CALCULUS CONCEPTS AND CONTEXTS 5TH EDITION BY JAMES STEWART, KOKOSKA (CHAPTER 1_13) SOLUTIONS MANUAL by QuentinAxel

TEST BANK

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

CALCULUS CONCEPTS AND CONTEXTS 5TH EDITION BY JAMES STEWART, KOKOSKA (CHAPTER 1_13) SOLUTIONS MANUAL Chapter 1-13

CHAPTER 1: SECTION 1.1 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1.1.1 (a) f (1)  3 (b) f (1)  0.2 (c) f (x)  1 when x = 0 and x = 3. (d) f (x)  0 when x ≈ –0.8. (e) The domain of f is 2  x  4. The range of f is 1  y  3. (f) f is increasing on the interval2  x  1. 1.1.2 (a) f (4)  2; g(3)  4 (b) f (x)  g(x) when x = –2 and x = 2. (c) f (x)  1 when x ≈ –3.4. (d) f is decreasing on the interval 0  x  4. (e) The domain of f is 4  x  4. The range of f is 2  y  3. (f) The domain of g is 4  x  4. The range of g is 0.5  y  4.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a) f (2)  12 (d) f (a)  3a2  a  2 (g)(g) f (2a)  12a2  2a  2



(i)  f (a)  3a 2  a  2 2

(b) f (2)  16 (e) f (a 1)  3a2  5a  4

(h) (h) f (a2)  3a4  a2  2

  9a4  6a3 13a2  4a  4 2

(j) f (a  h)  3  a  h    a  h   2  3a 2  3h 2  6ah  a  h  2 2

1.1.4

f (3  h)  f (3) (4  3(3  h)  (3  h)2 )  4 9  3h  9  6h  h2) 3h  h2       (3  h) h h h h

1.1.5



f (a  h)  f (a)  a  3a h  3ah  h  a  h 3a  3ah  h  h h h 3

 3a2  3ah  h2

1.1.6

1 1 a x   1 f (x)  f (a)  x a  ax ax  a  x    xa xa xa ax(x  a) ax 



1.1.7

x  3  2x  2 x 1 x 1 x  3 1 3 x  3 2  x 1   1 f (x)  f (1)  x 1 11  x 1 x 1    x 1   x 1 x 1 x 1 x 1 x 1 x 1 x 1

1.1.8



x4

The domain of f (x)  2

x 9

1.1.9



is x  | x  3,3.



2x3  5 The domain of f (x)  2 is x  | x  3, 2. x  x6

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 

1.1.10

The domain of f (t)  3 2t 1 is all real numbers. 1.1.11 g t   3  t  2  t is defined when 3  t  0  t  3 and 2  t  0  t  2. Thus, the domain is t  2,

or , 2. 1.1.12 The domain of h(x)  

1 4

x  5x 2

is , 0 5, .

1.1.13 The domain of F( p)  2 

p is0  p  4.

1.1.14 The domain of f (u)  u 1

1

is u  | u  2, 1.

u 1

1.1.15 (a) This function shifts the graph of y = |x| down two units and to the left one unit. (b) This function shifts the graph of y = |x| down two units (c) This function reflects the graph of y = |x| about the x-axis, shifts it up 3 units and then to the left 2 units. (d) This function reflects the graph of y = |x| about the x-axis and then shifts it up 4 units. (e) This function reflects the graph of y = |x| about the x-axis, shifts it up 2 units then four units to the left. (f) This function is a parabola that opens up with vertex at (0, 5). It is not a transformation of y = |x|.

1.1.16 (a)

g  f  x    g  x 2 1 10  x 2 1

(b)

f  g  4    f 104  402 1  1601

(c)

g  g 1  g 101  1010  100

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(d) (e)



 



f g  f 2  f g 22 1

1 f  g  x



1 f 10x 



  f 105  f 50  502 1  2501

10x  1 2



1 100x2  1

1.1.17 The domain of h(x)  4  x is 2  x  2, and the range is 2

0  y  2. The graph is the top half of a circle of radius 2 with center at the origin. 1.1.18 The domain of f (x)  1.6x  2.4 is all real numbers.

1.1.19



t 2 1 The domain of g(t)  ist  | t  1. t 1

1.1.20

 



x 1 f (x)  2 x 1 isx  | x  1,1. The domain of

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.1.21 The domain of f (x)  x3 1 is all real numbers.

1.1.22



x2 The domain of h(x)  isx  | x  1. x 1

1.1.23 The domain of g(x) 

x2 isx  | x  0. x3

1.1.24 f 3  3  2  1 f 0  1 0  1 f 2  1  2  1

1.1.25

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

f 3  3  12 3  92 f 0  3  12 0  3 f 2  2  2   5  1

1.1.26 f 3  3 1  2 f 0   0 2  0 f 2   2 2  4

1.1.27 f 3  1 f 0  1 f 2   7  2  2   3

1.1.28 Recall that the slope m of a line between the two points  x , y  and  x , y  is m  1

y2  y1 x x 2

and an

equation of the line connecting those two points is y  y1  m  x  x1  . The slope of the line segment joining the points 1, 3 and 5, 7 is function is f  x 

5 2

x

7   3  5 1

5 5  , so an equation is y 3   x 1. The 2 2

,1  x  5

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.1.29 The slope of the line segment joining the points 5,10 and 7, 10 is 10 10

7  5



5 , so an 3

5 f  x   x  , 5  x  7 equation is y 10   x    5 .  The function is 3

1.1.30 We need to solve the given equation for y. x  ( y 1)2  0  ( y 1)2  x  y 1   x 

y  1 x . The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we want f  x  1

x . Note that the

domain is x  0 .



1.1.31



x2  ( y  2)2  4  ( y  2)2  4  x2  y  2   4  x2  y  2  4  x2 . The top half is given by the function f  x  2  4  x , 2  x  2 2

1.1.32 For 0  x  3 , the graph is the line with slope 1 and y -intercept 3 , that is, y  x  3 . For 3  x  5 , the graph is the line with slope 2 passing through 3, 0 ; that is, y  0  2  x  3 , or y  2x  6 . So the function is

x  3 f  x   2x  6

if0  x  3 if3  x  5



1.1.33 For 4  x  2 , the graph is the line with slope 

passing through 2, 0 ; that is, 2 3 3 y  0   x   2  , or y   x  3. For 2  x  2 , the graph is the top half of the circle with 2 2 2 2 center 0, 0 and radius 2 . An equation of the circle is x  y  4 , so an equation of the top half is 3 y  4  x2 . For 2  x  4 , the graph is the line with slope passing through 2, 0 ; that is, 2 3 3 y  0   x  2  , or y  x  3. So the function is 2 2

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 3  x3  2  f  x   4  x2 3  x  3  2

if  4  x  2 if  2  x  2 if 2  x  4

1.1.34 From Figure 1.1 in the text, the lowest point before the end of the Super Bowl occurs at about t, r   145,300, where t is measured in minutes. The highest point occurs at about 20,360. Thus, the range of the rate of water usage is 300  t  360. Written in interval notation, we get 300,360.

1.1.35 Example 1: A car is driven at 60 mph for 2 hours. The distance d traveled by the car is a function of the time t. The domain of the function is {t | 0  t  2}, where t is measured in hours. The range of the function is {d | 0  d  120}, where d is measured in miles. Example 2: At a certain university, the number of students N on campus at any time on a particular day is a function of the time t after midnight. The domain of the function is {t | 0  t  24}, where t is measured in hours. The range of the function is {N | 0  N  k}, where N is an integer and k is the largest number of students on campus at one time. Example 3: A certain employee is paid $8.00 per hour and works a maximum of 30 hours per week. The number of hours worked is rounded down to the nearest quarter hour. This employee’s gross weekly pay P is a function of the number of hours worked h. The domain of the function is [0, 30], and the range of the function is {0, 2.00, 4.00, . . . , 238.00, 240.00}.

1.1.36 This is not the graph of a function because it does not pass the vertical line test.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.1.37 This is the graph of a function. The domain is 2  x  2 and the range is1  y  3. 1.1.38 This is the graph of a function. The domain is

3  x  2 and the range is 3  y  2

1  y  3.

1.1.39 This is not the graph of a function because it does not pass the vertical line test. 1.1.40 (a) When t = 1950, T ≈ 13.8℃, so the global average temperature in 1950 was about 13.8℃. (b) When T = 14.2℃, t ≈ 1990. (c) The average global temperature was smallest in 1910 (the year corresponding to the lowest point on the graph) and the largest in year 2016 (the year corresponding to the highest point on the graph). (d) When t = 1910, T ≈ 13.5℃, and when t = 2016, T ≈ 14.9℃. Thus, the range of T is about [13.5, 14.9].

1.1.41

(a)

The width varies from near 0 mm to about 1.6 mm, so the range of the ring width function is approximately [0, 1.6].

(b)

According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the late 1800s, and has been steadily warming since then. In the mid-19th century, there was variation that could have been associated with volcanic eruptions.

1.1.42

The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the water will slowly warm up to room temperature.

1.1.43

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

The graph indicates that runner A won the race, reaching the finish line at 100 meters in about 15 seconds, followed by runner B with a time of about 19 seconds, and then by runner C who finished in around 23 seconds. Runner B initially led the race, followed by C, and then A. Runner C passed runner B to lead for a while. Then runner A passed first runner B, then passed runner C to take the lead and finish first. All three runners finished the race. 1.1.44 (a)

The power consumption at 6AM is 500MW , which is obtained by reading the value of power P when t  6 from the graph. At 6 PM we read the value of P when t  18 , obtaining approximately 730MW.

(b)

The time of lowest power consumption is determined by finding the time for the lowest point on the graph, t  4, or 4 AM. The time of highest power consumption corresponds to the highest point on the graph, which occurs just before t  12, or right before noon. These times are reasonable, considering the power-consumption schedules of most individuals and businesses.

1.1.45 The summer solstice (longest day of the year) is around June 21, and the winter solstice (shortest day) is around December 22 (in the northern hemisphere). Therefore a reasonable graph for the number of hours of daylight vs. time of year is here:

1.1.46 The graph will depend upon geographical location, but here is one graph of the outdoor temperature vs. time on a spring day:

1.1.47 The value of the car will decrease rapidly initially, then somewhat less rapidly.

1.1.48

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

As the price increases, the amount of coffee sold will decrease:

1.1.49 The temperature of the pie would increase rapidly, level-off to oven temperature, decrease rapidly when removed from the oven, and then level-off to room temperature:

1.1.50 Here is a rough graph of the height of the grass on a lawn that is mown every Wednesday:

1.1.51 (a)

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b)

(c)(c)

(d)(d)

1.1.52 (a)

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b)

The temperature at 9 AM appears to have been roughly 51oF.

1.1.53 (a)

(b)

The rate or BAC jumps rapidly from zero to the maximum of 0.41 mg/mL. The concentration then gradually decreases over the next four hours to near zero.

1.1.54 With radius r + 1, the balloon has volumeV (r 1)  43  (r 1)3  43  (r3  3r2  3r 1). We wish to find the amount of air required to inflate the balloon from a radius of r to r+1 inches. Hence we need to find the difference V (r 1) V (r)  43 (3r3  3r2  3r 1)  43 r 3  43 (3r2  3r 1). 1.1.55 Let the length and width of the rectangle be l and w. Then the perimeter is 2l  2w  20 and the area is

A  lw. Solving the first equation for win terms of l gives w 

20  2l

 10  l. Thus

2 A(l)  l(10  l)  10l  l 2. Since the length must be positive, the domain of A is 0 < l < 10. If we further restrict l to be larger than w, the 5 < l < 10 would be the domain. 1.1.56 Let the length and width of the rectangle be l and w. Then the area is lw = 16 so that w = 16/l. The

perimenter is P  2l  2w, so P(l)  2l  216 / l   2l  32 / l, and the domain of P is l > 0 since the lengths must be positive. If we further require l to be larger than w, then the domain would be l > 4. 1.1.57 Let the length of a side of the triangle be x. Then by the Pythagorean Theorem, the height y of the 2 2 2 2 triangle satistfies y 2   12 x  , so that y  x  14 x  34 x and y  2 x. Using the formula for the area A 2

 x  x , with domain x > 0.

of a triangle, A = ½(base)(height), we find A(x)  12x

3 2

1.1.58

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Let the length, width, and height of the closed rectangular box be denoted by l, w, and h respectively. The length is twice the width, so l = 2w. The volume V of the box is V = lwh. Since V = 8, we have

8  (2w)wh  8  2w2h  h 



8  4 , so h  f (w)  4 . w2 2w2 w2

1.1.59 Let each side of the base of the box have length x, and let the height of the box be h. Since the volume is 2, we know that 2 = hx2 so that h = 2/x2, and the surface area is S  x2  4xh. Thus,





S (x)  x2  4x 2 / x2  x2  8 / x, with domain x > 0. 1.1.60 The area of the window is A  xh  2   x   xh  1

 x2

, where h is the height of the rectangular 8 portion of the window. The perimeter is P  2h  x  12  x  30  2h  30  x  12  x 2  60  2x   x  x   h  12 60  2x   x. Thus, 4 8 A(x)  x     4 15x  1 x2   x2   x2  15x  4 x2   x2  15x  x2 . Since the lengths x and h must be  8  2 4 8 8 8   positive, we have x > 0 and h > 0. For h > 0, we have 2h  0  30  x  21  x  0  60  2x   x 60 60 . . Therefore the domain of A is 0  x   x  2 2   1 2

1.1.61 The height of the box is x and the length and width are l  20  2x, w  12  2x. Then V = lwx so

V (x)  (20  2x)(12  2x)x  4(10  x)(6  x)x  4x(60 16x  x 2 )  4x3  64x2  240x. Because the sides must have positive lengths, l  0  20  2x  0  x  10; w  0  x  6; and x > 0. Combining these restrictions indicates the domain is 0 < x < 6. 1.1.62

We can model the monthly cost with a piecewise function: if 0  x  2 35 C(x)   . 35  5(x  2) if x  2

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.1.63 We can model the amount of the fine with a piecewise function:

15(40  x), if 0  x  40  F (x)  0, if 40  x  65 . 15(x  65), if 65  x  100 

1.1.64

if 0  x  1200 $10  0.06x, E(x)   $10  0.06(1200)  0.07(x 1200), if x  400 $10  0.06x, $82  0.07(x 1200),

equivalently, E(x)  

if 0  x  1200 if x  400

1.1.65 (a) Here is the graph:

(b) On $14,000 tax is assessed on $4000, and 10% of $4000 is $400. On $26,000, tax is assessed on $16,000 and 10%($10,000) + 15%($6000) = $1000 + $900 = $1900.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph of T is a line segment from (10,000, 0) to (20,000, 1000). The tax on $30,000 is $2500, so the graph of T for x > 20,000 is the ray with the initial point (20,000, 1000) that passes through (30,000, 2500). 1.1.66

One example is the amount paid for cable or appliance repair in the home, usually measured to the nearest quarter hour. Another example is the amount paid by a student in college tuition fees, if the fees vary according to the number of credits for which the student registers. 1.1.67 The function f is odd because its graph is symmetric about the point (0, 0). The function g is even because its graph is symmetric across the y-axis. 1.1.68 The function f is even because its graph is symmetric across the y-axis. The function g is neither even nor odd because it is not symmetric about the y-axis or the origin. 1.1.69 (a) If the point (5, 3) is on the graph of an even function then (–5, 3) must also be on the graph. (b) If the point (5, 3) is on the graph of an odd function then (–5, –3) must also be on the graph. 1.1.70 (a) Here is the complete graph if f is even:

(b) Here is the complete graph if f is odd:

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



1.1.71

x x is an odd function because f (x)    f (x). f (x)  2 x 1 (x)2 1

1.1.72

f (x) 

x2 x4 1

 f (x) is an even function.

1.1.73 

f (x) 

x x 1 is neither even nor odd.

1.1.74

f (x)  x x

is an odd function because

f (x)  x x   f (x).

1.1.75

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

f (x)  1 3x2  x4  1 3(x)2  (x)4  f (x) is an even function.

1.1.76

f (x)  1 3x3  x5 is neither even nor odd.

1.1.77 (i) If f and g are both even functions, then f  x   f  x and g  x   g  x  . Now

 f  g  x   f  x   g  x   f  x   g  x    f  g  x  , so f  g is an even function. (ii) If f and g are both odd functions, then f  x    f  x  and g  x   g  x  . Now

 f  g  x   f  x   g  x    f  x   g   x     f  x   g  x     f  g  x  , so f  g is an odd function. (iii) If f is an even function and g is an odd function, then

 f  g  x   f  x   g  x   f  x   g  x   f  x   g  x  , which is not  f  g  x  nor   f  g  x , so f  g is neither even nor odd. (Exception: if f is the zero function, then f  g will be odd. If g is the zero function, then f  g will be even.) 1.1.78 (i) If f and g are both even functions, then f  x   f  x and g  x   g  x  . Now

 fg  x   f  x  g  x   f  x  g  x    fg  x  , so fg is an even function (ii) If f and g are both odd functions, then f  x    f  x  and g  x   g  x  . Now

 fg  x   f  x  g  x     f  x   g  x   f  x  g  x    fg  x  , so fg is an even function. (iii) If f is an even function and g is an odd function, then

 fg  x   f  x  g  x   f  x  g   x     f  x  g  x     fg  x  , so fg is an odd function. © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION 1.2 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1.2.1 (a)

f  x  log2 x is a logarithmic function. 2x3

(b)

g  x 

(c)

u t   1  2t  3t2 is a polynomial of degree 2 (also called a quadratic function).

(d)

u t   11.1t  2.54t2 is a polynomial of degree 2.

(e)

h x  x2/7 is a power function.

(f)

k  x    x  3  x  2  is a polynomial of degree 6.

1  x2

is a rational function because it is a ratio of polynomials.

(g)(g)

v t   5t is an exponential function.

(h) (h)

wt   sin t cos2 t is a trigonometric function.

1.2.2 (a)

y  x4 is a power function and a polynomial of degree 4.

(b)

y  x2 2  x3   2x2  x5 is a polynomial of degree 5.

(c)

x y  is a rational function because it is a ratio of polynomials. 1 x

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x3 1

(d)

y 

(e)

y  x3/5  7 is an algebraic function.

(f)

y

(g)(g)

y

(h) (h)

y  tan t  cost is a trigonometric function.

1 3 x

is an algebraic function because it involves polynomials and roots of polynomials.

x2 is an algebraic function. x2  1 x2  7x  5 x3  x 1

is a rational function.

1.2.3 (a) y  x2 is the blue curve (function h) because it is a parabola. (b) y  x5 is the red curve (function f ) because it is an odd power function. (c) y  x8 is the green curve (function g) because it is an even power function whose degree is more than 2.

1.2.4 (a) y  3x is the blue curve (function f ) because it is a linear function. (b) y  x3 is the red curve (function g) because it is an odd power function. (c) y  3 x is the green curve (function h) because it is the cube-root function.

1.2.5 The domain and range of f (x)  2x  3 are all real numbers. 1.2.6 The domain of f (x)  x2  4 is all real numbers. The range is  y  | y  4. 

1.2.7 The domain of g(x)  x 1 is all real numbers. The range is also all real numbers. 3

1.2.8

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x2 is x  | x  2, 2. The range is  y  | y  0. x 4

The domain of g(x)  2 1.2.9 

The domain of h(x)  x3  2x  2 is all real numbers. The range is also all real numbers. 1.2.10

| x  0. The range is  y  | y  2.

x  2 is x 

The domain of h(x)  



1.2.11

The domain of f (x)  x  4 is all real numbers. The range is  y  2

1

1.2.12 The domain of f (x) 



is x  | x  2. The range is  y  | y  0.

1.2.13 The graph of f (x) 



x has a vertical asymptote at x  4 and a horizontal asymptote at y  1. x4

1.2.14 The graph of f (x) 



x2

| y  2.

x2 has a vertical asymptote at x  3 and no horizontal asymptotes. x3

1.2.15 The graph of g(x) 

x2 has a vertical asymptote at x = –2 and a horizontal asymptote at (x  2)(x  2)

y  0. 1.2.16

x has a vertical asymptote at x = 1 and a horizontal asymptote at y  0. x 1

The graph of g(x)  3 1.2.17 The graph of h(x) 



x2 has a vertical asymptote at x = 2 and a horizontal asymptote at y  0. x2

1.2.18



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x 1

The graph of h(x)  3

x 1

has a vertical asymptote at x 1 and a horizontal asymptote at y  0.

1.2.19 By the Vertical Line Test, the given graph is the graph of y as a function of x. 1.2.20 By the Vertical Line Test, the given graph is the graph of y as a function of x. 1.2.21 By the Vertical Line Test, the given graph is not the graph of y as a function of x. 1.2.22 By the Vertical Line Text, the given graph is not the graph of y as a function of x. 1.2.23 The linear function such that f (3)  11 and f (7)  9 must have slope = line is y  2(x  3) 11  2x  5. 1.2.24 If f (x)  3x  5 then

f (b)  f (a)



(3b  5)  (3a  5)



3b  3a

19 11 73

 2. An equation for this

ba  3  3. This will always be

  ba ba b  a  the value because the rate of change (i.e. ―slope‖) of a linear function is constant.

ba

1.2.25 If f (x)  5x  2 then

f (x  h)  f (x)



5(x  h)  2   5x  2



5x  5h  2  5x  2



 5.

1.2.26 The domain of f (x)  2(x  3)2  5 is all real numbers. The range is  y 

| y  5.



1.2.27 2 A quadratic function with range{y | y  6} such that f (4  d )  f (4  d) is f (x)  6  (x  4) or

f (x)  x2  8x 10. This function has a maximum of 6 that occurs when x = 4 and for all d, f (4  d )  f (4  d )  6  d 2. 1.2.28 A root function for which g(64) = 4 is g(x)  3 x.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.2.29 A rational function with a linear denominator which has an x-intercept at 4 and a vertical asymptote at x = 3 is f (x) 

x4 . x3

1.2.30 A rational function that is undefined at x = 2 but does not have a vertical asymptote at x  2 is

f (x) 



x2  4 x2

1.2.31 (a) An equation for the family of linear functions with slope 2 is y  2x  b. Below is a graph of several members of this family.

(b) An equation for the family of linear functions such that f (2)  1 is y  1b 2 x  b. Below is a graph of several members of this family.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.2.32 All members of the family of linear functions f (x)  1 m(x  3) go through the point (3,1). Below is a graph of several members of this family.

1.2.33 All members of the family of linear functions f (x)  c  x. have a common slope of –1. Each has a yintercept of (0, c). Below is a graph of several members of this family.

1.2.34 (a) The vertex of the parabola is 3, 0 , so an equation is y  a(x  3)2  0 . Since the point 4, 2 is on the parabola, we'll substitute 4 for x and 2 for y to find a.2  a(4  3)  a  2 , so an equation is 2

f  x  2(x  3)2 .

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b) The y -intercept of the parabola is 0,1 , so an equation is y  ax  bx 1 . Since the points 2

2, 2 and 1, 2.5 are on the parabola, we'll substitute 2 for x and 2 for y as well as 1 for x and 2.5 for y to obtain two equations with the unknowns a and b .

2, 2 : 2  4a  2b 1  4a  2b  1 1, 2.5 : 2.5  a  b 1  a  b  3.5 2   2   1 gives us 6a  6  a  1 . From (2), 1 b  3.5  b  2.5, so an equation is g  x  x2  2.5x 1. 1.2.35 . One expression for a cubic function f such that f (1)  6 , and f (1)  f (0)  f (2)  0 is

f (x)  3x(x  2)(x 1). 1.2.36 One cubic function that is increasing on the interval (–∞, ∞) is f (x)  x3 :

1.2.37 One quadratic function that intersects the x-axis at –2 is f (x)  (x  2)(x 1) :

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.2.38 One fourth degree polynomial that has zeros –2, 0, 1 and 6 is f (x)  x(x  2)(x 1)(x  6).

1.2.39 (a) In the equation T  0.02t  8.50, the slope of 0.02 indicates that for each additional year after 1900 the temperature is predicted to rise 0.02 ℃. This is the rate of change in temperature with respect to time. The T-intercept of 8.5 ℃ represents the average temperature of the earth’s surface in 1900. (b) The predicted average global surface temperature in 2100 is 0.02(200)  8.50  12.5 ℃. 1.2.40 (a) If c  0.0417  200 (a 1) then the slope of the graph of c is 8.34 mg/year . The slope is the increase in the mg of dose for each additional year in the age of a child and represents the rate of change in a child’s dose with respect to age. (b) The dosage for a newborn child would be c  0.0417  200(0 1) = 8.34 mg. 1.2.41 From d  kv2 we get 28  k  202  k  0.07, so d  0.07v2. A car traveling 40 mi/h would require d 40  0.07  402  112 ft to stop. 1.2.42 The slope for this linear model would be

y

$3.9

$7.25 $3.35 2015 1981



$3.9

. A linear model for these data would be

 x 1981  $3.35 . The estimated minimum wage in 1996 would be

34 $3.9

1996 1981 $3.35  $5.07 . This is $0.32 more than the actual minimum wage in 1996.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.2.43 (a) A graph of the model y  200  4x is given here:

(b) The slope of this line, 4, indicates that each additional dollar charged for rent is predicted to result in a decrease of four spaces rented. This is the rate of change in the number of spaces with respect to the rent charged. The y-intercept 0, 200 indicates that if nothing is charged, 200 spaces will be rented. The xintercept 50, 0 indicates that if $50 per space is charged, no spaces will be rented. 1.2.44 (a) Here is a graph of the function: (b) The slope of this graph is 9/5. The slope indicates that each increase of one degree in the Celsius temperature will result in a 9/5 degree increase in the Fahrenheit temperature. The slope represents the rate of change in the Fahrenheit temperature with respect to the Celsius temperature. The Fintercept is (0,32) and it tells us that a temperature of 0 ℃ is equivalent to 32 ℉.

1.2.45 (a)

The driver is driving at a constant rate of 40 miles / 50 minutes  4 / 5 mi/min or 48 mi/hr. We can write d  48t, where d is the distance traveled (in miles) and t is the time (in hours).

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b)

A graph of the model d  48t is given here:

(c)

The slope of this line is 48. At this rate, each hour the driver will travel 48 miles.

1.2.46 (a) Using N in place of x and T in place of y , we find the slope to be 80  70  10  1 . So a linear equation is T2  T1  N2  N1 173 113 60 6

T  80 

 N 173  T  80 

6 (b) The slope of

N

173

T 

N

307 307

  51.16 .  6  6

1 6

means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the

number of cricket chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 F . (c) When N 150 , the temperature is given approximately by T 

1 6

150 

307

 76.16 F  76 F .

1.2.47 (a) Let x denote the number of chairs produced in one day and y the associated cost. Using the points

100, 2200 and 300, 4800 , we get the slope 4800  2200 300 100



2600

 13. So y  2200  13 x 100 

200 y  13x  900.

(b) The slope of the line in part (a) is 13 and it represents the cost (in dollars) of producing each additional chair.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.2.48 (a) p  0.434d 15, where p = pressure and d = depth below the surface. (b) The pressure is 100 lb/in2 at a depth of 195.85 ft. 1.2.49 (a) Using d in place of x and C in place of y , we find the slope to be C2  C1 460  380 80  1 .   d2  d1 800  480 320 4 So a linear equation is C  460 

d  800  C  460 

4 (b) Letting d  1500 we get C 

1 4

d  200  C 

d  260 .

1500  260  635 .

4 The cost of driving 1500 miles is $635 . (c) The slope of the line represents the cost per mile, $0.25 .

(d) The y -intercept represents the fixed cost, $260. 1.2.50

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a)

The data appear to be increasing similarly to the values of a cubic function. A model of the form f  x   a  x  b   c seems appropriate. 3

The data appear to be decreasing in a linear fashion. A model of the form f  x  mx  b

(b)

seems appropriate. 1.2.51 The data are in the shape of a parabola. A model of the form f  x   a  x  b   c seems 2

(a)

appropriate. (b)

The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form f  x  a / x seems appropriate.

1.2.52 (a) The scatterplot indicates that a linear model would be appropriate. (b) Using the first and last points we would compute a slope of

8.2 14.1 60,000  4000

 0.0001.

A linear equation for this model would be Ulcer Rate = –0.0001(Income –4000) + 14.1. (c) The least squares regression equation is Ulcer Rate = –0.0001×Income + 13.95. (d) The ulcer rate for an income of $25,000 is predicted to be 11.45%. (e) A person with an income of $80,000 is predicted to have a 5.95% chance of peptic ulcers. (f) An income of $200,000 is quite far from the given income values so it would not be appropriate to apply the model in this case. If the model were applied, the resulting rate would be negative which makes no sense. 1.2.53 a) Graph (b) The regression equation is Chirp rate = 4.857×Temperature – 221. (c) The predicted chirping rate for an outdoor temperature of 100 ℉ is 48.57 – 221 = 264.7 chirps per minute.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.2.54 (a) Graph (b) The regression equation is Height = 1.881 × Femur Length + 82.65. (c) A person with a femur 53 cm long is predicted to have been 182.343 cm tall.

1.2.55 (a)

A linear model seems appropriate over the time interval considered.

(b)

Using a computing device, we obtain the least squares regression line y  0.025t  43.649.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(c)

When t  2020, y  6.44, which is higher than the actual winning height of 6.02 m.

(d)

No, since the times appear to be leveling off (or at least increasing at a lower rate) and getting further away from the model.

1.2.56 (a) The regression equation is Percent Tumors = 0.01879×Asbestos Exposure + 0.305. (b) The regression line does appear to be a suitable model for the data as the data is fairly linear. (c) The y-intercept, (0, 0.305), indicates that if there is no asbestos exposure then 0.305% of the mice are predicted to develop lung tumors.

1.2.57 (a)

A linear model seems appropriate over the time interval considered.

(b)

Using a computing device, we obtain the least squares regression line y  1.100t  60.109, where t represents years since 1985.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(c)

The year 2002 is 17 years after 1985. When t 17, y  78.80.

(d)

The year 2020 is 35 years after 1985. When t  35, y  98.60, which is higher than the 92.2 million barrels per day that the U.S. Energy Information Administration estimates the world consumed.

1.2.58 (a)

A linear model seems appropriate over the time interval considered.

(b)

Using a computing device, we obtain the least squares regression line y  0.277t  8.369.



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(c)

The year 2005 is 5 years after 2000. When t  5, y  9.75. The year 2017 is 17 years after 2000. When t 17, y  13.07. Using the linear model, we estimate the average retail price of electricity in 2005 as 9.75 cents/kWh and in 2017 as 13.07 cents/kWh. Each estimate is higher than the actual average retail price.

1.2.59 The light would be  12  = 4 times brighter. 2

1.2.60 (a) S(60)  0.7  60

 2.39 so you would expect to find 2 species of bats living in that cave. ln 4     4  ln  4   0.7   0.3 0.7  ln A  e 0.3 A 333.585 m2. (b) 4 S(A) 0.7 A0.3 0.7 A . Then 0.3 0.3



1.2.61 (a) T  1.000431227d1.499528750 (b) The power model in part (a) is approximately T  d1.5 . Squaring both sides gives us T 2  d 3 , so the model matches Kepler's Third Law, T 2  kd 3 . 1.2.62 Note that x  x for x  0 and x  x for x  0. Also, x  3    x  3 for x  3 and x  3  x  3 for x  3. Combining these, we have: x  x  3   x     x  3   2x  3 for x  3, x  x  3   x    x  3  3 for 3  x  0; and x  x  3  x   x  3  2x  3 for x  0.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Finally, adding 2 to each resulting expression above gives: 2x 1,  f (x)  5, 2x  3, 

x  3 3  x  0. x0

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION 1.3 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1.3.1 (a) To shift the graph of f up 3 units, graph y  f (x)  3 (b) To shift the graph of f down 3 units, graph y  f (x)  3. (c) To shift the graph of f 3 units to the right, graph y  f (x  3). (d) To shift the graph of f 3 units to the left, graph y  f (x  3). (e) To reflect the graph of f about the x-axis, graph y   f (x). (f) To reflect the graph of f about the y-axis, graph y  f (x).

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(g) To stretch the graph of f vertically by a factor of 3, graph y  3 f (x). (h) To shrink the graph of f vertically by a factor of 3, graph y  13 f (x). 1.3.2 (a) To obtain the graph of y  f (x)  8 from the graph of y  f (x), shift the graph up 8 units. (b) To obtain the graph of y  f (x  8) from the graph of y  f (x), shift the graph 8 units to the left. (c) To obtain the graph of y  8 f (x) from the graph of y  f (x), stretch the graph vertically by a factor of 8. (d) To obtain the graph of y  f (8x) from the graph of y  f (x), shrink the graph horizontally by a factor of 8. (e) To obtain the graph of y   f (x) 1 from the graph of y  f (x), first reflect the graph about the x-axis, then shift it down 1 unit. (f) To obtain the graph of y  8 f  18 x from the graph of y  f (x), stretch the graph horizontally and vertically by a factor of 8. 1.3.3 (a) y  f (x  4) is curve (3) because the graph of f has been shifted 4 units to the right. (b) y  f (x)  3 is curve (1) because the graph of f has been shifted up 3 units. (c) y 

f (x) is curve (4) because the graph of f has been shrunk vertically by a factor of 3.

(d) y   f (x  4) is curve (5) because the graph of f has been shifted 4 units to the left and reflected about then x-axis. (e) y  2 f (x  6) is curve (2) because the graph of f has been shifted 6 units to the left and stretched vertically by a factor of 2. 1.3.4 (a) y  f (x)  3 : Shift the graph of f down 3 units:

(b) y  f (x 1) : Shift the graph of f 1 unit to the left:

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(d) y   f (x) :

Shrink the graph of f vertically by a factor of

Reflect the graph of f about the x-axis.

1.3.5 (a) y  f (2x) :

(b) y  f  12 x :

Shrink the graph of f horizontally by a

Stretch the graph of f horizontally by a factor of

factor of 2:

(d) y   f (x) :

Reflect the graph of f about the y-axis:

Reflect the graph of f about the y-axis and then about the x-axis.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



1.3.6

2 The graph of y  f (x)  3x  x has been shifted 2 units to the right and stretched vertically by a

factor of 2. Thus a function describing the graph is

y  2 f (x  2)  2 3(x  2)  (x  2)2  2 x 2  7x 10. 1.3.7

 

2 The graph of y  f (x)  3x  x has been shifted 4 units to the left, reflected about the x-axis, and

shifted down 1 unit Thus a function describing the graph is

y   f (x  4) 1   3(x  4)  (x  4)2 1   x 2  5x  4 1 1.3.8 (a) The range of y  f  x  shown is {0,1, 2,3} (b) The range of the function y  f (x) is{1, 0,1, 2,3} (c) The range of the function y  f ( f (x)) is{1,0,1, 2,} 1.3.9

(a)

(b)

1.3.10 (a) The domain of y  g(x) is (0,) (b) The domain of y   f  g  (x) is the domain of f which is 2,3. (c) The domain of f (g(x)) is roughly 2, 2.25.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(d) The domain of y  f  x  is the domain of f which is 2,3. (e) The domain of y  f (2x) is [1, 2 ). 3

1.3.11  f  (a) The domain of y  (x) is 1,1 1,3.

   g 

(b) The domain of y   g

f  x is 7, 6 3,3.

1.3.12 The graphs of f (x)  2 x and g (x)  2x are reflections through the x-axis.



1.3.13

g1(x)  x , then shift to the left one unit  g2 (x)  x 1 . Finally, shift up 2 units to plot f (x)  x 1  2.

First graph the function





1.3.14 (a) The graph of y  2sin x is the graph of y  sin x stretched vertically by a factor of 2.

(b) The graph of y  1

x is the graph of y  x shifted up 1 unit.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.15 Start with the graph of y  x3 and reflect about the x-axis.

1.3.16 Start with the graph of y  x and shift 3 units to the right. 2

1.3.17 Start with the graph of y  x 1and shift up 1 unit: 3

1.3.18 Start with the graph of y 

1 x

reflect about the x-axis, and then shift

up 1 unit.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.19 Start with the graph of y  cos x, compress horizontally by a factor of 3, then stretch vertically by a factor of 2:

1.3.20



 Start with the graph of y  x, shift 1 unit to the left, and then stretch vertically by a factor of 2:

1.3.21 First note that y  x2  4x  5  (x  2)2 1. Now start with the graph of y  x2, shift 2 units to the right, then up 1 unit:

1.3.22 Start with the graph of y  sin x, compress horizontally by a factor of π, then shift up 1 unit.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.23 Start with the graph of y  x, reflect about the x-axis, and then shift up 2 units:

1.3.24 Start with the graph of y  cos x, stretch vertically by a factor of 2, reflect about the x-axis, and then shift up 3 units:

1.3.25 Start with the graph of y  sin x and stretch horizontally by a factor of 2:

1.3.26

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Start with the graph of y  x and shift 2 units down:

1.3.27 Start with the graph of y  x and shift 2 units to the right:

1.3.28 Start with the graph of y  tan x, shift π/4 units to the right, then compress vertically by a factor of 4.

1.3.29 Start with the graph of y 

x , shift down 1 unit, and then reflect the portion from x  0 to x 1 across

the x –axis to obtain the graph of y 

x 1.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.30 Start with the graph of y  cos x, shrink it horizontally by a factor of π, then reflect all the parts of the graph below the x-axis about the x-axis.

1.3.31 This is just like the solution to Example 4 except the amplitude of the curve (the 30 N curve in Figure 9

 2   t  80 . March 31 is the 90th day   365 of the year, so the model gives L 90  12.34 h . The daylight time (5:51 AM to 6 :18PM) is 12 hours on June 21) is 14 12  2 . So the function is L t   12  2sin

and 27 minutes, or 12.45 h . The model value differs from the actual value with a relative error of

12.45 12.34

 0.009 , less than 1%

12.45 1.3.32 Using a sine function to model the brightness of Delta Cephei as a function of time, we take its period to be 5.4 days, its amplitude to be 0.35 (on the scale of magnitude), and its average magnitude to be 4.0. If we let t = 0 at the time of average brightness, then the magnitude (brightness) as a function of time t in days can be modeled by M (t)  4.0  0.34sin 25.4t . 1.3.33

The water depth D(t) can be modeled by a cosine function with amplitude (12 – 2)/2 = 5 m, average magnitude (12 + 2)/2 = 7 m, and period 12 hours. High tide occurred at 1:02 PM t 13 2 h  391 h , so the curve begins a cycle at time t  391 h (we need to shift 391 units to the 60

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check 391  where D is measured in meters right). Thus D(t)  5 cos  212  t  391 30    7  5 cos  6  t  30    7,

and t is the number of hours after midnight. 1.3.34 The total volume of air V(t) in the lungs can be modeled by a sine function with amplitude (2500 – 200)/2 = 250 mL, average volume = (2500 + 2000)/2 = 2250 mL, and period 4 seconds. Thus

V (t)  250sin 24 t  2250  250sin 2t  2250, where V is in mL and t is in seconds. 1.3.35 (a) To obtain y  f  x  , the portion of the graph of y  f  x  to the right of the y -axis is reflected about the y -axis. (b) y  sin x



 1.3.36 The most important features of the given graph are the x-incepts and the maximum and minimum points. The graph of y  1/ f (x) has vertical asymptotes at the xvalues where the x-intercepts are on the graph of y  f (x). The maximum of 1 on the graph of y  f (x) corresponds to a minimum of 1/1 = 1 on Similarly, the minimum on the graph of y  f (x) corresponds to a maximum on the graph of

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

y  1/ f (x). As the values of y get large (positively or negatively) on the graph of y  f (x), the values of y get close to zero on the graph of y  1/ f (x).

1.3.37 (a)  f  g  (x)  x  5x 1. Its domain is all real numbers. 3

(b)  f  g  (x)  x  x 1. Its domain is all real numbers. 3

 3   (d)  f / g (x)  . Its domain is  x  | x   . 3x2 1 3   x3  2x2

  



1.3.38

(a)  f  g  (x)  3  x  x 1. Its domain is , 11,3. 2

2 (b)  f  g  (x)  3  x  x 1. Its domain is , 11,3. 2 (c)  fg  (x)  3  x  x 1. Its domain is , 11,3.

3 x

. Its domain is , 1 1,3. x2 1

(d)  f / g  (x) 

1.3.39 2 (a)  f g  (x)  3x  3x  5. Its domain is all real numbers.

(b)  g

f  (x)   3x  5   3x  5  9x 2  32x  30. Its domain is all real numbers.

f  (x)  3(3x  5)  5  9x  20. Its domain is all real numbers.



(d)  g g  (x)  x2  x

  x2  x  x4  2x3  2x2  x. Its domain is all real numbers. 2

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.40 3 3 2 (a)  f g  (x)  (1 4x)  2  64x  48x 12x 1. Its domain is all real numbers.





(b)  g f  (x)  1 x3  2 . Its domain is all real numbers. (c)  f

f  (x)  (x3  2)3  2  x9  6x6 12x3 10. Its domain is all real numbers.

(d)  g g  (x)  1 4(1 4x)  16x  3. Its domain is all real numbers. 1.3.41



(a)  f

g (x)  4x  2. Its domain is x  | x   12.

(b)  g

f  (x)  4 x 1  3. Its domain is x  | x  1.

f (x) 

x 11. Its domain is x  | x  1.

(d)  g g  (x)  4  4x  3  3  16x 15. Its domain is all real numbers.

1.3.42





(a)  f g  (x)  sin x 2 1 . Its domain is all real numbers. (b)  g

f  (x)  sin2 x 1. Its domain is all real numbers.

f (x)  sin(sin x) . Its domain is all real numbers.

2 

(d)  g g  (x)  x 1 1  x  2x  2. Its domain is all real numbers.

1.3.43



x 1 x  2 2x2  6x  5   (a)  f g  (x)  . Its domain is x  | x  2, 1. x  2 x 1 (x 1)(x  2) 1 x  1 2 x  x 1 x (b)  g f  (x)   . Its domain is x  | x  1, 0. 2 1 x   2 x  2x 1 x x4  3x2 1 1 1  (c)  f f  (x) x   . Its domain is x  | x  0. x x 1 x(x2 1) x © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x 1

(d)  g g  (x)  x  2

1



x 1 2 x2

2x  3

. Its domain is x  3x  5

| x  2,  5. 3

 | x 



1.3.44



sin 2x . Its domain is x   34  n , n  . 1 sin 2x  x  . Its domain is x  | x  1. (b)  g f  (x)  sin  (a)  f g (x) 



   1 x   x    (1 x) x

1 f  (x)  1 x   1 x  x1 Its domain is x  | x  1,  2. x x   2x  1  (1 x) 1 x 1 1 x  





(d)  g g  (x)  sin(2sin 2x).Its domain is all real numbers. 1.3.45

f





g h(x)  f g  x 2   f sin(x2   3sin x2  2

  

1.3.46

f

g h  (x) 



 x   4  x  4 2

1.3.47

f









g h  (x)  f g  x3  2   f  x 3  2   2

 x3  2   3  x6  4x3 1 2

1.3.48 3

  x 

 f g h  (x)  f g

 3 x   3 x   tan  f 3   3 x 1    x 1   

1.3.49 If f (x)  x4, g(x)  2x  x 2 , then  f

g  (x)   2x  x2   F(x). 4

1.3.50

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

If f (x)  x2, g(x) cos x. then  f

g  (x)   cos x   cos 2 x  F (x). 2

1.3.51 3 x x , g(x)  3 x, then  f g  (x)  3  F (x). 1 x 1 x

If f (x) 

 1.3.52

If f (x)  3 x, g(x) 

x x , then f g  (x)  3  F (x). 1 x 1 x

1.3.53

 

If f (t)  sec t tan t, g(t)  t , then f

   

g (t)  f t 2  sec t 2 tan t 2  F (t).

1.3.54

t tan t , g(t)  tan t, then f g (t)  f tan t    F(t). 1 t 1 tan t

If f (t) 

 1.3.55

Let f (x)  x, g(x)  x 1, h(x)  x. Then 

f

  x   f  x 1 

g h  (x)  f g



1.3.56 Let f (x) 

f

x 1 F (x).



x, g(x)  2  x, h(x)  x . Then







g h(x)  f g  x   f  x  2  8 x  2  F(x).

1.3.57 Let f (t)  t2, g(t)  sin t, h(t)  cos t. Then

f

g h  (t)  f  g cost   f sin(cost)  (sin(cost))2  sin2(cost)  F (t).

1.3.58 Let f (t)  tan t, g(t) 

1 t

, h(t)  t3. Then f

 g h  (t) 



f g t3   f

 1  1  tan  F (t).  t3  t3  



1.3.59 (a) f (g(1))  f (6)  5

(b) g( f (1))  g(3)  2

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(d) g(g(1))  g(6)  3

(e)  g

(f)  f

f (3)  g( f (3))  g(4)  1

g (6)  f (g(6))  f (3)  4

1.3.60 (a) f (g(2))  f (5)  4

(b) g( f (0))  g(0)  3

(d)  g

(e)  g

g (0)  f (g(0))  f (3)  0

g (2)  g(g(2))  g(1)  4 (f) f

f (6)  g( f (6))  g(6)  undefined

f (4)  f ( f (4))  f (2)  2

1.3.61 1. To find a particular value of f (g(x)), say for x = 0, we note from the graph that g(0) ≈ 2.8 and f(2.8) ≈ –0.5. Thus f (g(0))  f (2.8)  0.5. The other values listed in the table are obtained in a similar fashion.

g(x)

f (g(x))

–0.2

–4

1.2

–3.3

–3

2.2

–1.7

–2

2.8

–0.5

–1

–0.2

x –5

x 0 1 2 3 4 5

g(x)

f (g(x))

2.8

–0.5

2.2

–1.7

1.2

–3.3

–0.2

–4

–1.9

–2.2

–4.1

1.9

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.62 (a) Using the relationship distance = rate⋅ time with the radius as the distance, we have r  60t. (b)(b) A   r 2   A r  (t)  A(r(t))    60t   3600 t 2 . The area of the circle is increasing at a 2

rate of 3600π cm/s2. 1.3.63 (a) The radius of the balloon is increasing at a rate of 2 cm/s, so r(t)  2t in cm (b) Using V  4  r 3 , we get V r  (t)  V (r(t))  V (2t)  4   2t   32  t 3. This tells us the 3

volume of the balloon (in cm3) as a function of time in seconds. 1.3.64 From the figure, we have a right triangle with legs of length 6 and d, and hypotenuse s. By the Pythagorean Theorem, d 2  62  s2  s  f (d ) 

d 2  36.

(b) Using d = rt, we find d = (30 km)×(t hours) = 30t (in km). Thus d  g(t)  30t. (c)  f

g (t)  f (g(t))  f (30t)  900t 2  36. This represents the distance between the ship

and the lighthouse as a function of the time elapsed since noon. 1.3.65 (a) d  rt  d (t)  350t. (b) There is a Pythagorean relationship involving the legs with length d and 1 and the hypotenuse with length s: d 2 12  s2. Thus, s(d )  d 2 1. (c)  s d (t)  s(d (t))  s(350t) 



350t)2 1

1.3.66

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

0, t  0 1, t  0

(a) Heaviside function: H (t)  

(b)

 0, t  0 so V(t) = 120H(t). V (t)   120, t  0 (c) Starting with the equation in part (b), we replace 120 with 240 to reflect the different voltage. Also, because we are starting 5 units to the right of t = 0, we replace t with t – 5. Thus the formula isV (t)  240H (t  5). 1.3.67 (a) R t   tH t 

(b)

 



0  t

if t  0

0 V t    2t

if t  0

if t  0

if0  t  60

so V t   2tH t  , t  60 .

if t  7 0 (c) V t     4 t  7 if7  t  32

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

so V t   4  t  7 H t  7  , t  32 .

1.3.68 If f  x  m1x  b1 and g  x  m2 x  b2 , then

 f g  x  f  g  x    f  m2 x  b2   m1  m2 x  b2   b1  m1m2 x  m1b2  b1. g is a linear function with slope m1m2 .

So f 1.3.69

2 3 If A(x) 1.042 x, then  A A  (x)  1.04 x , and A A A  (x)  1.04 x , and

 A A A A(x)  1.044 x. These compositions represent the investment amounts after 2, 3 and 4 years respectively. After n compositions, the investment amount would be 1.04n x.

1.3.70 (a) If g(x)  2x 1 and f (x)  x2  6 then

f g  f  2x 1  (2x 1)2  6  4x2  4x  7  h(x). (b) If f (x)  x  4 and g(x)  x2  x 1then





g  f x 2  x 1  3(x2  x 1)  5  3x2  3x  3  5  3x2  3x  2  h(x).

1.3.71 If f (x)  x  4 and h(x)  4x 1 then h  g

f if g(x)  4x 17 .

1.3.72 Suppose g is an even function and h  f

 f (g(x)  h(x) so h  f

g. Then h(x)   f

g  (x)  f (g(x))

g is an even function.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.3.73 Suppose g is an odd function and h  f

h(x)   f

g. First suppose f is an odd function. Then

g  (x)  f (g(x))  f (g(x))   f (g(x))  h(x) so h  f

g is an odd

function. Now assume f is an even function. Then h(x)   f

g  (x)

 f (g(x))  f (g(x))  f (g(x))  h(x) so h  f

g is an even function.

If f is neither either nor odd, h  f

g will be neither even nor odd.

1.3.74

(a)

We need to examine h  x  . h  x    f

g  x   f  sin  x    f sin x  f sin x [because f is even]  f  g  x   h  x 



Because h  x   h  x  , h is an even function. (b)

We need to examine h  x  . h  x    g

f  x   cos f  x    cos f  x [because f is odd]

 cos f  x  g  f  x  h x

Because h  x   h  x  , h is an even function. (c)

We need to examine k  x  . k  x    f







g h  x   f g h  x    f g h  x  

 [since h is even]  k  x

Because k  x   k  x  , k is an even function. Note that the result does not depend on whether the functions f and g are each even, odd, or neither.

Solution and Answer Guide

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION 1.4 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

End of Section Exercise Solutions

END OF SECTION EXERCISE SOLUTIONS



1.4.1 28 3 8 8  86  2  (a) 4  2  2   

28

2 



(b) 3



 x4/3

4/3

1.4.2

   x33 x2  27x  x6  27x7

(a) 84/3  24  16

(b)(b) x 3x 2

1.4.3 3 12 6 y 3   1296 y  648 y  648 y7 (b)  4

(a) b8  2b   b8  24  b 4  16  b12 4

2 y5

1.4.4 1 7

32 7



25 7





52 7

5 2 52 7 25 52 7 5 n2 2n 4n 82 16n 22n  26  24n 26n  26 23n 3n6 (b) 4 8     2 162n 42 8n 162 24  23n  28 23n  212 26  25

(a) 5



1.4.5 (a)  20  2  5 2  4 2 5 2  5 2  2 5 2  55 2  7 5 2  7 5 1

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check 1

(b) 8 2  2 2  2 2  2 2  2  2 2  2 2 1.4.6

3x1 92 x  3x1  32

(a)

  

 3x1 34 x  35 x1

 

 3 3  3

(b)

3 3 2

1.4.7 (a)

x2n  x3n1  x2n3n1(n2)  x4n3 1 x n2 1 2

(b)

a b 3

 

a b 

2 1

3 1



a 231b 431 ab

 a 2 3  b 4 3  a 6  b  12 1 1

1 1

1.4.8 is in the interval [4, 5]

(b) 3

(d)

2 is in the interval [1, 2]

(a) 2

is in the interval [4, 5] 2

1.4.9 x a) f (x)  b , b  0 (b) The domain is

1.4.10 x (a) The number e is the value of a such that the slop of the tangent line at x = 0 on the graph of y  a is exactly 1. (b) e  2.71828 (c) f (x)  e

1.4.11

icly accessible

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

All of these graphs approach 0 as x → –∞, all of them pass through the point (0, 1), and all of them are increasing and approach ∞ as x → ∞. The larger the base, the faster the function increases for x > 0, and the faster it approaches 0 as x → –∞.

1.4.12 The graph of e–x is the reflection of the graph of ex about the y-axis, and the graph of is the reflection of that of 8x about the y-axis. The graph of 8x increases more quickly than that of ex for x > 0, and approaches 0 faster as x → –∞.

1.4.13 The functions with bases greater than 1 (3x and 10x) are increasing, whereas



those with bases less than 1  1  and  1  x

 are decreasing. The graph of

 1  is the reflection of 3x about the y-axis, and the graph of  1  is the x

reflection of that of 10 about the y-axis. The graph of 10 increases more quickly than that of 3x for x > 0, and approaches 0 faster as x → –∞.

1.4.14 Each of these graphs approaches ∞ as x → –∞, and each approaches 0 as x → ∞. The smaller the base, the faster the function grows as x → –∞, and the faster it approaches 0 as x → ∞.

1.4.15 If we start with the graph of y  4 and shift it 1 unit down to obtain the graph of y  4 1. x

1.4.16

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

If we start with the graph of y  (0.5) and shift it 1 unit to the right we obtain the graph of y  (0.5) x

x1

1.4.17 If we start with the graph of y  2 and reflect it about the y-axis, then about the x-axis, we obtain the x

x graph of y  2 . In each case, y = 0 is the horizontal asymptote.

1.4.18 If we start with the graph of y  e and reflect the portion of the graph in the first quadrant about the yx

axis, we obtain the graph of y  e x .

1.4.19 x x We start with the graph of y  e and reflect about the y-axis to get the graph of y  e . Then we

compress the graph vertically by a factor of 2 to obtain the graph of y  12 ex. and then reflect about the xaxis to get the graph of y   12 ex. Finally we shift the graph up one unit to obtain the graph of

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.20 We start with the graph of y  ex and reflect about the x-axis to get the graph of y  ex. Then shift the graph up one unit to get the graph of y  1 ex. Finally, we stretch the graph vertically by a factor of 2 to





obtain the graph of y  2 1 ex .

1.4.21 We start with the graph of

y  2x and then reflect the portion of the graph where x > 0 across the y-axis

and add this to the original graph. Then we flip the coordinates of every point on the graph of y  2 to x

x

get the corresponding values of y  2

. Finally, shift the graph up one unit to obtain the graph of

y  2 x 1.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.22 We start with the graph of y  e and flip across the x-axis to obtain the graph of y  e . Then we shift x

the graph up 2 units to graph axis to obtain the graph

y  ex  2. Finally, we flip the part to the right of x > ln(2) across the x-

y  ex  2 .

1.4.23 (a) To find the equation of the graph that results from shifting the graph of y = ex 2 units downward, we subtract 2 from the original function to get y  e  2. x

(b) To find the equation of the graph that results from shifting the graph of y = ex two units to the  x2

right, we replace x with x – 2 in the original function to get y  e

(c) To find the equation of the graph that results from reflecting the graph of y = ex about the x-axis, we multiply the original function by –1 to get y  e . x

(d) To find the equation of the graph that results from reflecting the graph of y = ex about the y-axis, x we replace x with –x in the original function to get y  e .

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(e) To find the equation of the graph that results from reflecting the graph of y = ex about the x-axis and then about the y-axis, we first multiply the original function by –1 (to get y  ex ) and then x

replace x with –x in this equation to get y  e . 1.4.24

(a)

This reflection consists of first reflecting the graph about the x-axis (giving the graph with equation y  ex ) and then shifting this graph 2  4  8 units upward. So the equation is y  ex  8.

(b)

This reflection consists of first reflecting the graph about the y-axis (giving the graph with equation y  ex ) and then shifting this graph 2  2  4 units to the right. So the equation is y  ex4.

1.4.25 1x 2

The denominator is zero when 1 e1x  0  e 2

 1  1 x 2  0  x  1. Thus, the function

has domainx | x  1. 1.4.26 The denominator is never equal to zero, so the function f (x) 

1 x ecos x

has domain .

1.4.27 The function g(t)  10 100 has domain t |10t 100  0  t |10t  102



 

  t | t  2.

1.4.28





 2x  3x  6  5x  6

 x

The sine and exponential functions have domain , so g(t)  sin et 1 also has domain 1.4.29 x2

1 9      27  x



   3  3

  32   3 x

 13

x2

3( x2)

6 5

1.4.30

 2    1 



1  1 1 (2 x1)  22( x5)  (2x 1)  2(x  5)  x   2x 10    2 2 2 2  4  19 19  3x    x   2 6 2 x1

x5



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 2 

1.4.31 x

8 



  23  

 8x  22 1

2 x2 4

2 x 4 2

 2x 2  23x  2x 2  3x  x  2  x  3x  2  0 2

 (x 1)(x  2)  0  x  1 or x  2 1.4.32

 5   5  5 3x

x6

 56 x 53x 5x6  56 x 53xx6  6x  4x  6  2x  6  x  3

1.4.33 x 4  2 x  6  4 x  8  0   3  2 x   2  2 x  4  0  3  2 x   2  2 x  4  0 Let u  2 , and the 2

equation becomes 3u 2  2u  4  0. Using the quadratic formula, we find that u 

1 13

, so

3 1 13 2x  1 13  x  . 3 6 1.4.34

4x4  4x3  96  4x44  4x43  96  4x43 4 1  96  4x 

 2x  1  x  

192

1 2

1.4.35 Using the points (1, 6) and (3, 24), we have 6  Cb  C  6band 1

24  Cb3  24   6b b3  4  b2. Then since b > 0, b = 2 and C = 6/2 = 3. The function is

f (x)  3 2x. 1.4.36 Using the points (–1, 3) and (1, 4/3), we have 3  Cb 4

 Cb  1

 3bb 

1

 C  3b and  b . Then since b > 0, b  2 and C  3 2  2. The function is 2

f (x)  2   23  . x

1.4.37 This function is y  e reflected through both the x- and y-axes, then shifted up 2 units, and then stretched x

vertically 3 units, because it goes through the point (0, 0) and has a horizontal asymptote at y = 3. The





equation of this graph is y  3 1 ex .

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.38 These graphs are reflections of each other through the y-axis because

y   12   2 x. x

1.4.39 2

x From the graph of f (x)  x we see that as x increases, the function 2 approaches zero.



1.4.40 If f (x)  5 we have x

f (x  h)  f (x)



5xh  5x



5x5h  5x h



5x 5h 1 h

 5h 1  5  . h   x

1.4.41 Suppose the month is February (in a non-leap year). Your payment on the 28th day under plan II would be 228–1 = 227 = 134,217,728 cent, or $1,342,177.28. Clearly the second method of payment results in a larger amount for any month.

1.4.42 We see from the graphs that for x less than about x 5 1.8, g(x)  5  f (x)  x , and then near the

point (1.765, 17.125), the curves intersect. Then

f (x)  g(x) from x ≈ 1.765 until x = 5. At (5, 3125) there is another point of intersection, and for x > 5 we see that g(x)  f (x). In fact, g increases much more rapidly than f beyond that point.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.43 The graph of g(x) finally surpasses that of f(x) at x ≈ 35.8.

1.4.44 We graph y = ex, and y = 1,000,000,000 and determine where ex  1109. This seems to be true at x ≈ 20.723, so ex  1109 for x > 20.723.

1.4.45 (a) Using the Rule of 72, if the interest rate is 10% your money should double in approximately 72/10 = 7.2 years. If your interest rate is 5%, your money should double in approximately72/5 = 14.4 years. If your interest rate is 2%, your money should double in approximately 72/2 = 36 years.

(b) Using technology to solve 0.95ert  2 when r = 10% suggests your money should double in 7.444 years. When r = 0.05, your money should double in 14.889 years, and when r = 0.02, your money should double in 37.222 years.

1.4.46 (a) Here is a scatterplot:

(b) Using technology, we obtain the exponential function

f (t)  36.89301(1.06614)t. (c) Using technology, we find that the bacteria count doubles form 37 to 74 in about 10.868 hours.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.47 (a) Three hours is 3 doubling periods (each doubling period is 1 hour) so there would be 4,000 bacteria after 3 hours.

(b) In t hours there will be t doubling periods. The initial population is 500 so at time t the number of bacteria will be y  500  2 . t

500 2(40/60)  500 2(2/3)  794 bacteria. (d) From the graphs of y  500 2t and y  100,000 we see that the curves intersect at about t ≈ 7.64, so 2

the population reaches 100,000 in about 7.64 hours. 1.4.48 (a) Fifteen days is 3 half-life periods (the half-life is 5 days). So 200  12  25 mg. 3

(b) There would be 1 5 doubling periods after t days. The initial population is 200 mg so after t days the amount of 210Bi is y  200  12 

t /5

 200  2t/5 mg.

(c) t = 3 weeks is 21 days ⇒ y  200  221/5  10.882 mg. (d) We graph y1  200 2 t/5 and y  1 . The two curves intersect at t ≈ 2

38.219, so the mass will be reduced to 1 mg in about 38.219 days. 1.4.49 (a) Sixty hours is 4 half-life periods. So 2   1   4

(b) In t hours, there will be t/15 half-life periods. The initial mass is 2 g, so the mass y at time t is

y  2   12 

t /15

96/15

 0.024 g.

(d) y = 0.01 ⇒ t = 114.658 hours.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.50 From Table 1.6 we see that V(1) = 76. Using the graph in Figure 1.87, we estimate that V = 38 (half of 76) when t ≈ 4.5. This gives us a half-life of 4.5 – 1 = 3.5 days. 1.4.51

Half the river’s original height is 12 13.07  6.535 ft. We graph y1  12.8626    0.890114  and t

y2  6.535. The two curves intersect at x  5.8, so we estimate that the river will be half its original height at about 5.8 days. Note that the river actually fell to this height between Day 4 and Day 5.

1.4.52 This is just like the solution to Example 4 except the amplitude of the curve (the 30 N curve in Figure 9  2  on June 21) is 14 12  2 . So the function is L t   12  2sin  t  80 . March 31 is the 90th day  365   of the year, so the model gives L 90  12.34 h . The daylight time (5:51 AM to 6 :18PM) is 12 hours and 27 minutes, or 12.45 h . The model value differs from the actual value with a relative error of

12.45 12.34

 0.009 , less than 1% .

12.45

1.4.53

Let t  0 correspond to 1900 to get the model P  abt , where a 1441.2575148222 and b 1.0139847176177. To estimate the population in 1997, let t  97 to obtain P  5544 million. To predict the population in 2022, let t  122 to obtain P  7845 million.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.4.54 Let t  0 correspond to 1900 to get the model P  abt , where a  681.873349517496 and b 1.0123458060325. To estimate the population in 1925, let t  25 to obtain P  111 million. To predict the population in 2025, let t  125 to obtain P  380 million.

1.4.55 From the graph it appears that f is an odd function (f is undefined for x = 0). To confirm this, we must show that f (x)   f (x) :

1 e

1/( x)

1

(1/ x)

1 1/ x

1/ x

1 e f (x)  e1/ x  e  e 1   1/( x) (1/ x) 1 e1/ x 1 e 1 e 1 e1/ x 1 e1/ x 

1 e1/ x   f (x), so f is an odd function. 1 e1/ x

1.4.56

 

If we start with b = –1 and graph f (x) 

1 for a = 0.1, 1, and 5, we 1 aebx

see there is a horizontal asymptote of y = 0 as x → –∞ and a horizontal asymptote of y = 1 as x → ∞. If a = 1, the y-intercept is As a gets smaller (closer to 0), the graph of f moves to the left. As a gets larger, the graph of f moves to the right.

As b changes from –1 to 0, the graph of f is stretched horizontally. As b changes through large negative values, the graph of f is compressed horizontally.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

If b is positive, the graph of f is reflected through the y-axis. Lastly, if b = 0, the graph of f is the horizontal line y = 1/ (1 + a).

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION 1.5 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1.5.1 (a)

A function is one-to-one if f  x1   f  x2  whenever x1  x2 .

(b)

The graph of a one-to-one function must pass the Horizontal Line Test.

 1.5.2



(a)

f 1  y   x  f  x  y for any y in B . The domain of f 1 is B and the range of f 1 is A .

(b)

Write y  f  x. Solve this equation for x in terms of y. Interchange the variables x and y to express f 1 as a function of x.

(c)

Reflect the graph of f about the line y  x.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



1.5.3

f is not one-to-one because 2  6 , but f 2  2.0  f 6 .

1.5.4 f is one-to-one because it never takes on the same value twice.

1.5.5 Because there are horizontal lines that intersect the graph in more than one point, this function is not oneto-one.

1.5.6 No horizontal line intersects the graph more than once, so this function is one-to-one.

1.5.7 No horizontal line intersects the graph more than once, so this function is one-to-one.

1.5.8 Because there are horizontal lines that intersect the graph in more than one point, this function is not oneto-one.

1.5.9 The graph of f (x)  2x  3 is a line with slope 2. It passes the horizontal line test, so f is one-to-one.

1.5.10 If f (x)  x 16, then f (1)  f (1)  15 so f is not one-to-one. 4

1.5.11 If g(x) 1 sin x, then g(0)  g( )  1, so g is not one-to-one.

1.5.12 The graph of g(x)  3 x passes the Horizontal Line Test so g is one-to-one.

1.5.13 A football will reach every height h up to its maximum height twice: once on the way up and once on the way down. Thus f is not a one-to-one function.

1.5.14

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Eventually, we all stop growing and remain at a fixed height for a while. Therefore the function f is not one-to-one.

1.5.15 1 (a) Because f is one-to-one, if f (6)  17, then f (17)  6. 1 (b) Because f is one-to-one, if f (3)  2, then f (2)  3.

1.5.16 Observe that f is a one-to-one function (f is an increasing function). By inspection, f(1) = 3, so

f 1(3)  1. Because f is one-to-one, f  f 1(2)  2. 1.5.17 Because g is an increasing function, g is one-to-one. By inspection g(0) = 4, so g 1(4)  0.

1.5.18 (a) f is 1-1 because it passes the Horizontal Line Test. (b) Domain of f  3, 3  Range of f (c) Since f 0  2, f

1

. Range of f  1, 3  Domain of f 1 .

2  0 . 1 (d) Since f 1.7  0, f 0  1.7 . 1

1.5.19 Solve C 

 F  32 for F: 9 C  F  32  F  9 C  32. This provides a formula for the Fahrenheit

temperature F as a function of the Celsius temperature C.

F  459.67  95 C  32  459.67  95 C  491.67  C  273.15, the domain of the inverse function.



1.5.20

 m2  2 m  1    1  v  c 1 0   v  c 1 m0 . 2 c2 m2 c2 m2 m2 1 v2 / c2  m  v2

m02

This provides a formula for the speed of the particle v in terms of its mass, m, i.e. v  f 1(m).

1.5.21

 



y  f (x)  1 2  3x  y 1  2  3x   y 1  2  3x   y 1  2  3x  2

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x  13  y 1  23 . 2

f 1(x)  31  x 1 2  23 .

The domain of f 1 is x ≥ 1.

1.5.22 y  f (x) 

4x 1

 y(2x  3)  4x 1  2xy  3y  4x 1  2  3x  3y 1  4x  2xy 

2x  3

3y 1  x(4  2 y)  x 

3y 1 4  2y

. So f 1(x) 

3x 1 4  2x

1.5.23 y  f (x)  e2x1  ln y  2x 1  1 ln y  2x  x  121 ln y .

So, f

1

(x) 

1 2

(1 ln x).

1.5.24 y  f (x)  x 2  x  y  x 2  x  14  14  y   x  12   14  y  14   x  12   x  12  2

f 1(x)  1  x  1 .

x  1  y  1. 2

y  14 

1.5.25 y  f (x)  ln(x  3)  x  3  e y  x  e y  3. So f 1(x)  ex  3. 1.5.26

1 ex

x

y  f (x)   y  ye  1 e x  y 1 e   1 e 1 e   1 y 1 y 1 x   ex   x  ln . So f 1(x)  ln .   1 y 

1 y 1.5.27



 2

y  f (x)  4x  3  y  4x  3  x 



1

f (x)  So

x

 ye  y  e 1  ye  e   y 1

  1 x 

y2  3 . 4

x2  3 . 4

1.5.28 y  f (x)  1 ex  y 1  ex  ln( y 1)  x  x  ln( y 1). So f 1(x)  ln(x 1).

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.5.29 Reflect the graph of f about the line y = x. The points (–1, –2) and (1, –1), (2, 2) and (3, 3) on f are reflected to (–2, –1), (–1, 1), (2, 2) 1 and (3, 3) on f .

1.5.30 Reflect the graph of f about the line y = x.

1.5.31



(a) y  f (x)  1 x

(0  x  1 and note that y  0)  y2  1 x2  x2  1 y2 



x  1 y2 . So f 1(x)  1 x2 , 0  x  1. In this case, f and f 1 are the same function. (b) The graph of f is the portion of the circle x  y  1with 0 ≤ x ≤ 1 and 0 ≤ 2

y ≤ 1 (the quarter-circle in the first quadrant). The graph of f is symmetric with respect to the line y = x, so its reflection about y = x is itself, that is, f 1  f .

1.5.32

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a)

y  g(x)  3 1 x3  y3  1 x3  x3  1 y3  x  3 1 y3 . So g 1(x)  3 1 y3 . In this case, g and g 1 are the same function.

(b) The graph of g is symmetric with respect to the line y = x, so its reflection about y = x is itself, that is,

g 1  g.

1.5.33 (a) The function logb x  y is defined as the inverse of the exponential function with base b, that is

logbx  y  by  x. (b) (0, ∞) (c) (d) Graph

1.5.34 (a) The natural logarithm is the logarithm with base e, denoted ln x. (b) The common logarithm is the logarithm with base 10, denoted log x. (c) Graph

1.5.35 (a) log 2 32  log 2 25  5.

(b) log 2  log 81/3  . 8

1.5.36 (a) log5

1 1  log 5 3  log 5 53  3. 125 5

 e

(b) ln 1 2  ln e2  2.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.5.37 (a) log 1040  log 102.5  log 10(40)(2.5)  log 10100  log 10102  2 (b) log 60  log 3  log 5  log 60  log 5  log 20 log 5  log 8

8 3

20 8 5

 log  log 82/3  4

2 3



1.5.38 1

(a) eln 2 





eln 2

(b) e ln 2 ln e3   eln3  3

1.5.39 ln10  2ln 5  ln10  ln 52  ln(10)(25)  ln 250 1.5.40 2

ln b  2ln c  3ln d  ln b  ln c  ln d  ln bc  ln d  ln

bc2 d3

1.5.41 1/3 2 x 2 3 ln  x  3  1 ln x  ln  x 2  3x  2    ln  x  2    1 ln x  ln(x  2)  ln 3 2  2 2 2      x2  3x  2  x2  3x  2

= ln (x  2) x

(x 1)(x  2)

 ln

x x 1

1.5.42 ln 1 e

2 x

  ln 1 e

1.5.43 ln plog  e

 1 e2x    ln  2 x   1 e 

   ln e   e   1 p 

 e    ln p  

 e2



ln p



 



 

p ln p

ln e ln p

ln p

p 1

1.5.44 log 10 

ln10

ln5

 1.430677

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.5.45 log 57  3

ln 57

 3.680144

ln3

1.5.46 To graph these functions, we use log1.5 x 

ln x ln x . and log50 x  ln 50 ln1.5

These graphs all approach –∞ as x → 0+, and they all pass through the point (1, 0). They are all increasing, and all approach ∞ as x→ ∞. The functions with larger bases do so somewhat more quickly than those with smaller bases. The functions with larger bases also approach the y-axis more closely as x → 0+.

1.5.47 The plot shows that the graph of ln x is the reflection of the graph of ex about the line y = x, and the graph of log10x is the reflection of the graph of 10x about the same line. The graph of 10x increases more quickly than that of ex. In addition, the graph of log10x approaches ∞ as x → ∞ more slowly than the graph of ln x.

1.5.48 ln(x  h)  ln x h

 xh  xh  ln   ln  h  h   h  1

1/h 1/h

 ln1 x   h 

1.5.49 3 ft = 36 in, so we need to find x so that log2 x  36  x  236  68,719476,736. In miles, this is

68,719476,736 in  1 ft  1 mi  1,084,587.7 mi. 12 in 5280 ft 1.5.50 The graphs overlap for x ≥ 1, and both have a vertical asymptote at y = 0. However, y = |ln x| is symmetric about the y-axis whereas y = |ln x| is not, and y = |ln x| approaches –∞ as x → 0 but y = |ln x|→ ∞ as x → 0.

1.5.51

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

From the graphs we see that f (x)  x0.1  g(x)  ln x for approximately 0  x  3.06, and then

g(x)  f (x) for roughly 3.06  x  3.431015. At that point, the graph of f finally surpasses the graph of g for good.

1.5.52 Shift the graph of y  log10 x five units to the left to obtain the graph of y  log10 (x  5). Note the vertical asymptote of x = –5.

1.5.53 Reflect the graph of y  ln x about the x-axis to obtain the graph of y  ln x.

1.5.54 Reflect the graph of y = ln x about the y-axis to obtain the graph of

y  ln(x).

1.5.55

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Reflect the portion of the graph y  ln x to the right of the y-axis about the y-axis. The graph of y  ln | x | is that reflection in addition to the original portion.



1.5.56

 1

Start with a graph of y  ln x and reflect across the x-axis to obtain the graph of y ln . Then

reflect the portion of this graph in Quadrant I acrros the y-axis and combine with that portion to obtain the graph of y  ln

1 . | x|

Finally, shift the graph 2 units to the left.

1.5.57 (a) The domain of f (x)  ln x 1 is x > 0 and the range is . 1

(b) y  0  ln x 1  0  ln x  1  x  e (c) Shift the graph of y  ln x up one unit.

1.5.58 (a) The domain of f (x)  ln(x 1) 1is x > 1 and the range is . (b) y  0  ln(x 1)  1  x 1  e1  x  e 1 (c) Shift the graph of y  ln x one unit to the right and down one unit. 1.5.59



e74 x  6  7  4x  ln 6  7  ln 6  4x  x  14 7  ln 6 1.5.60



ln 3x 10  2  3x 10  e2  3x  e2 10  x  31 e2 10



1.5.61



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check 2 3 2 3 3 ln  x 2 1  3  x 1  e  x  e 1  x   e 1

1.5.62







e2x  3ex  2  0  e x 1 ex  2  0  ex  1 or ex  2  x  ln1 or x  ln 2, so x = 0 or x = ln 2.

1.5.63 x5



 3  x  5  log2 3  x  log2 3  5 or



   ln 3  (x  5)ln 2  ln 3  x  5  ln 3  x  5  ln 3 x5

2x5  3  ln 2

ln 2



1.5.64



ln 2



ln x  ln(x 1)  1  ln(x  (x 1))  1  x(x 1)  e1  x2  x  e  0. The quadratic

  since the natural log is not defined for x < 0. So x  1 1 4e .

formula (with a = 1, b = –1, and c = –e) gives x  12 1 1 4e , but we reject the negative root 1



1.5.65



ln x  ln(x 1)  1  ln(ln x)  1  eln(ln x)  e  ln x  e  eln x  ee  x  ee.



1.5.66

e  Ce ax

 ln  e

  ln C e   ax  ln C  ln ebx  ax  ln C  bx  ax  bx  ln C

 (a  b)x  ln C  x 



ln C a b

1.5.67

log(x  5)  ln 6  0  log(x  5)  ln 6  x  5  10ln6  x  5 10ln6

1.5.68



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

log4 (x  6)  log4 (x  3)  1  log4  (x  6)(x  3)  1  (x  6)(x  3)  41    3  97   3  97  2 x  0  x2  3x 18  4  x  3x  22  0  x   2       2        x 

3  97 . 2

1.5.69

2ln x   ln x  1  2ln x ln x 1  0  (2ln x 1)(ln x 1)  0  ln x   1

x  e1/2 or x  e

1 2

or ln x 1 



1.5.70

ln(cos x)  1  cos x  e, but since | cos(x) | 1  e  2.71828, this inequality is true for all given x.



1.5.71

| ln x | 1  1  ln x  1  e1  x  e1 or x   1e, e



1.5.72

ln x  0  x  e0  x  1. Since the domain of f (x)  ln x is x > 0, the solution of the original inequality is 0 < x < 1.



1.5.73

e  5  ln ex  ln 5  x  ln 5 x

1.5.74

1 e

3x1



 2  ln  3x 1  ln 2  0  3x 1  ln 2  1  3x  1 ln 2  13  x  13 1 ln 2



1.5.75



1 2ln x  3   2ln x  2  ln x  1  x  e1



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



1.5.76







(a) We need ex  3  0  ex  3  x  ln 3. Thus the domain of f (x)  ln e x  3 is ln 3,.













(b)(b) y  ln e x  3  e y  ex  3  ex  e y  3  x  ln e y  3 , so f 1(x)  ln e x  3 . Now e  3  0  e  3, which is true for any real x, so the domain of f x

1

1.5.77

   300

(a) eln300  300; ln e300

ln300

(b) A calculator indicates that e

 300 but gives an error message for ln e300  because e300 is

larger than most calculators can evaluate. 1.5.78 The graph of the function y  f (x)  x3  x2  x 1 is increasing, so f is one-to-one. Maple gives two complex expressions as well as

y  f 1 (x)  1 M

2/3

 8  2M 1/3 , where 2M 1/3

M  108x2 12 48 120x2  81x4  80. 1.5.79 (a) Here is the graph of g(x)  x6  x4, x  0. (b) If we use Maple to solve x  y  y for y, we obtain two real solutions: 6

1/3 2/3 2C1/3  4 , where C  108x 12 3 x(27x  4), and  6 C  C  1/3 6 C

the inverse for y  x6  x4 (x  0) is the positive solution, whose domain is

 274 , .



1.5.80



  (a) n  f (t)  100  2t/3  n  2t /3  log 2  n   t  t  3log 2  n . We could 100 3 100  100     1

rewrite this as t  f (n)  3

ln  n /100  . This function tells us how long it will take to obtain n ln 2

bacteria.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b) n  50,000  t  f 1(50,000)  3

  ln 500   26.9 hours ln 50000 100   3 

ln 2 

1.5.81



(a) Q  Q0 1 et /a

   ln 2 



Q t ln  1 Q   Q  1 et /a  et/a  1   Q Q   a   Q  0 0  0 

t  a ln 1 Q / Q0 . This gives us the time t necessary to obtain a given charge. (b) Q  0.9Q0 and a = 2 ⇒ t  2ln1 0.9Q0 / Q0   2ln 0.1  4.6 seconds. 1.5.82 (a) To find the equation of the graph that results from shifting the graph of y  lnx3 units upward, we add 3 to the original function to get y  lnx  3. (b) To find the equation of the graph that results from shifting the graph of y  lnx3 units to the left, we replace x with x  3 in the original function to get y  ln  x  3 . (c) To find the equation of the graph that results from reflecting the graph of y  lnx about the x -axis, we multiply the original equation by 1 to get y  lnx . (d) To find the equation of the graph that results from reflecting the graph of y  lnx about the y -axis, we replace x with x in the original equation to get y  ln  x  . (e) To find the equation of the graph that results from reflecting the graph of y  lnx about the line

y  x , we interchange x and y in the original equation to get x  lny  y  ex . (f) To find the equation of the graph that results from reflecting the graph of y  lnx about the x -axis and then about the line y  x , we first multiply the original equation by 1 [to get y  lnx ] and then interchange x and y in this equation to get x  lny  lny  x  y  ex . (g) To find the equation of the graph that results from reflecting the graph of y  lnx about the y -axis and then about the line y  x , we first replace x with x in the original equation [to get y  ln  x  ] and then interchange x and y to get x  ln  y    y  ex  y  e . x

(h) To find the equation of the graph that results from shifting the graph of y  lnx3 units to the left and then reflecting it about the line y  x , we first replace x with x  3 in the original equation [to get

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

y  ln  x  3 ] and then interchange x and y in this equation to get x  ln  y  3  y  3  ex  y  ex  3 . 1.5.83 (a) If the point (x, y) is on the graph of y  f (x), then the point (x  c, y) is that point shifted c units to the left. Since f is one-to-one, the point ( y, x) is on the graph of y  f

1

(x) and the point

corresponding to (x  c, y) on the graph of f is ( y, x  c) on the graph of f 1. Thus, the curve’s reflection is shifted down the same number of units as the curve itself is shifted to the left. So an 1

expression for the inverse function is g (x)  f

1

(x)  c.

(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line y = x is compressed (or stretched) vertically by the same factor. Using this geometric principle, we see that 1 1 the inverse of h(x)  f (cx) can be expressed as h (x)  1/ c f (x).

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION 1.6 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

End of Section Exercise Solutions

END OF SECTION EXERCISE SOLUTIONS 1.6.1

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x  t 2  t, y  t 2  t,  2  t  2

2

1

1.6.2

x  t 2 , y  t3  4t,  3  t  3

3

t x

2 4

15

1

1.6.3

x  cos2t, y  1sint, 0  t   / 2

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 /6

 /3

3/ 4

1/ 4

3  0.13 2

1/ 2

1

 /2

1.6.4

x  et  t, y  et  t,  2  t  2

2

1

e2  2

e 1

e1 1

e2  2

5.39

1.72

e2  2

e1 1

2.14

1.37

2.14

e 1

e2  2

1.72

5.39

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.6.5 (a) x  2t 1, y  21 t 1 t

–4

–2

–9

–5

–1

(b) x  2t 1  2t  x 1  t  12 x  12 , so

y  1 t 1  1  1 x  1  1  1 x  1 1  y  1 x  5 2

1.6.6 (a) x  3t  2, y  2t  3 t

–4

–2

–10

–4

–5

–1

(b) x  3t  2 3t  x  2  t  13 x  213 , so

y  2t  3  2  1 x  2   3  2 x   4  3  y  2 x  5 3

1.6.7

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a) x  t2  3, y  t  2,  3  t  3 t

–3

–1

–2

–1

(b) y  t  2  t  y  2, so

x  t 2  3  ( y  2)3  y2  4 y  4  3  x  y2  4 y 1, 1  y  5 1.6.8

(a) y  sin t, y  1cos t, 0  t  2 t

/2

3 / 2

2

–1

(b) y  sin t, y  1 cos t [ or y 1  cos t] 

x 2  ( y 1)2  (sin t)2  (cos t) 2  x 2   y 1  1. As t varies from 0 to 2 , the circle with 2

center 0,1 and radius 1 is traced out.



1.6.9

(a) x  t , y  1 t

1.414 1.732

–1

–3

–2

(b) x  t  t  x2  y  1 t  1 x2. Since t  0, x  0. So the curve is the right half of the parabola y  1 x2 . 1.6.10 © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a) y  t2 , y  t3 t

–2

–1

–8

–1

(b) y  t 3  t  3 y  x  t 2 

 y  y 2

2/3

t  , y  , x  0. 1.6.11 (a) x  sin 12  , y  cos 12  ,       .

x2  y2  sin2 12   cos2 21  1. For     0, we have 1  x  0 and 0  y  1. For

0     , we have 0  x  1 and 1  y  0. The graph is a semicircle. (b)

1.6.12 (a) x  12 cos , y  2sin , 0     .

 12 x    12 y   cos2   sin 2   1  4x 2  14 y 2  1  2



y2 x2  1, which is an equation of an ellipse with x-intercepts  12 and y-intercepts 2. For (1/ 2)2 22 0     / 2, we have 1  x  0 and 0  y  2. For  / 2     , we have 0  x   1 and 2

2  y  0. So the graph is the top half of an ellipse.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b)

1.6.13 (a) x  sin t, y  csct, 0  t   . y  csct  2

1  . sin t x

For 0  t  2 , we have 0  x  1 and y  1. Thus, the curve is the portion of the hyperbola y  1/ x with y  1. (b)

1.6.14

   x  x for x  0 since x  e .

(a) y  e2t  et

2

(b)

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.6.15

   e2 y

(a) y  ln t  t  e y , so x  t 2  e y



1.6.16



(a) x  t 1  x2  t 1  t  x2 1.

y  t 1  (x 2 1) 1 

x2  2. The curve is the part of the hyperbola x2  y2  2, with

x  2 and y  0. (b)

1.6.17 (a) x  tan2  , y  sec ,   / 2     / 2

1 tan  sec2   1 x  y2  x  y2 1. x  y2 1. For  / 2    0, we have

x  0 and y  1. For 0     / 2, we have 0  x and 1  y. Thus, the curve is the portion of the 

parabola in the first quadrant. As increases from the point  x, y  approaches0,1 along the parabola. As increases from the point  x, y  retreats from0,1 along the parabola. (b)

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.6.18

x  3  2cost, y  1 2sint, / 2  t  3 / 2 . By Example 4 with r  2, h  3 , and k 1 , the motion  3 of the particle takes place on a circle centered at 3,1 with a radius of 2. As t goes from to , the 2 2 particle starts at the point 3, 3 and moves counterclockwise to 3, 1 [one-half of a circle]. 1.6.19



 x 2  x  2sint, y  4  cost  sint  , cost  y  4. sin t  cos t  1     ( y  4)2  1 . The motion 2  2  3 of the particle takes place on an ellipse centered at 0, 4 . As t goes from 0 to , the particle starts at 2 the point 0, 5 and moves clockwise to 2, 4 [three-quarters of an ellipse]. x

1.6.20

x  5sin t, y  2cos t  sin x5 , cos t  y2 . sin 2 t  cos2 t  1   x5    y2   1. The motion of the 2

particle takes place on an ellipse centered at 0, 0. As t goes from  to 5 , the particle starts at the point (3, 0, 2 1) and moves clockwise around the ellipse three times.

1.6.21

y  cos2 t  1 sin2 t  1 x2. The motion of the particle takes place on the parabola y  1 x2 . As

t goes from 2 to   , the particle starts at the point 0,1 moves to 1, 0 and goes back to 0,1. As t goes from  to 0, the particle moves to 1, 0 and goes back to 0,1. The particle repeats this motion as t goes from 0 to 2 . 1.6.22

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a) From the first graph, we have 1  x  2 . From the second graph, we have 1  y  1 . The only choice that satisfies either of those conditions is III. (b) From the first graph, the values of x cycle through the values from 2 to 2 four times. From the second graph, the values of y cycle through the values from 2 to 2 six times. Choice I satisfies these conditions. (c) From the first graph, the values of x cycle through the values from 2 to 2 three times. From the second graph, we have 0  y  2 . Choice IV satisfies these conditions. (d) From the first graph, the values of x cycle through the values from 2 to 2 two times. The values of x then cycle through the values from approximately 1.7 to approximately 2.1 two times. From the second graph, the values of y cycle through the values from 2 to 2 two times. Choice II satisfies these conditions. 1.6.23 When t  1,  x, y   1,1. As t increases to 0, x and y both decrease to 0. As t increases from 0 to 1, x increases from 0 to 1 and y decreases from 0 to –1. At t increases beyond 1, x continues to increase and y continues to decrease. For t  1, x and y are both positive and decreasing. We could achieve greater accuracy by estimating x- and y-values from selected values of t from the given graphs and plotting corresponding points.

1.6.24 When t  1,  x, y   0, 0. As t increases to 0, x increases from 0 to 1, while y first decreases to –1 and then increases to 0. As t increases from 0 to 1, x increases from 0 to 1 and y decreases from 0 to –1. At t increases beyond 1, x decreases from 1 to 0 while y first increases to 1 and then decreases to 0. We could achieve greater accuracy by estimating x- and y-values from selected values of t from the given graphs and plotting corresponding points.

1.6.25 When t  1,  x, y   0,1. As t increases to 0, x increases from 0 to 1, and y decreases from 1 and to 0. As t increases from 0 to 1, the curve is retraced in the opposite direction with x decreasing from 1 to 0 and y increasing from 0 to 1. We

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

could achieve greater accuracy by estimating x- and y-values from selected values of t from the given graphs and plotting corresponding points.

1.6.26 (a) x  t 4  t 1  (t 4 1)  t  0 [think of the graphs of and y  t] and y  t 2  0, so these equations are matched with graph V. (b) y  t  0. x  t 2  2t  t(t  2) is negative for 0  t  2, so these equations are matched with graph I. (c) x  sin 2t has period 2 / 2  .Note that y(t  2 )  sin[t  2  sin 2(t  2 )]

 sin(t  2  sin 2t)  sin(t  sin 2t)  y(t) so y has period 2 . These equations match graph II since x cycles through the values –1 to 1 twice as y cycles through those values once. (d) x  cos5t has period a 2 / 5 and y  sin 2t has period  , so x will take on values –1 to 1, and then 1 to –1, before y takes on the values –1 to 1. Note that when t  0,  x, y   1, 0. These equations are matched with graph VI. (e) x  t  sin 4t, y  t 2  cos 3t. As t becomes large, t and t2 become the dominant terms in the expressions for x and y, so the graph will look like the graph of y  x2 , but with oscillations. These equations are matched with graph IV. (f) x  sin 2t , y  cos 2t . As t  , x and y both approach 0. These equations are matched

4  t2

with graph III. 1.6.27

x  t 1, y  1 2t2 

dx dt

 1,

 4t 

dy dx



dy / dt dx / dt



4t

 4t. Critical points occur

d  dy  4 d  dy  dt dx        4  0. Because the  0   4t  0  t  0. when 2 dx dx dx dx  dx  1 dt dy

d2y

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

curve is concave down when t  0, the point when t  0 is a maximum on the curve. At this point,

x  0 1  1 and y  1 2(0)  1, so the x- and y-coordinates of this point are (A), 1,1. 1.6.28  /3

2 2   dx   dy  Looking ahead to the next section ,we know L    dt    dt  dt. In this case, we have 0     dx dy  2cos 2t. This means  dx 2  dy 2 x  cos 2t  dt  2sin 2t and y  sin 2t  dt      dt dt    





  2sin 2t    2cos 2t   4 sin 2 2t  cos 2 2t  4, so the length of the path is 2  /3  /3  /3 L  0 4 dt  0 2dt  2t 0  , option (B).   3 2

 

1.6.29

(a) The circle x2  ( y 1)2  4 has center 0,1 and radius 2, so it can be represented by

x  2cos t, y  1 2sin t, 0  t  2. This representation gives us the circle with a counterclockwise orientation starting at (a) To get a clockwise orientation, we could change the equations to x  2cos t, y  1 2sin t,

0  t  2. (b) To get three times around in the counterclockwise direction, we use the original equations

x  2cos t, y  1 2sin t, with the domain expanded to 0  t  6. (c) To start at 0, 3 using the original equations, we must have x1  0 that is, 2cos t  0. Hence,

t   . So we use x  2 cos t, y  1 2sin t,   t  3 . 2

Alternatively, if we want t to start at 0, we could change the equations of the curve. For example, we could use x  2sin t, y  1 2cos t, 0  t  . 1.6.30 (a) Let x2 / a2  sin2 t and y2 / b2  cos2 t to obtain x  a sin t and

y  b cos t with 0  t  2 as possible parametric equations for the ellipse x2 / a2  y2 / b2  1. (b) The equations are x  3sin t and y  b cos t for b {1, 2, 4,8}.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(c) As b increases, the ellipse stretches vertically. 1.6.31 Big circle: It is centered at (2, 2) with a radius of 2, so parametric equations are

x  2  2cos t, y  2  2sin t, 0  t  2. Small circles: They are centered at (1, 3) and (3, 3) with a radius of 0.1. The parametric equations are (left): x  1 0.1cos t, y  3  0.1sin t, 0  t  2 and (right): x  3  0.1cos t, y  3  0.1sin t, 0  t  2 /

Semicircle: Is the lower half of a circle centered at (2 2) with radius 1. The parametric equations are: x  2  cos t, y  2  sin t,   t  2 .

To get all four graphs on the same screen with a typical graphing calculator, we need to change the last t-interval to 0  t  2 in order to match the others. We can do this by changing t to 0.5t. This change gives us the upper half. There are several ways to get the lower half – one is to change the ―+‖ to a ―–‖ in the y-assignment, giving us x  2  cos(0.5t), y  2 sin(0.5t), 0  t  2. 1.6.32 If you are using a calculator or computer that can overlay graphs (using multiple t-intervals), the following is appropriate. Left side: x = 1 and y goes from 1.5 to 4, so use

x  1, y  t,

1.5  t  4

Right side: x = 10 and y goes from 1.5 to 4, so use

x  10, y  t,

1.5  t  4

Bottom: x goes from 1 to 10 and y = 1.5, so use

x  t, y  1.5,

1  t  10

Handle: It starts at 10, 4 and ends at 13, 7  , so use

x  10  t, y  4  t,

0t 3

Left wheel: It’s centered at 3,1, has a radius of 1, and appears to go about

 6

radians above the

horizontal, so use

x  3  cost, y  1 sin t,

5 6

 t  136

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Right wheel: Similar to the left wheel with center 8,1, so use

x  8  cos t, y  1 sin t, 56  t  136 If you are using technology that cannot overlay graphs (using one t-interval) the following is appropriate: We’ll start by picking the t-interval0, 2.5 since it easily matches the t-values for the two sides. We now need to find parametric equations for all graphs with 0  t  2.5. Left side: x = 1 and y goes from 1.5 to 4, so use

x  1, y  1.5  t,

0  t  2.5

Right side: x = 10 and y goes from 1.5 to 4, so use

x 10, y 1.5  t,

0  t  2.5

Bottom: x goes from 1 to 10 and y = 1.5, so use

x  1 3.6t, y 1.5,

0  t  2.5

To get the x-assignment, think of creating a linear function such that when t  0, x  1 and when

t  2.5, x  10. We can use the point-slope form of a line witht1, x1   0,1 and

t2, x2   2.5,10. x 1 

10 1 2.5  0

t  0  x  1 3.6t.

Handle: It starts a 10, 4 and ends at 13, 7  , so use

x  10 1.2t, y  4 1.2t,

0  t  2.5

t1, x1   0,10 andt2, x2   2.5,13 gives us x 10 

13 10 2.5  0

t  0  1 10 1.2t.

Left wheel: It’s centered at 3,1, has a radius of 1, and appears to go about

 6

radians above the

horizontal, so use

x  3  cos 815t  56 , y  1 sin 815t  56, 0  t  2.5

t ,   0, 5  andt ,    5 , 13  gives us  5  136  56 t  0    5  8 t. 1

0

Right wheel: Similar to the left wheel with center 8,1, so use

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x  8  cos 8 t  5 , y  1 sin 8 t  5 , 15

0  t  2.5

1.6.33 Big circle: It's centered at 2, 2 with a radius of 2 , so by Example 4, parametric equations are

x  2  2cost, y  2  2sint, 0  t  2 Small circles: They are centered at 1, 3 and 3, 3 with a radius of 0.1 . By Example 4, parametric equations are and

(left)

x  1 0.1cost,

y  3  0.1sint, 0  t  2

(right)

x  3  0.1cost,

y  3  0.1sint, 0  t  2

Semicircle: It's the lower half of a circle centered at 2, 2 with radius 1 . By Example 4, parametric equations are

x  2 1cost, y  2 1sint,   t  2 To get all four graphs on the same screen with a typical graphing calculator, we need to change the last t -

interval to 0, 2  in order to match the others. We can do this by changing t to 0.5t . This change gives us the upper half. There are several ways to get the lower half - one is to change the "+" to a "-" in the y assignment, giving us

x  2 1cos 0.5t  , y  2 1sin 0.5t , 0  t  2   

1.6.34 34 If you are using a calculator or computer that can overlay graphs (using multiple t -intervals), the following is appropriate. Left side: x 1 and y goes from 1.5 to 4 , so use

x  1, y  t, 1.5  t  4 Right side: x 10 and y goes from 1.5 to 4 , so use

x  10, y  t, 1.5  t  4 Bottom: x goes from 1 to 10 and y  1.5 , so use

x  t, y  1.5, 1  t  10

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Handle: It starts at 10, 4 and ends at 13, 7 , so use

x  10  t, y  4  t, 0  t  3 Left wheel: It's centered at 3,1 , has a radius of 1 , and appears to go about 30 above the horizontal, so use

x  3 1cost, y  11sint,

5

t 

13  6

Right wheel: Similar to the left wheel with center 8,1 , so use

x  8 1cost, y  11sint,

5

t 

13  6

If you are using a calculator or computer that cannot overlay graphs (using one t -interval), the following is appropriate. We'll start by picking the t -interval 0, 2.5 since it easily matches the t -values for the two sides. We now need to find parametric equations for all graphs with 0  t  2.5 . Left side: x 1 and y goes from 1.5 to 4 , so use

x  1, y  1.5  t, 0  t  2.5 Right side: x 10 and y goes from 1.5 to 4 , so use

x  10, y  1.5  t, 0  t  2.5 Bottom: x goes from 1 to 10 and y  1.5 , so use

x  1 3.6t, y  1.5, 0  t  2.5 To get the x -assignment, think of creating a linear function such that when t  0, x  1 and when

t  2.5 , x 10 . We can use the point-slope form of a line with t1, x1   0,1 and t2 , x2   2.5,10 . x 1 

10 1 2.5  0

t  0  x  1 3.6t .

Handle: It starts at 10, 4 and ends at 13, 7 , so use

x  10 1.2t, y  4 1.2t, 0  t  2.5

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

t , x   0,10 and t , x   2.5,13 gives us x 10  1

13 10

t  0  x  10 1.2t .

2.5  0

t , y   0, 4 and t , y   2.5, 7gives us y  4  7  4 t  0  y  4 1.2t . 1

2.5  0

Left wheel: It's centered at 3,1 , has a radius of 1 , and appears to go about 30 above the horizontal, so use

5  5   8  8 x  3 1cos  t  , y  11sin t , 0  t  2.5 15 6   15  6     13  5 5 5 13 5 t ,    0,  and t ,    ,  gives us    6  6 t  0    5  8 t. 1 1  6  2 2 2 6  6 15 5 6     0 2 Right wheel: Similar to the left wheel with center 8,1 , so use

5  5   8  8 x  8 1cos  t  , y  11sin t , 0  t  2.5 15 6   15  6     

1.6.35 The first two diagrams depict the case     32 , d  r. C has coordinatesr , r . Now Q (in the second diagram) has coordinates (r , r  d cos(   ))  (r , r  d cos ), so a typical point P of the trochoid has coordinates (r  d sin(   ), r  d cos ). That is, P has coordinates  x, y , where

x  r  d sin and y  r  d cos. When d  r, these equations agree with those of the cycloid.

1.6.36 It is apparent that x  OQ and y  QP  ST . From the diagram,

x  OQ  a cos and y  ST  b sin. Thus, the parametric equations are

x  a cos and y  b sin. To eliminate we rearrange: sin  y / b  © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

sin2   ( y / b)2 and cos  x / a cos2   (x / a)2. Adding the two equations: sin2   cos2   1  x2 / a2  y2 / b2. Thus, we have an ellipse.

1.6.37 A has coordinates  a cos , a sin . Since OA is perpendicular to AB, OAB is a right triangle and B has coordinates  a sec , 0. It follows that P has coordinates

 a sec , b sin . Thus, the parametric equations are x  a sec , y  b sin.

1.6.38

C  2a cot , 2a  , so the x-coordinate of P is 2a cot. Let B  0, 2a. Then OAB is a right angle and OBA   , so OA  2a sin and A  (2a sin ) cos , (2a sin ) sin  Thus, the y-coordinate of P is y  2a sin2 . 1.6.39 (a) Let  be the angle of inclination of segment OP. Then

OB 

2a . Let C  2a, 0. Then by use of right triangle OAC we see that cos

OA  2a cos. Now OP  AB  OB  OA 1 cos2  sin2   1   2a  2a   cos   2a  2a sin tan. So P has coordinates cos cos  cos  x  2a sin tan cos  2asin2  and y  2a sin tan sin  2a sin2  tan. x  2a sin2  , y  2a sin2  tan (b)

100



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.6.40 40 (a) C  2acot , 2a , so the x -coordinate of P is x  2acot . Let B  0, 2a . 

Then OAB is a right angle and OBA   , so OA  2asin and

A  2asin cos , 2asin sin  . Thus, the y -coordinate of P is y  2asin2 . (c) x  v0cos  t  t 

x v0cos

. 2

  1 x  g  x  g 2 y  v0sin  t  2 gt  y  v sin v cos 2  v cos   tan  x   2v cos  x 2 , 0 0  0   02 2  which is the equation of a parabola (quadratic in x ).

 



 

(a) If   30 and v0  500 m/s, then the equations become x  (500cos30)t  250 3t and

y  (500sin 30)t  12 (9.8)t2  250t  4.9t2.



t  250 4.9  51 s. Then x  250 3



250 4.9

y  0 when t  0 (when the gun is fired) and again when

 22, 092 m, so the bullet hits the ground about 22 km from the

gun. The formula for y is quadratic in t. To find the maximum y-value, we will complete the square:

y  4.9  t 2  250 t   4.9 t 2  250 t   125    125  4.99  t  125   125  125 4.9 4.9 4.9 4.9 4.9 4.9   4.9 2

with equality when t  125 sm so the maximum height attained is 1254.9  3189 m. 4.9 2

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(b)

1.6.41

x  t 2 , y  t3  ct. We use a graphing device to produce the graphs for various values of c with   t   . Note that all the members of the family are symmetric about the x-axis. For c  0, the graph does not cross itself, but for c  0 it has a cusp at 0, 0 and for the graph crosses itself at x  c, so the loop grows larger as c increases.

1.6.42

(a)

If   30 and v0  40 m/s, then the equations become x  40cos30t  20 3t and y  40sin 30t  12 9.8t2  20t  4.9t2. y  0 when t  0 (when the baseball is thrown) and





20 again when t  4.9  4.1s. Then x  20 3 4.920   141 m, so the baseball hits the ground

about 141 m from the outfielder. The formula for y is quadratic in t. To find the maximum y-value, we will complete the square:





2 2 2 2 2 y  4.9 t 2  20 t  4.9 t 2  20 t   10    10  4.9  t  10   10  10 4.9 4.9 4.9 4.9 4.9 4.9   4.9

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with equality when t  10 s, so the maximum height attained is 104.9  20.4 m. 4.9 2

As  0    90 increases up to 45, the baseball attains a greater height and a greater

(b)

range. As  increases past 45, the baseball attains a greater height, but its range decreases.

x v0cos

  1 x  g  x  g y  v0sin  t  2 gt2  y  v sin v cos 2  v cos   tan  x   2v cos  x 2 , 0 0  0   02 2  which is the equation of a parabola (quadratic in x ).



1.6.43 From the figure, we see that the curves roughly follow the line y  x, and they start having loops when a is between 1.4 and 1.6. The loops increase in size as a increases.

1.6.44 Consider the curves x  sin t  sin nt, y  cos t  cos nt, where n is a positive integer. For n  1, we get a circle of radius 2 centered at the origin. For n  1, we get a curve lying on or inside that circle that traces out n 1 loops as t ranges from 0 to 2. Note:

x2  y2  (sin t  sin nt)2  (cos t  cos nt)2

 sin2 t  2sin t sin nt  sin2 nt  cos2 t  2cos t cos nnt  cos2 nt  (sin2 t  cos2 t)  (sin2 nt  cos2 nt)  2(cos t cos nt  sin t sin nt)

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 11 2 cos(t  nt)  2  2 cos((1 n)t)  4  22, with equality for n  1. This shows that each curve lies on or inside the curve for n  1, which is a circle of radius 2 centered at the origin.

1.6.45 Note that all the Lissajous figures are symmetric about the x-axis. The parameters a and b simply stretch the graph in the x- and y-directions respectively. For a  b  n  1 the graph is a circle with radius 1. For n  2 the graph crosses itself at the origin and there are loops above and below the xaxis.

In general, the figures have n 1 points of intersection, all of which are on the y-axis, and a total of n closed loops. 1.6.46

x  cos t, y  sin t sin ct. If c  1, then y  0, and the curve is simply the line segment from 1, 0 to

1, 0. The graphs are shown for c  2,3, 4 and 5.

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1.6.47 Note that all the Lissajous figures are symmetric about the x -axis. The parameters a and b simply stretch the graph in the x- and y -directions respectively. For a  b  n  1 the graph is a circle with radius 1 . For n  2 the graph crosses itself at the origin and there are loops above and below the x axis. In general, the figures have n 1 points of intersection, all of which are on the y -axis, and a total of n closed loops.

1.6.48

x  cost, y  sint sinct . If c  1, then y  0 , and the curve is simply the line segment from 1, 0 to

1, 0 . The graphs are shown for c  2,3, 4 and 5 .

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Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION CONCEPT CHECK TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1. CV.1

(a) A function f is a rule that assigns to each element x in a set D (the domain) exactly one element, called f (x), in a set E. The set of inputs for the function, D, is the domain. The range of f is the set of all possible values of f (x) as x varies throughout the domain. (b) To obtain the graph of the function f, plot the ordered pairs of points,

 x, f (x).

(c) A curve is the graph of a function if it passes the Vertical Line Test – that is, if no vertical line intersects the curve more than once. 1. CV.2

There are four ways to represent a function:  Verbally: C(w) is the cost of mailing a large envelope of weight w.

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 Numerically (with a table of values):

 Visually (with a graph):  Algebraically: y  f (x)  x2 1 1. CV.3

(a) f is an even function if for every x in its domain, f (x)  f (x). The graph of an even function is symmetric with respect to the y-axis. Examples of even functions include y  x 2 , y  x 2 , and y  cos x. (b) f is an odd function if for every x in its domain, f (x)   f (x). The graph of an even function is symmetric about the origin. Examples of even functions include y  x , y  x  x, and y  sin x. 3

1.CV.4 A function f is increasing on an interval I if f  x1   f  x2  whenever x1  x2 in I. 1.CV.5 A mathematical model is a mathematical description of a real-world phenomenon. The purpose of the model is to understand the phenomenon and perhaps to make predictions about future behavior. 1.CV.6 (a) y  2x  5 is a linear function. (b) y  x7 is a power function. (c) y  e is an exponential function. x

(d) y  x  3x 13 is a quadratic function. 2

(e) y  x5  3x4  7x 12 is a polynomial of degree 5. (f) y 

x 3

x7

is a rational function.

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1. CV.7

1.CV.8 (a) y  sin x

(b) y  tan x

(d) y  ln x

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(e) y  1/ x

(f) y  x

(g) y 

(h) y  tan x

1

CV.9 (a) The domain of f  g is A  B. (b) The domain of fg is A  B. (c) The domain of f / g is{x | x A  B and g(x)  0}. 1.CV.10 The composite of f and g is ( f g)(x)  f (g(x)) and its domain is the set of all x in the domain of g such that g(x) is in the domain of f. 1.CV.11 (a) If the graph of f is shifted 2 units upward, its equation becomes y  f  x  2 . (b) If the graph of f is shifted 2 units downward, its equation becomes y  f  x   2 . (c) If the graph of f is shifted 2 units to the right, its equation becomes y  f  x  2  . (d) If the graph of f is shifted 2 units to the left, its equation becomes y  f  x  2 .

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(e) If the graph of f is reflected about the x-axis, its equation becomes y   f  x. If the graph of f is reflected about the y-axis, its equation becomes y  f  x  . (f) If the graph of f is stretched vertically by a factor of 2 , its equation becomes y  2 f  x . (g) If the graph of f is shrunk vertically by a factor of 2 , its equation becomes y 

(h) If the graph of f is stretched horizontally by a factor of 2 , its equation becomes y  f

 

f  x.

2 1  x .  2   

(i) If the graph of f is shrunk horizontally by a factor of 2 , its equation becomes y  f  2x  . 1.CV.12

(a)

If f  x  x2 is shifted 1 unit upward, the function would become f  x  x2 1.

(b)

If f  x  x2 is shifted 1 unit to the left, the function would become f  x    x  1 .

(c)

If f  x  x2 is shifted 2 units downward and 1 unit to the right, the function would

become f  x    x 1  2  x 2  2x 1. 2

(d)

If f  x  x2 is shifted 2 units upward, then reflected across the x-axis, the function would become f  x    x 2  2  x2  2.

(e)

If f  x  x2 is shifted 1 unit downward, w units to the right  w  0, and then reflected across the y-axis, the function would become f  x    x  w 1  x 2  2wx  w2 1. 2

1.CV.13 (a) A function is one-to-one if it never takes on the same value twice; if no horizontal line intersects the graph more than once the function is one-to-one. (b) If f is a one-to-one function with domain A and range B, then its inverse function has domain B 1

and range A and is defined by f ( y)  x  f (x)  y for any y in B. The graph of f 1 is obtained by reflecting the graph of f about the line y = x. 1.CV.14 (a) For 1  x  1, sin1 x is an angle between –π/2 and π/2 whose sine is x. The inverse sine function has domain [–1, 1] and range [–π/2, π/2].

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(b) For 1  x  1, cos1 x is an angle between 0 and π whose cosine is x. The inverse cosine function has domain [–1, 1] and range [0, π]. (c) tan1 x is an angle between –π/2 and π/2 whose tangent is x. The inverse cosine function has domain and range [–π/2, π/2].

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION PRINCIPLES OF PROBLEM SOLVING TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1.PofPS.1 By using the area formula for a triangle,

1 2

 4 y   h  a  , so a 

1 2

(base) (height), in two ways, we see that 2

. Since 4  y  h , y  h2 16 , and a 

4 h2 16 . h

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1.PofPS.2 Refer to Example 1, where we obtained h 

. The 100 came from 4 times the area of the

2P triangle. In this case, the area of the triangle is

h12  6h. Thus, h  P  4 6h 2Ph  P2  24h  2

2P 2

2Ph  24h  P  h  2P  24  P  h 

2P  24

PofPS.3



P2 100

 2x 1  2x 1   1 2x 

if x 

2 x  5 1 and x  5  x  5 if x   2

if x  5 if x  5

Therefore, we consider the three cases x  5, 5  x 

, and x 

If x  5 , we must have 1 2x   x  5  3  x  3 , which is false, since we are considering x  5 . If 5  x 

, we must have 1 2x   x  5  3  x  

2 If x 

, we must have 2x 1  x  5  3  x  9 .

2 So the two solutions of the equation are x  

7 3

and x  9 .

1.PofPS.4

x 1 x 1   1  x

if x  1 and if x  1

x  3 x  3   3  x

if x  3 if x  3

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Therefore, we consider the three cases x  1,1  x  3 , and x  3 . If x  1, we must have 1 x  3  x  5  0  7 , which is false. If 1  x  3 , we must have x 1  3  x  5  x 

9 2

, which is false because x  3 .

If x  3 , we must have x 1  x  3  5  2  5 , which is false. All three cases lead to falsehoods, so the inequality has no solution.

1. PofPS.5

f  x   x2  4 x 3 . If x  0 , then f  x  x2  4x  3   x 1 x  3 . Case (i): If 0  x  1, then f  x  x2  4x  3 .





Case (ii): If 1  x  3 , then f  x   x 2  4x  3  x 2  4x  3 Case (iii): If x  3 , then f  x  x  4x  3 . 2

This enables us to sketch the graph for x  0 . Then we use the fact that f is an even function to reflect this part of the graph about the y -axis to obtain the entire graph. Or, we could consider also the cases x  3, 3  x  1 , and 1  x  0 .

1. PofPS.6

g  x  x2 1  x2  4 .

 x2 1 x2 1   2 1 x 

 x2  4 and x2  4   2 if x  1  4  x if x  1



if x  2 if x  2



So for 0  x  1, g  x  1 x2  4  x2  3 , for





1  x  2, g  x  x2 1 4  x2  2x2  5 , and for

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



x  2, g  x  x2 1 x 2  4  3.

1.PofPS.7 Remember that a  a if a  0 and that a  a if a  0 . Thus,

x  x  2x  0

if x  0 and y  y 2 y  if x  0 0

if y  0 if y  0

We will consider the equation x  x  y  y in four cases.

(1) x  0, y  0 (2) x  0, y  0 (3) x  0, y  0 (4) x  0, y  0 2x  2 y 2x  0 0  2y 00 x  y x  00  y Case 1 gives us the line y  x with nonnegative x and y . Case 2 gives us the portion of the y -axis with y negative. Case 3 gives us the portion of the x -axis with x negative. Case 4 gives us the entire third quadrant.

1. PofPS.8

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







x4  4x2  x2 y2  4 y2  0  x2 x 2  4  y2 x 2  4  0  

 x 2  y2  x 2  4  0   x  y  x  y  x  2 x  2  0. So the graph of the equation consists of the graphs of the four lines y  x , y  x, x  2 , and x  2 .

1.PofPS.9

x  y  1 . The boundary of the region has equation x  y  1 . In quadrants I, II, III, and IV, this becomes the lines x  y  1, x  y  1, x  y  1 , and x  y  1 respectively.

1. PofPS.10



x  y  x  y  2 Case (i): x  y  0  x  y  x  y  2  x  y  1  y  x 1 Case (ii): y  x  0  y  x  x  y  2  0  2 (true) Case (iii): x  0 and y  0  x  y  x  y  2  2x  2  x  1

Case (iv): x 0 and y 0  y  x  x  y  2   2x  2  x  1 Case (v): y  x  0  x  y  x  y  2  0  2 (true) Case (vi): x  y  0  y  x  x  y  2  y  x  1  y  x 1

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Note: Instead of considering cases (iv), (v), and (vi), we could have noted that the region is unchanged if x and y are replaced by x and  y , so the region is symmetric about the origin. Therefore, we need only draw cases (i), (ii), and (iii), and rotate through 180 about the origin.



1.PofPS.11



5  ln3  ln4  ln5   ln32  ln32 ln2  5ln2  5  log 3132      ln2 ln2  ln2  ln3  ln4  ln31  ln2

log23log3 4log45

1.PofPS.12

 x  x2 1  2 (a) f  x   ln x  (x) 1  ln x  x 1   x  x2 1   











 x2  x2 1   ln   1 1    ln   ln      x  x2 1   x  x2 1   x  x2 1   









2 2  ln1 ln x  x 1  ln x  x 1   f  x 









e2x 1

(b) y  ln x  x2 1 . Interchanging x and y , we get x  ln y  y 2 1  e x  y 

 

e  y  y2 1  e  2 ye  y  y 1 e 1  2ye  y 

2ex

f

1

y2 1 

 x 



1. PofPS.13





ln x 2  2x  2  0  x2  2x  2  e0  1  x2  2x  3  0   x  3 x 1  0  x 1, 3 .

 

Since the argument must be positive, x  2x  2  0  x  1





  

3  x  1 3   0    x  ,1 3  1 3, . The intersection of these intervals is 1,1 3  1 3,3 .   2



 



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1.PofPS.14 Assume that log2 5 is rational. Then log2 5  m / n for natural numbers m and n . Changing to exponential form gives us 2m/n  5 and then raising both sides to the n th power gives 2m  5n . But 2m is even and 5n is odd. We have arrived at a contradiction, so we conclude that our hypothesis, that log2 5 is rational, is false. Thus, log2 5 is irrational.

1.PofPS.15 Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the first and second halves of the trip. For the first half of the trip we have t1  d / 30 and for the second half we have

t2  d / 60 . Thus, the average speed for the entire trip is total distance 2d  2d 60  120d  120d  40 . The average speed for the entire trip is     d d 60 2d  d total time 3d t1  t2  30 60 40mi / h . 



1. PofPS.16

Let f  sin, g  x , and h  x . Then the left-hand side of the equation is

 g  h  sin  x  x  sin2x  2sinxcosx ; and the right-hand side is

f g  f h  sinx  sinx  2sinx . The two sides are not equal, so the given statement is false. 1. PofPS.17

Let S n be the statement that 7n 1 is divisible by 6 . 

S1 is true because 71 1  6 is divisible by 6 .



Assume Sk is true, that is, 7k 1 is divisible by 6 . In other words, 7k 1  6m for some k1

positive integer m . Then 7

1  7k  7 1  6m 1 7 1  42m  6  6 7m 1 , which is

divisible by 6 , so Sk 1 is true. 

Therefore, by mathematical induction, 7n 1 is divisible by 6 for every positive integer n .

1. PofPS.18

Let S n be the statement that 1 3  5  

 2n 1  n2 .

S is true because 2 1 1  1  12 . 1  

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

 2k 1  k 2 . Then

Assume Sk is true, that is, 1 3  5 

  2k 1  2  k 1 1  1 3  5 

1 3  5 

  2k 1   2k 1  k 2   2k 1  (k 1) 2

which shows that Sk 1 is true.

 2n 1  n2 for every positive integer n .

Therefore, by mathematical induction, 1 3  5 



1.PofPS.19

f 0  x  x2 and fn1  x  f0  fn  x   for n  0,1, 2, f  x   f  f  x    f  x 2    x 2   x 4 , f  x   f  f  x    f  x 4    x 4   x8 , 2

f 3  x   f 0  f 2  x    f 0  x8    x8   x16 , Thus, a general formula is f n  x   x 2 . 2

n1

1.PofPS.20 (a) f0  x  1 / 2  x and fn1  f0

fn for n  0,1, 2,.

 f  x  f  1  1 2 x 2 x   2  2  x 1  3  2x , 1 0  1 2 x   2   2 x



f 2  x  f 0  2  x   3  2x  







3  2x 3  2x 1  2  x  2  3  2x   2  x  4  3x , 2    3  2x



f 3  x  f 0  3  2x   4  3x  4  3x 1  4  3x  3  2x  2  4  3x   3  2x  5  4x ,   2  4  3x 

Thus, we conjecture that the general formula is f n  x 



n 1 nx . n  2   n 1x

To prove this, we use the Principle of Mathematical Induction. We have already verified that fn is true for n 1 . Assume that the formula is true for n  k ; that is, f k  x 

k 1 kx k  2  k 1 x

. Then

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 x   f

f k 1

f  f  x   f  k 1 kx   fk  x  0 k 0   k  2   k 1 x  2  

k  2   k 1 x 2  k  2   k 1 x   k 1 kx 



1 k 1 kx k  2   k 1 x

k  2   k 1 x k  3  k  2 x

This shows that the formula for fn is true for n  k 1 . Therefore, by mathematical induction, the formula is true for all positive integers n . (b) From the graph, we can make several observations: 

The values at each fixed x  a keep increasing as n increases.



The vertical asymptote gets closer to x 1 as n increases.



The horizontal asymptote gets closer to y  1 as n increases.



The x -intercept for fn1 is the value of the vertical asymptote for fn .



The y -intercept for fn is the value of the horizontal asymptote for fn1 .

Solution and Answer Guide

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STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION REVIEW TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1.R.1 (a) When x  2, y  2.7, thus f (2)  2.7. (b) f (x)  3  x  2.3, 5.6

6, 6.

4, 4.

(e) f is increasing on

4, 4.

(f) f is not one-to-one because it fails the Horizontal Line Test. (g) f is odd because its graph is symmetric about the origin. 1.R.2 (a) When x  2, y  3. Thus g(2)  3. (b) g is one-to-one because it passes the Horizontal Line Test. 1 (c) When y  2, x  0.2, so g (2)  0.2.

(d) The range of g is1,3.5, which is the same as the domain of g 1. (e) We reflect the graph of g through the line y = x to obtain the graph of g 1. 1.R.3

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f (a  h)  f (a)



a2  2ah  h2  2a  2h  3   a2  2a  3 h 2a  h  2 

h 1.R.4

 

f (x) 

2 3x 1

Domain: {x | x  13} Range: {y | y  0}

1.R.5 Domain: 2, 2

g(x)  16  x4



Range: 0, 4



1.R.6

h(x)  ln  x  6 Domain: 6, 



 2a  h  2

Range: 1.R.7

F (t)  3  cos 2t

Domain: Range: 2, 4

1.R.8

f (x) 



3 x2



Domain:{x | x  2} Range: {y | y  0}

1.R.9

f (x)  3x  2

Domain: Range: {y | y  0}

1.R.10

f (x) 

sin x x

1.R.11

Domain: {x | x  0} 2 Range:  3 ,1



f (x)  tan x 1 Domain: {x | x  (k 1)   2}for k  Range:

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1.R.12

(a)

To obtain the graph of y  f  x  8, we shift the graph of y  f  x  down 8 units.

(b)

To obtain the graph of y  f  x  8, we shift the graph of y  f  x  right 8 units.

(c) To obtain the graph of y  1 2 f  x  , we stretch the graph of y  f  x  vertically by a factor of 2 , and then shift the resulting graph 1 unit upward. (d) To obtain the graph of y  f  x  2  2 , we shift the graph of y  f  x  right 2 units (for the "-2" inside the parentheses), and then shift the resulting graph 2 units downward. (e) To obtain the graph of y   f  x  , we reflect the graph of y  f  x  about the x -axis. (f) To obtain the graph of y  f

1

 x , we reflect the graph of y  f  x about the line y  x

(assuming f is one-to-one). 1.R.13 (a) To obtain the graph of y  f (x  8),

(b) To obtain the graph of y  21 f (x) 1, we

we shift the graph of y  f (x) right 8 units.

reflect the graph of y  f (x) about the x-axis.

(d) To obtain the graph of y  21 f (x) 1, we shrink

reflect the graph of y  f (x) about the x-axis, and then shift the resulting graph up 2 units.

the graph of y  f (x) by a factor of 2, and then shift the resulting graph down 1 unit.

(e) To obtain the graph of y  f 1(x), reflect

(f) To obtain the graph of y  f 1(x  3), we

the graph of y  f (x) about the line y = x.

reflect the graph of y  f (x) about the line y = x, and then shift the resulting graph left 3 units.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.R.14 Start with the graph of y  x3 and shift 2 units to the right.

1.R.15 Start with the graph of y 

x and stretch vertically by a factor of 2.

1.R.16





First note that y  x 2  2x  2  x 2  2x 1 1   x 1 1. Then start with the graph of y  x 2 , 2

shift 1 unit to the right, and shift 1 unit up.

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1.R.17 Start with the graph of y  ln x and shift left 1 unit.

1.R.18 Start with the graph of y  cos x, shrink horizontally by a factor of 2, and reflect about the x-axis.

1.R.19

f (x) 

 x

if x  0 . On (–∞, 0), graph y  x (the line with  x e 1 if x  0

slope –1 and y-intercept 0) with an open endpoint (0,0). On [0, ∞), graph y  ex 1 (shift the graph of y  ex down 1 unit) with closed endpoint (0, 0). 1.R.20 (a) The terms of f are a mixture of odd and even powers of x, so f is neither even nor odd. (b) The terms of f are all odd powers of x, so f is odd. 2

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(d) f (x)  1 sin(x)  1sin x. Because f (x)  f (x) and f (x)   f (x), f is neither even nor odd. 1.R.21

2a2b3a  a a b a  3 2 2 2  3a2 2ab a 2 x ln x y  (b) eln xln y  e y (a)



 a2b3  (c) 2 ln a  3ln b  4 ln c  ln a  ln b  ln c  ln  c4    2

1.R.22 (a) is a function

(b) not a function

(d) is a function

(e) is a function

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1.R.23 (b) 2a  2

(a) 2a  3

(d) 2a  2

1.R.24

02 For the line segment from (–2, 2) to (–1, 0), the slope is

1 2

 2, and an equation is y  0  2(x 1)

or equivalently, y  2x  2. The circle has equation x2  y2  1; the top half has equation y  1 x

 2x  2 if  2  x  1.

(we have solved for positive y). Thus f (x)  

if 1  x  1

 1 x2 

1.R.25









(a)  f

g  (x)  f (g(x))  f x 2  9  ln x 2  9 ; The domain is , 3 3, .

(b)  g

f  (x)  g( f (x))  g  ln x    ln x   9; The domain is 0,   .

f  (x)  f ( f (x))  f ln x   ln(ln x); The domain is 1,   .



 



(d)  g g  (x)  g(g(x))  g x 2  9  x 2  9  9  x 4 18x 2  72; The domain is 2

1.R.26 One possible combination is f (x) 

1 , g(x)  1 x, h(x)  x. Then f x

g h  (x) 

  x   f  1 x   11 x  F (x).

f (g(h(x)))  f g

1.R.27 (a) Let x denote the number of toaster ovens produced in one week and y the associated cost. Using the points 1000, 9000 and

1500,12, 000 , we get an equation of a line: y  9000 

12, 000  9000

 x 1000 

1500 1000 y  6  x 1000  9000  y  6x  3000.

126



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b)

The slope of the graph is 6. This means that each additional toaster oven produced adds $6 to the weekly production cost.

(c)

The y-intercept of the graph is 3000. This represents the overhead cost—the cost incurred without producing anything.

1.R.28 We need to know the value of x such that f  x  2x  lnx  2 . Since x 1 gives us

y  2, f 1  2   1. 1.R.29 We need to know the value of x such that f (x)  2x  ln x  2. Since x 1 gives us y  2, f 1.R.30



1

(2)  1.



y 1 x 1 . Interchanging x and y gives us x   2xy  x  y 1  2xy  y  1 x  2x 1 2 y 1 1 x y(2x 1)  1 x  y   f 1  x. 2x 1 y 

1.R.31 x (a) e  5  x  ln5

(b) lnx  2  x  e (c) e

(d)

 2  e x  ln2  x  ln  ln2 

 

e3x  2ex  ln e3x  ln 2ex  3x  ln 2  ln ex  3x  ln 2  x  2x  ln 2  x  12 ln 2

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ln x

(e)

 3 has no real solutions. Since ln x  x on the entire domain of ln x,

x ln x  3. x

 1. Thus,

ln  ln x   5  ln x  e5  x  ee

(f)

ln x

1.R.32 (a) After 4 days,

gram remains; after 8 days,

2 (b) m  4  

1 

, m8 

g ; after 12 days,

4 1 22



, m12 

g ; after 16 days,

, m16  1 . From the pattern, we see that m  t   1 , or 23 24 2t /4 



2t /4 . (c) m  2t/4  log 2m  t / 4  t  4log 2m .This is the time elapsed when there are m grams of 100

Pd . 31. (b) x  et  t  lnx; y  t so y  lnx . 0  t  1  0  y  1 and1  x  e

1.R.33 (a) After 4 days, ½ gram remains; after 8 days, ¼ g remains, after 12 days, ⅛ g remains; and after 16 days, 161 g remains.

(b) m(4)  , m(8) 

, m(12) 

, m(16) 

. From the pattern, we see that m(t)  2t/4.

23 24 2 22 (c) m(t)  2t/4  log 2m  t / 4  t  4log m; This is the time elapsed when there are m 2 grams of 100 Pd.  ln 0.01  (d) m  0.01  t  4log 20.01  4  ln 2  26.6 days.





1.R.34

(a)

f  x   6e x  6ex  f  x  , so f is an even function.

(b)

The average rate of change of f over 1, 3 is

f 3  f 1 6e9  6e1  9   3e 3e1  1.103 3 1 2

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 

1.R.35 2

(a) f (x)  6e(x)  6e(x)  f (x) so f is an even function

f (3)  f (1)



(b) The average rate of change of f over [1, 3] is

3 1





 

6e9  6e1  9  1     3e 6e 1.103 2 

(c)

1.R.36

As x increases without bound, y  ax has the largest y-values and y  loga x has the smallest yvalues. This makes sense because these functions are inverses of each other. 1.R.37 (a)

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

0  t  1  0  y  1 and1  x  e (b) x  et  t  lnx; y  t so y  lnx . 1.R.38 (a)

0  t  1  0  y  1 and1  x  e 1.R.39 2

(a) (x  2)  y  4 

(x  2)2



 1 . Let

(x  2)2

 sin t and

 cos 2t (since

4 4 4 4 sin2t  cos2t  1 ). Solving for x and y gives x  2  2sint and y  2cost . We want to move from

2, 2 to 2, 2 and pass through 0, 0 . When t  0 , we want y  2 , so choose y  2cost . When t



, we want x  0 , so choose x  2  2sint . Thus, parametric equations are

x  2  2sint, y  2cost, 0  t   . Another possibility is x  2  2cost, y  2sint,

 2

t 

3

130



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1.R.40 We sketch x  t, y  2t  lnt (the function) and x  2t  lnt, y  t (its inverse) for t  0 .



1.R.41

(a) Let  be the angle of inclination of segment OP . Then OB 

cos

Let C  2a, 0 . Then by use of right triangle OAC we see that OA  2acos . Now

OP 

 1



2a   cos   2a  cos 

 AB  OB  OA sin2   2a  2asin tan cos cos

1 cos2

So P has coordinates x  2asin tan cos  2asin2 and y  2asin tan sin  2asin2 tan .

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 1: SECTION TRUE/FALSE TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 1. TF.1

False

Let f (x)  x , s  1, and t  1. Then f (s  t)  (11)  0, but f (s)  f (t) 2

1.TF.2 Let f (x)  x . Then f (2)  4  f (2), but 2  2. 2

False 1.TF.3

Let f (x)  x . Then f (3x)   3x   9x 2 and 3 f (x)  3x . So f (3x)  3 f (x). 2

False

1.TF.4 True

If and f is a decreasing function, then the y-values get smaller as we move from left to right.

1.TF.5 Let f (x)  x , and g(x)  2x. Then f 2

False

g  (x)  f (g(x))  f (2x)  (2x)2  4x2

and  g f  (x)  g  f (x)   g x 2  2x2. So f g  g f .

 

1.TF.6

132

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check 1 3 Let f (x)  x . Then f is one-to-one and f (x)  x. But1/ f (x)  1/ x , which is not equal 3

False to f

1

(x).

1.TF.7 We can always divide by ex because ex  0 for every x.

True 1.TF.8

The function ln x is increasing on (0,).

True 1.TF.9

Let x  e. Then  ln x    ln e   1  1, but 6ln x  6ln e  61  6  1. It is true, however, 6

False

   6ln x for x  0.

that ln x 6 1.TF.10

Let x  e2 and a  e. Then ln x 

False

ln e2



2ln e

2 and ln

ln e a x the statement is false. What is true however, is that ln  ln x  ln a. a ln a

ln e

ln

 ln e  1, so in general

1.TF.11 For example, if x  3, then

False

 3  9  3, not –3. 2

1.TF.12

False Let f  x  x2 . Then the graph of y  f  x  x2 is a parabola opening downward, but the graph of y  f  x  x2 is a parabola opening upward.

Solution and Answer Guide

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.1 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.1.1 (a) msec 

f (b)  f (a) ba f (a 1)  f (a)

 f (a 1)  f (a) a 1 a (c) m  f (a  h)  f (a)  f (a  h)  f (a) sec aha h f (a  h)  f (a  h)  f (a  h)  f (a  h)  (d) msec  2h a  h   a  h (b) msec 



2.1.2



f (a  h)  f (a  h) k(a  h)  b(a  h)  k (a  h)  b(a  h)  (a) msec  2h 2h 2





k  a 2  2ah  h2   k  a 2  2ah  h2   2bh 2h

(b) msec 





4kah  2bh  2ka  b 2h

f (a  2h)  f (a  2h) k(a  2h)2  b(a  2h)   k(a  2h)2  b(a  2h)  2h 4h 

k  a 2  4ah  4h2   k  a 2  4ah  4h2   4bh 4h



8kah  4bh  2ka  b 4h

Q1   a  h, f  a  h   is parallel to the line containing P2  a  2h, f a  2h and

Q2   a  2h, f a  2h . Therefore, both lines have the same slope. 2.1.3

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

1. (a) y  f (x)  3x . Using P  (1,3) we construct the following table: 2

(1.1, 3.63)

3.633 0.1

 6.3

(1.01, 3.0603)

3.06033 0.01

 6.03

(1.001, 3.006003)

3.0060033 0.001

 6.003

slope = mPQi

Based on these values, the slope of the tangent line to the graph of f at P is 6. (b) y  f (x)  4x  3x 1. Using P  (1, 6) we construct the following table: 2

Based on these values, the slope of the tangent line to the graph of f at P is 11. (c) y  f (x)  x  x. Using P  (1, 0) we construct the following table: 3

Based on these values, the slope of the tangent line to the graph of f at P is 2.

2.1.4 (a) Using P  (15, 250) we construct the following table:

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(5, 694)

slope = mPQ 694250 515

  444 10  44.4

(10, 444)

444250 1015

  1945  38.8

(20, 111)

111250 2015

  1395  27.8

(25, 28)

28250 2515

  272 10  22.2

(30, 0)

0250 3015

  250 15  33.3

(b) Using the values of t that correspond to the points closest to P (t = 10 and t = 20), we have

33.8  (27.8)  33.3. 2 (c) From the graph, we can estimate the slope of the tangent line at P to be 33.3. (d) The slope of the tangent line indicates that the volume of the water in the tank is decreasing at a rate of 33.3 gallons per minute at time t = 15 minutes. 2.1.5

 4186  69.667 (a) Slope = 29482530 4236

 2874  71.75 (b) Slope = 29482661 4238

 1422  71 (c) Slope = 29482806 4240

 1322  66 (d) Slope = 30802948 4442

From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeats per minute after 42 minutes. After being stable for a while, the patient’s heart rate is dropping. 2.1.6 (a) y 

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

1 , P  (2, 1) 1 x x

Q  x,1/ (1 x)

mPQ

1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.001

(1.5, –2) (1.9, –1.111111) (1.99, –1.010101) (1.999, –1.001001) (2.5, –0.666667) (2.1, –0.909091) (2.01, –0.990099 (2.001, –0.999001)

2 1.111 111 1.010 101 1.001 001 0.666 667 0.909 091 0.990 099 0.999 001

(b) The slope appears to be 1.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

y  x  3. 2.1.7

y  ln x, P 1, 0

(a)   

(i) 0.5 (ii) 0.9 (iii) 0.99 (iv) 0.999 (v) 1.5 (vi) 1.1 (vii) 1.01 (viii) 1.001

mPQ

0.5, 0.693147 0.9, 0.105361 0.99, 0.010050 0.999, 0.001001 1.5, 0.405465 1.1, 0.095310 1.01, 0.009950 1.001, 0.001000

(b)

The slope appears to be 1.

(c)

y  0  1 x 1 or y  x 1

1.386294 1.053605 1.005034 1.000500 0.810930 0.953102 0.995033 0.999500

2.1.8 2. (a) y  cos x, P  (0.5,0) (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

Q  x,1/ (1 x)

mPQ

0 0.4 0.49 0.499 1 0.6 0.51 0.501

(0, 1) (0.4, –3.09017) (0.49, –3.141076) (0.499, –3.131587) (1, –1) (0.6, –3.090170) (0.51, –3.141076) (0.501, –3.131587)

–2 –3.090170 –3.141076 –3.131587 –2 –3.090170 –3.141076 –3.131587

(b) The slope appears to be –π.

(d)

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.1.9 (a) y  y(t)  40t 16t2. At t  2, y  40(2) 16(2)2 16. The average velocity between times 2 and 2 + h is

y(2  h)  y(2)   40(2  h) 16(2  h)  16 24h 16h2    24 16h, if h  0. vavg  h (2  h)  2 h 2

(i) 2, 2.5: h  0.5, vavg  32 ft/s

(ii) 2, 2.1: h  0.1, vavg  25.6 ft/s

(iii) 0: h  0.05, vavg  24.8 ft/s

(iv) 2, 2.01: h  0.01, vavg  24.16 ft/s

(b) The instantaneous velocity when t = 2 (h approaches 0) is –24 ft / s. 2.1.10 (a) y  y(t)  10t 1.86t . At t  1, y  10(1) 1.86(1)  8.14. The average velocity between times 1 and 1 + h is 2

10(1 h) 1.86(1 h) 2   8.14 

y(1 h)  y(1)

vavg 

(1 h) 1



(i) 1, 2: h  1, vavg  4.42 m/s

(iii) 1,1.1: h  0.1, vavg  6.094 m/s



6.28h 1.86h2 h

 6.28 1.86h, if h  0.

(ii) 1,1.5: h  0.5, vavg  5.35 m/s

(iv) 1,1.01: h  0.01, vavg  6.2614 m/s

(v) 1,1.001: h  0.001, vavg  6.27814 m/s (b) The instantaneous velocity when t = 1 (h approaches 0) is 6.28 m / s. 2.1.11

s(4)  s(2) 79.2  20.6   29.3 ft/s. 42 2 s(4)  s(3) 79.2  46.5   32.7 ft/s. (ii) On the interval [3, 4], vavg  43 1 (iii) On the interval [4, 5], v  s(5)  s(4)  124.8  79.2  45.6 ft/s. avg 54 1

3. (a) (i) On the interval [2, 4], vavg 

(iv) On the interval [4, 6],

vavg 

s(6)  s(4) 176.7  79.2   48.75 ft/s. 64 1

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b) Using the points (2, 16) and (5, 105) from the approximate tangent line, the instantaneous velocity at t = 3 is about

105 16 89   29.7ft/s. 52 3

2.1.12 (a) (i) s  s(t)  2sint  3cost.

s(2)  s(1) 3 (3)   6 cm/s. 2 1 1 s(1.1)  s(1) 3.471(3)   4.71 cm/s. (ii) On the interval [1, 1.1], vavg  1.11 0.1 (iii) On the interval [1, 1.01], v  s(1.01)  s(1)  3.0613 (3)  6.13 cm/s. avg 1.011 0.11 s(1.001)  s(1) 3.00627 (3)   6.27 cm/s. (iv) On the interval [1, 1.001], vavg  1.0011 0.001 On the interval [1, 2], vavg 

(b) The instantaneous velocity of the particle when t = 1 appears to be about – 6.3 cm/s.

2.1.13 (a)

h(2.1)  h(2) 369.4747  370.04475   5.70 ft/s. 2.1 2 0.1 h(2.01)  h(2) 370.0008194  370.04475   4.39292 ft/s. On the interval [2, 2.01], vavg  2.01 2 0.01 On the interval [2, 2.001], v  h(2.001)  h(2)  370.040489  370.04475  4.2596 ft/s. avg 2.001 2 0.001 (b) On the interval [2, 2.1], vavg 

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

h(2.0001)  h(2) 370.044324  370.04475   4.246 ft/s. 2.0001 2 0.0001 On the interval [2, 2.00001], v  h(2.00001)  h(2)  370.0445062  370.04475  4.24 ft/s. avg 2.000011 0.00001 On the interval [2, 2.0001], vavg 

2.1.14 (a) For the curve y  sin 10 / x and the point P  (1,0) : x

mPQ

2 1.5 1.4 1.3 1.2 1.1

(2, 0) (1.5, 0.86600 (1.4, –0.4339) (1.3, –0.8230) (1.2, 0.8660) (1.1, –0.2817)

0 1.7321 –1.0847 –2.7433 4.3301 –2.8173

0.5 0.6 0.7 0.8 0.9

(0.5, 0) (0.6, 0.8660)) (0.7, 0.7818) (0.8, 1) (0.9, –0.3420)

0 –2.1651 –2.6061 –5 3.4202

As x approaches 1, the slopes do not appear to be approaching any particular value. (b) We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at P that we need to take x-values much closer to 1 in order to get accurate estimates of its slope. (c) If we choose x = 1.001, then the point Q is (1.001, –0.0314) and mPQ  31.3794. If x = 0.999, then Q is (0.999, 0.0314) and

mPQ  31.4422. The average of these slopes is –31.4108. So we estimate that the slope of the tangent line at P is about –31.412.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.2 TABLE OF CONTENTS

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS

2.2.1 As x approaches 2, f(x) approaches 5. (Or, the values of f (x) can be made as close to 5 as we like by taking x sufficiently close to 2 (but not equal to 2). Yes, it is possible for lim f (x)  5 while f (2)  3 x2

if f has a hole at the point (2, 3). 2.2.2 As x approaches 1 from the left, f (x) approaches 3; and as x approaches 1 from the right, f (x) approaches 7. No, the limit does not exist because the left- and right-hand limits are different. 2.2.3 (a) As x approaches 2 from the left, the values of f (x) approach 3, so lim f (x)  3. x2

(b) As x approaches 2 from the right, the values of f (x) approach 1, so lim f (x)  1. x2



(d) When x = 2, y = 3, so f (2)  3. (e) As x approaches 4, the values of f (x) approach 4, so lim f (x)  4. x4

(f) f (4)  4

2.2.4 (a) As x approaches 1, the values of f (x) approach 1.7, so lim f (x)  1.7. x1

(b) As x approaches 3 from the left, the values of f (x) approach 1, so lim f (x)  1. x3

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

(d) lim f (x) does not exist since the left-hand limit does not equal the right-hand limit. x3

(e) When x = 3, y = 3, so f (3)  3.

2.2.5 (a) h(x) approaches 4 as x approaches –3 from the left, so lim h(x)  4. x3 

(b) h(x) approaches 4 as x approaches –3 from the right, so lim h(x)  4. x3 



(c)(c) lim h(x)  4 because the limits in part(a) and part (b) are equal. x3

(d) h(3) is not defined, so it doesn’t exist. (e) h(x) approaches 1 as x approaches 0 from the left, so lim h(x)  1. x0 



(f) h(x) approaches –1 as x approaches 0 from the right, so lim h(x)  1. x0 



(g) lim h(x) does not exist because the limits in part(a) and part(b) are not equal. x0

(h) h(0)  1since the point (0,1) is on the graph of h.

lim h(x)  2 lim h(x)  2, lim h(x)  2. (i) Since x2 we have x2 and x2 (j) h(2) is not defined, so it doesn’t exist. (k) h(x) approaches 3 as x approaches 5 from the right, so lim h(x)  3. x5 



(l) h(x) does not approach any one number as x approaches 5 from the left, so lim h(x) does not exist. x5 



 

2.2.6 (a) lim g(x)  1 x0 



(b) (b) lim g(x)  2 x0 

(d) (d) lim g(x)  2 x2 

(e) lim g(x)  0 x2 

(f) lim g(x) does not exist because the limits in part (d) and part (e) are not equal. x2

(g)(g) g(2)  1

(h) lim g(x)  3 x4

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.7 𝑙i𝑚 ƒ(𝑡) = 150 mg and 𝑙i𝑚 ƒ(𝑡) = 300 mg. These limits show that there is an abrupt change in the

𝑥→−12

amount of drug in the patient’s bloodstream at t = 12 h. The left-hand limit represents the amount of the drug just before the fourth injection. The right-hand limit represents the amount of the drug just after the fourth injection.

2.2.8 From the graph of

1 x, if x  1  2 f (x)  x , if 1  x  1, 2  x, if x  1 

(a) lim f (x)  0

(b) lim f (x)  1

x1 

(d) lim f (x)  1

(e) lim f (x)  1

x1

x1

(f) lim f (x)  1 x1

2.2.9

1 sin x, if x  0  f (x)  cos x, if 0  x   , sin x, if x    (a) lim f (x)  1

(b) lim f (x)  1

(d) lim f (x)  1

x0 

x0

(e) lim f (x)  0 x 

x0 

x 

(f) lim f (x)  0 does not exist because the limits in part (d) and part (e) are not x

equal. 

 2.2.10 (a) lim f (x)  1 x0

(b)(b) lim f (x)  0 x0 

part (b) are not equal.

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2.2.11

lim f (x)  1 (a) x0 lim f (x)  1 (b) x0 (c) lim f (x) does not exist because the limits in part (a) x0

2.2.12 (a) lim x0  f  x  2 (b) (c)(c)

lim x0  f  x  2 limx0 f  x does not exist because the limits

in part (a) and part (b) are not equal.

2.2.13

lim f (x)  2 lim f (x)  1

x3

x3



 2.2.14

(0,3) lies on the graph of f and lim f (x)  2 x0

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.15

lim f (x)   lim f (x)  

x0

x0 

       

2.2.16

lim f (x)  1 lim f (x)  2 f (0)  1

x0

x0

       

2.2.17

lim f (x)  1, lim f (x)  2, lim f (x)  2, f (0)  1, f (3)  1 x0

x3

x3

        

2.2.18

lim f (x)  4, lim f (x)  2, lim f (x)  2, f (3)  3, f (2)  1 x3

x3

x2

2.2.19

lim f (x)  2, lim f (x)  0, lim f (x)  3, f (0)  2, f (4)  1 x0

x0

x4

145



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.20 For f (x) 

x2  3x

x 3.1 3.05 3.01 3.001 3.001

 

x2  9

f (x) 0.508 197 0.504 132 0.599 832 0.500 083 0.4500 008

x 2.9 2.95 2.99 2.999 2.9999

f (x) 0.491 525 0.495 798 0.499 165 0.499 917 0.499 992

x –3.5 –3.1 –3.05 –3.01 –3.001 –3.0001

f (x) 7 31 61 301 3001 30,001

x –0.5 –0.1 –0.01 –0.001 –0.0001

f (x) 1.042 915 1.001 669 1.000 017 1.000 000 1.000 000

It appears that 2

x  3x

1  . x3 x2  9 2

lim



2.2.21

f (x) 

x2  3x x2  9

x –2.5 –2.9 –2.95 –2.99 –2.999 –2.9999

  

f (x) –5 –29 –59 –299 –2999 –29,999

x2  3x   and that 2 x3  x  9 x2  3x x2  3x lim lim 2 does not  so x3 x2  9 x3  x  9

It appears that lim 

exist.



2.2.22

x : sin x

For f (x)  x 0.5 0.1 0.01 0.001 0.0001

f (x) 1.042 915 1.001 669 1.000 017 1.000 000 1.000 000

It appears that lim x0

 1.

sin x

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.23 For f (x)  ln x : x 1.5 1.1 1.01 1.001 1.0001

f (x) 0.405 465 0.095 310 0.009 950 0.001 000 0.000 100

f (x) –0.693 147 –0.105 361 –0. 010 050 –0.001 001 –0.000 100

t –0.5 –0.1 –0.01 –0.001 –0.0001

f (t) 1.835 830 3.934 693 4.877 058 4.987 521 4.998 750

h –0.5 –0.1 –0.01 –0.001 –0.0001

f (h) 48.812 500 72.390 199 79.203 990 79.920 040 79.992 000

It appears that limln x  0. x1



2.2.24

e5t 1

For f (t)  t 0.5 0.1 0.01 0.001 0.0001

x 0.5 0.9 0.99 0.999 0.9999

f (t) 22.364 988 6.487 213 5.127 110 5.012 521 5.001 250

   

2.2.25

e5t 1  5. t0 t

It appears that lim



 2  h   32 5

For f (t)  h 0.5 0.1 0.01 0.001 0.0001

f (h) 131.312 500 88.410 100 80.804 010 80.080 040 80.008 000

It appears that

 2  h   32  80. 5

lim h0



2.2.26 For f (x) 

ln x  ln 4 : x4

x 3.9 3.99 3.999

f (x) 0.253 178 0.250 313 0.250 031

x 4.1 4.01 4.001

f (x) 0.246 926 0.249 688 0.249 969

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

3.9999

0.250 003

4.0001

0.249 997

ln x  ln 4  0.25. The graph confirms that result. x4 x4

It appears that lim



2.2.27 For f ( p)  p –1.1 –1.01 –1.001 –1.0001

1 p9

: 1 p15 f ( p)

p –0.9 –0.99 –0.999 –0.9999

0.427 397 0.582 008 0.598 200 0.599 820

f ( p) 0.771 405 0.617 992 0.601 800 0.600 180

It appears that lim 1 p  0.6. The graph confirms that result. 

1 p15

p1

2.2.28



sin 3 : tan 2 f ( )

±0.1 ±0.01 ±0.001 ±0.0001

1.457 847 1.499 575 1.499 575 1.499 9960

For f ( ) 



 It appears that lim sin 3  1.5. The graph confirms that result.



 0 tan 2



2.2.29 For f (t)  t 0.1 0.01 0.001 0.0001

5t 1

t f (t) 1.746 189 1.622 459 1.622 459 1.610 734

t –0.1 –0.01 –0.001 –0.0001

f (t) 1.486 601 1.596 556 1.622 459 1.610 734

5t 1  1.6094 (which is ln 5). The graph t0 t

It appears that lim

confirms that result.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.14 9 21 For f  x 

x4 2 : x x

It appears that lim x0

f  x

0.236068

0.5

0.242641

0.1

0.248457

0.05

0.249224

0.01

0.249844

f  x

1

0.267949

0.5

0.258343

0.1

0.251582

0.05

0.250786

0.01

0.250156

x  4  2  0.25  1 . x 4

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.15 0 For f  x

9x  5x

It appears that limx0

f  x

0.5

1.527864

0.1

0.711120

0.05

0.646496

0.01

0.599082

0.001

0.588906

f  x

0.5

0.227761

0.1

0.485984

0.05

0.534447

0.01

0.576706

0.001

0.586669

9x  5x  0.59 . Later we will be able to show that the exact value is ln 9 / 5 . x

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.32 For f (x)  xx : x 0.1 0.01 0.001 0.0001

f (x) 0.794 328 0.954 993 0.993 116 0.999 079

It appears that lim xx  1. The graph confirms that result. x0



2.2.33 For f (x)  x2 ln x : x 0.1 0.01 0.001 0.0001

f (x) –0.023 026 –0.000461 –0.000 007 –0.000 000

It appears that lim x2 ln x  0. The graph confirms that result. x0



2.2.34

2.2.35 (a) From the graphs, it seems that lim sin x  0.32. x0

sin x

(b)(b)

151

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

x ±0.1 ±0.01 ±0.001 ±0.0001

f (x) 0.323 068 0.318 357 0.318 310 0.318 310

(The exact value is 1/π.)

2.2.36 (a) Let f (x)  1 x  . 1/ x

x –0.001 –0.0001 –0.00001 –0.000001 0.000001 0.00001 0.0001 0.001

h(x) 2.71964 2.71842 2.71830 2.71828 2.71828 2.71878 2.71815 2.71692

It appears that lim h(x)  2.71828, which is approximately e. x0

2.2.37



For the curve y  2 and the points P 0,1 and Q x,2 x x

:

mPQ

0.1

0.1,1.0717735

0.71773

0.01

0.01,1.0069556

0.69556

0.001

0.001,1.0006934

0.69339

0.0001

0.0001,1.0000693

0.69317

The slope appears to be about 0.693.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.38





For f (x)  x2  2 x / 1000 : (a) It appears that lim f (x)  0. x0

x 1 0.8 0.6 0.4 0.1 0.05

(b) It appears that lim f (x)  0.001  f (0). x0

f (x) 0.998 000 0.638 259 0.358 484 0.158 680 0.008 928 0.001 465

x 0.04 0.02 0.01 0.005 0.003 0.001

f (x) 0.000 572 –0.000 614 –0.000 907 –0.000 978 –0.000 993 –0.001 000

2.2.39 For h(x)  (a)

tan x  x : x3 x 1.0 0.5 0.1 0.05 0.01 0.005

h(x) 0.557 407 73 0.370 419 92 0.334 672 09 0.333 667 00 0.333 346 67 0.333 336 67

x 0.001 0.005 0.0001 0.0005 0.00001 0.000001

h(x) 0.333 333 50 0.333 333 44 0.333 330 00 0.333 336 00 0.333 000 00 0.000 000 00

It seems that lim h(x)  1 . x0

(b) Here the values will vary from one calculator to another. Every calculator will eventually give false values.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.40 No matter how many times we zoom in toward the origin, the graphs of appear to consist of almostvertical lines. This indicates more and more frequent oscillations as x → 0.

2.2.41 We need to have 5.8  x3  3x  4  6.2 . From the graph we obtain the approximate points of

intersection P 1.9774, 5.8 and Q 2.0219, 6.2 ). So if x is within 0.021 of 2 , then y will be within

0.2 of 6 . If we must have x3  3x  4 within 0.1 of 6 , we get P 1.9888, 5.9 and Q 2.0110, 6.1 . We would then need x to be within 0.011 of 2 .

154

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

2.2.42

(a) Let y  x 0.99 0.999 0.9999 1.01 1.001 1.0001

x3 1 . From the table and the graph, we guess that the limit of y as x approaches 1 is 6. x 1 h(x) 5.925 31 5.992 50 5.999 25 6.075 331 6.007 50 6.000 75

(b) We need to have 5.5 

x3 1 x 1

 6.5. From the graph

obtain the approximate points of intersection P  (0.9314,5.5) and P  (1.0649,6.5). Now

1 0.9314  0.0696 and1.0649 1  0.0649, so by requiring that x be within 0.0649 of 1, we ensure that y is within 0.5 of 6.

2.2.43 (a) For any positive integer n, if x 

, then f (x)  tan

n tangent function has period π.) (b) For any nonnegative number n, if x 

 tann   0. (Remember that the x

, then

4n 1 4n 1 1     4n    f (x)  tan  tan  tan     tan n    tan   1. x 4 4 4 4 4     (c) From part (a), f (x)  0 infinitely often as x → 0. From part (b), f (x)  1 infinitely often as x → 1 0. Thus limtan does not exist since f (x) does not get close to a fixed number as x → 0. x0 x

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.44

(a)

(b)

f  x

0.77 0.78 0.785 0.7853 0.78539

5.541794 6.033899 6.311042 6.328415 6.333644

f  x

0.77 0.78 0.785 0.7853 0.78539

5.541794 6.033899 6.311042 6.328415 6.333644

It appears that

lim



x /4 

f  x  6.33.

It appears that

lim f  x  6.33. 

x /4 

Later we will be able to show that the exact values of the limits are  tan 2 and tan 2, respectively.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.3 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

156

     

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

END OF SECTION EXERCISE SOLUTIONS 2.3.1 (a) lim f (x)  5g(x)  lim f (x)  lim5g(x) x2

x2

Limit Law 1 Limit Law 3

x2

 lim f (x)  5limg(x) x2

 4  5(2)  6



(b) lim  g(x)   lim g(x) 3

x2



Limit Law 6

 (2)3  8 (c) lim x2

Limit Law 6

f (x)  lim f (x) x2



 4 2 lim3 f (x) (d) lim 3 f (x)  x2 x2 g(x) lim g(x)

Limit Law 5

x2



3lim f (x) x2

lim g(x)

Limit Law 3

x2



3 4 2

 6

(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim

g(x)

h(x) does not exist because the denominator approaches 0 while the numerator approaches a nonzero number. x2

limg (x)h(x)

(f) lim g(x)h(x)  x2 x2

f (x)

Limit Law 5

lim f (x) x2

lim g(x)  lim h(x)  x2

x2

lim f (x)

Limit Law 4

x2





2  0 0 4

2.3.2 

(a) lim f (x)  g(x)  lim f (x)  lim g(x) x2

x2

Limit Law 1

x2

 1 2  1

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b) lim f (x) exists, but lim g(x) does not exists (the left and right limits are not equal) so we cannot x0

x0

apply Limit Law 2 to lim f (x)  g(x).The limit does not exist. x 0

(c)(c) lim f (x)g(x)  lim f (x)  lim g(x) x1

x1

Limit Law 4

x1

 1 2  2 (d) lim f (x)  1, but lim g(x)  0, so we cannot apply Limit Law 5 to lim x3

exist. Note: lim x3

x3

f (x)

 since g(x)  0 as x  3 and lim

f (x)

g(x)

x3

g(x)

f (x)

. The limit does not

g(x)

  since g(x)  0 as x  3.

Therefore, the limit does not exist, even as a infinite limit. (e) lim  x2 f (x)   lim x2 lim f (x) Limit Law 4 x2

x2

 2  (1)  4 2

(f)(f) f (1)  lim g (x)  3  2  5 x1

2.3.3

lim5x3  3x2  x  6  lim5x3 lim3x2   lim x lim6 x3

x3

Limit Laws 2 and 1

 5lim  x 3   3lim  x 2   lim x  lim 6

3

 533   332   3  6

9,8, and 7

x3

 105     

2.3.4

limx4  3x  x 2  5x  3  limx4  3x  lim  x 2  5x  3

x1

Limit Law 4

   2, 1   lim x  3 lim x  lim x  5 lim x  lim 3  3 x1

x1

 lim x 4  lim 3x lim x 2  lim 5x  lim 3 x1

x1

 1 31 5  3

x1

9,8, and 7

 41  4 2.3.5

158

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

limt4  2 t4  2  t2 2 t2 2t2  3t  2 lim 2t  3t  2  lim

Limit Law 5



t2

lim t 4  lim 2

1,2, and 3

t2  2 limt2 t 2  3lim t  lim 2 t2

t2

16  2 2(4)  3(2)  2 14 7   16 8

9,8, and 7





 2.3.6

Limit Law 11

lim u4  3u  6  lim u4  3u  6 u2

u2

 lim u4  3 lim u  lim 6

1,2 and 3

 (2)4  3(2)  6

9, 8 and 7

u2

 16  6  6  16  4  

2.3.7



lim 1 3 x

2  6x  x   lim1 x lim2  6x  x  2



x8



x8

 lim1 lim 3 x lim 2  6lim x 2  lim x3 x8

x8



x8



 1 3 8   2  6 82  83 

Limit Law 4 1,2 and 3 7, 10,11 and 9

 (3)(130)  390

2.3.8

 2

 t  2   t 2  2  lim 3   lim 3  t2 t   3t  5   t2 t  3t  5  2



Limit Law 6

159

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 limt3  2    limt2t 3  3t  5    t2  2   lim t 2  lim 2 t2  3   lim tt2  3lim t  lim 5  t2  t2  t2  2  4  2     8  3(2)  5  2

5 1, 2 and 3 9, 7 and 8

4 2      7  49

2.3.9 2 2 lim 2x 1  lim 2x 1 x2 3x  2 x2 3x  2

 

Limit Law 11



lim2x2 1

5

x2

lim3x  2 x2







2 lim x2  lim1 x2

1, 2 and 3

x2

3lim x  lim 2 x2

x2

2 3  2(2) 1  9  4 2 3(2)  2

 9, 8 and 7



2.3.10 (a) The left-hand side of the equation is not defined for x = 2, but the right-hand side is. (b) Since the equation holds for all x ≠ 2, it follows that both sides of the equation approach the same limit as x  2. Remember that in finding lim f (x) we never consider x = a. xa

2.3.11

lim x0



 x  2 x  2   0  2 0  2  2(2)  4

2.3.12

160

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

lim 3 x  6  3 2  6  3 8  2 x2

2.3.13

x 1

lim x1

x2 1

1 1 1 x 1  lim   x1 (x 1)(x 1) x1 (x 1) 11 2

 lim



2.3.14    lim x sin x  sin       1   x2 2 2 2 2 2.3.15



lim x2 cos x  lim x  lim cos x  2

x /2

0  0

x /2

2.3.16

lim x  6x  5 lim (x  5)(x 1)  lim x 1  5 1  4 x5 x5 x5 x5 x5 2

2.3.17

lim x3

x2  3x x2  x 12

 lim

x3

x(x  3) (x  4)(x  3)

 lim

x3

x(x  3) (x  4)(x  3)

 lim

x3

x (x  4)



3 3  4



3 7

2.3.18

x2  5x  6 2 does not exist since x  5  0, but x  5x  6  6, as x  5. x5 x5

lim

2.3.19

2x2  3x 1 lim x1

x  2x  3 2

 lim x1

(2x 1)(x 1) (x  3)(x 1)

2(1) 1 1 1    x1 x  3  1 3 4 4

 lim

2x 1

2.3.20 2 25 10h  h2   25  lim h 10h lim h(h 10)  limh 10  10 lim  5  h   25  lim  h0 h0 h0 h0 h0 h h h h 2

161

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 2  h   8  lim 8 12h  6h2  h3   8  lim h3  6h2 12h  lim h2  6h 12  12 lim 3

h0



h0



2.3.22

lim x  2  lim x2

x3  8

x2

x2

 lim

(x  2)  x 2  2x  4

x2

x2  2x  4



1 444



2.3.23 Use the difference of squares in the numerator and the difference of cubes in the denominator:

t 1t 1t2 1 t 1t2 1 2(2) 4 t2 1t2 1  lim  lim lim  lim   2 1 t 1t  t 1 t 1t2  t 1 t  t 1 3 3 t t 4 1

t1

2.3.24





9  h  32 9  h  9  lim lim 9  h  3  lim 9  h  3  9  h  3  lim  lim h0 h0 h0 h0 h h 9  h  3 h0 h 9  h  3 h 9h 3 h  lim h0







 9  h  3

1 1 1   9  h  3 33 6

2.3.25

 4u 1   3 2

lim 4u 1  3 u2 u2 u2  lim

 lim u2

 

4u 1  3  lim 4u 1  3  u  2  4u 1  3 u2 u  2

4(u  2)



 lim

u  2 4u 1  3 u2

 lim

 4u 1  3

u2

4u 1 9 u  2 4u 1  3





2 4  4  4u 1  3  9  3 3



2.3.26

1 1 1 1 3x 1 1   lim x 3  lim x 3   lim 3  x  lim   x3 x  3 x3 x  3 3x x3 3x  x  3 x3 3x 9

2.3.27

162

 

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

lim  3  h   3 h0 h 1

1

1 3  lim 3  (3  h)  lim  lim 3  h h

h0

h  3  h3

h0

h h  3  h3

 lim h0

1 33  h



2.3.28



  2



1 1  3(3  0) 9

 

lim 1 t  1 t  lim 1 t  1 t 1  t  1  t  lim 1 t  1 t  lim 1 t   1 t   t0 t0 t0 t t 1 t  1 t t0 t 1 t  1 t t 1 t  1 t 2 2t  t 2   1  lim  lim t0 t0 1 1 2 t 1 t  1 t 1 t  1 t













2.3.29

1  1 1  t 11 1 1 1 lim   2   lim   lim  lim  1  t0 t t0 t0 t(t 1) t0 t 1 t 1 t t(t 1) 0 1     

2.3.30



 

 

 

4 x 4 x 4 x 16  x 1 1  lim lim  lim  2  lim  x16 x  x16 16x  x2  4  x x16 16x x 16  x 4  x x16 16 4  16 x 4  x 1 1   16(8) 128













2.3.31 2 (x  2)2 (x  2)2 (x  2) 0 lim x  4x  4 lim  lim  lim  0  4 2 2 x2 x  3x  4 x2 x 2  4 x2 (x  2)(x  2) x 2 1  x2 (x  2) x 2 1  4 5 x 1        

2.3.32



 

 t  lim t 1 t 1 1 t  

1 1 t 1 1 t 1 1  1 1 t    lim  lim lim t0 t0 t 1 t t  t0 t 1 t  t 1 t 1 1 t

t0

163



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



 lim



t0

2.3.34

lim

4(x  h)  4x



1 0 1 1 0

1 2

(x  4)(x  4)

 x  4

x4

 lim

h0

 lim

 x  9  5

4x  4h  4x

x4

 lim

h0







5

 lim x4



1

 x  9  x  9  5 lim x  9  25  lim x 16  x  4   x  9  5  x  4  x  9  5  x  4   x  9  5  4

x2  9  5  lim x4 x4

x4



1 t 1 1 t

2.3.33

lim

1

x4

(x  4)

4  4



16  9  5

x2  9  5



8  

55

 lim 4  4 h0

2.3.35

x 3  3x2h  3xh2  h3   x3 x  h   x3  3x2h  3xh2  h3   lim  lim  lim 3x2  3xh  h2  3x2 lim 3

h0

2.3.36

 x  h   x2 2

lim h0

 lim

x 2   x 2 h   x  h  x2

h0

 x2   x 2  2xh  h2 

 lim h0

h0



2x x2  x  h 



hx2  x  h 

h0







 lim h0

h  2x  h hx2  x  h

  2x  h

 lim 2 2 h0 x  x  h

2 x3

2.3.37 (a) Based on the graph, lim x0

x 1 3x 1



(b)(b) x –0.001 –0.0001 –0.000001 –0.0000001

f(x) 0.666 166 0.666 617 0.666 662 0.666 667

x 0.001 0.0001 0.000001 0.0000001

f(x) 0.667 166 0.666 717 0.666 672 0.666 667

164

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2  . 1 3x 1 3 x

Based on the table it appears that lim x0















x 1 3x 1  lim x 1 3x 1 x 1 3x 1 (c) lim   lim x0 1 3x 1 x0 3x 1 3x 1 x0 1 3x 1 1  lim 1 3x 1 3 x0 1  lim1 3lim x 1 x0 3 x0 1 1 2  1  (11)  1 3 0 3 3 3







Limit Law 3 1 and 11 1,3 and 7 7 and 8

2.3.38 (a) Based on the graph, lim

3 x  3

 0.29 .

x0

(b) x –0.001 –0.0001 –0.000001 –0.0000001

f(x) 0.288 699 0.288 678 0.288 675 0.288 675

x 0.001 0.0001 0.000001 0.0000001

Based on the table it appears that lim x0

3  x  3  3  x  3  lim x 3  x  3 x0

f(x) 0.288 651 0.288 673 0.288 673 0.288 675

3 x  3

 0.29.

x (3  x)  3

 lim

 3  x  3  lim1



x0

lim 3  x  lim 3 x0



x0

1 3 x  3

Limit Laws 5 and 1

x0

1 lim 3  x  3

7 and 11

1 1  3 0  3 2 3

Limit Laws 5 and 1

x0



165

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.2.39 Let f (x)  x , g(x)  x cos 20 x, and h(x)  x . Then 2

1  cos 20 x  1   x2  x2 cos 20 x  x2  f (x)  g(x)  h(x). So since lim f (x)  lim h(x)  0, by the Squeeze Theorem we have lim g(x)  0. x0

x0



2.3.40



Let f (x)   x  x , g(x)  x  x sin / x  , and h(x)  x3  x2 . Then 3

1  sin / x  1   x3  x2  x3  x2 sin / x  x3  x2  f (x)  g(x)  h(x). So since lim f (x)  lim h(x)  0, by the Squeeze Theorem we have lim g(x)  0. x0

x0

2.3.41

2



We have lim4x  9  4(4)  9  7 and lim x  4x  7  4  4(4)  7  7. Since x4

x4

4x  9  f (x)  x  4x  7 for x  0,lim f (x)  7 by the Squeeze Theorem. 2

x4

2.3.42





We have lim2x  2(1)  2 and lim x4  x2  2 14 12  2  2. Since for all x, x1

x1

2x  g(x)  x4  x2  2, lim g(x)  2 by the Squeeze Theorem. x1

2.3.43

1  cos2 / x  1   x4  x4 cos2 / x  x4. Since lim x4  0 and lim x4  0, we have x0

x0

lim x4 cos2 / x  0 by the Squeeze Theorem. x0



2.3.44



1  sin / x  1  e  e 1

sin( /x)

 e  1



  sin( / x0  x /e. Since lim xe  x / e 

x0

x /e0

sin( /x)   0 by the Squeeze Theorem. and lim xe  0, we have lim xe x0 

2.3.45

x0



 x  3   x  3 (x  3) 





if x  3  0 x  3 if x  3 if x  3  0   x if x  3 3

166

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check





Thus, lim 2x  x  3  lim2x  x  3  lim 3x  3  3(3)  6 and x3

x3

lim2x  x  3   lim2x  3  x  lim  x  3  3  3  6. Since the left and right limits are

x3



x3



equal, lim 2x  x  3  6. x3



2.3.46

 

 if x  6  0  if x  6 x  6   x  6  x6 (x  6) if x  6  0  (x  6) if x  6   2(x  6) 2x 12  2 and lim 2x 12  lim 2(x  6)  lim 2(x  6)  2.  lim Then lim x6 (x  6) x6  x  6 x6  x  6 x6  x  6 x6  x  6 





2x 12 does not exist. x6 x  6

The left and right limits are not the same, so lim

 

2.3.47

  2x 1   2x 1

2 2x  x  x  2x 1  x2  2x 1  x2 2x 1 ; 3



 (2x 1)

So 2x  x  x (2x 1) for x  0.5. Thus 3

if x  0.5 if x  0.5

1 2x 1  lim 2x 1  4.  lim 1  1  2 2 3 2  2 x0.5  2x  x x0.5 x (2x 1) x0.5 x (0.5) 0.25 lim









2x 1  lim 2x 1  lim 1 x0.5  2x3  x2 x0.5 x2 (2x 1) x0.5 x2

And 2x3  x2  x2 (2x 1) for x  0.5. Thus lim







2x 1 1 1 does not exist.   4. The left and right limits are not the same, so lim 2 x0.5 (0.5) 0.25 2x3  x2

2.3.48 Since x  x for x  0, we have lim x2

2 | x | 2 x

 lim x2

2 (x) 2 x

lim x2

2 x 2 x

2.3.49 1 1  1 1    lim  Since x  x for x  0, we have lim 

x0

   x | x | 



x0

 x

 lim1  1. x2

2 lim , which does not exist

   x  x0 x

since the denominator approaches 0 and the numerator does not.

167

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.3.50 1 1  1 1   lim   lim 0  0. Since x  x for x  0, we have lim 



x0

 x

 | x | 



x0

 x

 x 



x0

2.3.51

1 if x  0  sgn x  0 if x  0.  1 if x  0

(a)

(b) (i) Since sgn x  1 for x  0, lim sgn x  lim 1  1. x0

x0

(ii) Since sgn x  1 for x  0, lim sgn x  lim1  1. x0

x0

(iii) Since lim sgn x  lim sgn x, limsgn x does not exist. x0

x0

x0

(iv) Since sgn x  1 for x  0,lim sgn x  lim1  1. x0

x0

2.3.52

1 if sin x  0  (a) g(x)  sgn(sin x)  0 if sin x  0.  1 if sin x  0 (i) lim g(x)  lim sgn(sin x)  1 since sin x is positive for small positive values of x. x0

x0

(ii) lim g(x)  lim sgn(sin x)  1sincesin x is negative for small negative values of x. x0

x0

(iii) lim g(x) does not exist since lim g(x)  lim g(x). x0

x0

x0

(iv) lim g(x)  lim sgn(sin x)  1 sincesin x is negative for values of x slightly greater than π. x 

x 

x 

x 

(v) lim g(x)  lim sgn(sin x)  1 since sin x is positive for values of x slightly less than π. (vi) lim g(x) does not exist since lim g(x)  lim g(x). x

x 

x 

(b) The sine function changes sign at every integer multiple of π, so the signum function equals 1 on one side and –1 on the other side of nπ, n an integer. Thus lim g(x) does not exist xa

for a = nπ, n an integer.

168

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.3.53 (a) (i) lim g(x)  lim x2

x2  x  6 x2

 lim

(x  3)(x  2)

x2 (x  3)(x  2) Since x  2  0 as x  2   lim   x2  x2

x2



 lim(x  3)  5 x2

(ii) The solution is similar to the solution in part (i), but now x  2  2  x since x  2  0 as x  2. Thus, lim g(x)  lim (x  3)  5. x2

x2

(b) Since the right-hand and left-hand limits of g at x = 2 are not equal, lim g(x) does not exist. x2

2.3.54

 x2 1 if x  1 f (x)   2 (x  2) if x  1 lim f (x)  limx2 1 12 1  2, x1

x1

lim f (x)  lim  x  2   (1) 2 1. 2

x1

x1

(b) Since the right-hand and left-hand limits of f at x = 1 are not equal, lim f (x) does not exist. x1

2.3.55 1 For lim B(t) to exist, the one-sided limits at t = 2 must be equal: lim B(t)  lim4  t   4 1  3 t2

t2

t2

and lim B(t)  lim t  c  2  c.   t2 

t2 

Now 3  2  c  9  2  c  c  7.

169

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.3.56

For lim g  x to exist, the one-sided limits at x  2 must be equal: x2

lim g  x  lim 4  x   4  2  3 must equal lim g  x  lim x2

x2

x2

x  c  lim 2  c. Now x2

3  2  c  9  2  c  c  7.

 

2.3.57 (a) (i) x  2 for





2  x  1, so lim x  lim (2)  2. x2

x2

 3. (ii) x  3 for  3  x  2, so lim x  lim (3)  x2 

x2 

The right and left limits are not equal, so lim x does not exist. x2

(iii) x  3 for  3  x  2, so lim x  lim (3)  3. x2.4

x2.4

1)  n 1. (b) (i) x  n 1 for n 1  x  n, so lim x  lim(n  xn 

xn 

(ii) x  n for n  x  n 1, so lim x  lim n  n. xn 

xn 



 2.3.58 (a) See the graph of y  cos x. Since 1  cos x  0 on ,  / 2, we have

y  f (x)  cos x  1 on ,  / 2.

Since 0  cos x 1 on / 2, 0, 0, / 2, we have f (x)  0 on

 / 2, 0 , 0, / 2. Note that f (0)  1. (b) (i) lim f (x)  0 and lim f (x)  0, so lim f (x)  0. x0

x0

(ii) As x   / 2  , f (x)  0, so

x0



(iii) As x   / 2  , f (x)  1, so 

lim f (x)  0.

x /2

lim f (x)  1.

x /2



(iv) Since the answers in parts (ii) and (iii) are not equal, lim f (x) does not exist. x /2

170

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



2.3.59

The graph of f (x)  x  x is the same as the graph of g(x)  1 with holes at each integer, since

f (a)  0 for any integer a. Thus, lim f (x)  1 and lim f (x)  1 so lim f (x)  1. However x2

x2

x2

f (2)  2  2  2  (2)  0, so lim f (x)  f (2). x2

2.3.60

ax2 if x  2  f (x)  2  bx if  2  x  2; 2bx  3a  2 if x  2  lim f (x)  lim ax2  a(2)2  4a and lim f (x)  lim 2  bx  2  2b. If lim f (x) exists,

x2

x2 

x2

then lim f (x)  lim so 4a  2  2b or a   b.  1 2

x2 

1 2

lim f (x)  lim 2  bx  2  2b and lim f (x)  lim 2bx  3a  2  4b  3a  2. If lim f (x) exists, x2

x2

x2

x2

then lim f (x)  lim so 2  2b  4b  3a  2  2b  3a. Combining these two equations we have x2

x2 

2b  3a   2b  3 12  21 b  4b  3(1 b)  3  3b   b  3  b  3 and 2x2 if x  2  if  2  x  2. a  12  12 b  12  12 (3)  12  32  2. Thus f (x)  2  3x 6x  6  2 if x  2 

2.3.61

x0

 a x  b  ax   a x  b    ax 

 lim

 a 2 x 2  b  ax  lim

a x  b  ax

x0

a x  b  ax

a x  b  ax

lim 2 2

2 2

 lim

x0



b b

2 2

x0

2 2

 a x  b  ax lim b  a x  b  ax   b  a  0  b  a  0 2

a2 x2  b  a 2 x2



x0

1

171

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.3.62

 lim

 a 2 x 2  b  ax  lim

a x  b  ax

x0

a x  b  ax

a x  b  ax

2 2

x0

 a x  b  ax   ax  a x  b   

lim

2 2

 lim

b b

x0

2 2

 a x  b  ax lim b  a x  b  ax   b  a  0  b  a  0 2

a2 x2  b  a 2 x2

x0



2 2



x0

1

2.3.63 (a) lim L0 vc

 

 v0  L0  0. As the velocity approaches the speed of light, the length approaches 1 2  11 c 

0. (b) A left-hand limit is necessary since L is not defined for v > c.



2.3.64 Since p(x) is a polynomial, p(x)  a  a x  a x2 



 an xn. Thus, by the Limit Laws,

lim p(x)  lima  a x  a x2   a x n  a  a lim x  a lim x 2   a lim xn  0 1 2 n 0 1 2 n 0

xa

a aaa a  0

xa

 a a  p(a). Thus, for any polynomial p, lim p(x)  p(a).

xa

2.3.65 Let r(x) 

q(x) p(x)

where p(x) and q(x) are any polynomials, and suppose q(a)  0. Then

lim p(x)  xa xa q(x) lim q(x)

lim r(x)  lim xa

p(x)

xa

2.3.66

lim f (x)  8  lim x1

x1

 f (x)  8  x 1

Limit Law 5  p(a) Exercise 70  r(a). q(a)

f (x)  8   (x 1)  lim  lim(x 1)  10  0  0  x1 x 1 x1

172

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Thus, lim f (x)  lim x1

 f (x) 8  8  lim f (x) 8lim8  0  8  8.

x1

Note: The value of lim

lim x1

f (x)  8

x1

does not affect the answer since it’s multiplied by 0. What’s important is

x1 x 1 f (x)  8 that exists.

x 1

2.3.67

f (x)   x2  lim  lim x2  5 0  0 x0 2 x0  x2 x0  fx(x)  x0 f (x) f (x) (b) lim  lim  x  lim  lim x  5 0  0 2 2   x0 x0  x x0 x  x0 x (a) lim f (x)  lim

 f (x)

2.3.68 Observe that 0  f (x)  x for all x, and lim 0  0  lim x . So, by the Squeeze Theorem, lim f (x)  0. 2

x0

2.3.69 Let ƒ(𝑥) = ⟦𝑥⟧ and 𝑔(𝑥) = −⟦𝑥⟧. Then lim f (x) and lim g(x) do not exist [Example 13] but x3

x3

𝑙i𝑚[ƒ(𝑥) + 𝑔(𝑥)] = 𝑙i𝑚(⟦𝑥⟧ − ⟦𝑥⟧) = 𝑙i𝑚0 = 0. 𝑥→3

𝑥→3

2.3.70 Let f (x)  H (x) and g(x)  1 H (x), where H is the Heaviside function defined in Exercise XX. Thus either f or g is 0 for any value of x. Then lim f (x) and lim g(x) do not exist, but

lim f (x)g(x)  lim0  0. x0

x0

2.3.71

173



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

lim x2

6  x  2 x2 6  x  2 6  x  2 3  x 1  1  lim 1  2 1  3 x 3  x 6  x 3  x    6 x 2 4   2  x  6  x  4 3  x 1  1  3  x 1  3 x  lim    lim      lim  2 x2  x2 6  x  2  x2  3  x 1 6  x  2    2  x  6  x  2   3  x 1  3  x 1 1  lim  x2 6 x  2 2











2.3.72 Since the denominator approaches 0 as x  2, the limit will exist only if the numerator also approaches





0 as x  2. In order for this to happen, we need lim 3x2  ax  a  3  0  x2

3(2)2  a(2)  a  3  0  12  2a  a  3  0   a  15. With a = 15, the limit becomes 3x2 15x 18 3(x  2)(x  3) 3(x  3) 3(2  3) 3  lim  lim    1. lim x2 x2 (x 1)(x  2) x2 x2  x  2 x 1 2 1 3

2.3.73 Solution 1: First, we find the coordinate of P and Q as functions of r. Then we can find the equation of the line through these two points, and thus find the x-intercept (the point R) and take the limit as r  0. The coordinates of P are (0, r). The point Q is the point of intersection of the two circles

x2  y2  r 2 and (x 1)2  y2  1. Eliminating y from these equations, we get r2  x2  1(x 1)2  r2 1 2x 1  x  12 r2. Substituting back into the equation of the

 2   y2  r 2  y2  r 2 1 r 2 

shrinking circle to find the y-coordinate, we get 12 r

 yr

1 14 r 2



1 4



(the positive value). So the coordinates of Q are 1 r 2 , r 1 1 r 2 . The

equation of the line joining P and Q is thus y  r  find the x-intercept, and get x  r

1 2

r 1 14r 2  r 1 2



r2  0

 12 r 2

(x  0). We set y = 0 in order to

 1 r 1 21 r 1 4

 1 r 1 1 r 1 Now we take the limit as r  0 : lim x  lim 2  1 r 1  lim 2  1 1  4. 1 4



r0

1 4



1 .

r0

So the limiting position of R is the point (4, 0). Solution 2: We add a few lines to the diagram, as shown. Note that PQS  90 (subtended by diameter PS). So

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SQR  90  OQT (subtended by diameter OT). It follows that OQS  TQR. Also  PSQ  90 SPQ  ORP. Since QOS is isosceles, so is QTR implying that QT  TR. As the circle C2 shrinks, the point Q plainly approaches the origin, so the point R must approach a point twice as far from the origin at T, that is, the point (4, 0) as above.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.4 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.4.1 The function f is continuous at x = 4 provided lim f (x)  f (4). x4

2.4.2 If f is continuous on,  then the graph of f has no holes, jumps or vertical asymptotes. 2.4.3 (a) f is discontinuous at –4 because f (4) is not defined. The function is also discontinuous at –2, 2 and 4 because the limit does not exist at these points (the left and right limits are not the same). (b) The function is continuous from the left at x = –2 because lim f (x)  f (2). The function is  x2 

continuous from the right at 2 and 4 because lim f (x)  f (2) and x2 

lim f (x)  f (4). It is not

x4

continuous from either side at –4 because f (4) is undefined. 2.4.4

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From the graph of g we see that g is continuous on the intervals 3, 2,2, 1,1, 0, 0,1, and1,3.

2.4.5 The graph of y  f (x) must have a discontinuity at x = 2 and must show that lim f (x)  f (2). x2 



2.4.6 The graph of y  f (x) must have discontinuities at x = –1 and x = 4. It must show that lim f (x)  f (1) and lim f (x)  f (4). x1 

x4

      

2.4.7 The graph of y  f (x) must have a removable discontinuity (a hole) at x = 3 and a jump discontinuity at x = 5.

2.4.8

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.4.9 The function f (x) 



2 if x  3 satisfies lim f (x)  lim f (x)  2  f (3)  4.  x3 x3  4 if x  3

2.4.10 (a) The toll is $7 between 7:00 AM and 10:00 AM and between 4:00 PM and 7:00 PM. (b) The function T has jump discontinuities at t  7,10,16 and 19. Their significance to someone who uses the road is that, because of the sudden jumps in the tolls, they may want to avoid the higher rates between t  7 and t  10 and between t  16 and t  19 if feasible.

2.4.11

(f) Continuous; as the elapsed time since the bus left the first station increases, the speed changes smoothly as time passes, without any instantaneous jumps from one speed to another. Note that even in the event of a sudden stop, the speed passes through every value between its traveling speed and 0. 2.4.12

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

lim f (x)  lim  x  2x3   1 2(1)3   (3)4  81  f (1); by the definition of continuity, f is  4

x1

continuous at a = –1. 2.4.13

2 2 lim g(t)  lim t  5t  2  5(2)  14  g(2); by the definition of continuity, g is continuous at a = 2. t2 t2 2t 1 2(2) 1 5

2.4.14

lim p(v)  lim 2 3v2 1  2 3(1)2 1  2 4  4  p(1); by the definition of continuity, p is v1

v1

continuous at a = 1. 2.4.15







3 2 lim f (x)  lim 3x  5x  x  4  3(2)  5(2)  2  4  48 10  2  40  f (2); by the x2 x2 4

definition of continuity, f is continuous at a = 2.

2.4.16  For a < –2, we have lim f (x)  lim x  x  4  a  a  4  f (a). So f is continuous at x  a for xa

xa





every a in 4,  . Furthermore, lim f (x)  4  f (4) so f is continuous from the right at 4. So f is x4 

continuous on 4,   . 2.4.17

For a > 4, we have lim g(x)  lim x 1  xa

xa

, 2. 2.4.18

3x  6

a 1 3x  6

 g(a). So g is continuous at x  a for every a in

 1

if x  2  . For this function, f (2)  1, but f (x)  x  2  1 if x  2 lim f (x)   and lim f (x)  , so lim f (x) does not exist

x2 

x2

and f is discontinuous at x  2. 2.4.19

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f (x) 

x  3 if x  1  x if x  1  2

1  . x1 x1 x1  x1 2 Since the left-hand and right-hand limits of f at –1 are not equal, lim f (x) 1

lim f (x)  lim (x  3)  1 3  2 and lim f (x)  lim 2  2 x



x1

does not exist, and f is discontinuous at –1. 2.4.20

 x2  x  f (x)  x2 1  1

if x  1

if x  1 x2  x x(x 1) x 1  , but lim f (x)  lim  lim  lim x1 x1 x2 1 x1 (x 1)(x 1) x1 x 1 2 1 f (1)  1  so f is discontinuous at 1. 2

2.4.21

cos x  f (x)  0 1 x2 

if x  0 if x  0 if x  0

lim f (x)  1, but f (0)  0  1, so f is not continuous at 0. x0

2.4.22

 2x2  5x  3  if x  3 f (x)   x  3  6 if x  3  2x2  5x  3 (2x 1)(x  3) lim f (x)  lim  lim x3 x3 x3 x 3 x 3  lim 2x 1  7, but x3

f (3)  6  7, so f is discontinuous at 3.

2.4.23

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f (x) 

x2  x  2



x2

(x  2)(x 1) x2

 x 1 for x  2. Since lim f (x)  2 1  3, define f (2)  3. Then x2

f is continuous at 2. 2.4.24



(x  2) x2  2x  4  x2  2x  4 4 44 f (x)  x  8   for x  2. Since lim f (x)   3, define 3



x 4 (x  2)(x  2) f (2)  3. Then f is continuous at 2.

x2

22

x2

2.4.25

  4  x2  x 10 2 if x  2  x2 . Then lim f (x)  lim 4  x  x 10 Let f (x)    5  x2 x2 x2  if x  2  8  2 16   x 2  x 10  lim 4  x2  x 10  4  x  x 10  lim  lim x2

4  x2  x 10





 



  x 2  x  6



(x  2) 4  x  x 10 (x  2) 4  x  x 10 (x  3) 5 (x  3)(x  2) 5  5   and so f is  lim  lim  x2 x2 8 4  x2  x 10 4  4  2 10 4  16 (x  2) 4  x 2  x 10 continuous at x = 2. x2

x2



x2





2.4.26 By Theorem 5, the polynomials x2 and 2x 1 are continuous on  ,  . By Theorem 7 , the root function x is continuous on 0,  . By Theorem 9 , the composite function 2x 1 is continuous on 1  1  its domain, , . By part 1 of Theorem 4 , the sum R  x  x2  is continuous on , .

2 



2x 1



 2

 

2.4.27 By Theorem 7, the root function 3 x and the polynomial function 1 x3 are continuous on R . By part 4



of Theorem 4, the product G  x  3 x 1 x3

 is continuous on its domain, R .

2.4.28

By Theorem 5, the rational function H  x 

x2 1 2x2  x 1

x2 1 

 2x 1 x 1

is continuous on its

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By Theorem 7, the exponential function ex , the trigonometric functions sin x and cos x, and the polynomial function 2 are continuous on their domains, all of which are . By Theorem 9, the composite function esin x is continuous on . By part 1 of Theorem 4, 2  cos x is continuous on . We have cos x  1 2  cos x  1, so 2  cos x  0. By part 5 of esin x is continuous on its domain, . Theorem 4, Q  x  2  cos x 2.4.30 By Theorem 7, the exponential function e5t and the trigonometric function cos2t are continuous on

 ,  . By part 4 of Theorem 4, L t   e5tcos2t is continuous on  ,  . 2.4.31

By Theorem 7, the trigonometric function tan x and the polynomial function 4  x2 are continuous on their domains. By Theorem 7, the root function x is continuous on 0,   . By Theorem 9, the composite function tan x

Theorem 4, h  x 



4 x

4  x2 is continuous on its domain, 2, 2. By part 5 of

is continuous on its domain, 2,   2    2, 2   2 , 2.

2.4.32

By Theorem 7, the polynomial function x4 1 is continuous on ,  and the logarithmic

function ln x is continuous on 0,   . By Theorem 9, the composite function g  x  ln  x 4 1 is continuous on its domain, which is x | x4 1  0  x | x4  1  x | x  1  , 1 1,   . 2.4.33 The sine and cosine functions are continuous everywhere by Theorem 7, so f  x  sin cossin x   , which is the composite of sine, cosine, and (once again) sine, is continuous everywhere by Theorem 9. 2.4.34 The function f  x 

x  0 are different.

1 e

is discontinuous at x  0 because the left- and right-hand limits at 1/ x

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2.4.35

The function f  x  tan2 x is discontinuous at x  2  k, where k is any integer. The function

  is also discontinuous where tan2 x is 0—that is, at x  k. So f  x   ln tan2 x  is discontinuous at x  n, n any integer. f  x   ln tan2 x

 2

2.4.36 Because x is continuous on

and 20  x2 is continuous on its domain,

 20  x  20, the product f (x)  x 20  x2 is continuous on

 20  x  20. The number 2 is in that domain, so f is continuous at 2, and lim f (x)  f (2)  2 16  8. x2

2.4.37

, andsin x is continuous on , the composite function f (x)  sin(x  sin x) is continuous on , so lim f (x)  f    sin  sin   sin  0. Because x is continuous on

x 

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2.4.38

 5  x2  The function f (x)  ln   is continuous throughout its domain because it is the composite of a  1 x  5  x2  logarithmic function and a rational function. For the domain of f, we must have 1 x  0, so the



numerator and denominator must have the same sign, that is, the domain is ,  5  1,  number 1 is in that domain, so f is continuous at 1, and lim f (x)  ln x1

5 1

11

. The 5

 ln 2.

2.4.39 2.4.40

1 x2 if x  1 f (x)    ln x if x  1 By Theorem 3, since f (x) equals the polynomial1 x2 on,1, f is continuous on ,1. By Theorem 3, since f (x) equals the logarithmic function ln x on 1, , f is continuous on 1, . 2.4.41

1 x2 f (x)    ln x

if x  1 if x  1

By Theorem 3, the trigonometric functions are continuous. Since f (x)  sin x on, / 4 and

f (x)  cos x on  / 4,   , f is continuous on , / 4    / 4, . In addition,  lim f (x)  lim sin x  sin  1/ 2 since the sine function is continuous at π/4. Similarly, x /4

x /4

x /4 

lim f (x)  lim cos x  cos   1/ 2 by the continuity of the cosine function at π/4. Thus 4  

lim f (x) exists and equals1/

x /4

2 which agrees with the value of f  / 4. Therefore, f is continuous

at π/4, so f is continuous on (–∞, ∞).

2.4.42

x2 if x  1  f  x  x if 1  x  1 1/ x if x  1 

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The function f is continuous on , 1, 1,1, and 1,   , where it is a polynomial, a 2 polynomial, and a rational function, respectively. Now lim f  x  lim x  1 and x1

x1

lim f  x  lim x  1, so f is discontinuous at 1. Since f 1  1, f is continuous from x1  1 the right at 1. Also, lim f  x  lim x  1 and lim f  x  lim  1  f 1, so f is continuous x1  x1  x1  x1 x  at 1. 

x1



2.4.43





              

2x  f  x  3  x   x

if x  1 if 1  x  4 if x  4

The function f is continuous on ,1, 1, 4  , and 4,   , where it is an exponential, a x polynomial, and a root function, respectively. Now lim f  x  lim 2  2 and x1

x1

lim f  x  lim 3  x  2. Since f 1  2, f is continuous at 1. Also, x1

x1

x4

lim f  x  lim 3  x  1 and lim f  x   lim x4

x4

x  2, so f is discontinuous at 4. However,

f  4   1, so f is continuous from the left at 4.

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2.4.44

2.4.45 By Theorem 3, each piece of F is continuous on its domain. We need to check for continuity at r  R.

lim F(r)  lim 

rR 

GMr



rR

Since F(R) 

GM R2



GM 2

and lim F(r)  lim 

rR 



rR

GM r



, so lim F(r)  rR

, F is continuous at R. Therefore, F is a continuous function of r.

2.4.46

cx2  2x if x  2 f (x)   3  x  cx if x  2 f is continuous on ,2 and 2,   . Now lim f (x)  lim cx2  2x  4c  4 and x2

x2

lim f (x)  lim  x  cx  8  2c. So f is continuous  4c  4  8  2c  6c  4  c  2 . 3

x2

x2

Thus, for f to be continuous on ,   , c  23 . 2.4.47

 x3  a3  f (x)   x  a  c

if x  a if x  a

 2

For x ≠ a, f (x)  x  ax  a , so lim f (x)  lim x  ax  a 2

xa

 a2  a2  a2  3a2. In order for

f to be continuous at a, we need f (a)  c  lim f (x)  3a2. Therefore, c  3a2 and xa

x a  f (x)   x  a  3a2 3

if x  a

if x  a

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2.4.48

 x2  4  x2 if x  2  f (x)  ax2  bx  3 if 2  x  3 2x  a  b if x  3   x2  4 (x  2)(x  2) lim f (x)  lim  lim  lim(x  2)  4 At x = 2: x2 x2  x2  x  2 x2  x2  2 lim f (x)  lim ax  bx  3  4a  2b  3. 



x2

x2



We must have 4a  2b  3  4 or 4a  2b  1 lim f (x)  lim ax2  bx  3  9a  3b  3

At x =3:



x2

x3

x3

x3

(1)



lim f (x)  lim 2x  a  b  6  a  b.

We must have 9a  3b  3  6  a  b or 10a  4b  3 (2). Now solve the system of equations by addition –2 times equation (1) to equation (2):

8a  4b  2 10a  4b  3 2a 1

so a  12 .

Substituting a  12 for a in (1) gives us 2b  1, so b  12 as well. Thus, for f to be continuous on

,   , a  b  12 . 

2.4.49



There cannot be a function that is continuous at x = a if lim f (x)  lim f (x) because if the left and right xa

xa

limits are not equal, lim f (x) does not exist and the function is therefore not continuous at x = a. xa

2.4.50 If f and g are continuous and g(2)  6, then lim3 f (x)  f (x)g(x)  36  x2

lim f (x)  lim f (x)  lim g(x)  36  3 f (2)  f (2) 6  36  9 f (2)  36  f (2)  4. x2

x2

2.4.51 (a) f (x) 

x4 1 x 1



 x 2 1  x 2 1  x 2 1(x 1)(x 1) 

x 1

  x 2 1  (x 1)  x3  x2  x 1for x ≠ 1.

x 1 2 The discontinuity is removable and g(x)  x  x  x 1 agrees with f for x ≠ 1and is continuous on . 3

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x  x 2  x  2 x(x  2)(x 1) (b) f (x)     x(x 1)  x2  x for x ≠ 2. The x2 x2 x2 2 discontinuity is removable and g(x)  x  x agrees with f for x ≠ 2and is continuous on . 

x3  x2  2x

x 



x 



x 



x 



x 



x 

x

exist. The discontinuity at x = π is a jump discontinuity. 2.4.52 Both graphs below show functions that are continuous on (0,1) except at x = 0.25, and in both graphs, f(0) = 1, f(1) = 3.Thus neither graph satisfies the conditions of the IVT on (0,1). In the first graph, there is no 𝑥 ∈ (0,1) with f(x) = 2. But in the second graph, there are two values of x for which f(x) = 2.

2.4.53 2 Observe that f (x)  x 10sin x is continuous on the interval [31, 32], f (31)  957, and

f (32)  1030. Since 957 < 1000 < 1030, there is a number c in 32,31 such that f (c)  1000 by the Intermediate Value Theorem. Note: there is also a number c in 32, 31 such that f (c)  1000. 2.4.54 Suppose that f (3)  6. By the Intermediate Value Theorem applied to the continuous function f on the closed interval [2, 3], the fact that f (2)  8  6 and f (3)  6 implies that there is a number c in (2, 3) such that f (c)  6. This contradicts the fact that the only solutions of the equation are x 1 and x  4. Hence our supposition that f (3)  6 was incorrect. It follows that f (3)  6. But f (3)  6 because the only solutions of f (x)  6 are x 1 and x  4.Therefore f (3)  6. 2.4.55

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The function f (x)  x  x  3 is continuous on the interval [1, 2], f (1)  1, and f (2)  15. Because 4

1  0  15, there is a number c in (1, 2) such that f (c)  0 by the Intermediate Value Theorem. Thus, there is a root of the equation x4  x  3  0 in the interval (1, 2). 2.4.56 The equation ln x  x  x is equivalent to the equation ln x  x 

x  0. The function f (x)  ln x  x  x is continuous on the interval [2, 3], f (2)  ln 2  2  2  0.107, and f (3)  ln 3  3  3  0.169. Since f (2)  0  f (3), there is a number c in (2, 3) such that f (c)  0

by the Intermediate Value Theorem. Thus, there is a root of the equation ln x  x 

x  0, or

ln x  x  x, in the interval (2, 3). 2.4.57 The equation ex  3  2x is equivalent to the equation ex  3  2x  0. The function f (x)  e  3  2x is x

continuous on the interval [0, 1], f (0)  e0  3  2, and f (1)  e  3  2  e 1. Since 2  0  e 1, there is a number c in (0, 1) such that f (c)  0 by the Intermediate Value Theorem. Thus, there is a root of the equation ex  3  2x  0, or ex  3  2x, in the interval (0, 1). 2.4.58 The equationsin x  x2  x is equivalent to the equationsin x  x2  x  0. The function

f (x)  sin x  x2  x is continuous on the interval [1, 2], f (1)  sin1  0.84, and f (2)  sin 2  2  1.09. Sincesin1  0  sin 2  2, there is a number c in (1, 2) such that f (c)  0 by 2 the Intermediate Value Theorem. Thus, there is a root of the equation sin x  x  x  0, or sin x  x2  x, in the interval (1, 2).

2.4.59

(a) f  x  cos x  x3 is continuous on the interval 0,1, f 0  1  0, and f 1  cos11  0.46  0. Since 1  0  0.46, there is a number c in 0,1 such that f  c   0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x  x3  0 in the interval 0,1.

2.4.60

188

(b)

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

f (x)  ln x  3  2x is continuous on the interval [1, 2], f (1)  1  0, and f (2)  ln 2 1  1.7  0. Since –1 < 0 < 1.7, there is a number c in (1, 2) such that f (c)  0 by the Intermediate Value Theorem. Thus there is a root of the equation ln x  3  2x  0 or ln x  3  2x, in the interval (1, 2). 2.4.61 (a) Let f (x)  100ex/100  0.01x2. Then f (0)  100  0 and

f (100)  100e1 100  63.2  0. So by the Intermediate Value Theorem, there is a number c in (0, 100) such that f (c)  0. This implies that100ec/100  0.01c2. (b) Using the intersect feature of the graphing device, we find that the root of the equation is x  70.347.



2.4.62

1 8 1 . Then f 5    0 and f 6   0 , and f is continuous on x3 8 9 5,  . So by the Intermediate Value Theorem, there is a number c in 5, 6 such that f c   0 . This 1 implies that  c5 . c3 (a) Let f  x  x  5 

(b) Using the intersect feature of a graphing device, we find that the root of the equation is x  5.016 , correct to three decimal places.

2.4.63 (a)

Let f  x  ln x  e2x. Then f 1  e2  0 and f  2   ln 2  e4  0.67  0, and f is continuous on 1,   . So by the Intermediate Value Theorem, there is a number c in 1, 2 such that f  c   0. This implies that ln c  e2c.

Using the intersect feature of a graphing device, we find that the root of the equation is x  1.114, correct to three decimal places.

189

(b)

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.4.64 (a)

Let f  x  ln x sin x  2 

 

1 . Then f 0.2  0.37  0 and f 0.3  0.17  0, and f 1 x2

is continuous on 0,   . So by the Intermediate Value Theorem, there is a number c in

0.2, 0.3 such that f c   0. This implies that ln c  sin c  2 

1 . 1 c2

Using the intersect feature of a graphing device, we find that the root of the equation is x  0.265, correct to three decimal places.

2.4.65 The function ƒ(𝑥) = sin(𝑥2) is continuous everywhere and is positive for 0 < 𝑥 < √𝜋 and for √2𝜋 < 𝑥 < √3𝜋. The function is negative for√𝜋 < 𝑥 < √2𝜋. We know that √𝜋 = √3.14159 ≈ 1.77 and√2𝜋 ≈ 2.51, so ƒ(1) > 0, ƒ(2) < 0, ƒ(3) > 0. Therefore, f has an x-intercept between x = 1 and x = 2, and another x-intercept between x = 2 and x = 3. 2.4.66

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check 1

The function ƒ(𝑥) = 𝑥2 − 3 + is continuous for all 𝑥 G 0.ƒ(0.2) = 2.04 > 0, and 𝑥

ƒ(1) = −1 < 0, so f has a root (x-intercept) between x = 0 and x = 1. In addition, ƒ(2) = 1.5 > 0, so f also has an x-intercept between x = 1 and x = 2.

2.4.67

   If f is continuous at a, then by Theorem 4 with g(h)  a  h, we have





lim f (a  h)  f lim(a  h)  f (a). h0

h0

   Let   0. Since lim f (a  h)  f (a), here exists a δ > 0 such that xa

0  x a , then So if  f (x)  f (a)  f (a  (x  a))  f (a)  . Thus lim f (x)  f (a) and so f is continuous at a.

0  h    f (a  h)  f (a)  .

xa

2.4.68 lim sin( 𝑎 + ℎ) = lim(sin 𝑎 cos ℎ + cos 𝑎 sin ℎ) = (sin 𝑎)(1) + (cos 𝑎)(0) = sin 𝑎 ℎ→0

ℎ→0

2.4.69 As in the previous exercise, we must show that limcos(a  h)  cos a to prove that the cosine function is xa

continuous.

limcos(a  h)  limcos a coshsin a sinh  (cos a)(1) sin(a)(0)  cos a. h0

h0

2.4.70

 

If there is such a number, it satisfies the equation x3 1  x  x3  x 1  0 . Let the left-hand side of this equation be called f  x . Now f 2  5  0 , and f 1  1  0 . Note also that f  x is a polynomial, and thus continuous. So by the Intermediate Value Theorem, there is a number c between

2 and 1 such that f  c   0 , so that c  c3 1 .

2.4.71

0 if x is rational f (x)   is continuous nowhere. For any given number a and δ > 0, the interval 1 if x is irrational

 a   , a    contains both infinitely many rational and irrational numbers. Since f (a)  0 or 1, there © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

191

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check 

are infinitely many numbers x with 0  x  a   and f (x)  f (a)  1. Thus lim f (x)  f (a). In fact, xa

lim f (x) does not even exist. xa

2.4.72

0 if x is rational is continuous at 0. To see why, note that  x  g(x)  x , so by the  x if x is irrational Squeeze Theorem lim g(x)  0. But g is continuous nowhere else. For if a ≠ 0 and δ > 0, the interval g(x) 

x0

a   , a    contains both infinitely many rational and infinitely many irrational numbers. Since g(a)  0 or a, there are infinitely many numbers x with 0  x  a   and g(x)  g(a)  a / 2. Thus lim g(x)  g(a). xa

2.4.73 Assume a and b are positive and







a  ba  3 2  0. 2 x  2x 1 x  x  2 3



3 3 2 Then a x  x  2  b x  2x 1  0. Let denote the left side of the last equation. Since p is

continuous on [–1, 1], p(1)  4a  0, and p(1)  2b  0, there exists a c in (–1, 1) such that p(c)  0 by the Intermediate Value Theorem. Note that the only root of either denominator that is in (–1, 1) is









r  1 5 / 2 but p(4)  3 5  9 a / 2  0. Thus, c is not a root of either denominator, so

p(c)  0  x  c is a root of the given equation.

2.4.74

f (x)  x4 sin(1/ x) is continuous on , 0 0,  since it is the product of a polynomial and a composite of a trigonometric function and a rational function. Now since 1  sin(1/ x)  1, we have x4  x4 sin(1/ x)  x4. Because limx4   0 and limx4   0, the Squeeze Theorem gives us that





x0

lim x sin(1/ x)  0 which equals Thus, f is continuous at 0, and hence, on ,   . x0

2.4.75

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(a) lim 𝐹(𝑥) = 0 and lim 𝐹(𝑥) = 0, so lim 𝐹(𝑥) = 0, which is F(0) and hence F is continuous at x  a. 𝑥→0+

𝑥→0−

𝑥→0

For a  0, lim F (x)  lim x  a  F (a). For a < 0, xa

xa

lim 𝐹(𝑥) = lim(−𝑥) = −𝑎 = 𝐹(𝑎). Thus, F is continuous at x = a; that is, continuous everywhere. 𝑥→𝑎

𝑥→𝑎

(b) For any a, lim𝑥→𝑎|ƒ(𝑥)| = |limƒ(𝑥)| = |ƒ(𝑎)|. Therefore, if f is continuous on I, then f(x) is 𝑥→𝑎

continuous on I.

if 𝑥 ≥ 0

which is not continuous at x = 0.

−1, if 𝑥 < 0 However, |f(x)| is continuous everywhere. 2.4.76

Define u t  to be the hiker’s distance from their starting point S, as a function of time, on the first day, and define d t  to be their distance from S, as a function of time, on the second day. Let D be the distance from S to the top of the mountain. From the given information we know that u 0  0, u 12  D, d  0   D, and d 12  0. Now consider the function u  d, which is clearly continuous. We calculate that u  d  0   D and u  d 12  D. So by the Intermediate Value Theorem, there must be some time t0 between 0 and 12 such that

u  d t0   0  u t0   d t0 . So at time t0 after 7:00 AM, the hiker will be at the same place on both days.

Solution and Answer Guide

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STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.5 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.5.1 (a) As x approaches 2 (from the right or the left), the values of f(x) become large. (b) As x approaches 1 from the right, the values of f(x) become large negative. (c) As x becomes large, the values of f(x) approach 5. (d) As x becomes large negative, the values of f(x) approach 3.

194

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.1 95 (a) The graph of a function can intersect a vertical asymptote in the sense that it can meet but not cross it.

The graph of a function can intersect a horizontal asymptote. It can even intersect its horizontal asymptote an infinite number of times.

(b) The graph of a function can have 0, 1, or 2 horizontal asymptotes. Representative examples are shown.

195



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.1 96 (a)

limx2 f  x  

(b)

lim x1  f  x  

(c)

lim x1 f  x  

(d)

limx f  x  1

(e)

limx f  x  2

(f)

Vertical: x  1, x  2 ; Horizontal: y  1, y  2

2.5.4 (a) limx g  x  2 (b) limx g  x  2 (c) limx3 g  x   (d) limx0 g  x   (e) lim x2  g  x   (f) Vertical: x  2, x  0, x  3 ; Horizontal: y  2, y  2 2.5.5

lim f (x)   x0

lim f (x)  5

x

lim f (x)  5 x

196



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2.5.6

lim f (x)   x2

lim f (x)  

lim f (x)  

x2

x2

lim f (x)  0

x

x

f (0)  0

2.5.7 lim f (x)   x2

lim f (x)   x

lim f (x)  0

x

lim f (x)  

x0

lim f (x)  

x0

 



2.5.8

lim f (x)  3 x

lim f (x)  

x2

lim f (x)  

x2

f is odd

2.5.9

f (0)  3 lim f (x)  4

x0

lim f (x)  2

x0

lim f (x)  

lim f (x)  3

x

x

lim f (x)  

lim f (x)  

x4

x4

197

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.10

lim f (x)   x3

lim f (x)  2 x

f (0)  0 f is even 2.5.11 If f (x)  x / 2 , then a calculator gives 2

f (0)  0, f (1)  0.5, f (2)  1, f (3)  1.125, f (4)  1, f (6)  0.5625, f (5)  0.78125, f (7)  0.3828125, f (8)  0.25, f (9)  0.158203125, f (10)  0.09765625, f (20)  0.00038147, f (50)  2.22041012, f (100)  7.88861027. Combining this information with a





graph, it appears that lim x2 / 2x  0. x



2.5.12 From a graph of f (x)  1 2 / x 

in a window of [0, 10,000] by [0, 0.2],

we estimate that lim f (x)  0.14. From the table below, we estimate that x

lim f (x)  0.1353. x

f (x)

10,000 100,000 1,000,000

0.135 308 0.135 333 0.135 335

198



 

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.19 9 1 (a) f  x  3

x 1

From these calculations, it seems that

f  x

0.5

1.14

1.5

0.42

0.9

3.69

1.1

3.02

0.99

33.7

1.01

33.0

0.999

333.7

1.001

333.0

0.9999

3333.7

1.0001

3333.0

0.99999

33, 333.7

1.00001

33, 333.3

(b) If x is slightly smaller than 1 , then x3 1 will be a negative number close to 0 , and the reciprocal of

x3 1 , that is, f  x , will be a negative number with large absolute value. So lim x1 f  x   .

If x is slightly larger than 1 , then x3 1 will be a small positive number, and its reciprocal, f  x , will be a large positive number. So lim x1 f  x   . (c) It appears from the graph of f that lim x1 f  x   and lim x1 f  x  



199

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.20 0 Vertical: x  1.62, x  0.62, x  1; Horizontal: y  1

2.5.15

lim

x 1

  since the numerator is positive and the denominator approaches 0 from the positive side

x 5 as x  5.



x5



2.5.16



x2

lim



x3

x3

   since the numerator is negative and the denominator approaches 0 from the negative

side as x  3 . 2.5.17

lim

2  x   since the numerator is positive and the denominator approaches 0 through positive (x 1)2 values as x 1 . x1

2.5.18

lim x3



 x  3

  since the numerator is positive and the denominator approaches 0 through positive

values as x  3. 2.5.19 









Let t  x  9 . Then as x  3 , t  0 , and lim x3 ln x  9  lim t 0  lnt   by (3). 2

200



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.20 Let t  3 /  2  x  . As x  2 , t   . So lim x  e 

3/2x

 lim t2 et  0 by (7).

2.5.21 Let t  sin x. As x  0 , t  0 , and lim ln sin x  lim ln t   by (4). x0

t 0

2.5.22

1   since the numerator is positive and the denominator approaches 0 lim 1 sec x  lim    x cos x x  x x 2

through negative values—the product of x, which is positive, and cos x, which is negative—as

x . 2 

2.5.23

cosx    since the numerator is negative and the denominator approaches 0 sinx  through positive values as x   .

limx cotx  lim x 

 



2.5.24



 x   since the numerator is positive and the denominator approaches 0 sinx through negative values as x  2  .

lim x2 xcscx  lim x2 



2.5.25

 



 

 x  x  2 x2  2x x since the numerator is positive and the lim x2  2  lim x2   lim    x2  x  4x  4 (x  2)2 x2 denominator approaches 0 through negative values as x  2 . 



  x  4  x  2

2.5.26

  since the numerator is negative and the denominator lim x  2x  8  lim 2 x2 x  5x  6 x2  x  3 x  2 2



approaches 0 through negative values—the product of x  3, which is negative, and x  2, which is positive—as x  2.

201

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.27

3x  2

lim

 lim

2x 1

x

3 2 2

x 1 x



3  2(0) 3  20 2

2.5.28

1 2 1 x x3 0  2(0) 1 x  lim lim 3  1 0  0  0 x x  x 1 x 1 1  1 2 3 x

2.5.29

1  2 x  2 lim  lim x x2  0  2(0)  0 x 11 x x2 1 1 0 x2

2.5.30 3

4x  6x2  2

4  6x  2x3  4  0  0  2  lim 3 x 2  4  5 200 x 2x  4x  5 2 3 lim

2.5.31





2 2 1 1 0 1 2 t t /t t  t t3/2 lim  lim  lim   1 t 2 t 2t  t 2 t 2t  t 2 / t 2  t 1 0 1  

2.5.32

1 1 0 1 1 t t t t1/2  lim    3/2 1 t 2t  3t  5 t 2  3 1/2  5 3/21 200 2

lim





2.5.33

 2x2 1  lim 2x2 1 / x4  lim 2  x   x2  2x 1 / x 2   x 2  x  / x 2   x 1  x2  x   x 1  x2  x  / x4    2    lim (1 0)  4 1  x 1   1 0(200)    2

lim x

x

 2.5.34

lim x

x2  lim x4 1 x

1 x4 x4

 4 1

 lim x

1 1 1



1 1 0

1

202

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.35 6 6 3 lim 1 4x  lim 1 4x / x  lim 2 x x 2  x3 x 1 x3

1 x6

4

2 x3

1



0  4  2  2  0 1 1



2.5.36

Since x3   6 for x  0 x  

6 6 3 lim 1 4x  lim 1 4x / x 2 x x 2  x3 1 x3

 lim  x



2 x3

1 x6

4



1

 0  4 2  2 0 1 1

2.5.37 1 2 2 lim x  3x  lim x  3x / x  lim x  3  0  3  3 x 4  1 x 4x 1 x  4x 1 / x 40 4 x

2.5.38 1  3x  lim x   x 4x 1 1 x 4  x since as x  , 1 3x   and 4  1  4.

lim

x  3x2



 9x  x  3x  9x  x  3x 9x  x  9x lim  9x  x  3x   lim   lim lim

2.5.39

x 9x2  x  3x 9x2  x  3x 1 1 1 1     lim x 9  1  3 90 3 33 6 x

x



2.5.40

lim



x

x

x 9x2  x  3x

 4x2  3x  2x  4x2  3x  4x2   lim 4x2  3x  2x  lim 4x2  3x  2x  2 x x 4x2  3x  2x  4x  3x  2x   3 3x 3  x   lim x 4 40 2  4xx2  3x 2 x2







203

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.41



 x  ax  x  bx  x  ax  x  bx  lim  x  ax  x  bx  lim x  ax  x  bx 2

x

x  ax  x  bx 2

 lim

(a  b)x

 lim

x2  ax  x2  bx x x2  ax  x2  bx a b a b ab  lim   x 1 a  1 b 2 1 0  1 0 x x x

2.5.42 Let t  x2 . As x   ,t   . So lim x ex  lim t et  0 by (7). 2

2.5.43 For x > 0,

x2 1  x2  x. So as x  , we have

x2 1  , that is lim x2 1  . x

2.5.44

limx cosx does not exist because as x increases cosx does not approach any one value, but oscillates between 1 and 1. 2.5.45

sin2 x

x2 x sin2 x Squeeze Theorem, limx 0 x2

Since 0  sin x  1, we have 0 

 2 . Now limx 0  0 and limx

 0 , so by the

2.5.46 Since 1  cos x  1 and e2x  0, we have e2x  e2x cos x  e2x. We know that

lime2x   lime2x   0, so by the Squeeze Theorem, lim e2x cos x  0. x

x

 

2.5.47 3x 3x 6 x 1 0 Divide the numerator and denominator by e3x : lim e  e  lim 1 e  1 x e3x  e3x x 1 e6 x 1 0

204

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.48

lim 

1  1  [factor out the largest power of x   because x5   x  x  x   and 1/ x 1 1 as x   .

 x 4  x5   lim 



Or: lim x x 4  x5  lim x x4 1 x   

2.5.49 If we let t  tanx , then as x  ( / 2) , t   . Thus, lim x(



etanx  lim t / 2) et  0 .

2.5.50

1  2  1 x2   x  1 x2    x lim ln 1 x   ln 1 x  limln  ln  lim   ln  lim 1   , since the limit in x x x 1 x 1 x    x x 1  

parentheses is ∞.



2.5.51



 2 1   2  x   1  x lim ln  2  x   ln 1 x    lim ln   lim ln  ln       ln1  0 1 1  x x x  1   1 x  x 

2.5.52

(a) (i) lim f (x)  lim x0



 0 since x  0 and ln x   as x  0 .

x0

ln x x (ii) lim f (x)  lim   since x 1 and ln x  0 as x 1. x1  x1 ln x x (iii) lim f (x)  lim  since x 1 and ln x  0 as x 1. x1  x1 ln x 





(b)(b) x

f (x)

10,000 1085.7 100,000 8685.9 1,000,000 72,382.4 It appears that lim f (x)  . x

205



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.53 1  2 2 (a) (i) lim f (x)  lim    0 since  0 and

1  0 as x  . ln x x x x ln x x 1   2 1  2 (ii) lim f (x)  lim     as   since   and x0 ln x ln x  x0  x x x  0. 1  2 1 2 (iii) lim f (x)  lim     as   since  2 and x1 ln x ln x  x1  x x x  1. 1  2 1 2 (iv) lim f (x)  lim    as x  1.   since  2 and x1 x ln x x1  x ln x 



2.5.54



2 (a) From the graph of f (x)  x  x 1  x we estimate the value lim f (x) to be –0.5.

x

(b) Using the table we estimate lim f (x)  0.5. x

f (x)

–10,000 –100,000 –1,000,000

–0.499 962 5 –0.499 996 2 –4.99 999 6

(c)(c) lim



x

 x2  x 1  x   x 2  x 1 x2    lim x2  x 1  x  lim x2  x 1  x  2 x x x2  x 1  x  x  x 1  x  (x 1)1/ x 1 1/ x  lim  lim x x2  x 1  x 1 / x  x  1 1x  x12 1











1 1 0  2  1 0  0 1

206

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.55 (a) From the graph of f (x)  3x  8x  6  3x2  3x 1, we estimate the value of lim f (x) to be –0.5. 2

x

From the table, we estimate the limit to be 1.4434. x f (x) 10,000 100,000 1,000,000 (b)

1.443 39 1.443 38 1.443 38

 3x  8x  6  3x  3x 1 3x  8x  6  3x  3x 1 lim f (x)  lim 2

3x2  8x  6  3x2  3x 1  3x2  8x  6   3x2  3x 1 lim  5x  51/ x  lim   x 3x2  8x  6  3x2  3x 1 x 3x2  8x  6  3x 2  3x 1 (1/ x)

x



 lim x



5 3 55/ x 5  5   1.443376  lim 8 6 3 1 x 6 3  x  x2  3  x  x2 3 3 2 3

2.5.56 5 5  4x  lim x  4  0  4  4, so y  4 is a horizontal x x  3 x 1 3x 1 0 5  4x , so lim f (x)   since asymptote. y  f (x)  x3 x3 5  4x  7 and x  3  0 as x  3. Thus, x  3 is a vertical asymptote. The graph confirms our work.

lim

2

2.5.57

2 2x 1  lim 2   , so y  is a horizontal  x3  2 1 x 3x2  2x 1 3 3 x x2 lim

1 x2

asymptote. y  f (x) 



2x2 1





3x  2x 1 2

denominator is zero when x 

1 3 and



2x2 1 

. The

3x 1 x 1

x  1 but the numerator is

nonzero, so x  13 and x  1 are vertical asymptotes. The graph confirms our work.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.58

2x2  x 1

1 12 200  lim 1 x  x   2, so y  2 is a horizontal x1 1 2 x x2  x  2 1 0  0 x x2

lim

asymptote. y  f (x) 

2x2  x 1



(2x 1)(x 1)

, so x  x2 (x  2)(x 1) lim f (x)  , lim f (x)  , lim f (x)  . Thus, x  2 and

x2

x1

x1

x 1 are vertical asymptotes. The graph confirms our work. 2.5.59

1 x4

x

x 1 x2

1



0 1

 1, so y  1 is a horizontal 0 1 x x 4 4 1 x4  asymptote. y  f (x)  1 x  1 x

lim

 lim

1 1 x4

x2 1 x2  x2 1 x1 x The denominator is zero when x  0, 1 and 1, so these are vertical asymptotes. Notice that as x  0, the numerator and denominator are both positive, so lim f (x)  . The graph confirms our work. x2  x4

x0



2.5.60

y  f (x) 

x3  x





x  x 2 1



x(x 1)(x 1)

x(x 1)  (x 1)(x  5)  x  5



x2  6x  5 (x 1)(x  5) for  g(x) for x  1. The graph of g is the same as the graph of f with the exception of a hole in the graph of f at x  1. By long division, x2  x 30 g(x)   x  6  . As x  , g(x)  , so there is x 5 x 5 no horizontal asymptote. The denominator of g is zero when x  5, and lim g(x)   and lim g(x)  , so x  5 is a vertical asymptote. The graph confirms our work. x5

x5

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.61

2 2 x x   2, so y  2 is a  lim 2e 1/ e lim   x ex  5 1/ ex x ex  5 x 1 5 / ex  1 0   x 2e 2 horizontal asymptote. lim   0, so is a horizontal x ex  5 05 x

lim 2e

asymptote. The denominator is zero (and the numerator isn’t) when ex  5  0  ex  5  x  ln 5. Thus, x  ln 5 is a vertical asymptote. The graph confirms our work. 2.5.62 From the graph, it appears that y = 1 is a horizontal asymptote. x 3x3  500x2 3  500  lim x x3  500x2 100x  2000 x 1 500  100  2000

lim



3 0  3, so y = 3 is a horizontal asymptote. 1 0  0  0

The discrepancy can be explained by the choice of the viewing window. Try [–100,000, 100,000] by [–4, 4] to get a graph that lends credibility to our calculation that y = 3 is a horizontal asymptote. 2.5.63 (a) From the graph, it appears at first that there is only one horizontal asymptote, at y ≈ 0, and a vertical asymptote at x ≈ 1.7. However, if we graph the function with a wider and shorter viewing rectangle, we see that in fact there seem to be two horizontal asymptotes: one at y ≈ 0.5 and one at y ≈ –0.5. So

2x2 1 2x2 1  0.5 and lim   0.5 x 3x  5 x 3x  5

we estimate that lim 

209



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2x2 1  0.47. x 3x  5

(b) f (1000)  0.4722 and f (10,000)  0.4715, so we estimate that lim 

2x2 1  0.47. (c) x 3x  5

f (1000)  0.4706 and f (10,000)  0.4713, so we estimate that lim 

1 2 2  0.471404. For x < 0, we have x2  x  x, so when we lim 2x 1 lim 2  x2  x 3x  5 x 3  5 3 x 1 1 divide the numerator by x, with x < 0, we get 2x2 1   2x2 1   2  x12 . Therefore, 2 x x 1 2 2  0.471404. lim 2x 1  lim  2  x2   x 3x  5 x 3  5x 3

2.5.64 From the sketches (below) we see that

(a) lim x  n

x0

1

if n  0 if n  0

0  if n  0 

1 if n  0 n (c) lim x   if n  0  0 if n  0 x 

if n  0 1 if n  0 0 (b) lim x    x0   if n  0, n odd  if n  0, n even n



1  n (d) lim x    x  0

if n  0 if n  0, n odd if n  0, n even if n  0

210

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.65 Let’s look for a rational function: (1) lim f (x)  0  degree of numerator < degree of denominator x

(2) lim f (x)    there is a factor of x2 in the denominator (not just x since that would produce a sign x0

change at x = 0), and the function is negative near x = 0. (3) lim f (x)   and lim f (x)    vertical asymptote at x = 3; there is a factor of (x  3) in the x3

x3

denominator. (4) f (2)  0  2 in an x-intercept; there is at least one factor of (x  2) in the numerator. Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us f (x) 

2 x as one possibility. x (x  3) 2

2.5.66 Since the function has vertical asymptotes at x = 1 and x = 3, the denominator of the rational function we are looking for must have factors (x 1) and (x  3). Because the horizontal asymptote is y = 1, the degree of the numerator must equal the degree of the denominator, and the ratio of the leading coefficients must be one. One possibility is f (x) 

x2 . (x 1)(x  3)

2.5.67 (a) We must first find the function f. Since f has a vertical asymptote x = 4 and x-intercept x  1, x  4 is a factor of the denominator and x 1 is a factor of the numerator. There is a hole at x = –1, so x  (1)  x 1 is a factor of both the numerator and denominator. Thus f now looks like this:

a(x 1)(x 1) , where a is still to be determined. Now (x  4)(x 1) a(x 1)(x 1) a(x 1) a(11) 2 2 lim f (x)  lim  lim   a, so a  2, and a = 5. Thus is a x1 x1 (x  4)(x 1) x1 x  4 (1 4) 5 5 5(1)(1) 5 ratio of quadratic functions satisfying all the given conditions and f (0)   . (4)(1) 4 1 0 1 x2 1   2 lim f (x)  5lim  5lim 1 x  5   5(1)  5 2 3 4 x x x x  3x  4 1 0  0    1 x  2 f (x) 

(b)(b)

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.68

y  f (x)  2x3  x4  x2 2  x. The y-intercept is f (0)  0. The x-intercepts are 0 and 2. [found by solving f (x)  0 for x]. There are sign changes at 0 and 2 (odd exponents on x and 2 – x). As x  , f (x)   because

x3   and 2  x  .As x  , f (x)   because x3   and 2  x  . Note that he graph of f near x = 0 flattens out – that is, looks like y  x3. 2.5.69





y  f (x)  x4  x6  x4 1 x2  x4 (1 x)(1 x). The y-intercept is f (0)  0. The x-intercepts are 0, –1, and 1 [found by solving f (x)  0 for x]. Because x  0 for x ≠ 0, f doesn’t change sign change at x = 0. The function does change sign at 4



x = –1 and x = 1. As x  , f (x)  x4 1 x2

 approaches –∞ because a

x4  and 1 x2   . because x3  ,(x  2)2  , and (x 1)  . 2.5.70

y  f (x)  x3(x  2)2(x 1). The y-intercept is f (0)  0. The x-intercepts are 0, –2, and 1. There are sign changes at 0 and 1 (odd exponents on x and x–1). There is no sign change at x = –2. Also, f (x)   as x   because all three factors are large. And f (x)   as x   because x3  ,(x  2)2  , and (x 1)  . Note that the graph of f at x = 0 flattens out – i.e looks like y  x . 3

2.5.71

y  f (x)  (3  x) 1 x  (1 x) 4 . The y-intercept is f (0)  3(1)2(1)4  3. The xintercepts are 3, –1, and 1. There is a sign change at 3 but not at –1 or 1. When x is large and positive, (3  x) is negative and the other factors are positive so lim f (x)  . 2

x

When x is large but negative, (3  x) is positive so lim f (x)  . x

212

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.5.72

y  f (x)  x2  x2 1 (x  2)  x2  x 1  x 1 (x  2). The y-intercept is f (0)  0. 2

The x-intercepts are 0, –1, 1, and 2. There is a sign change at –2 but not at 0, –1 or 1. When x is large and positive, all the factors are positive so lim f (x)  . When x is x

large but negative, only is negative so lim f (x)  . x

2.5.73 (a) Since 1  sin x  1 for all x, 



sin x



for x > 0. As x  , 1/ x  0 and1/ x  0, so by

x x x sin x the Squeeze Theorem (sin x) / x  0. , Thus lim  0. x x

(b) From part (a), the horizontal asymptote is y = 0. The function y  (sin x) / x crosses the horizontal asymptote whenever sin x  0; that is, at x  n for every integer n. Thus, the graph crosses the asymptote an infinite number of times.

2.5.74 (a) In both viewing rectangles,

lim P(x)  lim Q(x)   and lim P(x)  lim Q(x)  . In the larger viewing rectangle, P and Q x

x

x

become less distinguishable. 5 2

P(x)

lim (b) x Q(x)

 lim x

3x  5x  2x 3x5

 lim x

3  x53  4x4 3

1

2.5.75

213

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No, because the calculator-produced graph of f  x  ex  ln x  4 looks like an exponential function, but the graph of f has an infinite discontinuity at x  4 . A second graph, obtained by increasing the numpoints option in Maple, begins to reveal the discontinuity at x  4 . (b) There isn't a single graph that shows all the features of f . Several graphs are needed since f looks like ln x  4 for large negative values of x and like ex for x  5 , but yet has the infinite discontiuity at x  4 .

A hand-drawn graph, though distorted, might be better at revealing the main features of this function.

2.5.76 (a) Divide the numerator and the denominator by the highest power of x in Q  x . (a) If degP  degQ , then the numerator  0 but the denominator doesn't. So

lim x P   x  / Q  x   0 . (b) If degP  degQ , then the numerator   but the denominator doesn't, so

lim x P   x  / Q  x    (depending on the ratio of the leading coefficients of P and Q ). 2.5.77

214



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

5 1/ x 5   5 and  lim lim 5 x  lim 5 x  x x 1 x x 1 1/ x x 1 1/ x 1 0

10   21/ e  10  0 10ex  21 1/ ex  lim   lim   5. Since lim x x 2ex 2ex 1/ ex x 2 2 10ex  21 5 x  f (x)  , we have lim f (x)  5 by the Squeeze Theorem. x x 2e x 1 x

10ex  21

2.5.78 (a) After t minutes, 25t liters of brine with 30 g of salt per liter has been pumped into the tank, so it contains 5000  25t  liters of water and 25t 30  750t grams of salt. Therefore, the salt concentration at time t will be C(t) 

750t 30t g  . 500  25t 200  t L 30 30  30. So the salt concentration approaches that of the (b) lim C(t)  lim 30t  lim  t t 200  t t 200 1 0 1 t 

brine being pumped into the tank.



2.5.79

limvc  m  lim vc 





2.5.80



. As v  c , 1 v / c  0 , and m   . 2

1 v / c





(a) limv(t)  limv 1 egt/ve  v (1 0)  v t

t

9.8t

(b) We graph v(t)  1 e

and v(t)  0.99v e, or in this case v(t)  0.99.

Using an intersect feature or zooming in on the point of intersection, we find that t ≈ 0.47 s.

2.5.81 (a) y  ex/10 and y  0.1 intersect at x1 ≈ 23.03. If x  x1 , then ex/10  0.1. (b) ex/10  0.1   x / 10  ln 0.1  x  10ln 110

 10ln101  10ln10  23.03

2.5.82

215

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(a) limx f  x  limx

4x2  5x 2x2 1

45/ x  limx

2 1/ x2

4 

2

(b) f  x  1.9  x  25.3744 , so f  x  1.9 when x  N  25.4 .

f  x  1.99  x  250.3974 , so f  x  1.99 when x  N  250.4 .

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.6 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.6.1

f (x)  f (3) . x3 f (x)  f (3) (b) The slope of the tangent line at point P is lim . x3 x3 (a) The slope of the secant line through points P and Q is

2.6.2

f (x)  f (1)

x1

x 1

x1

x 1

 lim

4x  4

 lim 4  4 x1 x 1  (b) lim f (x)  f (2)  lim 4x  3  11  lim 4x  8  lim 4( x  2)  4 x2 x2 x2 x  2 x1 x  (2) x2 x2 (a) lim

 lim

 4x  3 1

x1



f (x)  f (a) 4x  3  4a  3 lim 4x  4a  lim 4(x  a)  4  lim xa xa xa xa xa xa xa 

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2.6.3 As we zoom in near the point (0, 1), the graph of f (x)  ex seems to become a line with slope 1.

2.6.4 Using Definition 2, the slope of the tangent line to the graph of y  4x  x at (1, 3) is 2

4x  x2   3   x 2  4x  3  x  1 x  3  f ( x)  f (1) m  lim  lim  lim  lim  x1 x1 x1 x1 x 1 x 1 x 1 x 1 

 lim  x  3  lim3  x  3  1  2. x1

x1

Using the alternative definition, the slope of the tangent line at (1, 3) is

4 1  h   1  h 2   3 2 f (1  h)  f (1)  m  lim  lim   lim 4  4h 1  2h  h  3 h0 h0 h h h0 h h h  2 h2  2h  limh  2  2.  lim h0 h0 h0 h h

 lim

(b) An equation for the line tangent to the graph of y  4x  x2 at (1, 3) is

y  f (1)  x  1  f (1)  2  x 1  3 or y  2x  1. (c) As you continue to zoom in near the point (1, 3) it becomes clear that the graph of the line y  2x 1 is tangent to the graph of y  4x  x2 at the point (1, 3). As you zoom in even further, the parabola and tangent line become indistinguishable.

2.6.5 Using Definition 2, the slope of the tangent line to the graph of y  x  x at (1, 0) is 3

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f ( x)  f (1)  x  x   0  lim x 1  x  lim x 1  x1  x  lim x 1  x  x  1  lim x1 x1 x1 x1 x1 x 1 x 1 x 1 x 1 x 1  lim x 1 x  1(2)  2. 3

m  lim



x  1

Using the alternative definition, the slope of the tangent line at (1, 3) is

1  h   1  h 3   0 1  h  1 3h  3h 2  h3      lim  lim h0 h0 h0 h h h 2 3 2 h h  3h  2 h  3h  2h  lim  lim  lim h2  3h  2  2. h0 h0 h0 h h

m  lim f (1 h)  f (1)

(b) An equation for the line tangent to the graph of y  x  x3 at (1, 0) is

y  f (1)  x 1  f (1)  2  x  1  0 or y  2x  2. (c) As you continue to zoom in hear the point (1, 0) it becomes clear that the graph of the line

y  2x  2 is tangent to the graph of y  x  x3 at the point (1, 0). As you zoom in even further, the parabola and tangent line become indistinguishable. 2.6.6 With f (x)  4x  3x2 and the point 2, 4, the slope of the tangent line is

4x  3x2  4 f (x)  f (2) 4x  3x2  4 (3x  2)(x  2)  lim  lim  lim x2 x2 x2 x2 x2 x2 x2 x2

m  lim



 lim (3x  2)  8; x2

The equation of the tangent line is y  8 x  2  4 or y  8x 12. 2.6.7 With f (x)  x3  3x  1 and the point2, 3, the slope of the tangent line is



m  lim f (x)  f (2)  lim x  3x  1 3  lim x  3x  2 lim (x 1) (x  2) x2 x2 x2 x2 x2 x2 x2 x2 3



 lim (3x  2)  9; The equation of the tangent line is y  9  x  2  3 or y  9x  15. x2

2.6.8



With f (x)  x and the point 1,1, the slope of the tangent line is

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



(b)

 x  1 lim x  1  x  1  x  1 x  1



m  lim f (x)  f (1)  lim x  1  lim x  1 x1 x1 x1 x 1 x 1  x  1  lim x1

x1

1 1  ; The equation of the tangent line is y  1  x  1  1 or y  1 x  1 . 2 2 2 x 1 2

2.6.9 With f (x) 

2x  1 x2

and the point 1,1, the slope of the tangent line is

2x  1   x  2 2x  1  1 f ( x)  f (1) x 1 x2 x2  lim  lim m  lim  lim x1 x1 x1 x1 x 1 x 1 x 1  x  1 x  2 1 1 1 1  lim   ; The equation of the tangent line is y   x  1  1 or y  1 x  2 . 3 3 3 x1 x  2 1 2 3 2.6.10 For f ( x)  3  4x  2x 2

(a) The slope of the tangent line at the point where x = a is f (a)  lim

f (a  h)  f (a) h

h0

 lim h0

3  4(a  h)2  2(a  h)3  3  4a2  2a3  

3  4(a2  2ah  h2 )  2(a3  3a2h  3ah2  h3)  3  4a2  2a3 h0 h

 lim

3  4a2  8ah  4h2  2a3  6a2h  6ah2  2h3  3  4a2  2a3 h0 h

 lim

h 8a  4h  6a  6ah  2h  8ah  4h2  6a2h  6ah2  2h3  lim h0 h0 h h 2

 lim



 lim 8a  4h  6a2  6ah  2h2  8a  6a2. h0

 

At (1,5) the slope of the tangent line is 8(1)  6 12  2; The equation of the tangent line is

y  2(x 1)  5 or y  2x  3.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

   8; The equation of the tangent line is

At (2, 3) the slope of the tangent line is 8(2)  6 22

(c)

y  8(x  2)  3 or y  8x 19. (d)(d)

2.6.11 For f (x) 

1 x

(a) The slope of the tangent line at the point where x = a is f (a)  lim f (a  h)  f (a) h0

 a  x  a  x  ax  lim  lim  lim  lim xa xa ax  x  a   a  x  ax  x  a   a  x  1

 1a

xa

a x

xa

1

 lim xa

1



 a  x 

xa



2 a 

1 1 or  a3/2 [a  0]. 3/2 2a 2

(b) At (1,1) the slope of the tangent line is  12 (1)3/2   21 ; The equation of the tangent line is

y   1 ( x 1)  1 or y   1 x  3 . 2

3/2

  1 ; The equation of the tangent line is 16

y   161 (x  4)  12 or y   161 x  34 . (d)(d)

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2.6.12 (a) The particle is moving to the right when s is increasing; that is, on the intervals 0,1 and 4, 6. The particle is moving to the left when s is decreasing; that is, on the interval 2, 3. The particle is standing still when s is constant; that is, on the intervals 1, 2 and 3, 4  . (b) The velocity of the particle is equal to the slope of the tangent line of the graph. Note that there is no slope at the corner points on the graph. On the interval 0,1, the slope is

2, 3, the slope is

1 3 3 2

3 0 1 0

 3. On the interval

 2. On the interval 4, 6  , the slope is

3 1 64

 1.

2.6.13 (a) Runner A runs the entire 100-meter race at the same velocity since the slope of the position function is constant. Runner B starts the race at a slower velocity than runner A, but finishes the race at a faster velocity. (b) The distance between the runners is greatest at the time when the largest vertical line segment fits between the two graphs – this appears to be somewhere between 9 and 10 seconds. (c) The runners had the same velocity when the slopes of their respective position functions are equal – this also appears to be about 9.5 seconds. Note that the answers for part (b) and (c) must be the same for these graphs because as soon as the velocity for runner B overtakes the velocity for runner A, the distance between the runners starts to decrease.

2.6.14 Let s(t)  40t 16t . 2

s(t)  s(2)  lim 40t  16t   16  lim 16t2  40t  16  lim 82t  5t  2 t2 t2 t2 t2 t2 t2 t2 t2 8t  22t 1  lim  8lim2t 1  8(3)  24. t2 t2 t2

v(2)  lim



Thus, the instantaneous velocity when t = 2 is –24 ft/s.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.6.15



(a) Let H (t)  10t 1.86t2.

v(1)  lim

 2 H 1  h  H 1  lim 10(1  h)  1.86(1  h)   (10  1.86) h

h0

 lim

h0

10  10h  1.86 1  2h  h2  10  1.86 

h0

10 10h 1.86  3.72h 1.86h2 10 1.86 h0 h 2 6.28h 1.86h  lim6.28 1.86h  6.28  lim h0 h0 h  lim

The velocity of the rock after one second is 6.28 m/s.

10(a  h)  1.86(a  h)   (10a  1.86a ) H a  h  H a  lim  h0 h0 h h 2

(b) v(a)  lim

 lim

10a  10h  1.86 1  2ah  h2  10a  1.86a2  

h0

10a 10h 1.86a2  3.72ah 1.86h2 10a  1.86a2 10h  3.72ah 1.86h2  lim h0 h0 h h h 10  3.72a 1.86h  lim  lim10  3.72a 1.86h  10  3.72a  lim

h0

The velocity of the rock when t = a is (10 – 3.72a) m/s. (c) The rock will hit the surface when H = 0. This is when 10t 1.86t2  0  t(10 1.86t)  0  t  0 or1.86t  10. The rock hits the surface when t  10 / 1.86  5.4 s. (d) The velocity of the rock when it hits the surface is 10 10 v  1.86   10  3.72  1.86   10  20  10 m/s.

2.6.16 𝑎2 − (𝑎 + ℎ)2 1 1 2 2 2 − 𝑠(𝑎 + ℎ) − 𝑠(𝑎) 2 𝑎2 lim 𝑎2(𝑎 + ℎ)2 = lim 𝑎 − (𝑎 + 2𝑎ℎ + ℎ ) 𝑣(𝑎) = lim = lim (𝑎 + ℎ) ℎ ℎ ℎ ℎ ℎ→0 ℎ→0 ℎ→0 ℎ→0 −(2𝑎ℎ + ℎ2) −ℎ(2𝑎 + ℎ) −(2𝑎 + ℎ) = − 2𝑎 = lim 2 2 2 2 2 = lim 2 2 = lim 2 ℎ→0 ℎ ℎ→0 𝑎 (𝑎 + ℎ) ℎ→0 ℎ𝑎 (𝑎 + ℎ) 𝑎 (𝑎 + ℎ) 𝑎 ·𝑎 1 = − 3 m/s 𝑎 © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

So 𝑣(1) = −

2 13

= −2 m/s, 𝑣(2) = −

2 23

= m/s and 𝑣(3) = −

=−

m/s.

2.6.17 (a) The average velocity between times t and t + h is

s(t  h)  s(t)  (t  h)  t 

 t  h   6  t  h   23  12 t 2  6t  23  1 t 2  th  1 h2  6t  6h  23  1 t 2  6t  23 2 2

2

th  12 h2  6h h

h t  12 h  6   t  1 h  6 ft/s. 2 h

(i) [4, 8]: t = 4, h = 8 – 4 = 4, so the average velocity is 4  12(4)  6  0 ft/s. (ii) [6, 8]: t = 6, h = 8 – 6 = 2, so the average velocity is 6  12(2)  6  1 ft/s. (iii) [8, 10]: t = 8, h = 10 – 8 = 2, so the average velocity is 8  21 (2)  6  3 ft/s. (iv) [8, 12]: t = 8, h = 12 – 8 = 4, so the average velocity is 8  21 (4)  6  4 ft/s. (b) v(t)  lim h0

s(t  h)  s(t) h

 limt  1 h  6  t  6, so v(8)  2 ft/s. h0

(c)

2.6.18

(a)

The average velocity between times 4 and t is

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



s t   s  4  t  t  4  4 t  4 t 2 t 4 t 2     t4 t4 t 4 t 4 t 4 t 2 t 2 t4   1  1 t4 t 2 t  4  t  2  1   1 m/s.   t  2  







(i) 

(ii) 

(iii) 

(iv)

4, 5: t  5, so the average velocity is 1

1    1.236 m/s. 52    1   4, 4.5: t  4.5, so the average velocity is 1 4.5  2  1.243 m/s.   1   4, 4.1: t  4.1, so the average velocity is 1 4.1  2  1.248 m/s.   1  4, 4.01: t  4.01, so the average velocity is 1 4.01   1.250 m/s. 2   

It appears that the instantaneous velocity at t  4 is 1.25 m/s. s t   s  4  1   1  5  v  4   lim  lim 1  1     4 m/s  1.25 m/s. t 4  2  2 t 4 t 4 t 4    

(b) (c)

2.6.19

g(0) is the only negative value. The slope at x = 4 is smaller than the slope at x = 2 and both are smaller than the slope at x = –2, Thus g(0)  0  g(4)  g(2)  g(2). 2.6.20 (a) On [20, 60]:

f (60)  f (20) 700  300 400    10 60  20 40 40

(b) Pick any interval that has the same y-value at its endpoints. [0, 57] is such an interval since

f (0)  f (57)  600. (c) On [40, 60]:

f (60)  f (40) 700  200 500    25 60  40 20 20

On [40, 70]: f (70)  f (40)  900  200  700  23 1

70  40

Since 25 > 23⅓, the average rate of change on [40, 60] is larger.

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2.6.21 (a) The tangent line at x = 50 appears to pass through the points 40, f (40) and 50, f (50), so

f (50) 

200  400 40  50



200

 20.

10

(b) The tangent line at x = 10 is steeper than the tangent line at x = 30, so it is larger in magnitude, but less in numerical value; that is, f (10)  f (30). (c) The slope of the tangent line at x = 60, f (60), is greater than the slope of the line through

40, f (40) and 80, f (80). So yes, f (60) 

f (80)  f (40) . 80  40

2.6.22 Since g(5)  3, the point 5, 3 is on the graph of g. Since g(5)  4, the slope of the tangent line at

x  5 is 4. Then the equation of the tangent line at this point is y  4(x  5)  3 or y  4x  23. 2.6.23

3x  7x  2  3a  7a  2 g( x)  g(a)  lim Let g(x)  3x  7x  2. Then lim xa xa xa xa 2



3  x  a   7  x  a 3 x  a  x  a   7  x  a  3x2  7x  3a2  7a  lim  lim xa xa xa xa xa xa (x  a)3(x  a)  7  lim  lim 3(x  a)  7  3(a  a)  7  6a  7. x x xa 2

 lim

Now set this quantity equal to 11 and solve for a: 6a  7  11  6a  18  a  3. 2.6.24

(a  h)4  2(a  h)2   a 4  2a2  f (a  h)  f (a)  lim Let f (t)  t  2t . Then for t > 0, lim h0 h0 h h 4



a4  4a3h  6a2h2  4ah3  h4  2a2  4ah  2h2  a4  2a2 h0 h

 lim

h 4a  6a h  4ah  h  4a  2h  4a3h  6a2h2  4ah3  h4  4ah  2h2  lim h0 h0 h h

 lim



 lim 4a3  6a2h  4ah2  h3  4a  2h  4a3  4a h0

225

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Now set this quantity equal to 0 and solve for a:

4a3  4a  0  4a(a2 1)  0  a  0, or a  1, or a  1. 2.6.25 (a) f (a)  lim

f (x)  f (a)

 lim

xa

xa

kx  ka

 lim

xa

xa

k  x  a xa

 lim k  k. xa

k  x 2  a2  k  x  a  x  a f ( x)  f (a) kx2  ka2  lim  lim  lim xa xa xa xa xa xa xa xa

(b)(b) f (a)  lim



 lim k  x  a   2ka. xa

3 3 k  x 3  a3  k  x  a   x 2  ax  a2  f ( x)  f (a) kx  ka  lim  lim  lim (c)(c) f (a)  lim xa xa xa xa xa xa xa xa 



 lim k  x 2  ax  a2   k a2  a  a  a2   3ka2. xa

k  x 4  a4  f ( x)  f (a) kx4  ka4  lim  lim xa xa xa xa xa xa

(d)(d) f (a)  lim



k  x  a   x 3  a 2 x  ax2  x3   lim k x3  a2 x  ax2  x3  xa xa xa

 lim



 k  a 3  a2  a  a  a2  a3   4ka3.

k  x n  an  f ( x)  f (a) kxn  kan  lim  lim xa xa xa xa xa xa

(e)(e) f (a)  lim  lim



k  x  a  xn1  an2 x  an3x2 

xa

xa

 lim k xn1  an2x  an3x2  xa

axn2  an1 

 lim k an1  an2  a  an3  a2  xa

axn2  an1 

a  an2  an1   nkan1.

2.6.26 For the tangent line y  4x  5 : when x = 2, y = 4(2) – 5 = 3 and its slope is 4 (the coefficient of x). At the point of tangency, these values are shared with the curve y  f (x); that is, f (2)  3and

f (2)  4.

226

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.6.27 Since 4, 3 is on y  f (x), f (4)  3. The slope of the tangent line between 0, 2 and 4, 3 is

1 f (4)  . 4

2.6.28 We begin by drawing a curve through the origin with a slope of 3 to satisfy f (0)  0 and f (0)  3. Since f (1)  0, we will round off our figure so that there is a horizontal tangent directly over x = 1. Last, we make sure that the curve has a slope of –1 as we pass over x = 2. Two of the many possibilities are shown.

2.6.29 We begin by drawing a curve through the origin with a slope of 1 to satisfy g(0)  0 and g(0)  1. We round off our figure at x = 1 to satisfy g(1)  0, and then pass through (2, 0) with slope – 1 to satisfy g(2)  0 and g(2)  1. We round the figure at x = 3 to satisfy g(3)  0, and then pass through (4, 0) with slope 1 to satisfy g(4)  0 and g(4)  1. Finally we extend the curve on both ends to satisfy lim   and lim  . x

x

2.6.30 We begin by drawing a curve through (0,1) with a slope of 1 to satisfy

g(0)  1 and g(0)  1. We round off our figure at x = –2 to satisfy g(2)  0. As x  5 , y  , so we draw a vertical asymptote at x  5. As x  5, y  3, so we draw a dot at (5, 3) (the dot could be open or closed).

227

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.6.31 We begin by drawing an odd function (symmetric with respect to the origin) through the origin with slope –2 to satisfy f (0)  2. Now draw a curve starting at x = 1 and increasing without bound as

x  2 since lim f ( x)  . Lastly, reflect the last curve through the origin (or reflect 180◦) since f x2 

is an odd function.

2.6.32 Using Definition 2 with f (x)  3x2  x3 and a  1:

 3(1  h)  1  h)   2 f (1  h)  f (1)  lim  h0 h0 h h 2

f (1)  lim

 lim

3  6h  3h2   1  3h  3h2  h3   2  lim 3h  h3  lim h 3  h2  

h0

 lim 3  h2  3  0  3. h0

The tangent line is y  3(x 1)  2 or y  3x 1. 2.6.33

 x  2  (1)  lim x4 1 g(x)  g(1)  lim x1 x1 x1 x 1 x 1 x 1 4

With g(x)  x4  2 and a  1, g(1)  lim

x 2 1 x2 1 x 2  1 x  1 x  1    lim  x2  1  x  1  2(2)  4.  lim  lim x 1 x 1 

x1



x1

The tangent line is y  4(x 1) 1 or y  4x  5. 2.6.34





(a) Using Definition 2 with F ( x)  5x / 1  x2 and the point at 2, 2  , we have

5(2  h) 2 F (2  h)  F(2) 1  (2  h)2 F(2)  lim  lim h0 h0 h h

5h  10  2 h  4h  5 5h  10 2 2 h2  4h  5  lim h  4h  5  lim h0 h0 h h 2



228

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 lim h0

2h2  3h

h h2  4h  5

 lim h 2h  3  lim h0

h h2  4h  5

So the equation of the tangent line 2, 2 is y  

h0

2h  3

h2  4h  5

3  . 5

 x  2  2 or y   3 x  16 .

(b)(b)

2.6.35 (a) Use Definition 2 with G(x)  4x  x : 2

4(a  h)  (a  h)   4a  a  G(a  h)  G(a) G(a)  lim  lim h0 h0 h h 2



4a2  8ah  4h2  a3  3a2h  3ah2  h3   4a2  a3

 lim

h0

8ah  4h2  3a2h  3ah2  h3 h0 h

 lim

h 8a  4h  3a2  3ah  h2   lim h0 h 



 lim 8a  4h  3a2  3ah  h2  8a  3a2. h0

So G(2)  8 2  3 4  4 and G(3)  83  39  3. Therefore, the equation of the tangent line at the point  2,8  is y  4(x  2)  8 or y  4x and the equation of the tangent line at the point 3, 9 is

y  3(x  3)  9 or y  3x 18. 2.6.36 Use Definition 2 with f (x)  3x  4x  1: 2

229

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

3(a  h)  4(a  h)  1  3a  4a  1 f (a  h)  f (a)  lim h0 h0 h h 2

f (a)  lim



3a2  6ah  3h2  4a  4h 1  3a2  4a 1

 lim h0

h 6a  3h  4

 lim

6ah  3h2  4h

h0

 lim6a  3h  4  6a  4 h0

h0

lim

2.6.37 Use Definition 2 with f (t)  2t3  t :

f (a)  lim

f (a  h)  f (a)

 lim

h0

2(a  h)3  (a  h)  2a3  a  h

h0

 2a  6a h  6ah  2h  a  h  2a  a 6a2h  6ah2  2h3  h  lim h0 h0 h h 3

 lim

h 6a2  6ah  2h2  1

 lim

 lim  6a2  6ah  2h2  1  6a2  1 h0

h0

2.6.38 Use Definition 2 with f (t)  (2t 1) / (t  3) :

f (a)  lim h0

 lim

f (a  h)  f (a) h

2(a  h) 1 2a 1   lim (a  h)  3 a  3 h0

(2a  2h 1)(a  3)  (2a 1)(a  h  3) h0 h(a  h  3)(a  3)

 lim

2a2  6a  2ah  6h  a  3  2a2  2ah  6a  a  h  3 

h(a  h  3)(a  3) 5h 5 5  lim  lim  h0 h(a  h  3)(a  3) h0 (a  h  3)(a  3) (a  3)2 h0

2.6.39 Use Definition 2 with f ( x)  x2  1 / x2 :

1 1 a2  (a  h)2  2 2 2 2 2 2 a2 f (a  h)  f (a)  lim a (a  h)  lim a  (a  2ah  h ) f (a)  lim  lim (a  h) h0 h0 h0 h0 h h ha2 (a  h)2 h

230

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



2ah  h2 lim h0

ha2 (a  h) 2

 lim h0

h 2a  h ha 2 (a  h) 2

 lim h0

2a  h a 2 (a  h) 2



2a



a 2 (a) 2

2 a3

2.6.40 Use Definition 2 with f (x)  1 2x :

f (a)  lim

f (a  h)  f (a) h

h0

 lim 1  2(a  h)  1  2a h0 h



 lim 1 2(a  h)  1 2a  1 2(a  h)  1 2a  lim h0 h 1 2(a  h)  1 2a h0



(1 2a  2h)  (1 2a)

 lim h0

 

 1 2(a  h)  1 2a 

  



2



 1  2a  1  2a 

h0



2h

 lim

 1 2(a  h)    1 2a   h  1 2(a  h)  1 2a   lim

 1 2(a  h)  1 2a 

h0

2

 1 2(a  h)  1 2a 

2 1  2 1  2a 1  2a

2.6.41 Use Definition 2 with f (x) 

4 : 1 x

f (a  h)  f (a)  lim h0 h h0

f (a)  lim 

1 a  1 a  h

 4 lim h0

 lim h0

4 1 a

h0

 1  a  h  1  a  h 

 1  a  h  1 a  h0

1  a  1  (a  h) 1 a  1 a  h

 lim

 1 a  h  1 a  1 a  1 a  h  4

 4 lim

1  a  1  (a  h)

 4 lim

 1  a  h  1  a 

1 a  1 a  h

h0

(1 a)  (1 a  h)

 4 lim h0

4  1  (a  h)

 1 a  h  1 a  1 a  1 a  h 



 1  a  h  1  a  1  a  1  a  h   1  a  1  a  1  a  1  a 

231

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



1  a   2 1  a  (1  a)(1  a)

1/2



2 (1  a)3/2

2.6.42

Use (4) with f  x   x  2  x  2  x2  4.  a  h  2  4    a 2  4   f a  h  f a      f '  a   lim  lim  h0

 lim

h0 h 2 a  2ah  h  4  a2  4

h0 h  lim2a  h  2a

lim

h 2ah  h2

h0

 lim h 2a  h h0 h

h 0

2.6.43 Use (4) with f  x   x  2  x  2  x2  4x  4.  a  h 2  4  a  h   4    a 2  4a  4  f a  h  f a      f '  a   lim  lim  h0

 lim h0

 lim

h0 h a2  2ah  h2  4a  4h  4  a2  4a  4

h 2a  h  4

lim

2ah  h2  4h h

h0

h  lim2a  h  4  2a  4

 

2.6.44

lim 9  h  3  f (9), h0 h where f (x)  x and a  9.



2.6.45



lim e

 e2  f (2), where f (x)  ex and a  2, or lim e2h  e2

h0

2h

h0





 f (2), where

x

f (x)  e and a  2.

232

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.6.46

x6  64   f (2), where f ( x)  x6 and a  2. x2 x  2

lim

2.6.47 1 1 1 4 lim x  f (4), where f (x)  and a  . x1/4 x  1 x 4 4

2.6.48

lim

cos(  h) 1

h0

Or: lim

 f ( ), where f (x)  cos(x) and a   .

h cos(  h)  1

 f ( ), where f (x)  cos(  x) and a  0.

h0

2.6.49

sin  1

lim   /6

6

 f   , where f    sin and a 



 . 6

2.6.50

By (4), lim

ln 2  h  ln 2

h0

Or: By (4), lim

 f   2  , where f  x  ln x and a  2.

h ln 2  h  ln 2

 f   0  , where f  x  ln  2  x and a  0.

h0

2.6.51

 tan1 x   4  f 1, where f  x  tan1 x and a  1. By (5), lim x1 x 1 2.6.52



80(4  h)  6(4  h)   80(4)  6(4)  f (4  h)  f (4)  lim  h0 h0 h h 2

v(4)  f (4)  lim  lim h0



320  80h  96  48h  6h2   320  96 lim 32h  6h2 

h0

233

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 lim h0

h 32  6h

 lim32  6h  32 m/s. h0

The speed when t = 4 is 32  32 m/s.



2.6.53

v(4) 

10  45   10  45  45  9 4  h 1  4  1  f (4  h)  f (4)     5  h  lim  lim f (4)  lim h0

 lim h0

h 45  9  5  h

9 5

h0



9h

 lim

h 5  h

The speed when t = 4 is 

h0

 lim

h(5  h)

h0

9



(5  h)

m/s.

2.6.54 The sketch shows the graph for a room temperature of 72 and a refrigerator temperature of 38. The initial rate of change is greater in magnitude than the rate of change after an hour.

2.6.55 The slope of the tangent line (that is, the rate of change of temperature with respect to time) at t = 1 hour seems to be about

75 168 132  0

 0.7o F/min.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.6.56

(a)

(b)

(i)

1, 5:

P 5  P 1 33 167 134    33.5 mg/h. 5 1 4 4

(ii)

1, 4:

P 4  P 1 50 167 117    39 mg/h. 4 1 3 3

(iii)

1, 3:

(iv)

1, 2:

P 3  P 1 3 1



75 167 92   46 mg/h. 2 2

P 2  P 1 112 167   55 mg/h. 2 1 1

From (iv), the average rate of change over 1, 2 is 55 mg/h. The average rate of change P 3  P 2 75 112   37 mg/h. Using these two average rates of over 2, 3 is 3 2 1 55  37  46 mg/h as an estimate of the instantaneous rate of change, we have 2 change at t  2.

2.6.57

(a)

(i)

2015, 2017:

N 2017  N 2015 27,339  23, 043 4296    2148 locations 2017  2015 2 2

per year (ii)

2017, 2019:

N 2019  N 2017 31, 256  27,339 3917    1958.5 locations 2019  2017 2 2

per year These answers suggest that the rate of growth in the number of locations decreased from 2015 to 2019. (b)

The average growth rate over 2017, 2019 gives an estimate of the instantaneous rate of growth in 2018: 1958.5 locations per year.

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2.6.58

2010, 2015:

93, 770  87,302

(b)

2015, 2020:

100,300  93, 770 2020  2015



6468

 1293.6 thousand barrels per day per year. This 2015  2010 5 means that oil consumption rose by an average of 1293.6 thousand barrels per day each year from 2010 to 2015.

(a)

 1306

An estimate of the instantaneous rate of change in 2017 is 1306 thousand barrels per day per year.

2.6.59 (a)

1, 3:

(i)

(ii)

2, 3:

(iii)

3, 5:

(iv)

4, 5:

V 3 V 1 48, 600  60, 000 11, 400    5700 dollars per year 3 1 2 2 V 3 V 2 3 2



48, 600  54, 000

 5400 dollars per year

V 5 V 3 39,366  48, 600 9234    4617 dollars per year 53 2 2 V 5 V 4 54



39,366  43, 740

 4374 dollars per year

Using the value from (iv), we have 4374 dollars per year. The value of the BMW 550i was decreasing at a rate of approximately $4374/year 5 years after the initial sale.

(b) 2.6.60

C

C(105)  C(100)

6601.25  6500

 $20.25 / unit.   x  105 100  5 C C(101)  C(100) 6520.05  6500 (ii)    $20.05 / unit. x 101100 1

(a) (i)

5000  10(100  h)  0.05(100  h)   6500 20h  0.05h2  (b) C(100  h)  C(100)    h h h 2

 20.0.05h, h  0. So the instantaneous rate of change is

lim h0

C(100  h)  C(100) h

 lim20  0.05h  $20 / unit. h0

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2.6.61 2

2 t    t  h  60   100, 000 1   V  V (t  h) V (t)  100, 000 1     60   t  h  t  h  2   t t2   h 2th h2   100, 000  1   3600  1       100, 000      30 30 3600     30 3600 3600   



100,000

h 120  2t  h 

3600

250

h 120  2t  h

Dividing V by h and then letting h  0, we see that the instantaneous rate of change is 500 9

t  60 gal/min. t

Flow rate (gal/min)

Water remaining V(t) (gal)

3333.3

100,000

2777.7

69, 444.4

2222.2

44, 444.4

16662.6

25,000

1111. 1

11,111. 1

555.5

2777.7

The magnitude of the flow rate is greatest at the beginning, and gradually decreases to 0.

2.6.62 (a) f ( x) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are dollars per ounce. (b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17 / ounce. So the cost of producing the 800th or (801st) ounce is about $17. (c) In the short term, the values of f ( x) will decrease because more efficient use is made of start-up costs as x increases. But eventually f ( x) might increase due to large-scale operations.

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2.6.63 (a) f (5) is the rate of growth of the bacteria population with t = 5 hours. Its units are bacteria per hour. (b) With unlimited space and nutrients, f  should increase as t increases; so f (5)  f (10). If the supply of nutrients is limited, the growth rate slows down at some point in time, and the opposite may be true. 2.6.64 (a) H (58) is the rate at which the daily heating cost changes with respect to temperature when the outside temperature is 58 ℉. The units are dollars / F. (b) If the outside temperature increases, the building should require less heating, so we would expect

H (58) to be negative.

2.6.65 (a) f (8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is $8 per pound. The units for f (8) are pounds / (dollars/pound). (b) f (8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally less willing to buy a product when its price increases. 2.6.66

T   4  is the rate at which the temperature is changing at 4:00 PM. To estimate the value of T   4  , we will average the difference quotients obtained using the times t  3 and t  5. Let T 5 T 4 77  78 T 3 T 4 80  78 A   1 and B    2. Then 54 1 3 4 1 T t  T 4 A  B 1 2 T  4   lim    1.5F/h. t4 t 4 2 2 2.6.67

(a)

M   t  is the rate at which the average amount of milk produced per cow changes with respect to time. Its units are gallons per year.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

To estimate the value of M 5, we will average the difference quotients obtained using

(b)

the times t  4 and t  6. Let A  T 4  T 5  1950 1970  20 and 45 1 T 6  T 5 2030 1970 B   60. Then 65 1 M t   M 5 A  B 20  60 M 5  lim    40 gallons/year. The average amount of t5 t 5 2 2 milk produced per cow was increasing at the rate of approximately 40 gallons per year in 2019 (5 years after 2014). 2.6.68 (a) S(T ) is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature. The units of S(T ) are (cm/s) C. (b) For T  15 C, it appears the tangent line to the curve goes through the points 10, 25 and

20, 32. So S(15) 

32  25 20 10

 0.7 (cm/s) C . This tells us that at 15 C, the maximum

sustainable speed of Coho salmon is changing at a rate of 0.7 (cm/s)

In a similar fashion for T  25 C, we can use the point (20, 35) and (25, 25) to obtain

S(25) 

25  35 25  20

 2 (cm/s) C. As it gets warmer than 20 C, the maximum sustainable speed

decreases rapidly. 2.6.69 Since f (x)  x sin

h sin  lim h0

1 h

when x ≠ 0 and f (0)  0, we have f (0)  lim f 0  h  f (0) h0 h x

 f (0)

1 1 takes the values –1 and 1 on  limsin . This limit does not exist sincesin h0 h h

any interval containing zero.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.6.70 Since f (x)  x sin 2

1 x

when x ≠ 0 and f (0)  0, we have

h2 sin

1  f (0) 1 1 h  lim h sin . Since 1  sin  1, we have

f (0)  lim f 0  h  f (0)  lim h0 h0 h0 h h h h 1 1  h  h sin  h   h h sin  h . Because lim h   0 and lim h  0, h0 h0 h h 1    we know that lim h0 h sin h  0 by the Squeeze Theorem. Thus f (0  0. 





 2.6.71 (a) The slope at (0,0) appears to be 1.

(b) The slope at (0,0) still appears to be 1.

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(d) Answers will vary.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION 2.7 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.7.1 It appears that f is an odd function, so f  will be an even function – that is, f (a)  f (a) (a) f (3)  0.2 (b) f (2)  0

(d) f (0)  2

(e) f (1)  1

(f) f (2)  0

(g) f (3)  0.2

2.7.2 Your answers may vary depending on your estimates. (a) Note: By estimating the slopes of tangent lines on the graph of g, it appears that g(0)  6. (b) g(1)  0

(d) g(3)  1.3

(e) g(4)  0.8

(f) g(5)  0.3 (g) g(6)  0

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(f) g(7)  0.2 2.7.3 (a)′ = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then negative again. The actual function in graph II follows the same pattern. (b)′ = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents. (c)′ = I, since the slopes of the tangents to graph (c) are negative for x < 0 and positive for x > 0, as are the function values of graph I. (d)′ = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then positive, then 0, then negative again, and the function values in graph III follow the same pattern. Hints for Exercises 12-19: First plot x-intercepts on the graph of f  for any horizontal tangents on the graph of f. Look for any corners on the graph of f –there will be a discontinuity on the graph of

f . On any interval where f has a tangent with positive/negative) slope, the graph of f  will be positive/negative. If the graph of the function is linear, the graph of f  will be a horizontal line whose height is of the slope of the function. 2.7.4

2.7.5

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.7.6

2.7.7

2.7.8

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.7.9

2.7.10

2.7.11

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.7.12

2.7.13 (a) C(t) is the instantaneous rate of change of percentage of full capacity with respect to elapsed time in hours. (b) The graph of C(t) tells us that the rate of change of percentage of full capacity is decreasing and approaching zero. 2.7.14 (a) F (v) is the instantaneous rate of change of fuel economy with respect to speed. (b) Graphs will vary depending on estimates of F but will change from positive to negative at about v = 50. (c) To save on gas, drive at the speed where F is a maximum and F is 0, which is about 50 mi/h. 2.7.15

4000  3(5  h)  4000  3 5  G(5  h)  G(5)  lim (a) G(5)  lim h0 h0 h h 2



h 30  3h 4000  75  30h  3h2  4000  75 30h  3h2  lim  lim h0 h0 h0 h h h

 lim

 lim30  3h  30 h0

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(b) G(5) is the instantaneous rate of change in the amount of oil in the tank at 5:00 AM. The units

G(t) of are gallons per hour. So at 5:00 AM, the oil is leaking from the tank at a rate of 30 gallons per hour.

2.7.16 (a) T (8) 

T (9)  T (7) 97



54  60 2



6

 3 degrees per cm.

(b) T (8) tells us that eight cm from the heated end of the rod, the temperature is decreasing at a rate of 3 degrees per cm. 2.7.17 The slope at x = –2π, 0 and 2π appears to be one. At x = –3π/2, –π/2, 3π/2, and 3π/2, the slope is zero. At x = ±π, the slope appears to be –1. Therefore, we might guess that f (x)  cos( x).

2.7.18 The slope at 0 appears to be 1 and the slope at 1 appears to be 2.7. As x decreases, the slope gets closer to 0. Since the graphs are so similar, we might guess that f ( x)  e . x

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2.7.19 As x increases toward 1, f ( x) decreases from very large numbers to 1. As x becomes

( large, f ( x) gets closer to 0. As a guess, f ( x)  2 or f x) 

makes sense.

2.7.20

This is same as our guess.

2.7.21

(c)

It appears that f  x is three times the square of x, so we guess that f   x  3x2.

(d)

This is the same as our guess.

2.7.22

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f (x  h)  f (x)

f (x)  lim

 lim

h0

 lim

h0

3(x  h)  8  (3x  8)

 lim

h0

3x  3h  8  3x  8 h

h0

 lim3  3. Domain of f   . h0

2.7.23

f (x  h)  f (x)

f (x)  lim

h0

 lim

mh h

h0

m(x  h)  b  (mx  b) h

h0

 lim

mx  mh  b  mx  b h

h0

 lim m  m. Domain of f and f  is . h0

2.7.24

2.5(t  h)  6(t  h)  2.5t  6t  f (t  h)  f (t) f (t)  lim  lim h0 h0 h h 2



2.5(t2  2th  h2  6t  6h  2.5t2  6t 2.5t2  5th  2.5h2  6h  2.5t2  lim h0 h0 h h

 lim

h 5t  2.5h  6 5th  2.5h2  6h  lim5t  2.5h  6  5t  6.  lim h0 h0 h0 h h

 lim

Domain of f and f  is

2.7.25

4  8( x  h)  5( x  h)  4  8x  5x  f ( x  h)  f ( x)  lim h0 h0 h h 2

f ( x)  lim  lim



4  8x  8h  5  x 2  2xh  h2   4  8x  5x2 h

h0

8h  5x2  10xh  5h2  5x2 h0 h

 lim

h 8 10x  5h 8h 10xh  5h2  lim8 10x  5h  8 10x.  lim h0 h0 h0 h h

 lim

Domain of f and f  is

2.7.26

( x  h)2  2( x  h)3   x 2  2x 3  f ( x  h)  f ( x) f ( x)  lim  lim h0 h0 h h 

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x2  2xh  h2  2x3  6x2h  6xh2  2h3  x2  2x3 h0 h

 lim

h  2x  h  6x  6xh  2h  2xh  h2  6x2h  6xh2  2h3  lim  lim h0 h0 h h 2



 lim  2x  h  6x2  6xh  2h2  2x  6x2.

Domain of f and f  is

h0

2.7.27

g(t)  lim

g(t  h)  g(t) h

h0

1 1  t  lim  lim t  h h0 h0 h

t  th th t h

 t  th t  th  t  (t  h) h  lim    lim  lim  h0 t  t  h  h0 h t  h t h0  h t  h t t  th t  th h th t





  lim h0

1

th t

1



 t  t  h 

t t



 t  t 



 

t 2 t





1 2t3/2

Domain of g = domain of g  0,   . 2.7.28

g(x)  lim

g(x  h)  g(x) h

h0

9  (x  h)  9  x

 lim h0

 lim 9  (x  h)  9  x  9  (x  h)  9  x    h0 h  9  (x  h)  9  x 

 9  ( x  h)  9  x  1

h

 lim



h0

 9  ( x  h)  9  x  Domain of g = , 9, domain of

1

 9  ( x  h)  9  x  2 9  x

g  , 9.



2.7.29



( x  h)  1  x2  1  f ( x  h)  f (x) 2( x  h)  3 2x  3  f ( x)  lim  lim h0 h0 h h 2



249

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 lim h0

h 2(x  h)  3  2x  3



2x3  4x2h  2xh2  2x  3x2  6xh  3h2  3  2x3  2x2h  3x2  2x  2h  3

 lim

h  2x  2h  3  2x  3

h0



h  2x 2  2xh  6x  3h  2 4x2h  2xh2  6xh  3h2  2x2h  2h  lim h0 h0 h 2x  2h  3  2x  3 h 2x  2h  3  2x  3

 lim



 x 2  2xh  h2 1(2x  3)  (2x  2h  3)  x 2 1



2 2x 2  2xh  6x  3h  2  2x  6x  2   lim  

h0

2x  2h  3  2x  3



 2x  3



Domain of f = domain of f   , 2   32,   . 3

2.7.30

1 2(t  h) 1 2t  G(t  h)  G(t) 1 2(t  h)(3  t) 3  (t  h)(1 2t) 3  (t  h) 3t G(t)  lim  lim  lim h0 h0 h0 h h h(3  (t  h))(3  t) 2 2 6h  h  lim 3  t  6t  2t  6h  2ht  3  6t  t  2t  h  2ht  lim h0 h(3  t  h)(3  t) h0 h(3  t  h)(3  t) 7 7 7h  lim  lim  h0 h(3  t  h)(3  t) h0 (3  t  h)(3  t) (3  t)2

Domain of g = domain of g  , 3  3,   . 2.7.31

f (x)  lim f (x  h)  f (x) lim (x  h)  x h0 h0 h h 3/2

3/2

3/2 3/2 3/2 3/2   h)  x  (x  h)  x   lim (x h0 h (x  h)3/2  x 3/2 



2 2 3 2 2 3 3 (x  h)3  x3  lim x  3x h  3xh  h  x  lim h 3x  3xh  h  h0 h0 h (x  h)3/2  x 3/2  h0 h (x  h)3/2  x 3/2  h (x  h)3/2  x 3/2     

 lim

 lim h0



3x2  3xh  h2 

3/2 3/2 (x   h)  x 

 lim h0

3x2

2x3/2  2 x

1/2

250

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Domain of f = domain of f   0,   . Strictly speaking, the domain of f  is 0,  because the limit that defines f (0) does not exist as a two-sided limit. But the right-hand derivative does exist at 0, so in that sense one could regard the domain of f  to be0,   .

2.7.32

f (x)  lim h0

f (x  h)  f (x) h

lim

(x  h)4  x4 h

h0

 lim x  4x h  6x h  4xh  h  x h0 h 4

2 2

h  4x 3  6x2h  4xh2  h3   lim  4x 3  6x2h  4xh2  h3   4x 3 h0 h0 h

 lim



Domain of f = domain of f   . 

2.7.33 If f (x)  2x3  3x2 then using the limit definition we find

2( x  h)  3( x  h)   2x  3x  f ( x  h)  f ( x)  lim h0 h0 h h 3

f ( x)  lim



2x3  6x2h  6xh2  2h3  3x2  6xh  3h2  2x3  3x2 h0 h

 lim

h  6x 2  6xh  2h2  6x  3h 6x2h  6xh2  2h3  6xh  3h2  lim  lim h0 h0 h h 

 lim 4x2  4xh  2h2  6x  3h  6x2  6x. h0

Then using the limit definition again, f (x)  lim f (x  h)  f (x) h0

 lim h0

6( x  h)2  6( x  h)   6x 2  6x  h

6x2  12xh  6h2  6x  6h  6x 2  6x h0 h

 lim

h 12x  6h  6 12xh  6h2  6h  lim12x  6h  6  12x  6 .  lim h0 h0 h0 h h

 lim



251

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

If f  x  x3, then f  x  3x2 , f (x)  6x, and f (x)  6, which is a constant function, so f 4  x  0. 2.7.35 (a)

(b) Note that the third graph in part (a) has small negative values for its slope,

f ; but as x  6 , f   .

 lim h0

h0

6 xh  h

6 x



h0



 lim 6  ( x  h)  6  x h0 h

6  x  h  6  x  lim 6  x  h  6  x 6  x  h  6  x h0 h 6  x  h  6  x



h

 lim 

f (x  h)  f (x)

 6  x  h  6  x 

 lim h0

1

 6  x  h  6  x 







1 2 6 x



 

 2.7.36

252

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.7.37

2.7.38 (a) In the viewing window [–5, 5] × [–3, 3], the graphs of f ( x)  x and

g(x) 

x2  0.0001  0.01 are indistinguishable.

(b) x

0.2

0.4

0.6

0.8

1.0

f(x)

1.000

0.966

0.939

0.919

0.902

0.889

g(x)

0.190

0.390

0.590

0.790

0.990

The values of f and g are not similar for small x near zero. As x → 1 however, the values of both functions become fairly close to each other. (c) As we zoom in on the graphs repeatedly, we see that the graphs are distinct. They intersect at the point (0, 0), but for all other x near zero, f (x)  g(x).

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

f (x  h)  f (x)  1 h0  h

(d) At x = 0, the graph of f has a sharp point (cusp) at x = 0, so lim

f (x  h)  f (x)

 lim

h0

 1 and f (0) does not exist.

f (g  h)  g(x)  lim (e) g( x)  lim h0 h0 h

 ( x  h)  0.0001  0.01   x  0.0001  0.01 2

(x  h)  0.0001    x  0.0001    (x  h)  0.0001    x  0.0001     lim    (x  h)  0.0001    x  0.0001  h 2

h0



(x  h)2  0.0001   x 2  0.0001

 lim h0

h0



2xh  h2

 lim

 (x  h)  0.0001   x  0.0001 2

h  2x  h

 lim



h0

 (x  h)  0.0001   x  0.0001 2

2x  h

 lim

 (x  h)  0.0001   x  0.0001  ( x  h)  0.0001   x  0.0001 2

2x x2  0.0001  x2  0.0001



h0

x x2  0.0001

. Therefore, g(0) 

0 02  0.0001

 0.

2.7.39 (a) U (t) is the rate at which the unemployment rate is changing with respect to time. Its units are percent unemployed per year. (b) To findU (t), we use lim h0

For 2004: U (2004) 

U (t  h) U (t)



U (t  h) U (t)

U (2005) U (2004) 2005  2004

for small values of h.

h 

5.1 6.0

 0.9

For 2005: We estimate by using h = –1 and h = 1, and then average the two results to obtain a final estimate.

254

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

h = –1 ⇒ U(2005)  h = 1 ⇒ U(2005) 

U (2004) U(2005) 6.0  5.1   0.9; 2004  2005 1

U(2006) U(2005) 4.5  5.1   0.5. 2006  2005 1

So we estimate that U 2005  12 0.9  0.5 0.7. We continuing in this manner through 2018. For 2019 we use only h  1 because the pandemic of 2020 effectively causes a jump discontinuity in the unemployment-rate graph, meaning that a one-sided derivative estimate may be best. (Not taking the pandemic into account would give us U   t   1.4 for 2019.) For 2020 we use only h  1 because we are not given data about the 2021 unemployment rate. U   t 

t 2004 2005 2006 2007 2008

0.9 0.7 0.25 0.60 2.35

t 2009 2010 2011 2012

U   t 

t 2013 2014 2015 2016

1.90 0.20 0.75 0.75

0.95 1.05 0.65 0.5

t 2017 2018 2019 2020

U   t  0.5 0.3 0.2 3.0

2.7.40 (a) N (t) is the rate at which the number of minimally invasive cosmetic surgery procedures performed in the United States is changing with respect to time. Its units are thousands of surgeries per year. (b) To find N (t), we use lim h0

N (t  h)  N (t)



N (t  h)  N (t)

for small values of h.

For 2000: N(2000) 

N (2002)  N (2000) 4897  5500   301.5 2002  2000 2

For 2002: N(2002) 

N(2004)  N (2000) 7470 5500   492.5 2000  2000 4

Note that this gives the same estimate that averaging two difference quotients would.

N  t 

2000 301.5

2002 492.5

2004 2006 2008 1060.25 856.75 605.75

2010 534.5

2012 596

2014 2016 594.25 938.75

2018 1144

255

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

We could obtain more-accurate estimates of N  t  by obtaining data for more values of

(d)

t. 2.7.41 We use one-sided difference quotients for the first and last values, and average two difference quotients for all other values. t

H(t)

H′(t)

13/7

23/14

18/14

14/14

11/14

5/7

2.7.42 We use one-sided difference quotients for the first and last values, and average two difference quotients for all other values. The units forW ( x) are grams per degree g/ C . x

15.5

17.7

20.0

22.4

24.4

W(x)

37.2

31.0

19.8

9.7

–9.8

W′(x)

–2.82

–3.87

–4.53

–6.73

–9.75

2.7.43 (a) dP dt is the rate at which the percentage of the city’s electrical power produced by solar panels changes with respect to time, t, measured in percentage points per year. (b) Two years after January 1, 2000 (January 1, 2002), the percentage of electrical power produced by solar panels was increasing at a rate of 3.5 percentage points per year. 2.7.44

256

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

dN dp is the rate at which the number of people who travel by car to another state for vacation changes with respect to the price of gasoline. If the price of gasoline goes up, we would expect fewer people to travel, so we would expect dN dp to be negative.

2.7.45 f is not differentiable at x = –4, because the graph has a corner there, at x = 0, because there is a discontinuity there, and at x  2 because there is a vertical tangent line there.

2.7.46 f is not differentiable at x = –1, because there is a discontinuity there, and at x = 2, because the graph has a corner there.

2.7.47 f is not differentiable at x = 1, because f is not defined there.

2.7.48 f is not differentiable at x = –2 and x = 3, because the graph has corners there, and at x = 1, because there is a discontinuity there.

2.7.49 As we zoom in toward (–1, 0), the curve appears more and more like a straight line, so f ( x)  x 

x is

differentiable at x = –1. But no matter how much we zoom in toward the origin, the curve does not straighten out – we can’t eliminate the sharp point (a cusp). So f is not differentiable at x = 0.

2.7.50 As we zoom in toward (0, 1), the curve appears more and more like a straight line, so f is differentiable at x = 0. But no matter how much we zoom in toward (1, 0), or (–1, 0), the curve doesn’t straighten out – we cannot eliminate the sharp point (a cusp). So f is not differentiable at x = ±1.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.7.51 Call the curve with the positive y-intercept g and the other curve h. Notice that g has a maximum (horizontal tangent) at x = 0, but h ≠ 0, so h cannot be the derivative of g. Also notice that where g is positive, h is increasing. Thus h = f and g  f . Now f (1) is negative since f  is below the x-axis there and f (1) is positive since f is concave upward at x = 1. Therefore, f (1) is greater than f (1). 2.7.52 Call the curve with the smallest positive x-intercept g and the other curve h. Notice that where g is positive in the first quadrant, h is increasing. Thus h = f and g  f . Now f (1) is positive since f  is above the x-axis there and f (1) appears to be zero since f has an inflection point at x = 1. Therefore,

f (1) is greater than f (1).

2.7.53

a  f ,b  f , c  f . We can see this because where a has a horizontal tangent, b = 0, and where b has a horizontal tangent, c = 0. We can immediately see that c can be neither f nor f  since at the points where c has a horizontal tangent, neither a nor b is equal to 0.

2.7.54 Where d has horizontal tangents, only c is 0 so d  c. c has negative tangents for x < 0 and b is the only graph that is negative for x < 0, so c  b. b has positive tangents on ℝ (except at x = 0), and the only graph that is positive on the same domain is a, so b  a. We conclude that

d  f ,c  f ,b  f , and a  f .

2.7.55 We can see immediately that a is the graph of the acceleration function, since at the points where a has a horizontal tangent, neither c nor b is equal to 0. Next, we note that a = 0 at the point where b has a horizontal tangent, so b must be the graph of the velocity function, and hence b  a. We conclude that c is the graph of the position function.

2.7.56

258

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

Curve a must be the jerk since none of the graphs are 0 at its high and low points. a is 0 where b has a maximum, so b  a. b is 0 where c has a maximum, so c  b. We conclude that d is the position function, c is the velocity, b is the acceleration, and a is the jerk.

2.7.57

2.7.58



2.7.59



If x  y  k is tangent to the graph of y  f  x  at x  2, then lim f  x  f 2  f 2 is equal x2 x2 to the slope of the line x  y  k, which is 1. 2.7.60

If f  x  x  5  3x  6 , then when x  1, f  x    x  5     3x  6   4x 1. In this case, f  x  4. 2.7.61 (a) True; if lim h0

f (5  h)  f (5x)

  then f (5) does not exist because f has a vertical tangent line

at x = 5.

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f (5  h)  f (5x) would have h0 h

(b) False. If f had a horizontal tangent line at the point (5, 7), then lim to be zero. (c) False. Because f (5)  lim f (5  h)  f (5x)  , f (5)  7. h0

(d) False. Because f (5)  lim

h f (5  h)  f (5x)

 , f (5)  0 .

h0



2.7.62

Because the line 2x  y  13 is tangent to y  f (x) at x = 3, the line and y intersect at the point

3, f (3). Therefore f (3)  13  2(3)  7. The slope of the line 2x  y  13is –2, and this slope must also be the value of the derivative of f at x = 3. So f (3)  2.

2.7.63

(a) f (x)  lim

a(x  h)  1 ax  1  b(x  h)  1 bx  1

f (x  h)  f (x)

 lim h0 h h (ax  ah  1)(bx  1)  (ax  1)(bx  bh  1)  lim h0 h(bx 1)(bx  bh  1) h0

 lim

abhx  abx2  ah  ax  bx  1  abhx  abx2  ax  bh  bx  1 

h(bx  1)(bx  bh  1)

h0

ah  bh h(a  b) a b a b  lim  lim  h0 h(bx 1)(bx  bh 1) h0 h(bx 1)(bx  bh 1) h0 (bx 1)(bx  bh 1) (bx 1)2

 lim

(b) For g(x) 

3x  1

2 , using part (a), g(x)  3  5   (5x  1)2 5x  1 (5x  1)2

a(x  h)  c ax  c  b(x  h)  c bx  c

h0 h h (ax  ah  c)(bx  c)  (ax  c)(bx  bh  c)  lim h0 h(bx  c)(bx  bh  c) h0

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 lim

abhx  abx2  ach  acx  bcx  c2  abhx  abx2  acx  bch  bcx  c2  

h(bx  c)(bx  bh  c)

h0

 lim h0

 lim

h ac  bc ach  bch  lim h0 h(bx  c)(bx  bh  c) h(bx  c)(bx  bh  c)

ac  bc



(a  b)c (bx  c)2

(bx  c)(bx  bh  c) 3x  5 5 , using part (c), g(x)  (3  4)5   (d) For g(x)  2 (4x  5) 4x  5 (4x  5)2 h0

2.7.64

2.7.65

(a)

We estimate the acceleration to be its most negative, 2, at t  15 s.

(b)

Since we estimate the acceleration to be a minimum at t  15 s, the jerk is 0 at t 15. The units of measure would be ft / s2  / s, or ft / s3.

2.7.66 (a) Note that we have factored x – a as the difference of cubes in the third step.

f (a)  lim f (x)  f (x)  lim x1/3  a1/3  lim xa xa xa xa xa

x1/3  a1/3

 x  a  x 1/3

1/3

2/3

 x1/3a1/3  a 2/3 

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

 lim xa

x  x a  a  2/3

1/3 1/3



1 2/3 .

or a

3a 2/3

2/3

f (0  h)  f (0)

3  lim h  0  lim 12/3 . This function increases without bound, h0 h0 h0 h h h so the limit does not exist, and therefore f (0) does not exist.

(b)(b) f (0)  lim

x0

  and f is continuous at x = 0 (it is a root function), so f has a vertical

3x2/3

tangent at x = 0. 2.7.67

g(x)  g(0) x2/3  0 1   lim 1/3 , which does not exist.  lim (a) g (0)  lim x0 x x0 x0 x0 x (b) g(a)  lim g(x)  g(a)  lim x2/3  a2/3  lim

xa

xa

x0

 2a 3a

1/3 1/3 xx aa  a

 lim x2/3

1/3 1/3

x a

x0





2/3

1/3

2/3

1/3



1/3

x0

2 1/3

1/3

 x  a  x  a   x  a  x  x a  a  

1/3 1/3

2/3

2  3a1/3 or 3 a 2/3

2.7.68

   



if x  6  0  x  6 f ( x)  x  6   x  6    x  6   if x  6  0 6  x 

if x  6 if x  6

So the right-hand limit is lim f (x)  f (6)  lim x  6  0  lim x  6  lim 1  1, and the leftx6

x 6

x6

x 6

x6

x6

x6

hand limit is lim f (x)  f (6)  lim x  6  0  lim 6  x  lim1  1. Since these limits are x6  x6  x6 x6 x6 x  6 f (x)  f (6) does not exist and f is not differentiable at 6. not equal, f (6)  lim x6 x6 x6



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1

(b) A formula for is f ( x)  

if x  6

1 if x  6

. Another way of writing the formula is f (x) 

x6 x6

A graph of the function and its derivative are shown here:

2.7.69 (a) f ( x)  x is not continuous at any integer n, so f is not differentiable at n by the contrapositive of Theorem 1. (b) If a is not an integer, then f is constant on an open interval containing a, so f (a)  0. Thus,

f (x)  0, x not an integer. Here is a plot of the derivative:

2.7.70

 x2 (a) f ( x)  x x     2 x

 if x  0 if x  0

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(b) Since f ( x)  x2 for x  0, we have f (x)  2x for x > 0. Similarly, f ( x)  x 2 for x  0, we have f (x)  2x for x < 0. At x = 0, we have f (0)  lim

f (x)  f (0)

x0

 lim

x0

 2x 2x

(a) x 

x0

 lim x  0. So f is differentiable at 0. Thus, f is differentiable for all x.

2.7.71

.

x

if x  0 so f  x  x 

  x if x  0

if x  0 2x. if x  0 2x   0

if x  0 . if x  0

Graph the line y = 2x for x ≥ 0 and graph y = 0 (the x-axis) for x < 0. (b) g is not differentiable at x 0 because the graph has a corner there, but g is differentiable at all other values; that is g is differentiable on , 0  0, .

2x 0 

2 if if x  0  g(x)   x0 if x  0  0 if x  0 

Another way of writing the formula is g(x)  1 sgn x, for x  0. 2.7.72 (a) If f is even, then

f (x)  lim

f (x  h)  f (x)

h f (x  h)  f (x)  lim h0 h h0

 lim

f (x  h)  f (x) h

h0

 lim

f (x  h)  f (x) h

h0

let x  h   lim f (x  x)  f (x)   f (x) x

x0

Therefore, f  is odd. (b) If f is odd, then

f (x)  lim

f (x  h)  f (x)

h0

f (x  h)  f (x) h0 h

 lim

f (x  h)  f (x) h

h0

 lim

 f (x  h)  f (x)

h0

let x  h  lim f (x  x)  f (x)  f (x) x0

x

Therefore, f  is even.

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2.7.73 (a) f (4)  lim f (4  h)  f (4)  lim 5  (4  h) 1  lim h  1 

h0

h0

h0

h 1

1

and f (4)  lim f (4  h)  f (4)  lim 5  (4  h) 





h0



h0

 lim

1  (1  h) h(1  h)



h0 

 lim 

h0

1

(1  h)

(b)

 0  (c) f ( x)  5  x 1  (5  x)

if x  0 if 0  x  4 if x  4

At 4 we have lim f ( x)  lim(5  x)  1 and lim f (x)  lim x4

x4

x4

x4



1 5 x



 1, so lim f (x)  1  f (4) and x4

f is continuous at 4. Since f(5) is not defined, f is discontinuous at 5. These expressions show that f is continuous on the intervals (–∞, 0), (0, 4), (4, 5) and (5, ∞). Since lim f ( x)  lim(5  x)  5  0  lim f ( x), lim f (x) does not exist, so f is discontinuous at 0.

x0

x0

x0

(d) f is not differentiable at 0. 2.7.74 These graphs are idealizations conveying the spirit of the problem. In reality, changes in speed are not instantaneous, so the graph in (a) would not have corners and the graph in (b) would be continuous.

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2.7.75 (a)

(c)

2.7.76 In the right triangle shown in the diagram, let y be the side opposite angle ϕ and

x the side adjacent to the angle ϕ. Then the slope of the tangent line ℓ is y m  tan . Note that 0 < ϕ < π/2. We know that the derivative of x f ( x)  x2 is f ( x)  2x. So the slope of the tangent to the curve at the point (1, 1) is 2. Thus, ϕ is the angle between 0 and π/2 whose tangent is 2; that is  tan1 2  63 .

Solution and Answer Guide

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION CONCEPTS CHECK TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.CV.1 (a) lim𝑥→𝑎 ƒ(𝑥) = 𝐿: See Definition: Limit of a Function and Figures 2.12 and 2.13 in Section 2.2. (b) lim𝑥→𝑎+ ƒ(𝑥) = 𝐿: See Definition: One-Sided Limits and Figure 2.21 in Section 2.2. (c) lim𝑥→𝑎− ƒ(𝑥) = 𝐿: See Definition: One-Sided Limits and Figure 2.20 in Section 2.2. (d) lim𝑥→𝑎 ƒ(𝑥) = ∞: See Definition: Intuitive Idea of an Infinite Limit and Figure 2.45 in Section 2.5. (e) lim𝑥→∞ ƒ(𝑥) = 𝐿: See Definition: Limit at Infinity and Figure 2.52 in Section 2.5.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

2.CV.2 In general, the limit of a function fails to exist when the function does not approach a fixed number. For each of the following functions, the limit fails to exist at x = 2.

The left- and right-hand limits are not equal.

There is an infinite discontinuity.

There are an infinite number of oscillations.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2.CV.3 See Figure 1.23 (b) and (c) in Section 2.2. 2.CV.4 If the two one-sided limits are the same value, then the overall (two-sided) limit is that same number. 2.CV.5 (a) – (g) See the statements of Limit Laws 1– 6 and 11 in Section 2.3. 2.CV.6 See The Squeeze Theorem in Section 2.3. 2.CV.7 (a) See Definition 2.5.3 and Figures 2.47 – 2.49 in Section 2.5. (b) See Definition 2.5.6 and Figures 2.53 and 2.54 in Section 2.5. 2.CV.8 (a) 𝑦 = 𝑥4: No asymptote (b) 𝑦 = sin 𝑥: No asymptote (c) 𝑦 = tan 𝑥: Vertical asymptotes 𝑥 = 𝜋2 + 𝜋𝑛 , n an integer (d) 𝑦 = 𝑒 𝑥 : Horizontal asymptote 𝑦 = 0 (lim𝑥→−∞ 𝑒𝑥 = 0) (e) 𝑦 = ln 𝑥: Vertical asymptote 𝑦 = 0 (lim𝑥→0+ ln 𝑥 = −∞) (f) 𝑦 = 1/𝑥: Vertical asymptote x = 0, horizontal asymptote y = 0 (g) 𝑦 = √𝑥: No asymptote 2.CV.9 (a) A function f is continuous at a number a if f(x) approaches f(a) as x approaches a; that is, lim𝑥→𝑎 ƒ(𝑥) = ƒ(𝑎). (b) A function f is continuous on the interval (−∞, ∞) if f is continuous at every real number a. The graph of such a function has no breaks and every vertical line crosses it. 2.CV.10 Answers will vary. 2.CV.11 See A Closer Look for The Intermediate Value Theorem (IVT) in Section 2.4. 2.CV.12 See Definition: Tangent Line in Section 2.6.

2.CV.13 See the paragraph containing Formula 2 in Section 2.6. 2.CV.14 (a) The average rate of change of y with respect to x over the interval  x1 , x2  is

f  x2   f  x1  x2  x1

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(b) The instantaneous rate of change of y with respect to x at x  x is lim 1

x2 x1

f  x2   f  x1  . x x 2

2.CV.15 See Definition 2.6.4. The pages following the definition discuss interpretations of ƒ′(𝑥) as the slope of a tangent line to the graph of f at x = a and as an instantaneous rate of change of f(x) with respect to x when x= a. 2.CV.16 See the paragraphs before and after Example 6 in Section 2.7. 2.CV.17 (a) A function f is differentiable at a number a if its derivative f  exists at x  a ; that is, if

f   a 

exists. (b) See Theorem 2.7.4. This theorem also tells us that if f is not continuous at a , then f is not differentiable at a . 2.CV.18 See the discussion and Figure 2.80 in Section 2.7.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION FOCUS ON PROBLEM SOLVING TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

END OF SECTION EXERCISE SOLUTIONS 2.FPS.1 Let t  6 x , so x  t6 . Then t 1 as x 1 , s

t 1t 1 t 2 1 t 1 11 2 x 1 lim   .  lim 3  lim  lim x1 x 1 t1 t 1 t 1 t 1t 2  t 1 t1 t 2  t 1 12 1 1 3 3

Another method: Multiply both the numerator and the denominator by

 x 1  x  x 1 . 3

2.FPS.2 First rationalize the numerator: lim x0

ax  b  4 ax  b  2 ax  b  2   lim x0 . Now since x ax  b  2 x ax  b  2





the denominator approaches 0 as x  0 , the limit will exist only if the numerator also approaches 0 as

x  0 . So we require that a 0  b  4  0  b  4 . So the equation becomes a a lim x0  1  1 a  4 . Therefore, a  b  4 . ax  4  2 4 2

2.FPS.3 For 

x

1 2

, we have 2x 1  0 and 2x 1  0 , so 2x 1    2x 1 and 2x 1  2x 1 .

Therefore, limx0

2x 1  2x 1

 limx0   2x 1   2x 1  limx0 4x  limx04  4 x x 

x 2.FPS.4

1 1 2 since the coordinates of P are  2 x, 2 x   x2 1 x, m (negative reciprocal). But  x  x

Let R be the midpoint of OP , so the coordinates of R are

 x, x2 . Let Q  0, a  . Since the slope m

1 2 OP 2 QR x  a  1 1 x  2a 2 , so we conclude that 1  x2  2a  2a  x2 1 a  x2  . As m   QR 1 x 2 2 x0 2 1  1 x  0, a  , and the limiting position of Q is 0, .  2  2   



2.FPS.5

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Since   x    x 

x  x 



 x    1 , we have  x  1 x   1

 1

 x 

for x  1. As x   ,   x     ,

 x   x   x  1 1 x  0 and 1  1. Thus, lim x  1 by the Squeeze Theorem. so  x   x    x           2.FPS.6

[ [𝑥] ]2 + [ [𝑦] ]2 = 1. Since [ [𝑥] ]2 and [ [𝑦] ]2 are positive integers or 0 , there are only 4 cases:

(a)

Case (i):   x    1,   y    0  1  x  2 and0  y  1 Case (ii):   x    1,   y    0  1  x  0 and0  y  1

Case (iii):  x    0,  y    1  0  x  1 and1  y  2

Case (iv):  x    0,  y  1  0  x  1 and 1  y  0 (b)  x    y   3 . The only integral solution of n  m  3 is n  2 and m  1 . So the graph 2

 x, y ∣  x   2,  y  1   x, y  21  yx  32 oror 12 yx  1,0 . 



or 1  x  y  0 (d) For n  x  n 1,   x    n . Then   x      y   1    y    1 n 

1 n  y  2  n . Choosing integer values for n produces the graph. 

2.FPS.7

f is continuous on  , a  and a,  . To make f continuous on R , we must have continuity at a .

2 2 2 Thus, lim xa f  x  lim xa f  x  lim xa x  lim xa  x 1  a  a 1  a  a 1  0 





[by the quadratic formula] a  1 5 / 2  1.618 or 0.618 . 2.FPS.8 (a) Here are a few possibilities:

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

(b) The "obstacle" is the line x  y (see diagram). Any intersection of the graph of f with the line y  x constitutes a fixed point, and if the graph of the function does not cross the line somewhere in

0,1 , then it must either start at 0, 0 (in which case 0 is a fixed point) or finish at 1,1 (in which case 1 is a fixed point). (c) Consider the function F  x  f  x  x , where f is any continuous function with domain 0,1 and range in 0,1. We shall prove that f has a fixed point. Now if f  0   0 then we are done: f has a fixed point (the number 0 ), which is what we are trying to prove. So assume f 0  0 . For the same reason we can assume that f 1  1. Then F 0  f 0  0 and F 1  f 1 1  0 . So by the Intermediate Value Theorem, there exists some number c in the interval 0,1 such that

F  c   f  c   c  0 . So f  c   c , and therefore f has a fixed point. 2.FPS.9





(a) Consider G  x  T x 180  T  x  . Fix any number a . If G  a   0 , we are done: Temperature at a  Temperature at a 180 . If G  a   0 , then







G a 180  T a  360

  T a 180   T a   T a 180   G a   0 . Also, G is

continuous since temperature varies continuously. So, by the Intermediate Value Theorem, G has a zero on the interval  a, a 180  . If G  a   0 , then a similar argument applies. (b) Yes. The same argument applies. (c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity. 2.FPS.10 (a) Solution 1: We introduce a coordinate system and drop a perpendicular from P , as shown. We see from  NCP that tan2 

y , and from  NBP that tan  y / x . Using the double-angle formula 1 x

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2  y / x 2tan  tan2   . After a bit of simplification, this becomes 1 x 1 tan2 1 ( y / x)2 y

for tangents, we get

1 2x  y2  x 3x  2 . As the altitude AM decreases in length, the point P will  2 1 x x  y2 approach the x -axis, that is, y  0 , so the limiting location of P must be one of the roots of the equation x 3x  2  0 . Obviously it is not x  0 (the point P can never be to the left of the altitude

AM , which it would have to be in order to approach 0 ) so it must be 3x  2  0 , that is, x 

Solution 2: We add a few lines to the original diagram, as shown. Now note that  BPQ   PBC (alternate angles; QP BC by symmetry) and similarly CQP  QCB . So

BPQ and CQP are

isosceles, and the line segments BQ,QP and PC are all of equal length. As AM  0 , P and Q approach points on the base, and the point P is seen to approach a position two-thirds of the way between B and C , as above.

(b) The equation y2  x 3x  2 calculated in part (a) is the equation of the curve traced out by P . Now as AM , 2 



, 



2 4 part of the curve with 0  y  1.

, x 1 , and since tan  y / x, y 1. Thus, P only traces out the

274



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2.FPS.11 Let a be the x -coordinate of Q . Since the derivative of y  1 x2 is y  2x , the slope at Q is 2a . But since the triangle is equilateral, AO / OC  3 / 1 , so the slope at Q is  3 . Therefore, we must

 3  3 2   1   3,  . Thus, the point Q has coordinates  ,1  have that 2a   3  a     2  2  2    2 4     3 1  and by symmetry, P has coordinates  2 , 4 .     3



2.FPS.12 (a) V   t  is the rate of change of the volume of the water with respect to time. H   t  is the rate of change of the height of the water with respect to time. Since the volume and the height are increasing,

V   t  and H   t  are positive.

(b) V   t  is constant, so V t  is zero (the slope of a constant function is 0 ). (c) At first, the height H of the water increases quickly because the tank is narrow. But as the sphere widens, the rate of increase of the height slows down, reaching a minimum at t  t2 . Thus, the height is increasing at a decreasing rate on 0, t2  , so its graph is concave downward and H t1   0 . As the sphere narrows for t  t2 , the rate of increase of the height begins to increase, and the graph of H is concave upward. Therefore, H t2   0 and H t3   0 2.FPS.13 2 2 (a) Put x  0 and y  0 in the equation: f 0  0  f 0  f 0  0  0  0  0  f 0  2 f 0 .

Subtracting f 0 from each side of this equation gives f  0   0 . (b) 2 2    f   0   limh0 f 0  h  f 0  limh0 f 0  f  h   0 h  0h   f 0  limh0 f h  limx0 f  x   1 h h h x 

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

  2 2   2 2 f   x  limh0 f  x  h  f  x  limh0  f  x  f h   x h  xh   f  x   limh0 f  h   x h  xh

(c)



 f h   x2  xh  1 x2   h0  h 

 lim 

2.FPS.14 We find the equation of the parabola by substituting the point 100,100 , at which the car is situated,

1 . Now we find the equation of a tangent 100 1 1 to the parabola at the point  x , y  . We can show that y  a  2x   2x  x , so an equation 0 0 100 50 1 x  x  x  . Since the point  x , y  is on the parabola, we must have of the tangent is y  y  0 0 0 0 50 0 1 2 y x , so our equation of the tangent can be simplified to y  1 x2  1 x  x  x  . We want 0 0 100 0 100 0 50 0 the statue to be located on the tangent line, so we substitute its coordinates 100, 50 into this equation: into the general equation y  ax2 :100  a(100)2  a 

2 50  1 x2  1 x 100  x   x  200x  5000  0  0 0 0 0 0 50 1100 2 x  200  200  45000  x  100  50 . But x  100 , so the car's headlights 2 0 0 0   2

illuminate the statue when it is located at the point

100  50 2,150 100 2   29.3,8.6 , that is,

about 29.3 m east and 8.6 m north of the origin. 2.FPS.15

g  x  h  g  x

g   x  lim h0

 

 lim h0

 x  h f  x  h  xf  x   lim

f  x  h  f  x x lim h0

h0

 xf  x  h   xf  x  hf  x  h     h h   

 lim f  x  h  xf  x  f  x h0

because f is differentiable and therefore continuous. 2.FPS.16

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

We are given that f  x   x2 for all x . In particular, f  0   0 , but a  0 for all a . The only conclusion is that f  0  0 . Now

f  x   f 0 x2 f  x  f  x x2 f  x   f  0  x  x   x.     x x0 x x x x0 f  x  f 0  0 . So by the x0 definition of a derivative, f is differentiable at 0 and, furthermore, f   0   0 . But lim x0   x   0  lim x0 x , so by the Squeeze Theorem, lim x0

2.FPS.17

f  x  lim

lim xa

1 1 1  1  f  x  g  x   f  x  g  x  lim  f  x   g  x  lim  f  x   g  x xa  xa  xa   2  2  2  2  



  





and xa

So lim

1 3  2  1  , 2 2 2

 f  x  g x  f x  lim

g  x  lim

lim



xa



3 1 f x  2   .  f  x   g  x    lim xa xa   2 2

3 1 3  f  x g  x  lim f  x   lim g  x      .  2 2 4 xa    xa   xa

Another solution: Since lim xa  f  x   g  x  and lim xa  f  x   g  x  exist, we must have



 and lim  f  x   g  x ]  lim f  x   g  x   , so

lim f  x   g  x ]2  lim f  x   g  x    xa  xa  

xa



xa





 







1 lim f  x  g  x    lim f  x   g  x ]2   f  x   g  x ]2 [because all of the f 2 and g 2 cancel] xa xa 4 1 1 3  lim f  x   g  x ]2  lim  f  x   g  x ]2   22 12   . xa  4 xa  4 4





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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION REVIEW TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.R.1 (a) (i) lim f (x)  3

(ii) lim f (x)  0

x2

x3 

(iii) lim f (x) does not exist because the left and right limits are not equal. x3

(iv) lim f (x)  2 x4

(vi) lim f (x)  

(v) lim f (x)  

x2 

x0

(vii) lim f (x)  4

(viii) lim f (x)  1

x

x

(b) The equations of the horizontal asymptotes are y  1 and y  4. (c) The equations of the vertical asymptotes are x  0 and x  2. (d) f is discontinuous at x  3,0,3 and 4. The discontinuities are jump, infinite, infinite, and removable, respectively. 2.R.2

lim f (x)  2,

lim f (x)  0,

x

x

x3

x3

lim f (x)  ,

lim f (x)  ,

lim f (x)  2

x3

f is continuous from the right at 3,

2.R.3 To meet the first two requirements and the discontinuity, the rational function may have the form (3𝑥2 + 𝑘)(𝑥 + 2) . ƒ(𝑥) = 2 (𝑥 + 4𝑥)(𝑥 + 2) 12+𝑘 = 3 ⇒ 12 + 𝑘 = 36 ⇒ 𝑘 = 24. Thus, Then lim 𝑥→2 ƒ(𝑥) = 4+8

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ƒ(𝑥) =

3(𝑥2 + 8)(𝑥 + 2) . 𝑥(𝑥 + 4)(𝑥 + 2)

2.R.4 To meet the first two requirements and the discontinuity, the rational function may have the form (2𝑥2 + 𝑘)(𝑥 − 1) . ƒ(𝑥) = 𝑥(𝑥 + 3)(𝑥 − 1) 2+𝑘 Then lim 𝑥→1 ƒ(𝑥) = 4 = 2 ⇒ 2 + 𝑘 = 8 ⇒ 𝑘 = 6. Thus, 2(𝑥2 + 3)(𝑥 − 1) . ƒ(𝑥) = 𝑥(𝑥 + 3)(𝑥 − 1) 2.R.5 Let (2𝑥2 − 8)(𝑥 − 3)

ƒ(𝑥) =

(𝑥 − 3)(𝑥 − 2)2

This function is rational, and f(0) = -2. (2𝑥2 − 8)

lim ƒ(𝑥) = lim

𝑛→3 (𝑥 − 2)

𝑛→3

2𝑥2 − 8

= 10, and lim ƒ(𝑥) = lim

𝑛→±∞ 𝑥

𝑛→±∞

= lim

− 4𝑥 + 4

𝑛→±∞

8 2− 2 =2 4𝑥4 1 − 𝑥 + 𝑥2

This is as close to the requested function as possible. 2.R.6 (𝑥−1)2 0 As x 1, the function ƒ(𝑥) = → . But lim (𝑥−1)4

𝑥→1

ƒ(𝑥) = lim

1 (𝑥−1)2 = ∞. 𝑥→1 (𝑥−1)4 = lim 𝑥→1 (𝑥−1)2

2.R.7 3 Because the exponential function is continuous, lim e x x  e11  e0  1. x1

2.R.8

x2  9 32  9 0     0. 2 2 x3 x  2x  3 3  2(3)  3 12

Because rational functions are continuous, lim 2.R.9

lim x3

x2  9

x2  2x  3

 lim

x3

(x  3)(x  3) (x 1)(x  3)

 lim

x3



(x  3) (x 1)



3  3 3 1





6 4





3 2



2.R.10

x  9   x2  9  0 for 1  x  3. 2 x1 x2  2x  3 since x2  2x  3  0 as x 1 and x  2x  3

lim



2.R.11 (h 1)3 1

 lim

lim h0

(h3  3h2  3h 1) 1

h0

h3  3h2  3h  lim

h0

 lim h  3h  3h  3 h0

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2.R.12 2

t 4 (t  2)(t  2) (t  2) 22 1  lim    lim lim 3 t2 (t  2) t 2  2t  4 t2 t  8   t2 t 2  2t  4  4  4  4 3

2.R.13

lim r9

   since  r  9   0 as r  9 and 4

r  9

r  0 for r  9. (r  9)4

2.R.14

1  1 lim 4  v  lim 4  v  lim v4  4  v v4 (4  v) v4 1 



u2 1u2 1 (u 1)(u 1)u2 1 (u 1) u2 1 2(2) 4   lim  lim  lim  lim u  5u  6u u(u 1)(u  6) u(u  6) 1(7) 7 u u  5u  6

2.R.15 u1

u4 1

u1



2.R.16

    x 2  x  x  x  6  x2 x6 x x6  x x6  x  lim 3  lim   lim  lim  x3  3x2  x3  x3 x  3x2 x3 2 x3 2 x  6  x   x (x  3) x6  x x (x  3) x  6  x 5 5 (x  3)(x  2) (x  2)    lim  lim x3 2 x3 9(3  3) 54 x6  x x (x  3) x  6  x x2















2.R.17

lim x0

tan x

1 sin x  lim sin x  lim  lim  11  1 x0 x cos x x0 cos x x0 x x

2.R.18

lim x1

ln x x



ln1

0

2.R.19

x  lim 1  1 x0 2 x x0 2 2

lim

2.R.20 Since x is positive,

x2  9

9  lim 1 x2  1 0  1 x 2x  6 x 2  6 20 2 x2

x2  x  x. Thus lim

280



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2.R.21 Since x is negative,

x2  x  x. Thus lim

x

x2  9 2x  6

 lim

1 x92

x

2  x62



1 0 1  2  0 2

2.R.22 Let t  sin x. Then as x   ,sin x  0, so t  0. Thus lim ln(sin x)  lim ln t  . t0

x 



2.R.23

lim

x

1  1 1 x4 x2

1 2x2  x4 5  x  3x



 lim 5 x

 3 3 



0  0 1 

003 



 x  4x 1  x   lim  x  4x1 1  x 

x

1 3



2.R.24

lim

x2  4x 1  x 

 x 2  4x 1  x2

  lim 2  x  4x 1  x  x x  4x 1  x 4x 1 4  1x 40 4 lim   lim   2 x 1 0  0 1 2 x2  4x 1  x x 1 4x  x12 1x 2

x

2.R.25 2 xx 2  lim et  0. Let t  x  x  x(1 x). Then as x  ,t  , and lim e x

t



 1   1 1     lim    lim  x2 1 lim   2   x1 x1 x1 x 1 x  3x  2 x 1 (x 1)(x  2)  (x 1)(x  2)  (x 1)(x  2)           1  1  lim   lim  1  x 1  x1  (x 1)(x  2)      x1  x  2  1 2

2.R.26 1











281

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2.R.27





From the graph of y  cos2 x / x2 it appears that y  0 is the horizontal asymptote and x  0 is the 2 vertical asymptote. Now 0  (cos x)2  1  0  (cos x)  1 2 2 2

 0 

(cos x)2

.  2 But lim 0  0 and lim 2  0, so by the Squeeze x x x x x cos2 x Theorem, lim  0. Thus y  0 is the horizontal asymptote. Now x x2 2 cos x lim   because cos2 x 1 and x2  0 as x  0, so x  0 is the vertical asymptote. 2 x0 x 2



2.R.28

From the graph of y  f (x)  x2  x 1  x2  x, it appears that there are two horizontal asymptotes and possibly 2 vertical asymptotes. To obtain a different form for f, let’s multiply and divide it by its conjugate:

f1(x) 



 x  x 1  x  x  x  x 1  x  x 2

x 2  x 1  x 2  x

x2  x 1  2x 1

x2  x

x  x 1  x  x



x

 lim

x  x 1  x  x 2

2x 1

Now lim f1(x)  lim x

x

2  1x 1  1 x

1 x2



 1

1 x

 2, so y  1 is a

11

horizontal asymptote. For x < 0, we have x2  x  x, so when we divide the denominator by x, with x

x2  x 1 

x2  x

< 0, we get

 Therefore lim f1(x)  lim x

x

  x2  x 1  x2  x      1  1 1 1 x  x2 1 x .  x2 2x 1

 lim

x  x 1  x  x 2

x

2  1x 1 1x  x12  1 1x



2 (11)

 1, so

y  1 is a horizontal asymptote. The domain of f is , 0 1,   . As x  0 , f (x) 1, so x  0 is not vertical asymptote. As 

x 1, f (x)  3, so x 1 is not a vertical asymptote and hence there are no vertical asymptotes.

2.R.29

282

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

Since 2x 1  f (x)  x for 0  x  3 and lim2x 1  1  lim x , we have lim f (x)  1 by the 2

x1

Squeeze Theorem. 2.R.30



Let f (x)  x 2 , g(x)  x2 cos 1 / x2

 and h(x)  x2. Then since cos1/ x2  1for x  0, we have

f (x)  g(x)  h(x) for x  0, and so lim f (x)  lim h(x)  0  lim g(x)  0 by the Squeeze x0

x0

Theorem.

2.R.31

4 1 sin x  1 cos x  1,       and   we have Since 1 4   10sin  6x2 cos  6 x  x x2  6 10 1 6x2 1  lim 10   3     lim  lim 2 2 3 2 x x x 2x  3x  2 2  x  x2 2x  3x  2 2 2.R.32



(a) f (x)  x if x  0, f (x)  3  x if 0  x  3, f (x)  (x  3) if x  3. (i) lim f (x)  lim(3  x)  3 (ii) lim f (x)  lim 2

x0

(iii) Because of (i) and (ii), lim f (x) does not exist. x0

0 x (iv) lim f (x)  lim(3  x)  0 x0

x0

x3

(v) lim f (x)  lim(x  3)  0 2

x3

x3

(vi) Because of (iv) and (v), lim f (x) does not exist x3

f (x) does not exist. (b) f is discontinuous at 0 since lim x0 f is discontinuous at 3 since lim f (x) does not exist. x3

2.R.33

283

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

(a) g(x)  2x  x2 if 0  x  2, g(x)  2  x if 2  x  3, g(x)  x  4 if 3  x  4, g(x)   , if x  4. Therefore, lim g(x)  lim 2x  x2  0 and lim g(x)  lim 2  x  0. Thus



x2



x2

x2

lim g(x)  g(2)  0, so g is continuous at 2. lim g(x)  lim 2  x  1, and x3

x2

x3

lim g(x)  lim  x  4  1. Thus, lim g(x)  1  g(3), so g is continuous at 3.

x3

x4

x3

lim g(x)  lim(x  4)  0 and lim g(x)     .Thus, lim g(x) does not exist, so g is x4

x4

discontinuous at 4. But lim g(x)    g(4), so g is continuous from the right at 4. x4

2.R.34

sin x and ex are continuous on

, esin x is continuous on by since it’s a polynomial and the product xesin x is continuous

by Theorem 3. Since is continuous on

Theorem 5. Lastly x is continuous on on its domain. 2.R.35 x2  9 is continuous on

because it is a polynomial and



x is continuous on0,   , so the

  , 3 3, . Note that x2  2  0 on

2 composition x  9 is continuous on x | x2  9  0

this set and so the quotient function f (x) 

x2  9 is continuous on its domain, , 33, . x2  2

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2.R.36

1 x Let f (x)  . The domain of f isx | x  0 and x  6. At all other values, f is a composite of x6 rational and polynomial functions, so it is continuous. Therefore, f has discontinuities at x = 0 and x = 6.

6 1 x  6 x  Now f (x)  x6

1 x6



for x ≠ 6, so the discontinuity at x = 6 is removable whereas the

discontinuity at x = 0 is not. 2.R.37



kx  k 5x

kx  k 5x Let f (x)  x  10. But x 2x  5 2x x5 k k kx  5x k  x5x k kx  k  5  lim  lim   k  20. We also need to find a k so that lim 1 x x 2x  5x 2  2 x 2x  1x 5 x5 x . We need to find k so that lim

k 5x kx  k 5 kx  k 5x lim k  k  10. Thus x  10. In this case, lim x  lim x 2x  5 x 2x  5 x 5 x k  20 or k  10. x

2.R.38

f (x) 

2  3 x1 73

(a) Let

1 x

lim f (x)  lim

. Then

x

2  3x1



x

7  3x

2 1 3  . 7 1 8

2  3t 3t 1  t  (b) Let and t  x . Then as x  0 ,t  , so lim f (x)  lim 7  3  lim 3t  1. t x x0 

 t



2  3 

t . lim f (x)  lim x Then as x  0  ,t  , and x0  t 7  3t (c) Let 



20

2  . x0 7  0 7 lim

2.R.39 First find the horizontal asymptote of f (x) 

x3  x2  2x x3  2x2  x

1 22 1 0  0 lim 1 x  x   1 so y  1 is the horizontal asymptote. Using the solve function, we find that x1 2 1 1 0  0 x x2

f (3)  1 so this function crosses its horizontal asymptote at the point (3,1).

2.R.40

285

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

f (x)  x5  x3  3x  5 is continuous on the interval [1, 2], f (1)  2, and f (2)  25. Since 2  0  25, there is a number c in (1, 2) such that f (c)  0 by the Intermediate Value Theorem. Thus, there is a root of the equation x5  x3  3x  5  0 in the interval (1, 2). 2.R.41

f (x)  cos x  ex  2 is continuous on the interval [0, 1], f (0)  2, and f (1)  0.2. Since 0.2  0  2, there is a number c in (0, 1) such that f (c)  0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x  ex  2  0, or cos x  ex  2, in the interval (0, 1). 2.R.42 (a) 𝑠 = 𝑠(𝑡) = 1 + 2𝑡 + 𝑡2/4. The average velocity over the time interval [1, 1+h] is (1 + ℎ)2 − 13 𝑠(1 + ℎ) − 𝑠(1) 1 + 2(1 + ℎ) + 10ℎ + ℎ2 10 + ℎ 4 4 = = 𝑣𝑎𝑣𝑒 = = ℎ 4ℎ 4 (1 + ℎ) − 1 So for the following intervals, the average velocities are: 10+2 = 3 m/s (i) [1,3]: ℎ = 2, 𝑣𝑎𝑣𝑒 = (ii) [1,2]: ℎ = 1, 𝑣𝑎𝑣𝑒 =

4 10+1 4

(iii) [1,1.5]: ℎ = 0.5, 𝑣𝑎𝑣𝑒 = (iv) [1,1.1]: ℎ = 0.1, 𝑣𝑎𝑣𝑒 =

= 2.75 m/s 10+0.5

4 10+0.1 4

= 2.625 m/s = 2.525 m/s

(b) When t = 1, the instantaneous velocity is lim

𝑠(1+ℎ)−𝑠(1) ℎ→0

ℎ

= lim ℎ→0

10+ℎ =

= 2.5 m/s.

2.R.43 (a) When V increases from 200 in3 to 250 in3, we have ∆𝑉 = 250 − 200 = 50 in3, since 𝑃 = 800/𝑉, 800 800 ∆𝑃 = 𝑃(250) − 𝑃(200) = − = 3.2 − 4 = −0.8 lb/in2. So the average rate of change is ∆𝑃 0.8 lb/in2 = − − −0.016 3 . ∆𝑉 50 in

250

200

(b) Since V = 800/P, the instantaneous rate of change of V with respect to P is (𝑃 + ℎ)] lim ∆𝑉 = lim 𝑉(𝑃 + ℎ) − 𝑉(𝑃) = lim 800/(𝑃 + ℎ) − 800/𝑃 = lim 800[𝑃 − ℎ→0 ℎ→0 ℎ→0 ℎ→0 ∆𝑃 ℎ ℎ ℎ(𝑃 + ℎ)𝑃 −800 800 = lim =− ℎ→0 (𝑃 + ℎ)𝑃 𝑃2 which is inversely proportional to the square of P. 2.R.44 Estimating the slopes of the tangent lines at x = 2, 3, and 5, we obtain approximate values 0.4, 2, and 0.1. Since the graph is concave downward at x = 5, ƒ′′(5) is negative. Arranging the numbers in increasing order, we have:

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

ƒ′′(5) < 0 < ƒ′(5) < ƒ′(2) < 1 < ƒ′(3)

2.R.45 (a) ƒ′′(2) = lim 𝑥→2

ƒ(𝑥)−ƒ(2) 𝑥3−2𝑥−4 (𝑥−2)(𝑥2+2𝑥+2) = lim 𝑥→2 = lim 𝑥→2 = lim 𝑥→2(𝑥2 + 2𝑥 + 2) = 10 𝑥−2 𝑥−2 𝑥−2

(b) 𝑦 − 4 = 10(𝑥 − 2) or 𝑦 = 10𝑥 − 16 (c)

2.R.46 (a) Estimating ƒ′(1) from the triangle in the graph below, we get ∆𝑦 ≈ ∆𝑥

−0.37

= −0.74 . To estimate ƒ′(1)

0.50

numerically, we have 2

𝑒−(1+ℎ) − 𝑒−1 ƒ(1 + ℎ) − ƒ(1) = lim =𝑦 ℎ ℎ ℎ→0 ℎ→0 From the table, we have ƒ′(1) ≈ −0.736. ƒ′(1) = lim

h 0.01 0.001 0.0001 -0.01 -0.001 -0.0001

y -0.732 -0.735 -0.736 -0.739 -0.736 -0.736

(b) 𝑦 − 𝑒−1 ≈ −0.736(𝑥 − 1) or 𝑦 ≈ −0.736𝑥 + 1.104 (c) See the graph in part (a).

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Review

2.R.47 26 = 64, so ƒ(𝑥) = 𝑥6 and 𝑎 = 2.

2.R.48 (a) ƒ′(𝑟) is the rate at which the total cost changes with respect to the interest rate. Its units are dollars/(percent per year). (b) The total cost of paying off the loan is increasing by $1200/(percent per year) as the interest rate reaches 10%. So if the interest rate goes up from 10% to 11%, the cost goes up approximately $1200. (c) As r increases, C increases. So ƒ′(𝑟) will always be positive.

2.R.49

2.R.50

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

2.R.51

2.R.52

2.R.53

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2.R.54

2.R.55 f is not differentiable: at x = -4 because f is not continuous, at x = -1 because f has a corner, at x = 2 because f is not continuous, and at x = 5 because f has a vertical tangent.

2.R.56 The graph of a has tangent lines with positive slope for x < 0 and negative slope for x > 0, and the values of c fit this pattern, so c must be the graph of the derivative of the function for a. The graph of c has horizontal tangent lines to the left and right of the x-axis and b has zeros at these points. Hence, b is the graph of the derivative of the function for c. Therefore, a is the graph of f, c is the graph of f´, and b is the graph of f´´.

2.R.57

C2017 is the rate at which the total value of U.S. currency in circulation is changing, in billions of dollars per year. To estimate the value of C2017, we will use the average rate of change from t  2016 to t  2018. C2017 

C 2018  C 2016



1671.9 1463.4

2018  2016



208.5

 104.25 billion dollars/year.

2.R.58

(a) Drawing slope triangles, we obtain the following estimates: F2000  0.1 10  0.01, 0.2 F2010  0.5 10  0.05, and F2015  10  0.02.

(b) The rate of change of the average number of children born to each woman was increasing by 0.01 in 2000, decreasing by 0.05 in 2010, and decreasing by 0.02 in 2015. (c) There are many possible reasons:

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

  

In the dot-com era (around 2000), there was perhaps economic optimism, with family size rising. During the economic downturn following 2008 (affecting 2010), there was perhaps economic distress, with family size falling. There was perhaps less economic distress by 2015, leading to family size falling at a lower rate.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 2: SECTION TRUE/FALSE TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 2.T.1 False. Limit Law 2 applies only if the individual limits exist (these don’t). 2.T.2 False. Limit Law 5 cannot be applied if the limit of the denominator is 0 (it is).

2.T.3 True.

Limit Law 5 applies.

2.T.4 False.

f (x) 

x2  9 x3

 g(x)  x  3 because f (3) is undefined but g(3)  6.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

2.T.5

x2  9 (x  3)(x  3)  lim(x  3)  6 lim x3 x  3 x3 x3 (x  3)

lim True.

2.T.6

False.

A limit must be unique. It cannot have two distinct values. lim 4 x  3  4 x4

2.T.7

 x x  16 x0 sin x0  sin 6xx  61 lim lim    False.   

 2.T.8

lim False.

4(x  h)  4x h

h0

 lim h0

 lim1  1

h0

2.T.9

sin 3x 3  x0 sin 2x 2

lim True.

2.T.10 True. The limit doesn’t exist since ƒ(𝑥)/𝑔(𝑥) doesn’t approach any real number as x approaches 5. (The denominator approaches 0 and the numerator doesn’t.) 2.T.11 𝑥(𝑥−5) . The first limit exists and is equal to 5. By Example 3 False. Consider lim 𝑥→5 𝑥−5 or lim 𝑥→5 sin(𝑥−5) 𝑥−5 in Section 2.2, we know that the latter limit exists (and it is equal to 1). 2.T.12 False.

If f (x)  1/ x, g(x)  1/ x, and a = 0, then lim f (x) does not exist and lim g(x) does not xa

xa

exist, but lim f (x)  g(x)  lim 0  0 does exist. xa

xa

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

2.T.13 True.

Assume lim f (x) exists and lim g(x) does not exist. Now suppose lim f (x)  g(x) exists. xa

xa

Then lim𝑔(𝑥) = 𝑥→𝑎 lim {[ƒ(𝑥) + 𝑔(𝑥)] − ƒ(𝑥)} = 𝑥→𝑎 lim[ƒ(𝑥) + 𝑔(𝑥)] − 𝑥→𝑎 limƒ(𝑥)

𝑥→𝑎

exists because both limits on the right exist. But this is a contradiction and therefore lim f (x)  g(x) xa

cannot exist. 2.T.14 1 False. Consider lim 𝑥→6[ƒ(𝑥)𝑔(𝑥)] = lim𝑥→6 *(𝑥 − 6) 𝑥−6+. It exists (its value is 1) but f(6) = 0 and g(6) does not exist, so f(6)g(6) ≠ 1. 2.T.15 True. A polynomial is continuous everywhere, so lim𝑥→𝑏 𝑝(𝑥) exists and is equal to 𝑝(𝑏). 2.T.16 False. Consider lim

𝑥→0

[ƒ(𝑥) − 𝑔(𝑥)] = lim

individual functions approaches ∞.

( 1 − 1 ). This limit is -∞ (not 0), but each of the 𝑥→0 𝑥2 𝑥4

2.T.17 False. See Figure 2.56 in Section 2.5. 2.T.18 False. Consider ƒ(𝑥) = sin 𝑥 for x > 0. lim𝑥→∞ ƒ(𝑥) G ±∞ and f has no horizontal asymptote. 2.T.19 1/(𝑥 − 1) False. Consider ƒ(𝑥) = , 2

if 𝑥 G 1 if 𝑥 = 1

2.T.20 False. The function f must be continuous in order to use the Intermediate Value Theorem. For example, let 1 if 0 < 𝑥 < 3 ƒ(𝑥) = , −1 if 𝑥 = 3 There is no number 𝑐 ∈ [0,3] with ƒ(𝑐) = 0. 2.T.21 True. Use Theorem 2.4.8 with a = 2, b = 5, and 𝑔(𝑥) = 4𝑥2 − 11. Note that f(4) = 3 is not needed. 2.T.22 True. Use the Intermediate Value Theorem with a = -1, b = 1, and 𝑁 = 𝜋, since 3 < 𝜋 < 4. 2.T.23 2 False. For example, let ƒ(𝑥) = {𝑥 + 1 if 𝑥 G 0 2 if 𝑥 = 0 Then ƒ(𝑥) > 1 for all x, but lim𝑥→0 ƒ(𝑥) = lim𝑥→0(𝑥2 + 1) = 1. © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

2.T.24 True.

f (x)  x10 10x2  5 is continuous on the interval[0, 2], f (0)  5, f (1)  4, and

f (2)  989. Since 4  0  5, there is a number c in (0, 1) such that f (c)  0 by the Intermediate Value Theorem. Thus there is a root of the equation x10 10x2  5  0. in the interval (0, 1). Similarly, there is a root in (1, 2).

2.T.25 False. f (x) 

x2  3x 10 x5



 x  2 if x  5  . Since f is a line for x ≠ –5, it takes on all real undefined if x  5

values except –5 – 2 = –7. But there is no c such that f (c)  7 because the only c that could produce –7 is c = –5, but f is undefined there.



2.T.26

True. For any a , we have limxa f  x  limxa f  x  f  a  



 1, x  0

2.T.27



. f  x is continuous, but f  x is not. False. Consider the function f  x   1, x  0  2.T.28



 x2  x   x 1  x 1  and the limit oscillates  lim cos (a) False. limx0 cos . As x  0,  x0  x2  2 x3 x     between 1 and 1.   x2  x   x 1   x 1   lim cos  coslim  cos0  1 (b) True. limx cos  x  x2  x  x3 x2       

2.T.29

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 dy 2  d2 y False. is the second derivative while   is the first derivative squared. For example, if y  x , dx2  dx   dy   d2 y then  0 , but    1. dx2  dx  2

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.1 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.1.1

eh 1  1. h0 h

(a) e is the number such that lim

lim 2.7 1  0.993 lim 2.8 1  1.03. h h (b) From the tables it appears that h0 and h0 h

lim h0

2.7h 1

lim h0

2.8h 1 h

–0.001

0.9928

–0.001

1.0291

–0.0001

0.9932

–0.0001

1.0296

0.001

0.9937

0.001

1.0301

0.0001

0.9933

0.0001

1.0297

Since 0.99 < 1 < 1.03, we conclude that 2.7 < e < 2.8. 3.1.2

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) Observe that the graph of never crosses the x-axis, and crosses the y-axis at the point (0,1). At this point, the slope of the tangent line appears to be roughly 1.0. (b) f (x)  ex is an exponential function and g(x)  xe is a power function.

f (x) 

e  e and g(x)  dx x  ex . dx x

e1

The function f (x)  ex grows more rapidly than g(x)  xe when x is large, because for large x,

(c)

f (x)  ex  g(x)  exe1. See the graph to the right. 3.1.3

f ( x)  240 is a constant function, so its derivative is 0, that is f (x)  0.

3.1.4

f (x)  e5 is a constant function, so its derivative is 0, that is f (x)  0.



3.1.5

f (x)  5.2x  2.2  f (x)  5.2(1)  0  5.2

3.1.6 g(x)  7 x2  3x 12  g(x)  7 (2x)  3(1)  0  7 x  3 4



3.1.7

f (t)  2t  3t  4t  f (t)  2 3t2   3(2t)  4(1)  6t2  6t  4 3

3.1.8 f (t)  1.45  2.5t2  6.7  f (t)  1.4 5t4   2.5(2t)  0  7t4  5t

3.1.9 g( x)  x2 (1  2x)  x2  2x 3  g( x)  2x  2 3x2   2x  6x 2

3.1.10

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

H (u)  (3u 1)(u  2)  3u2  5u  2  H (u)  3(2u)  5(1)  0  6u  5

3.1.11 g(t)  2t3/4  g(t)  2  34t7/4    32 t7/4

3.1.12 B( y)  cy6  B( y)  c 6 y 7   6cy 7

3.1.13 F(r) 

5 r

 5r3  F(r)  5 3r 3

4

 15r   r 4



3.1.14

y  x5/3  x2/3  y  35 x2/3  23 x1/3

3.1.15 R(a)   3a  1  9a 2  6a  1  R(a)  9(2a)  6(1)  0  18a  6 2

3.1.16 h(t)  4 t  4et  t1/4  4et  h(t)  1 4t 3/4  4 et   14 t 3/4  4et

3.1.17 S( p)  3.1.18

p  p  p1/2  p  S( p)  12 p1/2  1 or 1  1 2 p



y  3 x (2  x)  2x1/3  x4/3  y  2  31 x2/3   34 x1/3  32 x2/3  43 x1/3

1 33 x2

 43 3 x

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.1.19

 

4 y  3ex  3  3ex  4x1/3  y  3 ex  4  31  x4/3  3ex  34 x4/3 x 3.1.20

S(R)  4R2  S(R)  r (2R)  8R 3.1.21 h(u)  Au3  Bu2  Cu  h(u)  A3u2   B 2u  C(1)  3Au2  2Bu  C

3.1.22 xx x x y    x3/2  x1  y   3 x5/2  1x2    3 x5/2  x2 2 2 x2 x2 x2 3.1.23 y

x2  4x  3 x



 



 x 3/2  4x1/2  3x1/2 

1/2

y 2x

 4 2  x

 

3.1.24 G(t) 

5t 

7 t





 3 2  x

 

3/2

2

x



3x2  4x  3

3 

2x x



2x x



1/2  7 1t 2   5t1/2  7t1  G(t)  5  21 t  

5 2 t



7 t2



3.1.25

j(x)  x

1/2

2.4

e

2.4

 j(x)  2.4x1.4  0  2.4x1.4

3.1.26

k(r)  er  re  k(r)  er  ere1 3.1.27

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

G(q)  1 q 1   1 2q1  q2  G(q)  0  2  1q 2    2q3   2q2  2q3 2

3.1.28 A  Bz  Cz2

F(r) 





Cz2

 Az

2

1



Bz  C 

F(r)  A2z3   B 1z2   0  2 Az3  Bz2  

2A z

f (v) 

v  2vev v

B z

or 

2A  Bz z3

 

3.1.29 3



2ve 2/3 v  v 2 5/3  v  v  2e  f (v)   3 v  2e v v 3

3.1.30 D(t) 

1  16t2 



(4t)

1  16t2 

64t

3

 1 t  1t

1



D(t)  1 3t4   1 1t2    3 t4  1 t2 or  3  1 64 4 64 4 64t4 4t2 3.1.31 10A 11 y y A z  10  Bey  Ay10  Bey  z  10Ay  Be   11  Be y y



3.1.32

ye

x1

 1  e e  1  e  e  1  y  e  ex  0  ex1 x 1



3.1.33

y  2x  x  2  y  6x2  2x. At (1,3), y  6(1)2  2(1)  4 and an equation of the tangent line 3

is y  4(x  1)  3or y  4x 1.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.1.34

y  2e  x  y  2ex  1. At (0, 2), y  2e0 1  3 and an equation of the tangent line is x

y  3(x  0)  2 or y  3x  2. 3.1.35 y  x

 x  2x1  y  1 2x2. At (2,3), y  1  2(2)2  1 and an equation of the tangent 2

x line is y  1 (x  2)  3 or y  1 x  2. 2



3.1.36



y  4 x  x  x1/4  x  y  14 x3/4 1 

1 4

4 x tangent line is y   3 (x 1)  0 or y   3 x  3 . 4

1. At (1, 0), y  14 1   34 and an equation of the

3.1.37

y  x4  2ex  y  4x3  2ex. At (0, 2), y  2, and an equation of the tangent line is y  2(x  0)  2 or y  2x  2. The slope of the normal line is  12 (the negative reciprocal of 2) and an equation of the normal line is y   12( x  0)  2 or y   1 2x  2.



3.1.38



y  x  y  x 2

3/2

[since x and y are positive at (1,1)]  y  32 x1/2. At (1,1), y  32

equation of the tangent line is

and an

y  32 ( x  1)  1 or y  32 x  12 .

The slope of the normal line is  23 (the negative reciprocal of 32 ) and an equation of the normal line is

y   23 (x 1)  1 or y   23x  53.

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

3.1.39

y  3x  x  y  6x  3x2. At (1, 2), y  6  3  3, so an equation of the tangent 2

line is y  3(x 1)  2 or y  3x 1.

3.1.40



 

1/2

y  x  x  y  1 x 1 2

 1



. At (1,0), y  12 , so an equation of the

2 x tangent line is y  (x 1)  0 or y  x  1 .

3.1.41

 

f (x)  x4  2x3  x2  f (x)  4x3  6x2  2x. Note that f ( x)  0 when f has a horizontal tangent, f  is positive when f is increasing, and f  is negative when f is decreasing. When f changes direction, f  crosses the x-axis.

3.1.42



f (x)  x5  2x3  x 1  f (x)  5x4  6x2 1. Note that f ( x)  0 when f has a horizontal tangent, f  is positive when f is increasing, and f  is negative when f is decreasing.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.1.43 To graphically estimate the value of f 1 for f  x  3x2  x3 , we'll graph f in the viewing rectangle

1 0.1,1 0.1 by

f  0.9  , f 1.1 , as shown in the figure. [When assigning values to the window

variables, it is convenient to use Y1 0.9 for Ymin and Y1 1.1 for Ymax .] If we have sufficiently

zoomed in on the graph of f , we should obtain a graph that looks like a diagonal line; if not, graph again with 1 0.01 and 1 0.01, etc.

Estimated value:

f 1 

2.299 1.701 0.589   2.99 1.1 0.9 0.2

Exact value: f  x  3x2  x3  f  x  6x  3x2 , so f 1  6  3  3 .

3.1.44 See the previous exercise. Since f is a decreasing function, assign Y1 3.9 to Ymax and Y1 4.1 to Ymin .

0.49386  0.50637 0.01251   0.06255 . 4.1 3.9 0.2 1 11 1 1 1/2    0.0625 .  f   x    x3/2 , so f   4    43/2   Exact value: f  x  x 2 2  8  16 2 Estimated value: f   4  





3.1.45 (a)

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(b) From the graph in part (a), it appears that f  is zero at x1  1.25, x2  0.5, and x3  3.05. The slopes are positive (so f  is positive) on

 x1, x2  and  x3,   . The slopes are negative (so f  is negative) on , x1  and  x2 , x3 . 4 3 2 3 2 (c) f (x)  x  3x  6x  7x  30  f (x)  4x  9x 12x  7

3.1.46 (a)

(b) From the graph in part (a), it appears that g is zero at x1  0.2 and x2  2.8. The slopes are positive (so g is positive) on , x1  and  x2,   . The slopes are negative (so g is negative) on  x1, x2 . (c) g(x)  e  3x x

 g(x)  ex  6x



3.1.47



f (x)  0.001x  0.02x  f ( x)  0.005x  0.06x  f (x)  0.02x3  0.12x 5

3.1.48

304

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

r  3 r  G(r)  2 r

G(r) 

1/2

 1 r2/3  G (r)   1 r3/2  2 r5/3 3

3.1.49 f (x)  2x  5x3/4  f (x)  2  15 x1/4  f (x)  15 x5/4 4

Note that f  is negative when f is decreasing and positive when f is increasing. f  is always positive since f  is always increasing.



3.1.50



f (x)  e  x  f (x)  e  3x  f (x)  ex  6x x

Note that f ( x)  0 when f has a horizontal tangent and that f (x)  0 when

f  has a horizontal tangent.



3.1.51

2 (a) s  t  3t  v(t)  s(t)  3t  3  a(t)  v(t)  6t 3

(b) a(2)  6(2)  12 m/s

3.1.52 3 2 2 (a) s  t  2t  t  t  v(t)  s(t)  4t  6t  2t 1  a(t)  v(t)  12t 12t  2 4

(b) a(1)  12(1) 12(1)  2  2 m/s 2

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(c)

3.1.53

The velocity of the particle is given by v t   st   3t 2 12t  9, and the acceleration is given by a t   vt   6t 12. Then acceleration is zero when t  2, and the velocity at this time is 3 inches per second.

3.1.54 The velocity of the particle is given by v t   st   6t 2  2t  4  2 3t  2  t 1, and the acceleration is given by a t   vt   12t  2. Then velocity is zero when t  2 3, and the acceleration at this time is 10 cm/sec2.

3.1.55 If f  x  3ex 1, then f  x  3ex and f   0   3e0  3. The slope of the tangent line at the point where x  0 is 3, and the point on the graph has coordinates 0, 3e0 1  0, 2. The equation of the tangent line at that point is y  3 x  0  2, or y  3x  2. 3.1.56

L  0.0155A3  0.372A2  3.95A 1.21 

dL dA

 0.0465A2  0.744 A  3.95, so

dL  0.0465(12)2  0.744(12)  3.95  1.718. The derivative is the instantaneous rate of dA A12 change of the length of an Alaskan rockfish with respect to its age when its age is 12 years.

3.1.57

306

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

L A  9.4 A  54, so L5  9.45  54  7, and the units mm/yr. A bluegill fish in this lake grows in length 7 mm per year.

3.1.58 (a) P 

and P  50 whenV  0.106, so k  PV  50(0.106)  5.3.Thus, P 

5.3

andV 

V 1

(b) V  5.3P



 0.5.31P2   

5.3

. When P  50,

dV dP



5.3

 0.00212.

The derivative is the instantaneous rate of change of the volume with respect to the pressure at 25 C. Its units are m3 / kPa (kilopascal). 3.1.59 (a) Using technology, we find the quadratic model L  aP  bP  c, where a  0.275428, 2

b  19.74853, and c  273.55234. (b)

 2aP  b. When P  30,

 3.2, and when P  40,

 2.3. The derivative is the

instantaneous rate of change of tire life with respect to pressure. Its units are (1000 miles) / (lb/in2). When

is positive, the life of the tire is increasing, and when

 0, the life of the tire is decreasing.

3.1.60 The curve y  2x  3x 12x  1 has a horizontal tangent when y  6x  6x 12  0  3

6  x 2  x  2  0  6( x  2)( x  1)  0  x  2 or x  1. The points on the curve are (2, 21) and (1, 6).



3.1.61



f (x)  e  2x  f (x)  e  2. f (x)  0  e  2  x  ln 2, so f has a horizontal tangent x

when x = ln 2.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.1.62

y  2ex  3x  5x3  y  2ex  3  15x2. Since 2ex  0 and15x2  0 we must have y  0  3  0  3, so no tangent line can have slope 2.



3.1.63

y  x4  1  y  4x3. The slope of the line 32x  y  15 or y  32x  15 is 32, so the slope of any line parallel top it is also 32. Thus, y  32  4x3  32  x3  8  x  2, which is the xcoordinate of the point on the curve at which the slope is 32. The y-coordinate is 24 + 1 = 17, so an equation of the tangent line is y  32(x  2)  17 or y  32x  47.

3.1.64 The slope of the line 3x  y  15 or y  3x  15 is 3, so the slope of both tangent lines to the curve is





3. y  x3  3x2  3x  3  y  3x 2  6x  3  3 x 2  2x 1  3  x 1 . Thus, 2

3  x  1  3   x  1  1  x  1  1  x  0 or 2, which are the x-coordinates at which the 2

tangent lines have slope 3. The points on the curve are and so the tangent line equations are

y  3(x  0)  3 or y  3x  3and y  3(x  2) 1 or y  3x  7.

3.1.65 The slope of y  1  2e  3x is given by m  y  2e  3. x

The slope of 3x  y  5  y  3x  5is 3.

m  3  2ex  3  3  ex  3  x  ln 3. This occurs at the point ln 3, 7  3ln 3  1.0986, 3.704.

3.1.66 The slope of y 



x is given by y  1 x1/2  1 . The slope of 2x  y  1 (or y  2x 1) 2 2 x

308

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

is –2, so the desired normal line must have slope –2, and hence, the tangent line to the curve must have slope 12 . This occurs if

1 1  2 x 2

x  1  x  1. When x = 1, y  1  1, and an equation of the

normal line is y  2(x 1)  1or y  2x  3. 3.1.67

y  f (x)  x2  1  f (x)  2x. So f (3)  6, and the slope of the normal line is  16. The equation of the normal line at (3, 1) is y   16 ( x  3)  1 or y   16x  32.

3.1.68

y  f (x)  x2 1  f (x)  2x. So f (1)  2, and the slope of the normal line is 1 . The equation of the normal line at (–1, 0) is y  1 ( x  (1))  0 or 2

y  12 x  12 . Substituting this into the equation of the parabola, we obtain 1 2

x  12  x2 1  x  1  2x 2  2  2x2  x  3  0 

(2x  3)(x 1)  0  x  3 or –1. Substituting 3 into the equation of the normal 2

line gives us y  . Thus, the second point of intersection is  3 ,  , as shown in the sketch. 5

2 4

3.1.69

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3.1.70





(a) If y  x2  x, then y  2x 1. If the point at which a tangent meets the parabola is a, a2  a then the slope of the tangent is 2a 1. But since it passes through 2, 3, the slope must also be

a2  a  3 y a2  a  3  . Therefore, 2a  1  . Solving this equation for a we get x a 2 a2 a2  a  3  2a2  3a  2  a2  4a  5  (a  5)(a  1)  0  a  5 or –1. If a = –1, the point is 1, 0 and the slope is –1, so the equation is y  1(x  1)  0 or

y  x 1. If a = 5, the point is 5, 30 and the slope is 11, so the equation is y  11(x  5)  30 or y  11x  25. (b) As in part (a), but using the point  2,7  , we get the equation 2a 1  

a 2 a7 a2



2a2  3a  2  a2  a  7  a2  4a  5  0. The last equation has no real solution (discriminant = – 16 < 0), so there is no line through the point 2,7 that is tangent to the parabola. The diagram shows that the point  2,7  is ―inside‖ the parabola, but lines tangent to the parabola do not pass through points inside the parabola.

3.1.71

310

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (x  h)  f (x)

f (x)  lim

1 1 x x  (x  h) x  h  lim  lim  lim

h0 hx(x  h) h0 h h n n1 (a) f (x)  x  f (x)  nx  f (x)  n(n 1)xn2 

h0

f (n) (x)  n(n 1)(n  2) 1

(b) f (x)  x

h

 lim

hx(x  h)

h0

1



x(x  h)

1 x2



2 1xnn  n!

 f (x)  (1)x2  f (x)  (1)(2)x3 





f (n) (x)  (1)(2)(3)

(n)x(n1)  (1)n n! x(n1) or

(1)n n! xn1

3.1.72 (a) f (x)  x

 f (x)  nxn1  f (x)  n(n 1)xn2 

f (n) (x)  n(n 1)(n  2) 1

(b) f (x)  x



2 1xnn  n!

 f (x)  (1)x2  f (x)  (1)(2)x3 





f (n ) (x)  (1)(2)(3)

(n)x(n1)  (1)n n! x (n1) or

(1)n n! xn1

3.1.73 Let P(x)  ax  bx  c. Then P(x)  2ax  b and P(x)  2  2a  2  a  1. 2

P(2)  3  2(1)(2)  b  3  4  b  3  b  1. P(2)  5  (1)(2)2  (1)(2)  c  5  2  c  5  c  3.

3.174





Since f '(x)  2x 1 and the tangent line passes through the points 1, 2 and x, x2  x , it follows that

 x2  x  (2)  2x 1  x 2  x  2  2x2  x 1  x2 2x  3  0  x  3, x  1. Each of these x 1 values of x represents a different tangent line. For x  3, y = 12, and the slope through3,12 and 1, 2 is 7. So y  7(x 1)  2 or y  7x  9, which has y-intercept k  9. For x  1, y  0 and the slope

311



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

through 1, 0 and1, 2 is –1. So the tangent line in this case is y  1(x 1)  0 or y  x 1. In this case, the y-intercept is k  1.



3.1.75

f (x)  2 x  2x1/2  f (x)  2  12 x1/2  and 0, b, and therefore has slope



1 . The tangent line goes through the points 9, 0 x

b0

b  . Let the point of intersection be 0  (9) 9



P   x0, y0   x0, 2 x0 . The slope of the tangent line at P must also be f  x 0  that

1 , which means x0

9 1 1 b   b (1). Then the tangent line has equation y   x  b and at the point of x0 9 x0 x0

intersection, 2 x0 

b  x0 

9  x0

1   x0   b  2 x0  x0  b  x0

x0  b. Substituting this in (1) we have

 x   x  9 . Thus the y-intercept is 0,b  0, 9   0,3. 2

3.1.76

312

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

1 3/2 1 1/2 p p q  .  q x  px1/2  qx1/2  f (x)   px  qx   3 2 2 x 2 x Then 2 x

f ( x) 

q p f (4)   p  q   which must be zero since the tangent line is horizontal at this point. In 2 43 2 4 4 16 addition, y  f (4)  12 

p p  q 4   2q. Solving both equations simultaneously, we find 2 4

p  12, and q  3.

3.1.77

f (x)  ax 

b x

f (1)  a(1)  f (1)  a 

b 12

 ax  bx1  f (x)  a  b(1)x2  a  bx2. When x = 1,

b  a  b. The slope of the line y  5x  6 is 5 so we must have (1)  a  b  5(1), and the point 1, f (1)  1, a  b must also be on this line. Thus,

a  b  5(1)  6  11  a  b  11. Now substituting into (1) we find, 5  a  b  (11 b)  b  11 2b   6  2b  b  3, and

a  b 11  6 11  5.

3.1.78



y  Ax  Bx  C  y  2 Ax  B  y   2A. We substitute these expressions into the equation 2

y   y  2 y  x2 to get (2 A)  (2 Ax  B)  2  Ax2  Bx  C   x2

2A  2Ax  B  2Ax2  2Bx  2C  x2 2 Ax2  2 A  2B x  2 A  B  2C   (1)x2  (0)x  (0) The coefficients of x2 on each side must be equal, so

2A  2B  0  A  B  

1 2 and

2A  1  A   12 .

Similarly,

2A  B  2C  0  1  2C  0  C   3 . 1

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3.1.79 Substituting x = 1 and y = 1 into 𝑦 = 𝑎𝑥2 + 𝑏𝑥 gives us 𝑎 + 𝑏 = 1 (1). The slope of the tangent line 𝑦 = 3𝑥 − 2 is 3 and the slope of the tangent to the parabola at (x, y) is 𝑦′ = 2𝑎𝑥 + 𝑏. At 𝑥 = 1, 𝑦′ = 3 ⟹ 3 = 2𝑎 + 𝑏 (2). Subtracting (1) from (2) gives us 2 = a and it follows that b = -1. The parabola has equation 𝑦 = 2𝑥2 − 𝑥.



3.1.80

y  x  ax  bx  cx  d  y(0)  d. Since the tangent line y  2x 1 is equal to 1 at x = 0, we 4

3 2 must have d = 1. y  4x  3ax  2bx  d  y(0)  c. Since the slope of the tangent line

y  2x 1 at x = 0 is 2, we must have c = 2. Now y(1)  1 a  b  c  d  a  b  4 and the tangent line y  2  3x at x = 1 has y-coordinate –1, so a  b  4  1 or a  b  5 (1). Also y(1)  4  3a  2b  6  3a  2b  6 and the slope of the tangent line y  2  3x at x = 1 is –3, so

3a  2b  6  3or 3a  2b  9 (2). Adding –2 times (1) to (2) gives us a = 1 and b = –6. The curve has equation y  x4  x3  6x2  2x  1. 3.1.81

y  f (x)  ax2  f (x)  2ax. So the slope of the tangent to the parabola at x = 2 is m  2a(2)  4a. The slope of the given line, 2x  y  b  y  2x  b, is seen to be –2, so we must have 4a  2  a   1 . So when x = 2, the point in question has y-coordinate  1  22  2. Now we 2

simply require that the given line, whose equation is 2x  y  b, pass through the point

2, 2 : 2(2)  (2)  b  b  2. So we must have a   21 and b  2.

3.1.82 The slope of the curve y  c x is y 



c 2 x

and the slope of the tangent line y  32 x  6 is 32 . These



must be equal at the point of tangency a, c a , so

c 2 a



3 2

 c  3 a.

314



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

The y-coordinates must be equal at x = a, so c a  32 a  6 

3 a  a  a  6  3 2

3a  3 a  6  3 a  6  a  4. Since c  3 a , we have c  3 4  6. 2



3.1.83

The line y  2x  3 has slope 2. The parabola y  cx  y  2cx has slope 2ca at x = a. Equating 2

slopes gives us 2ca  2 or ca  1. Equating y-coordinates at x = a gives us

ca2  2a  3  (ca)a  2a  3  1a  2a  3  a  3. Thus c 



3.1.84

f (x) 

1  . a 3

6  6x1/2  f (x)  6   12  x3/2  3x3/2. x 3/2

The instantaneous rate of change at c is f (c)  3c



The average rate of change over the interval1, 4 is f (4)  f (1) 

64  6 1



4 1 They are equal when f (c)  

3 3

 1 

3 3

 1  3 

3  6   3  1. 3

 c  32/3  2.080.

3.1.85



If P  a, a

 is a point of intersection of the graphs of f and the tangent line, then the slope of the

tangent line must be m 

y

1 2 a

(x  8)  3 or y 

its equation as y 

1 2 a

x  3 

4 a

 x  a  a or y 

2 a

equations we find 3 

, and an equation of the tangent line (which goes through the point 8, 3 is

. Since the tangent line also goes through P, we can write



2 a

 a

x  a 

. Equating the y-intercepts of both

4  a  a  a  3 a   a  8 2 2 2 a a 2 a.

315

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Solving, we find a  4 or a  16. Therefore, the possible slopes of such tangent lines are m  and m 

1 2 4



1 1  . 2 16 8

Alternative method: Set the slope of the line through P and 8, 3 equal to



1 2 a

a  3  1  , and solve for a. 2 a 

i.e. set a  8  





3.1.86





line that goes through the origin has equation y  mx. Let P  a, a2  5 be the point where the tangent line intersection the graph of f. Then f (x)  x  5  f (x)  2x, the slope of the tangent line is 2

f (a)  2a, and the equation of the tangent line is y  2ax. At the point of tangency we have

y  a2  5  2a(a)  2a2  5  a2 

5  a. Therefore the slope of the tangent line is 2 5.

3.1.87 Let  B,0  and 0, A be the x- and y-intercepts of a tangent line. At P a, 6 a , the point of tangency, the tangent line has slope f (a) 

6 2

, and the equation of the

a 6 12  12  y  2 x  . Thus 0, A  0, . tangent line is y  6 2 (x  a)  6 or a a a a  a    If we let y = 0 in the linear equation, we find

0

6

12 6 12 B   B   6B  12a  B  2a. The area of the triangle formed by the a2 a a2 a

tangent line and the coordinate axes is quadrant.

1 12 AB   2a  12, for any tangent line in the first  2   2 a 1

3.1.88

316

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

The area of the triangle is 12 AB because the intercepts of the tangent line are 0, A and  B, 0  . The slope of the tangent line is f (4)  2(4)  3  5, and because the tangent line contains the point 4, 3, and an equation for the line is y  5 x  4  3 or y  5x  23. Thus A  23. Letting y = 0 we find

B  23 . Then the area of the triangle is 1 AB  1  23   23   529 . 5



3.1.89

f (x)  ax  bx  c  f (x)  2ax  b. The slope of the tangent line at x  p is 2ap  b, the slope of 2

the tangent line at x  q is 2aq  b, and the average of those slopes is

2ap  b  2aq  b

 ap  aq  b. The midpoint of the interval  p, q is

pq

and the slope of the

2 pq   b  a  p  q  b. This is equal to ap  aq  b, as required. tangent line at the midpoint is 2a  2    3.1.90 f is clearly differentiable for x < 2 and for x > 2. For x < 2, f (x)  2x, so f(2)  4. For x > 2, f (x)  m, so f(2)  m. For f to be differentiable at x = 2, we need 4  f(2)  f(2)  m. So f (x)  4x  b. We must also have continuity at x = 2, so 4  f (2)  lim f ( x)  lim(4x  b)  8  b. Hence b  4. x2

  

x2

 3.1.91 Solution 1: Let f (x)  x1000. Then, by the definition of a derivative,

f (x)  f (1) x1000 1  lim  1000. But this is just the limit we want to find, and we know x1 x1 x 1 x 1

f (1)  lim

(from the Power Rule) that f (x)  1000x

, so f (1)  1000(1)999  1000. Therefore,

999

x1000 1  1000. x1 x 1

lim

Solution 2: Note that  x1000  1  ( x  1)  x999  x998  x997 

 x2  x  1.

317



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(x 1) x x1000 1  lim So lim x1 x1 x 1

999

 x998  x997 

 x2  x 1



x 1

 lim x999  x998  x997  x1

 x2  x 1   1  1  1 

 1  1  1  1000, as above.

1000 ones

3.1.92 In order for the two tangents to intersect on the y-axis, the points of tangency must be at equal distances from the y-axis, since the parabola y  x2 is symmetric about the y-axis. Say the points of tangency are a, a2  and a, a2  for some a > 0. Then since the derivative of y  x2 is dy / dx  2x, the left-hand tangent has slope –2a and equation y  2a(x  a)  a 2 , or y  2ax  a 2 , and similarly the right-hand tangent line has equation y  2a(x  a)  a , or y  2ax  a . So the two lines 2

intersect at 0, a2 . Now if the lines are perpendicular, then the product of their slopes is –1, so

 2a 2a  1  a2  1 1  a  1 . So the lines intersect at 0,  1 . 4

3.1.93

y  x2  y  2x, so the slope of a tangent line at the point a, a2  is y  2a and the slope of a normal 1 line is 

, for a ≠ 0. The slope of the normal line through the points a, a2  and 0, c is

a2  c a 0

, so

 1 a2  c   1  a2  c   1  a2  c  . The last equation has two solutions if c  1 , one 2 a 2a 2 2 solution if c  1 , and no solution if c  1 . Since the y-axis is normal to the curve regardless of the value 



of c (this is the case for a = 0), we have three normal lines if c > ½ and one normal line if c  12 .

3.1.94

318

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

From the sketch, it appears that there may be a line that is tangent to both curves. The slope of the line through the points P a, a2  and Q b, b2  2b  2 is

b2  2b  2  a2 ba

. The slope of the tangent line at P is 2a  y  2x  and at Q is

2b–2  y  2x  2. All three slopes are equal so 2a  2b  2  a  b 1. Also, 2b  2 

b2  2b  2  a2 ba

 2b  2 

b2  2b  2   b 1 2 b  (b 1)



2b  2  b2  2b  2  b2  2b 1  2b  3  b  3 and a  3 1  1 . 2

Thus an equation of the tangent line at P is y  2  1  x  1     or y  x  1 . 1 2

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.2 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.2.1 Product Rule:

f ( x)  1  2x 2  x  x2  

f ( x)  1  2x 2 1  2x    x  x2   4x   1  2x  2x 2  4x 3  4x 2  4x 3  1  2x  6x 2  8x3. 2 3 Multiplying first: f ( x)  1  2x 2  x  x2   x  x2  2x 3  2x 4  f (x)  1  2x  6x  8x

(equivalent)

319



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.2.2 For this problem, simplifying first seems to be the better method, as there is no longer a quotient and there are thus fewer operations. 3.2.3 2 h( x)  x  x 1 Quotient Rule:

 h (x) 

(x 1)(2x)  x2(1)

(x 1)(2x)  x2 

(x 1)2

(x 1)

2x2  2x  x2 

(x 1)



x2  2x (x 1)2

(b) Using the Product Rule: h(x)  x  x 2 (x 1)1  x 1

  2  2 1  h (x) x 1(x 1)   (x 1) (2x)  

x 2 (x 1) 2



2x (x 1)



x 2  2x(x 1) (x 1)2

2x2  2x  x2 x2  2x x(x  2)   (x 1)2 (x 1)2 ( x 1)2

x(x  2) (x 1)2

3.2.4 By the Product Rule,

f (x) 

f ( x)   2x 2  5x  e x 







 



 



d  2x 2  5x e x   2x 2  5x e x   e x 2x 2  5x   2x 2  5x e x  e x  4x  5  dx 

 e x  2x2  5x    4x  5  e x  2x 2  x  5 

3.2.5

   2   g( x)  2  2 x   e   e 2  2 x   2  2 x  e  e   

By the Product Rule,

g(x)  2  2 x e x  x

 ex





 



22 x 

  1 1    ex 2  2 x      x   x  



x 



3.2.6

320

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y

 x

ex (1)  xex

e 



By the Quotient Rule,

ex (1  x)

e 

1 x





3.2.7 x

y  e 1  ex  

By the Quotient Rule,

1  ex e x  ex ex  e  e  e e   . x  1  e  1  ex  1  e x  x



The notations  and  indicate the use of the Product and Quotient Rules, respectively.

3.2.8

g(x) 

1 2x QR  (3  4x)(2)  (1  2x)(4) 6  8x  4  8x 10 3  4x g( x)    2 2 2  3  4x   3  4x   3  4x 

3.2.9

x2  2 QR G( x)  2x  1 

3.2.10



(2x  1)(2x)  ( x  2)(2) 4 x  2 x  2 x  4 2 x  2 x  4    G ( x)  2 2 2  2x  1  2x  1  2x  1 2

  u  u   1   1  H (u)  u  u  1  u  u  1 u    

H (u)  u  u





u  u 



1 2



 1 u  1 2



 1  2u 1. 2



An easier method is to simplify first and then differentiate as follows:



H (u)  u  u

u  u   u   u   u  u  H  u   2u 1. 2

3.2.11

J (v)  v3  2v  v 4  v2   PR

J (v)  v3  2v 4v4  2v3   v4  v2 3v2  2 

 4v2  2v0  8v4  4v2  3v2  2v4  3v0  2v2  1 v2  6v4

321



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.2.12

F ( y) 

 1  y2 

3 



 y  5y 3    y2  3y4  y  5y 3   PR

y4  



F ( y)   y2  3y4 1  15 y2    y  5y 3  2 y 3  12 y 5    y2  15  3y4  45 y2   2 y2  12 y4  10  60 y2 

 5  14 y2  9 y4 or 5  14 / y2  9 / y4

3.2.13

f (z)  1  ez  z  ez   PR

f (z)  1 e z 1 ez    z  e z  ez   1  e z   ze z   e z   1 ze z  2e2z 2

3.2.14

x2  1 QR y  x3 1

y 

 x3 1  2x    x2 1 3x2   x 1 3



 

  x  y  2 x  

3.2.15



x  x3 1  2    x2 1  3x 

y 

2  x  

1   x (1) 2 x   2 2  x

 x 1 3



x  2x3  2  3x3  3x  x x3  3x  2    2 2 3  x 1  x3 1 

   

1 x  2  x  2x   x 2 x 2 x 2 x   2 2 2 2 x 2  x  2  x 2  x

3.2.16

322



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

t3  3t QR  y 2  t  4t  3



t2  4t  33t2  3  t3  3t 2t  4  t 2  4t  3 3t4  3t2 12t3 12t  9t2  9  2t4  4t3  6t2 12t  t  8t  6t  9   2 t  4t  3  t  4t  3   y 



3.2.17

t3  2t2 1(0) 13t2  4t  3t2  4t y     t 3  2t 2 1 t  2t 1 

QR 1  y 3 t  2t2 1 

3.2.18





y  e p p  p p  e p  p  p3/2   PR



y  e p 1 32 p1/2    p  p3/2  e p  e p 1 32 p  p  p p



 3.2.19





 e r  aber  ae2r  ae2r



QR b  er aer  aer h(r)  ae  r be h(r)  



b  e 





b  e 



r 2



aber

r 2



 b  er  

3.2.20

1 3 2 5/2    3  2 s y  s  s  s  s  s1  s3/2  y  s  3 s 2 2 2 2 2 5/2 s s s s 2s 2s5/2 





3.2.21

y   z 2  ez  z  PR





ez  2z z  ze z     2 z 2 z z2

2 1 z  z 2z  ez 

y  z2  e  z

z2  ez  4z2  2zez 2 z

5z2  ez  2zez  2 z

3.2.22

323

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (t) 



QR t  t3  3

 t  3 

f (t) 

1 2/3 t

  t (1)

 t  3

1 1/3





1/3

t

2/3

 t  3

t

2t

1/3

 t

2 1/3



t

2/3

3 

3t 2  t  3

2t  3

2/3

 3t

 t  3



2/3

3t 2/3  t  3

3.2.23

4  t QR  V (t)  t  te 

tet (1)  4  t tet  et (1) V (t)  tet 

tet  4tet  4et  t 2et  tet



4tet  4et  t 2et

t 2e2t

3.2.24



t 2e2t



et  t 2  4t  4 



t 2e2t

 t  2 2 t 2et



x2ex QR  f (x)  2 x  ex

x 2  e x   x 2 e x  e x (2x)   x 2 e x  2x  e x   f ( x)   x2  e x  2



x4ex  2x3ex  x2e2x  2xe22x  2x3ex  x2e2x

x  e  2



x4ex  2xe22x  xe  x  2e2  x



 x 2  e x 

 x 2  ex 



3.2.25

F (t) 

At Bt 2  Ct3



A Bt  Ct 2



Bt  Ct 2 (0)  A  B  2Ct   A B  2Ct  A  B  2Ct   F (t)   2  2 2 2 2 2 t B  Ct t  B  Ct     Bt  Ct  QR

3.2.26

f (x) 

x xc/ x





  x  c / x(1)  x  x  c / x    f (x)    2 dx  x  c / x    x  cx  d 



324

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 2 x  c / x  x  c / x  2c / x  x  2cx   2 2 2 2  x2  c  x  c  x2  x 2  c    x  x2  



3.2.27

f (x) 

ax  b QR  cx  d  a  ax  b c  acx  ad  acx  bc  ad  bc cx  d f ( x)  2 2 2  cx  d   cx  d   cx  d 

3.2.28





 









f ( x)   x 3  1ex  f (x)  x 3  1 e x  e x 3x 2  e x  x 3  1  3x 2   e x x 3  3x 2  1  PR

f ( x)  ex 3x2  6x    x 3  3x2  1  e x  ex  x 3  6x 2  6x  1      



3.2.29

f (x)  xex

f ( x) 



 xex  ex  1    x 2x  1 x   e  e. x    1   2 x   2 x 2 x

Using the Product Rule and f ( x)   x1/2  12 x1/2  e x , we get





f (x)  x1/2  12 x1/2 e x  ex

x 1 2

1/2

1 4

x

3/2

  x  x 1/2

1

1/2

x

3/2

e  x

4x2  4x 1 4x

3/2

3.2.30

f (x) 

1  e x  2x  x2  e x 

1 e

x 

f ( x) 

1  e 

x 2



x 1  e x  2  xe x

Using the Quotient and Product Rules and f ( x) 



1  e 

x 2

2x  2xex  x2ex

1  e x 

x 2  2ex  xex  

1  e x 

, we get

1  e x  2  2  xe x  e x    x 2 e x  2xe x    2x  2xe x  x 2e x  1  e x  e x  1  e x  e x   f (x)  1  ex  2



325

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

1  e x  1  2e x  2  2xe x  2e x  x2e x  2xe x    2x  2xe x  x 2e x  2e x   1  e x  1  ex  2  2ex  x2ex   4xex  4xe2x  2x2e2x  1  e x  4

2  2ex  x2ex  2ex  2e2x  x2e2x  4xex  4xe2x  2x 2 e 2x



1  e x 

2  4ex  x2ex  4xex  2e2x  x2e2x  4xe2x



1  e x 

3.2.31

f (x) 

x x 1



f (x) 



  x 2 1(1)  x 2x x2 1  2x2

 x  1 2





 x 1





x 2 1

 x  1 



2 2 3 x 2 1  2x    x 2  1 x 4  2x 2  1  x  1  2x    x  1 4x  4x   (x) f    x2 1  x2 1  x2  1  2x    x2  1  4x   x2 1  x2  1  x2  1  2x    x2  1  4x     x2  1  x2  1 2





2x3  2x  4x 3  4x

 x 2  1





2x 3  6x

 x 2  1

3.2.32

y

x2 1 x2  x  1



2 2 3 2 3 2 x2  4x  12 y   x  x  1   2x    x 21   2x  1 2x  2x  2x  2x 2 x  2x  1    x 2  x  1  x 2  x  1  x 2  x  1 6 2 At 1,0, y   , and an equation of the tangent line is y  2 (x 1)  0 or y  2 x  2 . 3 3 3 32 3

326



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



 x x 1  xex . 1  ex  ex  xex y  1 x  y  1  e (1)  1  x e   x x 1 e 1  e 1  e     1  ex 

3.2.33



At 0,

1  , and an equation of the tangent line is y  1 (x  0)  1 or y  1 x  1 . 2 4 2 4 2 1  1 4

 , y 



3.2.34

y  2xex  y  2  xe x  ex 1  2ex ( x  1). PR

At 0, 0  , y  2 e (0  1)  2 11  2, and an equation of the tangent line is y  2(x  0)  0 or 0 0

y  2x. The slope of the normal line is  12 , so an equation of the normal line is y   12 ( x  0)  0 or y   12 x.



3.2.35

2 2  2x . At 1,1, y  0, and an equation of the tangent 2x  y   x  1(2)  2x  2x  y x 1  x 2  1  x 2  1  2



line is y  0(x 1) 1 or y  1. The slope of the normal line is undefined, so an equation of the normal line is x  1. 3.2.36

f  x  xex  f  x  xex  ex 1  ex  x 1. The tangent line is horizontal when f  x  0,

which happens when x  1. 3.2.37 2 4x 2 2 4(1  x2 ) . The tangent line is 4  4x  8x f (x)  1  x  4  (4x)(2x) f (x)   2 2 2  2  2 2 2 1 x 1  x  1  x  1  x  



horizontal when f ( x)  0 which happens when x  1, 1. 3.2.38

(4x 1)2x  x2 (4) 8x2  2x  4x2 4x2  2x 2x(2x 1)  x2 f (x)   f (x)     0  2  (4x 1)2 (4x 1)2 4x 1 (4x 1) 2 (4x 1) x  0,  12 . Therefore, f has a horizontal tangent line. 

3.2.39

327

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

s(t)  

t2  5  2t2 2 5  t2 2 . The particle is at rest when its  s(t)  t  5  1  (t)(2t) 2   2 t2  5 t2  5 t2  5 t 5 t

velocity, s(t) is zero. This happens when t  5, and at that time, the particle’s position is

 5 



 5   5 10 2

3.2.40 

 (a) y 

f (x) 



1 

1 x

2







 2 f ( x)  1  x (0)  2x  2x   2x 2 



1  x  2

1  x 

So the slope of the tangent line at the point y  1, 2 is 1

f (1) 

2 2

 12 and the equation of that tangent line is

y  12  x  1  12 or y  12 x  1. 3.2.41 (a) y  f (x) 

x   2 1 x

f ( x)  1  x (1)  2x  2x   1  x 2 . So the slope of the 2



1  x2 



1  x2 

tangent line at the point y  3, 0.3 is f (3)   100 and the 8

equation of that tangent line is y  0.08 x  3  0.3 or y  0.08x  0.54. 3.2.42 (a)

328

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f ( x)   x 3  x  e x  f ( x)   x 3  x  e x  ex 3x2  1  ex  x 3  3x2  x  1 (b) f   0 when f has a horizontal tangent line, f  is negative when f is decreasing, and f  is positive when f is increasing.

3.2.43 (a) f (x) 

ex  2x2  x  1

x 2 x 2 2x 2  x  1ex  ex  4x  1 e  2x  x  1  4x  1 e  2x  3x   f (x)    2 2 2x  x  1 2x  x  1    2x2  x  1   2

(b) f   0 when f has a horizontal tangent line, f  is negative when f is decreasing, and f  is positive when f is increasing.

3.2.44 (a) f (x) 

f ( x) 



x2 1

 x2 1 x 2  1 (2x)  x 2 1 (2x)





x  1





(2x)  x 2  1   x 2  1

x  1



x  2

f (x) 



    

(2x)2



 x  1



1 (4)  4x  x  2x  1



2 

x  1 2



4  x 1  4x 4x  4x  

 x 2  1  x2  1 4  x 2 1 16x 2  x 2 1 4(x 2 1)  x 2 1  4x 2  4 1 3x2       2 2   x 1  x 1  x2 1 4



(b)

f   0 when f has a horizontal tangent line and f   0 when f  has a horizontal tangent. f  is negative when f is decreasing, and

f  is positive when f is increasing. f  is negative when f  is decreasing,

329

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

and f  is positive when f  is increasing. f  is negative when f is concave down, and f  is positive when f is concave up.



3.2.45

(a) f ( x)   x2  1  e x  f ( x)   x 2  1  e x  ex  2x   ex  x 2  2x  1  f ( x)  ex  2x  2   x2  2x  1  e x  ex  x 2  4x  1 We can see that our answers are plausible, since f has horizontal

(b)

tangents where f (x)  0, and has horizontal tangents where

f (x)  0.

3.2.46

1  x (2x)  x2 1

x2  f ( x)   f ( x)  1 x

f ( x) 

1  x 

2x  2x 2  x2 

1  x 

 x 2  2x  1(2x  2)   x 2  2x   (2x  2)  x2  2x  1

  







x2  2x x2  2x  1



(2x  2)  x 2  2x  1  x2  2x  2

 x  1 2   

2  2  1. 2(x 1)(1)  2 , so (1)  f  (x 1)4 (x 1)3 (1  1)3 8 4

3.2.47

g( x) 

x e

g( x) 

 g( x)  x

ex 1  x  ex

  ex

ex  (1)  (1  x)ex

 g(x) ) 

e 

ex 1  ( x  2)ex

g (4) ( x) 

 



e 

 



ex 1  (1  x) x  2    x  x 2 e e

ex 1  ( x  2) 3  x   x  2  e ex

ex  (1)  (3  x)ex x 2

ex 1  x 1  x   x  2  e ex

 

x 2



 





ex 1  (3  x)

e 

x 2



x4 e

330

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

The pattern suggests that g (n) (x) 

(x  n)(1)n ex

. (We could use mathematical induction to prove this

formula.) 3.2.48 We are given that f (5)  1, f (5)  6, g(5)  3, and g(5)  2. (a) ( fg)(5)  f (5)g(5)  g(5) f (5)  (1)(2)  (3)(6)  2 18  16 

 f  g(5) f (5)  f (5)g(5) (3)(6)  (1)(2) 20   (b)   (5)  2 (3)2 9  g(5)  g    g  f (5)g(5)  g(5) f (5) (1)(2)  (3)(6)  20 (c)   (5)   2 (1)2  f (5)  f  3.2.49 We are given that f (4)  2, g(4)  5, f (4)  6, and g(4)  3.

(a) h(x)  3 f (x)  8g(x)  h(4)  3 f (4)  8g(4)  3(6)  8(3)  18  24  6 (b) h(x)  f (x)g(x)  h(4)  f (4)g(4)  g(4) f (4)  2(3)  5(6)  6  30  24 (c) h(x)  f (x)  h(4)  g(4) f (4)  f (4)g(4)  5(6)  2(3)  30  6  36 2 2 

 g(4)

g(x)

(d) h( x) 

(2  5)(3)  5(6  (3)) 2115 36   2 49 49 (2  5)



3.2.50

1  x  f (x)  x  f (x)  f (x)   dx  2 x 1 1 x  f (x)  4  f (4)  f (4)  2  (1)  16   2  4  2. x4 2 4 2 4

(a) Using the Product Rule,

g( x)  f (4)  g(4) g(4)  g(4) f (4)  g(4)  h(4)  2 f ( x)  g( x)  f (4x)  g(4) 

(b) Using the Quotient Rule, d  g( x)   f ( x)g( x)  g( x) f ( x)  2  dx  f ( x)   f ( x) 



331

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (2)g(2)  g(2) f (2) 3  (1)  (16)  (5) 3  80 83 d  g( x)      2   dx  f ( x) x2 32 9 9  f (2) 3.2.51 (a) h(x)  2  g(x)  h(1)  2  g(1)  2 1  3 (b) q( x) 

f ( x)g( x)  g( x) f ( x)   2  f ( x)

q(1)  f (1)g(1)  g(1) f (1)  (3)(1)  (8)(5)  3  40  37 2 (3)2 9 9  f (1) 3.2.52

d  h( x)  xh( x)  h(x) 1    dx  x  x2 d  h(x)  2  h(2)  h(2) 1 2  (2)  7 1 4  7 11     2 2 4 4 4 dx  x x2 3.2.53

d  h(x)  dx  x 



xh(x)  h(x) 1

 d  h(x)  2h(2)  h(2) 2(3)  4 10      2.5 2 2 4 4 dx  x x2 



3.2.54

f (x)  e x g(x)  f (x)  exg(x)  g(x)ex  ex g(x)  g(x). f (0)  e0 g(0)  g(0)  1(5  2)  7 3.2.55

g(x)  xf (x)  g(x)  xf (x)  f (x) 1. Now g(3)  3 f (3)  3 4  12 and g(3)  3 f (3)  f (3)  3(2)  4  2. Thus, an equation of the tangent line to the graph of g at the point where x  3 is y  2(x  3)  12, or y  2x 18.



3.2.56

f (x)  x f (x)  f (x)  x2 f (x)  f (x)  2x. Now f (2)  22 f (2)  4(10)  40, so 2

f (2)  22  (40) 10(4)  200.

332

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.2.57 (a) From the graphs of f and g, we obtain the following values: f (1)  2 since the point 1, 2 is on the graph of f ; g(1)  1since the point 1,1 is on the graph of g; f (1)  2 since the slope of the line segment between 0, 0 and 2, 4 is

2, 4 and 0, 0 is

04 2  2

40 20

 2; g(1)  1since the slope of the line segment between

  1. Now u(x)  f (x)g(x), so

u(1)  f (1)g(1)  g(1) f (1)  2  (1) 1 2  0. 2 (b)(b) v(x)  f (x) / g(x) so v(5)  g(5) f (5)  f (5)g(5)  2  ( 13)  3 23   3  . 2 2 8

 g(5)

3.2.58 (a) P(x)  F(x)G(x), so P(2)  F (2)G(2)  G(2)F(2)  3 2 4 2  0  32. (b) Q(x)  F(x) / G(x), so Q(7)  G(7)F (7)  F (7)G(7)  1 4  5    3   1  10  43 . 2 1 4 3 12 G(7) 1



3.2.59

g  x   ex  ex g x ex y  y  2 g  x  g  x  

(d)



3.2.60 (a) y  x f (x)  y  x f ( x)  f ( x)  2x  2

f (x)



(b) y 



 x2 f ( x)  f ( x)(2x) xf ( x)  2 f ( x)    y  2 x3 x2



 

x  y  f ( x)(2x)  x2 f ( x) (c) y  2 f ( x)  f ( x) 2

1  xf (x)   y  (d)(d) y  x

x xf ( x)  f ( x)  1  xf (x) 

1 2 x

 x   

333

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

x3/2 f (x)  x1/2 f (x)  1 x1/2  1 x1/2 f (x) 2x1/2 xf (x)  2x2 f (x) 1 2 2    1/2  x 2x 2x3/2 3.2.61 The average rate of change of f over the interval1, 4 is

2 4 3 4 2 12 5   2 2 2 f (4)  f (1)  4  4 1  4   . 20 5     4 1 3 3 3 15 



 x2  4    4x  x 3/2 x  2  2x   2 x    . 2 2 2 x  4   x2  4 

 x  4  2 1 x  2

f ( x) 





Using a CAS, we find that the derivative is equal to the average rate of change when x  0.964.



3.2.62 If y  f (x) 

x (x 1)(1)  x(1)  1 , then f (x)   . When x  a, the equation of the tangent (x 1)2 (x 1)2 x 1

(x  a)  a . This line passes through 1, 2 when a 1 (a 1) a 1 2  (1 a)  2(a  1)2  a(a  1)  1 a  2 a  1 (a 1) 2 2a  4a  2  a2  a 1 a  0  a2  4a  1  0. The quadratic formula

line is y 





4  42  4(1)(1) 2(1)

gives the roots of this equation as a 

 4  12   2  3, so there are two such tangent lines. 2



3 1

2    22 3 31  1 3 1

Since f 2  3 



1 3 the lines touch the curve at A 2  3, 2



3 2  3 3  3 1 3 1 3    , 1 3 2 2 3

  0.27, 0.37 and



B 2  3, 1 23  3.73,1.37.

334

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.2.63

y

x 1 (x  1)(1)  (x 1)(1)  2 . If the tangent intersects the curve when x  a,  y   x 1 (x 1)2 (x 1)2

then its slope is 2 / (a  1)2. But if the tangent is parallel to x  2 y  2, that 1 is, y  1 x  1, then its slope is . Thus,

2 (a  1)2

 

 (a  1)2  4 



a  1  2  a  1, or a  3. When a  1, y  0 and the equation of the tangent is y  1 ( x 1)  0 or y  1 x  1 . When a  3, y  2 and the 2

equation of the tangent is y  (x  3)  2 or y  1 x  7 . 1

3.2.64

R

f g

 R 

gf   fg . 2 4 3 5 g2 For f (x)  x  3x  5x , f (x)  1  9x  25x , and for

g(x)  1  3x3  6x6  9x 9 , g(x)  9x 2  36x5  81x8. Thus R(0) 

g(0) f (0)  f (0)g(0) 11  0  (0) 1    1. 2 12 1  g(0)

3.2.65

Q

f g

 Q 

gf   fg . x x 2 x g2 For f (x)  1  x  x  xe , f (x)  1  2x  xe  e , and for

g(x)  1  x  x2  xex , g(x)  1  2x  xex  e x . Thus Q(0) 

g(0) f (0)  f (0)g(0) 1 2 1(2) 4    4. 2 12 1  g(0)

3.2.66

335



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) Let h(x) 

f (x)

. Then h(x)  g(x) 

g(x)

f (x) g(x)

 g(x)  f (x) 

  f ( x)  g( x)  h( x)g( x)  f ( x)  h(x)g(x)  h(x)g(x)  f (x)     g( x)   f ( x)  f ( x)  g( x)    f ( x)    g( x) g( x)  h( x)   h( x)g( x)  f ( x)    g( x)  g( x)  h( x)  f ( x)  f ( x)g( x)  h( x)  f ( x)g( x)  f ( x)g( x)  g( x) f ( x)  f ( x)g( x) . 2 2 2 2 g( x)  g( x)  g( x)  g( x)  g( x) 

(b) Let h(x) 

3x  2



3x  2 4x  5



. Then h(x)  (4x  5)  3x  2  h(x)(4)  h(x)(4x  5)  3 



(4)  h(x)(4x  5)  3  h(x)(4x  5)  3 

4x  5 h( x) 

12x  8 4x  5

 h( x) 

12x  8 4x  5  4x  5

3



3 12x  8 3(4x  5)  12x  8 12x  15  12x  8  23    2  2 2 2 . 4x  5  4x  5  4x  5  4x  5  4x  5 



3.2.67

If P t  denotes the population at time t and At  the average annual income, then T t   P t  A  t  is the total personal income. The rate at which T t  is rising is given by T   t   P t  At   A  t  Pt   T 2018  P 2018 A2018  A2018 P2018  694, 583$1200/yr  $41, 7949624/yr  $833, 499, 600/yr  $402, 225, 456/yr  $1, 235, 725, 056/yr. So the total personal income was rising by about $1.236 billion per year in 2018. The term P t  At   $833 million represents the portion of the rate of change of total income due to the existing population’s increasing income. The term A  t  Pt   $402 million represents the portion of the rate of change of total income due to increasing population. 3.2.68 (a) f (20)  10, 000 means that when the price of the fabric is $20/yard, 10,000 yards will be sold.

336



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (20)  350 means that as the price of the fabric increases past $20/yard, the amount of fabric which will be sold is decreasing at a rate of 350 yards per (dollar per yard). (b) R( p)  pf ( p)  R( p)  pf ( p)  f ( p) 1 

R(20)  20 f (20)  f (20)  20(350) 10,000  3000. This means that as the price of the fabric increases past $20/yard, the total revenue is increasing at $3000/($/yard). Note that the Product Rule indicates that we will lose $7000/($/yard) due to selling less fabric, but this loss is more than made up for by the additional revenue due to the increase in price. 3.2.69

v

0.14[S] dv 0.15  [S] (0.14)  0.14[S] (1)  0.0021 .   2 2 d[S] 0.15  [S]  0.15  [S]  0.15  [S] dv d[S] represents the rate of change of the rate of an enzymatic reaction with respect to the concentration of a substrate S.

3.2.70

B(t)  N (t)M (t)  B(t)  N (t)M (t)  M (t)N (t), so B(4)  N (4)M (4)  M (4)N(4)  820(0.14) 1.2(50)  174.8 g/week.



3.2.71



  fg h   fg  h  f g h  fgh  fgh (a)  fgh    fg  h    fg  h   fg  h   f g (b) Putting f = g = h, in part (a), we have

d  f (x)3   fff   f  ff  ff  f  fff   3 fff   3 f (x) 2  f (x). dx (c) 3.2.72

3 2 d 3x d e    e x   3  e x  e x  3e2x e x  3e3x  dx dx



 

  fg  (a) We use the Product Rule repeatedly: F  fg  F  f g

337



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

F    f g  f g    f g   fg   f g  2 f g   fg . (b)

F   f g  f g   2  f g   f g   f g   fg   f g  3 f g   3 f g   fg    F (4)  f (4) g  f g   3  f g   f g   3  f g   f g   f g   fg (4)  f (4) g  4 f g   6 f g   4 f g   fg (4)

n F ( n )  f ( n ) g  nf ( n1) g    f ( n2) g   k  where

n    f ( nk ) g ( k )   k

 f g ( n1)  fg ( n) ,

n n! n(n  1)(n  2) (n  k  1)  .  k  k!   k !(n  k )!  

3.2.73





For f ( x)  x2ex , f ( x)  x2ex  ex (2x)  ex x 2  2x . Similarly, we have

f ( x)  ex  x 2  4x  2

  f (4) ( x)  ex  x 2  8x  12 f ( x)  ex x 2  6x  6



f (5) ( x)  ex x 2  10x  20



It appears that the coefficient of x in the quadratic term increases by 2 with each differentiation. The pattern for the constant terms seems to be 0  1 0, 2  2 1, 6  3 2, 12  4  3, 20  5 4. So a





reasonable guess is that f (n) ( x)  ex x 2  2nx  n(n  1) .









Proof: Let Sn be the statement that f (n) ( x)  ex x 2  2nx  n(n  1) .





1. S1 is true because f ( x)  ex x 2  2x . 2. Assume that Sk is true; that is f (k ) ( x)  ex x 2  2kx  k (k  1) . Then

f (k1) (x) 

d  f (k ) (x)  e x  2x  2k    x 2  2kx  k(k 1) e x dx

 e x  x 2   2k  2 x   k 2  k  e x  e x  x 2  2  k 1 x  (k 1)k  .

338



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

This shows that Sk+1 is true. 3. Therefore, by mathematical induction, Sn is true for all n; that is, for every positive





integer n, f (n) ( x)  ex x 2  2nx  n(n  1) . 3.2.74



(a) d  1  

dx  g( x) 

g( x) 



d d   (1)  1  g( x) dx dx [Quotient Rule]  g( x)  0  1 g( x)   g( x) 2 2 2  g( x)   g( x)  g( x)



3 2  3t2  4t d 1  t  2t 1  (b)   dt t 3  2t2 1    t 3  2t 2 1 2  t 3  2t 2 12  xn   d d  1  nxn1  n n12n (c) x   nxn1   n    n 2 [by the Reciprocal Rule]   x2n  nx dx dx  x  x  

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.3 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.3.1 PR

f (x)  x2 sin x  f (x)  x2 cos x  (sin x)(2x)  x2 cos x  2x sin x

339

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.3.2

f (x)  x cos x  2 tan x  f (x)  x(sin x)  (cos x)(1)  2sec2 x  cos x  x sin x  2sec2 x



3.3.3

f (x)  ex cos x  f (x)  ex (sin x)  (cos x)ex  ex (cos x  sin x) 3.3.4

y  2sec x  csc x  y  2(sec x tan x)  (csc x cot x)  2sec x tan x  csc x cot x 3.3.5 y  sec tan  y  sec sec2    tan (sec tan )  sec (sec2   tan2  ). Using the identity

1 tan2   sec2  , we can write the alternative forms of the answer as sec (1  tan2  ) or sec (2sec2  1). 3.3.6

g(x)  ex (tan x  x)  g(x)  ex (sec2 x  1)  (tan x  x)ex  ex (sec2 x  1 tan x  x)

3.3.7

y  t cos t  t 2 sin t  y  t(sin t)  cos t 1 t 2 cos t  sint(2t)  t sin t  cos t  t 2 cos t 3.3.8

f (t) 

3.3.9

y

cot t et x

2  tan x

 f (t) 

 y  

et (csc2 t)  (cot t)et (et )2



et (csc2 t  cot t) (et )2

(2  tan x)(1)  x(sec2 x) (2  tan x)2



csc2 t  cot t et

2  tan x  x sec2 x 

(2  tan x)2

3.3.10

y  sin cos  y  sin sin   cos (cos )  cos2   sin2   cos 2



3.3.11

340

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f ( ) 



sin

 (1  cos ) cos  (sin )(sin ) cos  cos2   sin2  )   f ( )  (1  cos )2 1  cos  (1  cos )2 cos  1  1   (1  cos )2 1  cos

3.3.12 

y  cos x  1 sin x (1  sin x)(sin x)  cos x(cos x) sin x  sin2 x  cos2 x sin x 1 1    y  2 2 2 (1  sin x) (1  sin x) (1  sin x) 1  sin x

3.3.13

t sin t

y



 y 

1 t

(1  t)(t cos t  sin t)  t sin t(1) (1  t)2 2 (t  t) cos t  sin t  (1  t)2

t cos t  sin t  t 2 cos t  t sin t  t sin t  

(1  t)2

3.3.14

 

sin t  y   1  tan t

cos t  sin t  sin t (1  tan t) cos t  (sin t) sec t  cos2 t cos t  sin t  tan t sec t y    (1  tan t)2 (1  tan t)2 (1  tan t)2 2

3.3.15 Using Exercise 3.4.78(a), f ( )   cos sin 

f ( )  1cos sin   (sin ) sin   cos (cos )  cos sin   sin2   cos2   sin cos  (cos2   sin2  )  12 sin 2  cos 2 [using double-angle formulas] 3.3.16 t Using 3.4.78(a), f (t)  te cot t 

f (t)  1et cot t  tet cot t  tet (csc2 t)  et (cot t  t cot t  t csc2 t) 3.3.17

341

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(sin x tan x)2  2x(sin x tan x) 2x   f (x)  sin x tan x (sin x tan x)2 2 sin x tan x  x   sin x sec x  tan x cos x 2 sin x tan x  2x sin x sec  tan x cos x   2  cos x   2 2 (sin x tan x) (sin x tan x) sin x tan x  x tan x(sec x  cos x) sin x  x(sec x  cos x) 2 2 2 (sin x tan x) sin2 x tan x f (x) 



3.3.18

f (x)  2x cos x  3sin x  f (x)  2x(sin x)  2cos x  3cos x  2xsin x cos x

3.3.19

cscx 

dx 3.3.20

secx 

d  1 

sinx0 1cosx

dx  sinx 



dx  cosx 



cotx 



sin2x

d  1 

3.3.21



d  cosx 

cosx



sin2x

cosx0 1sinx



sinx

cos2 x

sinxsinx

cosxcosx

  dx  sinx  

sin2x



cosx

 cscxcotx

sinx sinx





sinx

 secxtanx

cosx cosx

sin2 x  cos2x 1 2     csc x sin2x sin2x

3.3.22

f  x  cosx  f  x  h  f  x  cos x  h  cosx cosxcosh  sinxsinh  cosx f   x  lim  lim  lim h0 h0 h0 h 1 h cosh sinh  cosh 1 sinhh   lim cosx  sinx  cosx lim  sinx lim  h0  h h  h0 h0 h h  cosx0  sinx1  sinx

3.3.23

342

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  sin x  cos x  y  cos x  sin x, so y(0)  cos 0  sin 0  1 0  1. An equation of the tangent line to the curve y  sin x  cos x at the point 0,1 is y  1(x  0)  1or y  x  1. 3.3.24

y  ex cos x  y  ex (sin x)  (cos x)ex  ex (cos x  sin x)  the slope of the tangent line at the point 0,1 is e0(cos 0  sin 0)  1(1 0)  1and an equation for that tangent line is

y  1(x  0) 1or y  x  1. 3.3.25

y  cos x  sin x  y  sin x  cos x, so y( )  sin  cos  0  (1)  1. An equation of the tangent line to the curve y  cos x  sin x at the point  , 1 is y  1(x   )  1or

y  x   1. 3.3.26

y  x  tan x  y  1  sec2 x, so y( )  1  (1)2  2. An equation of the tangent line to the curve y  x  tan x at the point  ,  is y  2(x   )   or y  2x   .

3.3.27

y  2x sin x  y  2x cos x  2sin x, so y( )  2  cos   2sin   2   2  2 2  4



An equation of the tangent line to the curve 2x sin x at the point  ,

y

 1(x   )  2 or y 

  4 x 

2 4

 is

2 4

 1.

 2.

3.3.28 



(a) y  2x sin x  y  2(x cos x  sin x 1). At  2 ,  , y  2( 2 cos 2  sin 2 )  2(0 1)  2, and an equation of the tangent line is y  2  x  2    , or y  2x.

343

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.3.29 (a) y  3x  6cos x  y  3  6sin x. At   ,  3, y  3  6sin   3  6 3  3  3 3, and an 3



equation of the tangent line is y  3  3 3

  x      3, or y  3  3 3 x  3   3. 



3.3.30



(a) f (x)  sec x  x  f (x)  sec x tan x 1 (b) Note that f   0 where f has a minimum. Also note that f  is negative when f is decreasing and f  is positive when f is increasing.



3.3.31

x x x (a) f (x)  e cos x  f (x)  e (sin x)  (cos x)e  e (cos x  sin x)  x

f (x)  ex (sin x  cos x)  (cos x  sin x)ex  ex (sin x  cos x  cos x  sin x)  2ex sin x. (b) Note that f   0 where f has a minimum and f   0 where f  has a minimum. Also note that f  is negative when f is decreasing and f  is negative when f  is decreasing.

3.3.32

H (x)  x sin x  H (x)  x(cos x)  (sin x) 1  x cos x sin x  H (x)  x( sin x)  cos x 1 cos x  x sin x  2cos x

344

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.3.33

f (x)  sin x cos x  f (x)  sin x(sin x)  cos x(cos x)  cos2 x sin2 x  f (x)  2cos x(sin x)  2sin x (cos x)  2(cos xsin x  cos xsin x)  4cos xsin x 3.3.34

f (t)  sec t  f (t)  sec t tan t  f (t)  (sec t) sec2 t  (tan t) sec t tan t  sec3 t  sec t tan2 t, so f (4 ) 

 2   2(1)  2 2  2  3 2. 3

3.3.35 (a) f (x) 

tan x 1

f (x) 

 sec s sec x(sec2 x)  (tan x 1)(sec x tan x) 2



sec x(sec2 x  tan2 x  tan x) 2

1  tan x  sec x

(sec x) (sec x) sin x  cos x sin x tan x  1 cos x  1   cos x  sin x  cos x  (b) f ( x)  1 1 sec s cos x cos x f (x)  cos x  (sin x)  cos x  sin x 1 tan x tan x  1   cos x  sin x, which is the expression for (c) From part (a), f (x)  sec x sec x sec x f ( x) in part (b). 



3.3.36 (a)

f (x)  tan x 

sin x cos x

 f (x) 

f (x)  tan x  sin x(cos x) 1 

cos2 x

cos x

(b)

sin2 x

cos x(cos x) sin x(sin x)

1



cos2 x sin2 x 2

 1 tan x  sec x 2

cos x 2

sin x  2 1  f (x)  sin x(1)(cos x) (sin x)  (cos x) (cos x)   1 cos2 x

 1tan 2 x  sec2 x

345

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.3.37 (a) g(x)  f (x)sin x  g(x)  f (x) cos x  sin x  f (x), so

g( )  f (  ) cos   sin   f ( )  4  1  3  (2)  2  3 3 3 3 3 3 2 2 cos x f (x)  (sin x)  cos x  f ( x) (b) h(x)   h(x)  , so f (x) [ f (x)]2 3 f ( )  (sin  )  cos   f ( ) 4    1 (2) 2 1 1  2 3 3 2 2   3 3 3   h( 3 )  3  2  2 4 1616 [ f ( 3 )]

3.3.38

f (x)  x  2sin x has a horizontal tangent when f ( x)  0  1  2 cos x  0  cos x   1  x  2   2n or 4   2n, where n is an integer. Note that 4 and 2 are   units 2

from π. This allows us to write the solutions in the more compact, equivalent form (2n  1)  3 , n an integer. 3.3.39

f (x)  ex cos x has a horizontal tangent when f (x)  0. f (x)  ex (sin x)  (cos x)ex  ex (cos x  sin x). f (x)  0  cos x  sin x  0  cos x  sin x  tan x  1  x  4  n , n an integer.

3.3.40

The particle’s velocity is v t   st   4 cos t cos t   4sin t sin t   4 cos t



 v t   0  0  cos2 t  sin2 t  cos t  0  cos2 t  1 cos2 t   cos t 

 4 cos2 t  4sin2 t  4 cos t  4 cos2 t  sin2 t  cos t . The particle is at rest when



0  2cos2 t cos t 1 0  2cos t 1cos t 1  cos t  

1 2

or cos t 1. The first time t  0

with velocity zero is t  0 ( cos 0  1). If we require t  0, then t  cos

2 3



2  is the first time (with 3

3.3.41

346

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 The particle’s velocity is 2 2 2  sin t  sin t   cos t cos t  2sin t sin t  cos t  1 2sin t v t   st      . The 2 2 

2  sin t 



2  sin t 

1 particle is at rest when v t   0 1 2sin t  0  sin t   . The first time t  0 for which 2 7 1 . sin t   is t  6 2 3.3.42 (a) x(t)  8sin t  v(t)  x(t)  8cos t  a(t)  x (t)  8sin t

   4 3, velocity v( )  8cos( )  8( )  4, and acceleration a( )  8sin( )  8   4 3. Since

(b) The mass at time t 

2 3

has position x( 23 )  8sin( 2 3)  8

3 2

2

v( 3 )  0, the particle is moving to the left. 2

3.3.43 (a) s(t)  2 cos t  3sin t  v(t)  2sin t  3cos t  a(t)  2cos t  3sin t. (b)

t2  2.55 s. (d) v  0  t1  0.98, s(t1)  3.61 cm. So the mass travels a maximum of about 3.6 cm (upward and downward) from its equilibrium position. (e) The speed is greatest when s = 0, that is, when t  t2  n , n a positive integer. 3.3.44 From the diagram we can see that sin  x / 10  x  10sin. We want to find the rate of change of x with respect to θ, that is, dx / d. Taking the derivative of x  10sin we  dx  get dx / d  10(cos ). So when  ,  10cos  10  1   5 ft/rad. 3

d

3.3.45

347

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) F 

dF (b)

d

W dF ( sin  cos )(0)  W ( cos sin ) W (sin cos )     sin  cos d ( sin  cos )2 ( sin  cos )2  0  W ( cos sin )  0  sin   cos  tan      tan1 

0.6(50) for 0    1, we see that 0.6sin  cos 

 0    0.54. Checking this with part (b) and μ = 0.6, we calculate d   tan1 0.6  0.54. So the value from the graph is consistent with the value in part (b).

3.3.46



d3 d4 d (sin x)  cos x  d 2 (sin x)  sin x  (sin x)  cos x  (sin x)  sin x. dx4 dx3 dx dx2 The derivatives of sin x occur in a cycle of four. Since 99 = 4(24) + 3, we have d 99 d3 (sin x)  (sin x)  cos x. dx99 dx3 3.3.47 Let f (x)  x sin x and h(x)  sin x, so f (x)  xh(x). Then f (x)  h(x)  xh(x), f (x)  h(x)  h(x)  xh (x)  2h(x)  xh (x),

f (x)  2h (x)  h (x)  xh (x)  3h (x)  xh ( x), , f (n) (x)  nh(n1)(x)  xh(n)(x). 35  and h (x)  cos x. Since 34 = 4(8) + 2, we have h(34) (x)  (2) d2 h (x)  2 (sin x) sin x dx 35 d (34) (35) Thus, (x sin x)  35h (x)  xh (x) )  35sin x  x cos x. dx35 3.3.48

y  Asin x  b cos x  y  Acos B sin x  y    Asin x  B cos x. Substituting these expressions for y, y, and y into the given differential equation y   y  2 y  sin x gives us ( Asin x  B cos x)  ( Acos x  B sin x)  2( Asin x  B cos)  sin x  3Asin x  B sin x  Acos x  3B cos x  sin x  (3A  B)sin x  ( A  3B) cos  sin x, so we must have 3A  B  1 and A  3B  0 (since 0 is the coefficient of cos x on the right side). Solving 1 for A and B, we add the first equation to three times the second to get B   10 and A   310. 3.3.49

348

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) If x  0 , then 5x  0 , and hence

  5x  0.lim x0

sin5x sin sin  lim 0  5lim 0  51  5 . 1  x   5

(b) Let f  x  sin5x . Then

d dx

 sin5x   f   x   lim

f  x  h  f  x

 lim

sin 5  x  h    sin5x

 lim

sin  5x  5h   sin5x

h0 h0 h0 h h h sin5xcos5h  cos5xsin5h sin5x cos5h 1 sin5h    h0 lim  lim sin5x   cos5x h  h0  h h    cos5h 1 sin5h  sin5x lim  cos5x lim h0 h0 h h

The first limit can be shown to equal 0 in a manner similar to part (a) and Equation 3. The second limit is

equal to 5 , by part (a). Thus, we have f   x  sin5x0  cos5x5  5cos5x . 3.3.50 (a) Let 

. Then as x  ,  0, and

x 1 sin  lim sin  lim  1. x  0  x  0  (b) Since 1  sin(1 / x)  1, we have (as illustrated in the figure) lim x sin



 x  x sin(1 / x)  x . We know that lim  x  0 and x0

lim x   0; so by the Squeeze Theorem, lim x sin(1 / x)  0. x0

x0

(c)(c)

3.3.51

d (a)

tan x 

d sin x

 sec x 

cos x cos x  sin x(sin x) cos2 x

dx cos x



cos2 x  sin2 x cos2 x

. So

1 sec2 x  2 . cos x d d 1 cos x(0) 1(sin x) sec x   sec2 x  . So sec x tan x  sin x . (b) 2 dx dx cos x cos x cos2 x 





349

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



d (sin x  cos x)  d  1 cot x  dx dx csc x    



csc x(csc2 x)  (1  cot x)(csc x cot x) csc x   csc x  (1  cot x) cot x  cos x  sin x  csc2 x csc2 x csc2 x  cot2 x  cot x 1 cot x cot x 1  . So cos x  sin x  .  2 csc x csc x csc x 2

3.3.52 𝑡 lim tan 6 = lim ( 𝑡→0 sin 𝑡2 𝑡→0

sin 6𝑡 𝑡

1 cos 𝑡6

𝑡

1 ) = lim 6 sin𝑡6 · lim · lim 2𝑡 𝑡→0 𝑡→0 cos 𝑡6 𝑡→0 2 sin 2𝑡 sin𝑡2 6𝑡

1 1 2𝑡 = 6(1) · 1 1 (1) = 6 lim sin 𝑡6 · lim · =3 · lim 𝑡→0 6𝑡 𝑡→0 cos 𝑡6 2 𝑡→0 2 sin 2𝑡 1 2

350

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.3.53 lim

sin 3𝑥 sin 5𝑥 𝑥2

𝑥→0

= lim (

3 sin 3𝑥 3𝑥

𝑥→0

5 sin 5𝑥

) = lim

5𝑥

3 sin 3𝑥

𝑥→0

· lim

3𝑥

5 sin 5𝑥

𝑥→0

5𝑥

sin 3𝑥 sin 5𝑥 · 5 lim = 3(1) · 5(1) = 15 𝑥→0 3𝑥 𝑥→0 25𝑥

= 3 lim

3.3.54 sin 𝜃 lim 1 1 1 sin 𝜃 𝜃 𝜃→0 lim = = = = 𝜃 + tan 𝜃 lim (1 + sin 𝜃 · 1 ) 1 + 1 · 1 2 𝜃→0 𝜃 + tan 𝜃 lim 𝜃 𝜃 cos 𝜃 𝜃→0 𝜃→0

3.3.55 We get the following formulas for r and h in terms of θ:

  h  10cos  2 10 2 1 2 10 2 Now A( )  1  r 2 and B( )  (2r)h  rh. So sin



lim



A( )

 r  10sin 2

 lim

2 r 1



and cos





 1  lim

B( )  0 rh 1   lim tan( / 2)  0.  0



 0

 1  lim

10sin( / 2)

 / 2)

 0 10cos(

 0

      

3.3.56 By definition of radian measure, s  r , where r is the radius of the circle. By drawing the bisector



  d  2r sin . 2 r 2 s r So lim  lim   lim 2  ( / 2)  lim  / 2  1. [This is just the reciprocal of   0  0 d 2r sin( / 2)  0 2sin( / 2)  0 sin( / 2) sin x the limit lim  1combined with the fact that as  0,   0 also.] of the angle θ, we can see that  

d /2





x0

351

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.3.57

x has a jump discontinuity at x  0. 1  cos 2x (b) Using the identity cos2x  1sin2 x, we have x x  x x    . 2 2 1  cos 2x 2 sin x 2 sin x 1  (1  2sin x) x x x  1 lim  1 lim  lim Thus, lim x0  1  cos 2x x0  2 sin x 2 x0  2 x0 (sin x) (a) It appears that f (x) 



1 1 2   1 lim  1   2 2 x0 sin x / x 2 1 Evaluating lim f ( x) is similar but sin x  sin x, so we get 12 x0

2. These values appear to be

reasonable values for the graph, so they confirm our answer to part (a). Another method: Multiply numerator and denominator by 1  cos 2x.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.4 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

352

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

END OF SECTION EXERCISE SOLUTIONS 3.4.1 Let u  g(x)  1 4x and y  f (u)  3 u. Then



dy du 4    1 u 2/3  (4)  . 3 2 3 du dx 3 (1 4x)

3.4.2 Let u  g(x)  2x  5 and y  f (u)  u . Then 3



3 dy du   4u3  6x2   24x2 2x3  5 . du dx

3.4.3 Let u  g(x)   x and y  f (u)  tan u. Then

dy dy du    (sec2 u)( )   sec2  x. dx du dx

3.4.4 Let u  g(x)  cot x and y  f (u)  sin u. Then

dy dy du    (cos u)(csc2 x)  cos(cot x) csc2 x. dx du dx

3.4.5 Let u  g(x)  2x3  5 and y  f (u)  u 4 . Then



3 dy du   4u3  6x2   24x2 2x3  5 . du dx

3.4.6

e x dy du 1/ 2 u 1     e  21 u   e x x and y  f (u)  e . Then  . dx du dx 2 x 2 x dy

Let u  g(x)  3.4.7

F(x)   5x6  2x3   F(x)  4  5x6  2x3   4

3 d 5x6  2x3   4 5x6  2x3  30x5  6x 2 .  dx

   5x3  2 6x2 5x3 1  24x11  5x3  2  5x3 1.

We can factor as follows: 4 x3

3.4.8

F(x)  1 x  x2   F(x)  99 1 x  x 2   99

98 d 1 x  x 2   99 1 x  x 2  1 2x  .  dx

353



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 

3.4.9

 

f (x)  5x 1  (5x 1)1/2  f (x)  12 (5x 1)1/2 (5) 

5 2 5x 1



3.4.10

f (x) 

1 3

x 1 2

 2 1/3  (x 1)  f (x)   13 (x2 1)4/3(2x) 

3.4.11

f ( )  cos 2   f ( )  sin   2



2x 3(x 1)4/3 2



 2   sin 2 (2 )  2 sin  2   d

3.4.12

g( )  cos2    f ( )  2 cos  (sin  )  2sin cos  2sin 2 3.4.13

y  x2e3x  y  x2e3x (3)  e3x (2x)  e3x (3x2  2x)  xe3x (2  3x) 3.4.14

f (t)  t sin t  f (t)  t(cost)  sin t 1  t cost sin t 3.4.15

f (t)  eat sin bt  f (t)  eat (cos bt) b  (sin bt)eat  a  eat (b cos bt  a sin bt)

3.4.16

g(x)  e x  x  g(x)  e x  x (2x 1) 2

3.4.17

f (x)  (2x  3)4 (x2  x 1)5  f (x)  (2x  3)4 5(x2  x 1)4 (2x 1)  (x2  x 1)5  4(2x  3)3  2  (2x  3)3(x2  x 1)4[(2x  3) 5(2x 1)  (x2  x 1)4 8]  (2x  3)3(x2  x 1)4 (20x2  20x 15  8x2  8x  8)  (2x  3)3(x2  x 1)4 (28x2 12x  7) 3.4.18

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

g(x)   x 2 1  x 2  2   3



        6x  x2 1  x2  2 2  x2 1   x2  2  6x  x2 1  x2  2   3x 2  4  3

g(x)  x 2 1  6 x 2  2  2x  3 x 2 1  2x  x 2  2 2



3.4.19

h(t)  (t 1)2/3(2t2 1)3  h(t)  (t 1)2/3 3(2t2 1)2  4t  (2t2 1)3  23(t 1)1/3  23 (t 1)1/3(2t2 1)2[18t(t 1)  (2t2 1)]  23 (t 1)1/3(2t2 1)2 (20t2 18t 1) 3.4.20

F (t)  (3t 1)4 (2t 1)3  F(t)  (3t 1)4 (3)(2t 1)4 (2)  (2t 1)3  4(3t 1)3(3)  6(3t 1)3(2t 1)4[(3t 1)  2(2t 1)]  6(3t 1)3(2t 1)4(t  3)

355

    



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.21 1/ 2

x  x  y    x 1  x 1  1/2 d  x   1 x1/21/2 (x 1)(1)  x(1) y  1  x  2 x 1 dx  x 1  2 (x 1) (x 1)2     1/2 1 1 (x 1) 1  2  2 x (x 1) 2 x (x 1)3/2

3.4.22

1 5 1 4 d  1 1 4  1     y   x    y  5 x   x   5 x  1      x2  x x dx x x           5(x2 1)4 (x2 1) Another form of the answer is y  . x6 3.4.23

y  etan  y  etan  

d d

(tan )  (sec2  )etan

3.4.24

f (t)  2t  f (t)  2 (ln 2) t3



(t 3 )  3(ln 2)t 2 2t

3.4.25 8

 u3 1   u3 1  d  u3 1   8 (u3 1)7 (u3 1)(3u2 )  (u3 1)(3u2 )  3  (u3 1)7 g(u)   3   g(u)  8 3  du u 1 (u3 1)2  u 1   u 1    3 7 2 3 3 3 (u 1) 3u [u 1 (u 1)] (u 1)7 3u2 (2) 48u2 (u3 1)7  8 3 8 3 (u3 1)9 7 3 2 7 3 2  (u 1) (u 1) (u 1) (u 1)  3.4.26

 1 sin t 1/2 1 sin t s(t)     1 cos t  1 cos t  

1  1 sin t 1/2 (1 cos t) cos t  (1 sin t)(sin t)  s(t)   2  1 cos t  (1 cos t)2



cos t  sin t 1 1 (1 sin t)1/2 cos t  cos2 t  sin t  sin2 t   1/2 2 2 (1 cos t) (1 cos t) 2 1 sin t (1 cos t)3/2

356

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.27

f (t)  102 t 



2 t

f (t)  10

(ln10)

1 1/ 2

(ln10) 2  2 t

(2 t )  10

102 t (ln10)

 

3.4.28  

f (z)  ez /( z1) 

z  z/(z1) (z 1)(1)  z(1) z /( z1) d   ez /( z1)    f (t)  e (z 1)2 (z 1)2   e   dz z 1 

3.4.29

H (r)  H (r) 

(r2 1)3

 (2r 1)5 (2r 1)5 3(r2 1)2 (2r)  (r2 1)3 5(2r 1)4 (2)

2(2r 1)4  r 2 1 3r(2r 1)  5(r 2 1)    2



(2r 1)5  

(2r 1)5 

2(r2 1)2 (6r2  3r  5r2  5) 2(r2 1)2 (r2  3r  5)  (2r 1)6 (2r 1)6

3.4.30

J ( )  tan2 (n )  [tan(n )]2  d

J ( )  2 tan(n )

d

tan(n )  2 tan(n ) sec2 (n )  n  2 tan(n ) sec2 (n )

3.4.31

F (t)  etsin2t  F(t)  etsin2t (t sin 2t)  etsin2t (t cos 2t  2  sin 2t 1)  etsin2t (2t cos 2t sin 2t) 3.4.32

F (t) 

t2 t3 1





  (t3 1)1/2 (2t)  t 2  12 (t3 1)1/2 (3t 2 ) F (t) 

 t 1 3

t(t3 1)1/2[2(t3 1)  3 t 3 ] (t3 1)

t  2 t  2 t(t  4)  3  3 (t 1)3/2 2(t 1)3/2 1 2



3.4.33

357



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

G(x)  4C/x  d d  C  C / x 1 C/x C/x 2 4C / x   G (x)  4 (ln 4)    4 (ln 4) Cx   4 (ln 4) Cx   C  (ln 4)  2 dx x dx  x  

3.4.34 4 2 3 4  y4 1   y4 1  ( y 1) 4 y   ( y 1)(2 y)  U ( y)   2   U ( y)  5 2  y 1 ( y2 1)2    y 1  5( y4 1)4 2 y[2 y2 ( y2 1)  ( y4 1)] 10 y( y4 1)( y4  2 y2 1)   ( y2 1)4 ( y2 1)2 ( y2 1)4 ( y2 1)2 5





10 y( y4 1)( y4  2 y2 1) ( y2 1)6

3.4.35

 1 e2x  y  cos 2 x    1 e   1 e2x  d  1 e2x   1 e2x  (1 e2x )(2e2x )  (1 e2x )(2e2x )  y  sin  (1 e2x )2     2 sin   1 e2 x  dx  1 e2 x  1 e2 x  

 1 e2x  2e2x[(1 e2x )  (1 e2x )]  sin  1 e2x  2e2x (2)  4e2x sin  1 e2x       (1 e2x )2 1 e2x (1 e2x )2 (1 e2x )2 1 e2x  sin        1 e2 x  

 



3.4.36

 y  x e

2 1/ x

yxe

1    e1/ x (2x)  e1/ x  2xe1/ x  e1/ x (1 2x)  2   x 

3.4.37

 y  cot2 (sin )  [cot(sin )]2  y  2cot2 (sin )  d (cot(sin )) d  2 cot(sin ) [csc2 (sin ) cos ]  2 cos cot(sin ) csc2 (sin ) 3.4.38 2x

y  1 xe



 y  (1 xe 1 2

2 x 1/2

)

[x(2e

2 x

)e

2 x

]

e2x (1 2x) 2 1 xe2x

3.4.39

358

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (t)  tan(sec(cos t)) 



d d sec(cos t)  sec2(sec(cos t))[sec(cost) tan(cos t)] cos t dt dt 2  sec (sec(cos t)) sec(cos t) tan(cos t) sin t

f (t)  sec2 (seccos t)



3.4.40



y  esin2x  sin(e2x )  y  esin 2 x

sin 2x  cos(e2x )

d dx

e2 x

 esin2x (cos 2x)  2  cos(e2x )e2x  2  2 cos 2xesin2x  2e2x cos(e2x )

3.4.41



   



f (t)  sin 2 esin t  sin esin t     2 2 2 2 2 d d sin 2 t f (t)  2 sin esin t   sin esin t  2sin esin t cos esin t  e   dt dt 2





     4sin  e   cos  e   e

 2sin esin t  cos esin t  esin t 2

sin 2 t

3.4.42

g

x x x



 





 



2 2 2 d sin 2 t   2sin esin t cos esin t  esin t  2sin t cos t  dt

sin t cos t





 



  y  21 x  x  x 1/2 1 x 1/2 1 21 x 1/2  21 x     





3.4.43

g(x)  (2rarx  n) p  g(x)  p(2rarx  n) p1  d (2rarx  n) dx  p(2rarx  n) p1  2rarx (ln a)  r  2r2 p(ln a)(2rarx  n) p1arx 3.4.44 4x

y  23

 x

y  23 (ln 2)

 

x d 4x d x 34 4x 3  2 (ln 2)3 (ln 3) 4 dx dx

 23 (ln 2)3 4 (ln 3)4 x(ln 4)  (ln 2)(ln 3)(ln 4)4 x3 4  2 3 4

3.4.45

359



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  cos sin(tan  x)  cos(sin(tan  x))1/2  d d y  sin(sin(tan x))1/2  (sin(tan x))1/2  sin(sin(tan  x))1/2  1 (sin(tan x))1/2  (sin(tan x)) 2 dx dx d sin (sin(tan  x)  sin (sin(tan  x) cos(tan  x)  tan  x  cos(tan  x) sec2 ( x)  dx 2 (sin(tan  x) 2 (sin(tan  x) 

 cos(tan  x) sec2 ( x) sin (sin(tan  x) 2 (sin(tan  x)



 

3.4.46 4

2 2 y  x  (x  sin 2 x)3   y  4  x  (x  sin 2 x)3   1  3(x  sin x)  (1 2sin x cos x)

3.4.47

y  cos(sin(3 ))  y  sin(sin 3 )  (cos 3 ) 3  3cos 3 sin(sin 3 )  y  3cos 3 cos(sin 3 )(cos 3 ) 3  sin(sin 3 )(sin 3 ) 3   9 cos2 (3 ) cos(sin 3 )  9(sin 3 ) sin(sin 3 )   

3.4.48

y

 (1 tan x) (1 tan x)2

3 2   y  2(1 tan x) sec

2

2sec2 x x

(1 tan x)3

Using the Product Rule with y  2(1 tan x)3(sec x)2, we get

y  2(1 tan x)3  2(sec x)(sec x tan x)  (sec x)2  6(1 tan x)4 sec2 x  2 sec 2 x(1 tan x) 4 2(1 tan x) tan x  3sec 2 x 

 

2 2  2 sec 2 x(1 tan x) 4 2  tan x  2 tan x  3(tan x 1) 



2sec2 x(tan2 x  2 tan x  3) (1 tan x)4

3.4.49

y  1sec t  y  21 (1 sec t)1/2 (sect tan t) 

2sec t tan t . 2 1sect

Using the Product Rule with y   21 (sect tan t)(1 sect)1/2 we get y   1 sec t tan t  1 (1 sec t)3/2 (sec t tan t)  (1 sec t)1/2 ( 1 )[sec t sec2 t  tan t sec t tan t]. 2 2  2  3/2

. Note that  3 is the lesser exponent on (1 – sec t). Continuing, 2 2 y   1 (1 sec t)3/2  1 sec t sec2 t  (1 sec t)(sec2 t  tan2 t  2 2 

Now factor out  1 (1 sect)

360

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1





  12 (1 sec t)3/2 21 sec t sec2 t  sec2 t  tan2 t  sec3 t  sec t tan2 t   1 (1 sec t)3/2  1 sec t(sec2 t 1)  sec2 t  (sec2 t 1)  sec3 t  2  2 





  12 (1 sec t)3/2  32 sec3 t  2 sec2 t  12 sec t 1

 sec t(1 sec t)3/2  34 sec3 t sec2 t  1 4sec t  1 2 

sec t 3sec3 t  4sec2 t sec t  2



There are many other correct forms of y, such as y 

4(1 sec t)3/2 sec t(3sec t  2) 1 sec t

. We chose to find

4 a factored form with only secants in the final form.

3.4.50

y  ee  y  ee  (e x )  ee  e x  x



   e  e  e  e  e  e  e (1 e ) or e

y   ee  (e x )  e x  ee x

ex  x

(1 e x )

3.4.51

y  2x  y  2x ln 2. At 0,1, y  20 ln 2  ln 2, and an equation of the tangent line is y  (ln 2)(x  0) 1 or y  (ln 2)x 1. 3.4.52 3 1/ 2

y  1 x  (1 x )



3 1/ 2

 y  2 (1 x )



3x2

 3x 

. 2(1 x3)1/2 At equation of the tangent line is y  2(x  2)  3 or y  2x 1.





2,3, y 

3 4 2 9  2, and an

3.4.53

y  sin(sin x)  y  cos(sin x) cos x. At  , 0 , y  cos(sin  ) cos  cos(0)  (1)  1, and an equation of the tangent line is y  1(x   )  0 or y  x  . 3.4.54

y  xe x  y  xe  x (2x)  e  x (1)  e  x (2x 2 1). At 0, 0, 2

y  e0 (1)  1, and an equation of the tangent line is y  1(x  0)  0 or y  x. 3.4.55

361

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 y  (a) y  1 ex  2e0

(1 ex )(0)  2(ex )

2ex

 . (1 ex )2 (1 ex )2 2(1) 2 1 1    At 0,1, y  (1 e 0 )2 (11)2 22 2 . So an equation of the tangent line is y  2 (x  0) 1or y  12 x 1.

3.4.56 a) For x  0, x  x, and y  f (x)  

 f (x) 

 2  x2 2 1/2 1 2 1/ 2 2  x2  (1)  x   2  (2  x )  (2x) (2  x )  2 (2  x2 )1/2 2 2 x





2 (2  x2 )  x2   2 3/2 (2  x ) (2  x2 )3/2 So at (1,1) the slope of the tangent line is f (1)  2 and the equation of the tangent line is y  2(x 1) 1or y  2x 1. 

3.4.57 (a) f (x)  x 2  x  x(2  x2 )1/2  2

2  2  2  2x (2  x )  (2  x f (x)  x  2 (2  x ) (2x)  (2  x ) 1 x ) .   2  x2 (b) f   0 when f has a horizontal tangent line, f  is negative when f is decreasing, and f  is positive when f is increasing.



2 1/2

2 1/2



2 1/ 2

362

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.58 (a) From the graph of f, we see that there are 5 horizontal tangents, so there must be 5 zeros on the graph of f . From the symmetry of the graph of f, we must have the graph of f  as high at x = 0 as it is low at x = π. The intervals of increase and decrease as well as the signs of f  are indicated in the figure. (b) f (x)  sin(x  sin 2x) 

f (x)  cos(x sin 2x) 

d dx

(x sin 2x)  cos(x sin 2x)(1 2cos 2x)

3.4.59 2 For the tangent line to be horizontal f (x)  0. f (x)  2sin x  sin x  f (x)2cos x  2sin x cos x  0  2cos x(1 sin x)  0  cos x  0 , or sin x  1, so x    2n or 3  2n , where n is any integer. Now f     3 and f  3   1, so the points on 2

the curve with a horizontal tangent are    2n , 3 and 3  2n , 1, where n is any integer. 2

3.4.60

y  sin 2x  2sin x  y  2 cos 2x  2 cos x  2 cos 2x  cos x  . The tangent line will be horizontal for y  0  cos 2x  cos x  2cos2 x 1 cos x  2cos2 x  cos x 1  0. We have 1 2 4 2cos x 1cos x 1  0  cos x   or cos x  1 x   2k  , x   2k  , or 2 3 3 x  2k  , k an integer.

363



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.61

 



1 . The line 6x  2 y  1 or y  3x  1  2 1 2x 1 1 has slope –3, so the tangent line perpendicular to it must have slope 1 . Thus   3 3 1 2x  1 1   1 2x  3  1 2x  9  2x  8  x  4. When x  4, y  1 2(4)  3, so 3 1 2x the point is 4, 3. y  1 2x  y  12 (1 2x)1/2  2 



3.4.62

F(x)  f (g(x))  F(x)  f (g(x))  g(x), so F(5)  f (g(5))  g(5)  f (2)  6  4  6  24. 3.4.63



h(x)  4  3 f (x)  h(x)  12 (4  3 f (x))1/2 3 f (x),

6 h(1)  1 (4  3 f (1))1/2 3 f (1)  1 (4  37)1/2 3 4  6  . 2 2 25 5 3.4.64 (a) h(x)  f (g(x))  h(x)  f (g(x))  g(x), so h(1)  f (g(1))  g(1)  f (2)  6  5 6  30. (b) H (x)  g( f (x))  H (x)  g( f (x))  f (x), so

H (1)  g( f (1))  f (1)  g(3)  4  9  4  36. 3.4.65 (a) F(x)  f ( f (x))  F(x)  f ( f (x))  f (x), so

F(2)  f ( f (2))  f (2)  f (1) 5  4 5  20. (b) G(x)  g(g(x))  G(x)  g(g(x))  g(x), so

G(3)  g(g(3))  g(3)  g(2) 9  7 9  63. 3.4.66

 q(x)   f (x)  p q(x)  (q(x))  q(x)  1 7 7 f (5)  p q(5)   (q(5))  q(5)  p 16    7  p(4)   8  7

(a) f (x)  p

1 2

1/2

(b) h(x) 

q(x)

 h(x)  

1/2

xq(x)  q(x) 

8 8 2 16  h(4)  4  q(4)  q(4)  4  2  4  4  1 42 16 16 4

364

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.4.67

f (x) g(x)  f (x)  f (x)g(x)   h(x)   g(x) (g(x))2 g(2)  f (2)  f (2)g(2) 5(1)  (3)(2) 5  6 1 h(2)     2 2 (g(2)) 5 25 25 (b) h(x)  f (g(x))  h(x)  f (g(x))  g(x)  h(2)  f (g(2))  g(2)  f (5) (5)  7  35 1 (c) h(x)   h(x)  f (x)  h(5)  f (5)  7  7  7 f (x) 2 f (x) 2 f (5) 2 4 2  2 4 (a) h(x) 

3.4.68

D(x)  D(4) 

[ f (x)]2

 D(x) 

x [2 f (x) f (x)] [ f (x)]2 1

x 4 [2 f (4) f (4)] [ f (4)]2 2



x2 4 [2  6(1/ 6)]  62



8  36 28 7    . 16 16 4

3.4.69 The particle changes direction from left to right when the velocity, v(t), changes from negative to positive. s(t)  cos t  cos2 t  v(t)  s(t)  (sin t)  2 cos t(sin t)  sin t(2 cos t 1).

v(t)  0  sin t(2cos t 1)  0  sin t  0  t  0, or t   , or (2 cos t 1)  0  t  3 . The velocity is negative for 3  t   , so the particle changes direction from left to right when t  .

3.4.70



3sin  2t  , is 0. That occurs when The particle is at rest when its velocity, v t   t t  t   t  3sin  0  sin 0  k  t  2k, k an integer. Ignoring interval endpoints,  2   2  2     t  4 is the only time of this form with 2  t  6. 

365

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.71 f  x  6x2  3  f   x 

1 2 6x2  3

12x. If the point of tangency is P a, f  a   , then the

slope of the tangent line at P is f   a   6a

y  2x  k is 2. Now

6a  3 2

6a 6a2  3

 2, since the slope of the tangent line

 2  36a 2  4  6a 2  3  a 2  1  a  1. Note that a  1 is

extraneous. The x-coordinate of P is thus 1, and the y-coordinate is f 1  6 1  3  3. The 2

point of tangency is 1, 3. 3.4.72







 2 x2x 3  cos  x  3  x x 3 . The slope  (1)   of the tangent line at the point where x  1 is f (1)  cos  1  3   cos  2    , f (x)  sin  x2  3  f (x)  cos  x 2  3 

12  3

3.4.73

y  g(sec2 x) 

 g(sec2 x)  2sec x sec tan x  2 2sec4 x 1sec2 x tan x  dx 4 2   y     2 2sec4    1 sec2    tan     2 2 2 1  2 (1)  2 8 1  2  4  4 3  12. 9 4 4 4 4    

 

 

3.4.74

h(x)  x2 f (x)g(x)  h(x)  x2[ f (x)g(x)]  2x  f (x)g(x)  x2[ f (x)g(x)  f (x)g(x)]  2x  f (x)g(x)  h(1)  12[ f (1)g(1)  f (1)g(1)]  2(1)  f (1)g(1)  2(0) 1(1)  2(2)(0)  1.



3.4.75

f (x)  (ax  b)  f (x)  n(ax  b)n1  a  an(ax  b)n1  f (x)  an(n 1)(ax  b)n2  a  a2n(n 1)(ax  b)n2  n

f (x)  a3n(n 1)(n  2)(ax  b)n3  f (n) (x)  ann(n 1)(n  2)

(n  (n 1))(ax  b)0  an  n!

366

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.76 (a) u(x)  f (g(x))  u(1)  f (g(1))  g(1)  f (3) 3. But f (3)  the slope of the graph of f at

3 4

1 1 3   . Therefore, u(1)   3   . 62 4 4 4 (b) v(x)  g( f (x))  v(1)  g( f (1))  f (1)  g(2)  f (1). But g(2) does not exist because the graph of g has a cusp at x = 2. Therefore v(1) does not exist. 03  3, and (c) w(x)  g(g(x))  w(1)  g(g(1))  g(1)  g(3)  g(1). Now g(1)  2 1 20 2 2 g(3)   . So w(1)  g(3)  g(1)  (3)  2. 3 5 2 3 x  3, which is

3.4.77

2.5 1.5 1.5  0.75  1; f (2)   0.75; 0.5 1.5 1.5  2.5 h(x)  f ( f (x))  h(x)  f ( f (x))  f (x)  h(2)  f ( f (2))  f (2)  f (1)  f (2)  (1)(0.75)  0.75 42 (b) f (4)   2.5; 4.3  3.5 g(x)  f (x2 )  g(x)  f (x2 )  2x  g(2)  f (22 )  2(2)  4  f (4)  4  2.5  10 (a) f (1) 

3.4.78



   1 f (3) f (3) 1/2 g(x)  f (x)  ( f (x))  g(x)  f (x)  g(3)   . Now f (3) = 2 f (x) 2 f (3) 2 2 04 2 the slope of the tangent line in the figure which is m   . 60 3 2 1 1  3 Thus g (3)    2. 3 2 6 2 2 3.4.79 ℎ(𝑥) = ƒ(𝑔(𝑥)) ⟹ ℎ′(𝑥) = ƒ′(𝑔(𝑥))𝑔′(𝑥). So ℎ′(0.5) = ƒ′(𝑔(0.5))𝑔′(0.5) = ƒ′(0.1)𝑔′(0.5). We can estimate the derivatives by taking the average of two secant slopes. For ƒ′(0.1): 𝑚 = 14.8−12.6 = 22, 𝑚 = 1 2 0.1−0

18.4−14.8

0.2−0.1 0.05−0.10

For 𝑔′(0.5): 𝑚 = 0.10−0.17 = −0.7, 𝑚 = 1 2 0.5−0.4

= 36. So ƒ′(0.1) ≈

𝑚1+𝑚2

22+36

= −0.5. So 𝑔′(0.5) ≈

0.6−0.5

= 29.

2 𝑚1+𝑚2

= −0.6.

Hence, ℎ′(0.5) = ƒ′(0.1)𝑔′(0.5) ≈ (29)(−0.6) = −17.4.

367

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.80 𝑔(𝑥) = ƒ(ƒ(𝑥)) ⟹ 𝑔′(𝑥) = ƒ′(ƒ(𝑥))ƒ′(𝑥). So 𝑔′(1) = ƒ′(ƒ(1))ƒ′(1) = ƒ′(2)ƒ′(1). We can estimate the derivatives by taking the average of two secant slopes. For ƒ′(2): 𝑚 = 3.1−2.4 = 1.4, 𝑚 = 1 2 2.0−1.5

For ƒ′(1): 𝑚 = 2.0−1.8 = 0.4, 𝑚 = 1 2 1.0−0.5

4.4−3.1 2.5−2.0 2.4−2.0

= 2.6. So ƒ′(2) ≈ = 0.8. So ƒ′(1) ≈

1.5−1.0

𝑚1+𝑚2 2 𝑚1+𝑚2

= 2. = 0.6.

Hence, 𝑔′(1) = ƒ′(2)ƒ′(1) ≈ (2)(0.6) = 1.2.

3.4.81 (a) F(x)  f (x )  F(x)  f (x )

(x )  f (x )x1

dx 1 (b) G(x)  [ f (x)]  G(x)  [ f (x)] f (x) 

3.4.82 (a) F(x)  f (ex )  F(x)  f (ex ) (b) G(x)  e

f (x)

 G(x)  e

f (x)

(ex )  f (ex )ex

dx d f (x)  e f (x) f (x) dx

3.4.83 cx cx 0 (a) g(x)  e  f (x)  g(x)  e  c  f (x)  g(0)  e  c  f (0)  c  5

g(x)  ecx  c  f (x)  g (x)  cecx  c  f (x)  g (0)  c2e0  f (0)  c2  2 kx kx kx 0 0 (b) h(x)  e f (x)  h(x)  e f (x)  f (x)  ke  h(0)  e f (0)  f (0)  ke  5  3k

An equation of the tangent line to the graph of h at the point 0, h(0)  (0, f (0)  (0, 3) is

y  (5  3k)(x  0)  3 or y  (5  3k)x  3. 3.4.84

r(x)  f (g(h(x)))  r(x)  f (g(h(x)))  g(h(x))  h(x), so r(1)  f (g(h(1)))  g(h(1))  h(1)  f (g(2))  g(2)  4  f (3) 5 4  65 4  120 3.4.85

f (x)  xg(x2 )  f (x)  xg(x2 ) 2x  g(x 2 )1  2x 2 g(x 2 )  g(x 2 )  f (x)  2x 2 g (x2 ) 2x  g(x2 )4x  g(x2 )2x  4x3g (x2 )  6xg(x2)

368

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.86

F (x)  f (3 f (4 f (x)))  d d F(x)  f (3 f (4 f (x)))  3 f (4 f (x))  f (3 f (4 f (x))) 3 f (4 f (x))  (4 f (x)) dx dx  f (3 f (4 f (x))) 3 f (4 f (x))  4 f (x), so F(0)  f (3 f (4 f (0))) 3 f (4 f (0))  4 f (0)  f (3 f (4  0)) 3 f (4 0)  4  2  f (3 0) 3 2 4 2  23 2 4 2  96 3.4.87

F(x)  f (xf (xf (x)))  d d   (xf (x))  f (xf (x)) 1  F(x)  f (xf (xf (x))) (xf (xf (x)))  f (xf (xf (x))) x  f (xf (x))   dx dx 



 f (x f (x f (x)))   x f (x f (x))  (x f (x)  f (x) 1)  f (x f (x)), so

F(1)  f ( f ( f (1)))  f ( f (1))  ( f (1)  f (1))  f ( f (1))  f ( f (2)) [ f (2) (4  2)  f (2)]  f (3) [5 6  3]  633  198. 3.4.88

y  e2x ( Acos 3x  B sin 3x)  y  e2x (3Asin 3x  3B cos 3x)  ( Acos 3x  B sin 3x)  2e2x  e2x (3Asin 3x  3B cos 3x  2Acos 3x  2B sin 3x)  e2x[(2A  3B) cos 3x  (2B  3A) sin 3x]  y  e2x[3(2A  3B) sin 3x  3(2B  3A) cos 3x]  (2A  3B) cos 3x  (2B  3A) sin 3x] 2e2x

 e2x [3(2 A  3B)  (2(2B  3A)]sin 3x [3(2B  3A)  2(2 A  3B)]cos 3x  e2x[(12A  5B) sin 3x  (5A 12B) cos 3x] Substitute the expressions for y, y and y into y   4 y 13y  0 to get

y   4 y 13y  e2x[(12 A  5B) sin 3x  (5A 12B) cos 3x]  4e2x[(2 A  3B) cos 3x  (2B  3A) sin 3x] 13e2x ( Acos 3x  B sin 3x)

 e2x[(12A  5B  8B 12A 13B) sin 3x  (5A 12B  8A 12B 13A) cos 3x]  e2x[(0) sin 3x  (0) cos 3x]  0. Thus, the function y satisfies the differential equation y   4 y 13y  0.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.89

y  erx  y  rerx  y   r2erx. Substituting y, y and y into y   4 y  y  0 gives us r2erx  4rerx  erx  0  erx  r 2  4r 1  0. Since rerx  0, we must have r 2  4r 1  0  r

4  16  4  2  3. 2

3.4.90

, Dn is just a derivative notation. In general, Df (2x)  2 f (2x),

The use of D, D ,

D2 f (2x)  4 f (2x),

, Dn f (2x)  2n f (n) (2x). Since f (x)  cos x and 50  4(12)  2, we have

f (50) (x)  f (2) (x)  cos x, so D50 cos 2x  250 cos 2x. 3.4.91 ƒ(𝑥) = 𝑥𝑒−𝑥, ƒ′(𝑥) = 𝑒−𝑥 − 𝑥𝑒−𝑥 = (1 − 𝑥)𝑒−𝑥, ƒ′′(𝑥) = −𝑒−𝑥 + (1 − 𝑥)(−𝑒−𝑥) = (𝑥 − 2)𝑒−𝑥. Similarly, ƒ′′′(𝑥) = (3 − 𝑥)𝑒−𝑥, ƒ−1; ƒ(4)(𝑥) = ƒ(𝑥 − 4)𝑒−𝑥,…, ƒ(1000) = (𝑥 − 1000)𝑒−𝑥.



3.4.92

f (x)  e

 f (x)  e x (2x)  2xe x . At the point 1, f (1)   1, e  , the tangent line has slope 2

f (1)  2(1)e1  2e, and the equation of the tangent line is y  2e(x 1)  e or y  2ex  e. 2

The tangent line intersects the x-axis at the point  12 ,0  and intersects the y-axis at the point 0, e. Therefore the base of the right triangle is 1 , the height is e, and the area of the triangle is 1  1 e  e . 2

3.4.93

370

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

s(t)  10  1 sin(10 t)  the velocity after t seconds is v(t)  s(t)  1 cos(10t)(10 ) 4



5 2

cos(10 t) cm/s.

3.4.94 (a) 𝑠 = 𝐴 cos(𝜔𝑡 + 𝛿) ⟹ velocity = 𝑠′ G −𝜔𝐴 sin(𝜔𝑡 + 𝛿) (b) If 𝐴 G 0 and 𝜔 G 0, then 𝑠′ = 0 - sin(𝜔𝑡 + 𝛿) = 0 - 𝜔𝑡 + 𝛿 = 𝑛𝜋 - 𝑡 =

𝑛𝜋−ð 𝜔

n an integer.

3.4.95

2t dB  2t 2t 7 2t    0.35sin  2  0.7 5.4 dt  5.4  5.4   5.4 cos 5.4  54 cos 5.4    dB 7 2 (b) At t  1,  cos   0.161. This means that the brightness on day 1 into the 5.4-day dt 54 5.4 (a) B(t)  4.0  0.35sin

cycle is increasing at a rate of 0.161 on the brightness scale. 3.4.96  L(t)  12  2.8sin  2365 (t  80)  L(t)  2.8cos 2365(t  80)  2365. On March 21, t  80, and L(80)  0.0482 hours per day. On May 21, t  141, and L(141)  0.02398, which is approximately one-half of L(80).

3.4.97

s(t)  2e1.5t sin 2t  v(t)  s(t)  2[e1.5t (cos 2t)(2 )  (sin 2t)e1.5t (1.5)]  2e1.5t (2 cos 2t 1.5sin 2t)

371

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

  kt

3.4.98

(a) lim p(t)  lim t

t

 kt

1 ae

1 a  0



 1, since k  0   kt    e

 0.

kaekt dp  kt 2 kt )  (b) p(t)  (1 ae )   (1 ae ) (kae (1 aekt )2 dt (c) From the graph of p(t)  (110e0.5t )1, it seems that p(t) = 0.8 kt 1

(indicating that 80% of the population has heard the rumor) when t ≈ 7.378 hours. 3.4.99 (a) Use C(t)  atebt with a  0.0225 and b  0.0467 to get

C(t)  a(tebt  b  ebt 1)  a(bt 1)ebt . C(10)  0.0225(0.533)e0.467  0.00752, so the BAC was increasing at approximately 0.00752 (mg/mL)/min after 10 minutes. (b) A half an hour later gives us t = 10 + 30 = 40. C(40)  0.0225(0.868)e1.88  0.003002, so the BAC was decreasing at approximately 0.003002 (mg/mL)/min after 40 minutes. 3.4.100

P  t   1368.59  1.01471  ln1.01471 . The rates of change (increase) in the world population t

for 1950, 1990, and 2020 are, respectively, are P50  41.477, P90  74.385, and P120  115.277 millions of people per year. 3.4.101 By the Chain Rule, 𝑎(𝑡) =

𝑑𝑣

𝑑𝑡

𝑑𝑣 𝑑𝑠 𝑑𝑠 𝑑𝑡

𝑑𝑣

𝑣(𝑡) = 𝑣(𝑡)

𝑑𝑠

𝑑𝑣

. The derivative dv/dt is the rate of change of

𝑑𝑠

the velocity with respect to time (in other words, the acceleration) whereas the derivative dv/ds is the rate of change of the velocity with respect to the displacement.

3.4.102 (a) The derivative dV/dr represents the rate of change of the volume with respect to the radius and the derivative dV/dt represents the rate of change of the volume with respect to time. (b) Since 𝑉 = 4 𝜋𝑟3, 3

𝑑𝑉 𝑑𝑟 𝑑𝑉 = 𝑑𝑟 𝑑𝑡 𝑑𝑡

= 4𝜋𝑟

3 𝑑𝑟

𝑑𝑡

3.4.103 (a) Using a calculator or CAS, we obtain the model Q  abt with a  100.0124369 and b  0.000045145933.

372

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(b) Use Q(t)  abt (ln b) with the values of a and b from part (a) to get Q(0.04)  670.629 A. 3.4.104

P  abt with a  1.0618795261015 and b  1.01892541, where P is measured in millions of people. The fit appears to be very good.

7.71 6.41 9.58  7.71  0.13 and m2   0.187. So 1960 1950 1950 1940 m  m2 P1950  1  0.1585 million people/year. 2 For 2000: m  20.85 16.99  3.86  0.386 and m  25.24  20.85  0.439. 1 2 2000 1990 10 2010  2000

For 1950: m1 

P2000 

m 1  m2

 0.4125 million people/year.

2 Using Pt   abt ln b with the values of a and b from part (a), we get

(c)

P1950  0.1502 million people/year and P2000  0.3836 million people/year. These values are relatively close to the values in part (b). P 2021  30.33 million. The actual population was 29.53 million. Growth may have

(d)

slowed due to the pandemic that began in 2020. 3.4.105 𝑥 = 𝑡4 + 1, 𝑦 = 𝑡3 + 𝑡; 𝑡 = −1. 𝑑𝑦 = 3𝑡2 + 1, (𝑥, 𝑦) = (2, −2) and 𝑑𝑦 = 𝑑𝑥

𝑑𝑡

−4

𝑑𝑥 𝑑𝑡

= 4𝑡3, and 𝑑𝑦 = 𝑑𝑥

𝑑𝑦/𝑑𝑡

𝑑𝑥/𝑑𝑡

3𝑡2+1

. When 𝑡 = −1,

4𝑡3

= −1, so an equation of the tangent to the curve at the point

corresponding to 𝑡 = −1 is 𝑦 − (−2) = (−1)(𝑥 − 2), or 𝑦 = −𝑥.

3.4.106

373

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 𝑑𝑦/𝑑𝜃

𝑥 = cos 𝜃 + sin 2𝜃 , 𝑦 = sin 𝜃 + cos 2𝜃 ; 𝜃 = 0. 𝑑𝑦 = 𝑑𝑥

𝑑𝑥/𝑑𝜃

cos 𝜃−2 sin 2𝜃

. When 𝜃 = 0,

− sin 𝜃+2 cos 2𝜃

(𝑥, 𝑦) = (1,1) and 𝑑𝑦 = , so an equation of the tangent to the curve is 𝑑𝑥

𝑦 − 1 = (𝑥 − 1), or 𝑦 = 1 𝑥 + . 2

3.4.107 𝑥 = 𝑒√𝑡, 𝑦 = 𝑡 − ln 𝑡2 ; 𝑡 = 1. 𝑑𝑦 = 1 − 𝑑𝑡

𝑡2

2 𝑑𝑥

=1− , 2

𝑡 𝑑𝑡

𝑒 √𝑡

, and

2√𝑡

𝑑𝑦 𝑑𝑥

𝑑𝑦/𝑑𝑡 𝑑𝑥/𝑑𝑡

1−2/𝑡

𝑒√𝑡/(2√𝑡)

2𝑡 2𝑡

2𝑡−4

√𝑡𝑒√𝑡

When 𝑡 = 1, (𝑥, 𝑦) = (𝑒, 1) and 𝑑𝑦 = − , so an equation of the tangent line is 2 𝑑𝑥 𝑒 𝑦 − 1 = − (𝑥 − 𝑒), or 2 𝑦 = − 𝑥 + 3. 𝑒 𝑒

3.4.108 𝑥 = 2𝑡3 + 3𝑡2 − 12𝑡, 𝑦 = 2𝑡3 + 3𝑡2 + 1. 𝑑𝑦 = 6𝑡2 + 6𝑡 = 6𝑡(𝑡 + 1), so 𝑑𝑦 = 0 - 𝑡 = 0 or − 1 𝑑𝑡

𝑑𝑡

(𝑥, 𝑦) = (0, 1) or (13, 2). 𝑑𝑥 = 6𝑡2 + 6𝑡 − 12 = 6(𝑡 + 2)(𝑡 − 1), so 𝑑𝑥 = 0 - 𝑡 = −2 or 1 𝑑𝑡

𝑑𝑡

(𝑥, 𝑦) = (20, −3) or (−7, 6). The curve has horizontal tangents at (0, 1) and (13, 2), and vertical tangents at (20, -3) and (-7, 6).

3.4.109 𝑥 = 10 − 𝑡2, 𝑦 = 𝑡3 − 12𝑡. 𝑑𝑦 = 3𝑡2 − 12 = 3(𝑡 + 2)(𝑡 − 2), so 𝑑𝑦 = 0 - 𝑡 = ±2 - (𝑥, 𝑦) = 𝑑𝑡

𝑑𝑡

(6, ±16). 𝑑𝑥 = −2𝑡, so 𝑑𝑥 = 0 - 𝑡 = 0 - (𝑥, 𝑦) = (10, 0). The curve has horizontal tangents at 𝑑𝑡

𝑑𝑡

(6, ±16) and a vertical tangent at (10, 0).

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.110

3.4.111

3.4.112

3.4.113

375

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

a) Using technology we find f (x) 

3x4 1



 

2 x  x 1 4 x 4  x 1 x  x 1 4 x  x 1 (b) Using technology, we find f (x)  0  x  0.7598 and x  0.7598. (c) From the graphs we see that lim f (x)  1 and lim f (x). The graph

x





 x 4  x 1

3/ 2

x

of f seems to have horizontal tangents when x is larger than about 5 or smaller than about –5. 3.4.114 (a) If f is even, then f (x)  f (x). Using the Chain Rule to differentiate this equation, we get

f (x)  f (x)

d dx

(x)   f (x). Thus f (x)   f (x), so f  is odd.

(b) If f is odd, then f (x)   f (x). Differentiating this equation, we get f (x)   f (x)(1)

 f (x), so f  is even. 3.4.115

 f (x)  1  1 2  g(x)    f (x)[g(x)]   f (x)[g(x)]  (1)[g(x)] g(x) f (x)   f (x) f (x)g(x) f (x)g(x)  f (x)g(x)    2 g(x) [g(x)] [g(x)]2 This is an alternative derivation of the formula in the Quotient Rule. But part of the purpose of the Quotient Rule is to show that if f and g are differentiable, so is f /g. The proof in Section 3.4 does that; this one doesn’t. 3.4.116

d (sinn x cos nx)  nsinn1 x cos x cos nx sinn x(nsin nx) dx  n sinn1 x(cos nx cos x sin nx sin x)

(a)

(b)

[Product Rule] [factor out n sinn1 x ]

 n sinn1 x cos(nx  x)

[Addition formula for cosine]

 n sinn1 x cos(n 1)x

[factor out x]

d (cosn x cos nx)  n cosn1 x(sin x) cos nx  cosn x(nsin nx) [Product Rule] dx  n cosn1 x(cos nx sin x sin nx cos x) [factor out n cosn1 x ]

 n cosn1 x sin(nx  x)

[Addition formula for sine]

 n cosn1 x sin(n 1)x

[factor out x]

376

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.117 ―The rate of change of y5 with respect to x is eighty times the rate of change of y with respect to x‖



y5  80

 5y 4

 80

 5y 4  80 (Note that dy/dx ≠ 0 since the curve never

dx dx  4 has a horizontal tangent)  y  16  y  2 (since y > 0 for all x). dx

3.4.118



d  d      we have  ( )  rad, (sin  )  (sin 180  )  180 cos 180   180 cos . Since 180 d  d 

3.4.119



(a) f (x)  x 



x  (x )

2 1/2

 f (x)  12 (x2 )1/2 (2x) 

x x2

x for x ≠ 0. x



f is not differentiable at x = 0. (b) f (x)  sin x 

 cot x  cot x

sin2 x  (sin x2 )1/2  f (x)  12 (sin2 x)1/2 (2sin x cos x) 

x cos x sin x

if sin x  0

; if sin x  0 f is not differentiable when x  n , n an integer



if x  0

 cos x x x  cos x   x x cos x g is not differentiable at 0. g(x)  cos x 

if x  0

;

3.4.120



dy du

, so dx du dx d 2 y  d  dy   d  dy du    d  dy  du  dy d  du    dx2   [Product Rule] dx  dx  dx  du dx   dx 2 du  2dx du2 dx  du 2 dy d u   d  dy  du  du  dy d 2u  d 2y  du  du du dx dx du dx dx dx  du dx2      

The Chain Rule says that







377

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.4.121 From the previous exercise, 2 d 2 y  d 2 2y  du 2 dy d 2u d 3 y d d 2 y d  d y  du 2  d  d d 2u2  2  du dx   du dx2  dx3  dx dx2  dx du dx  2  dx dx  dx  du       

2  dy    du 2   d  du 2  d 2 y   d  dy  d 2u   d  d 2u    d  d 2y  dx du dx  dx dx du2  dx du dx2 dx dx2 du                 2 2 2 2 3 2 du d u d y  d  dy  du   d u  d u dy   d  d 2y  du   du  du du dx dx   2 dx dx2 du2   du  du  dx   dx2   dx3 du            3 3 2 2 3 d y  dy  du d u d y dy d u  3   3  . du  dx  dx dx2 du2 du dx3 

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.5 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.5.1 (a)

d dx

(9x2  y2 ) 

(1)  18x  2 yy  0  2 yy  18x  y 

9x y

(b) 9x2  y2  1  y2  9x2 1  y   9x2 1, so y   12 (9x2 1)1/2 (18x)   (c) From part (a), y 

9x y



9x  9x2 1

9x 9x2 1

, which agrees with part (b).

378

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.2 (a)

(2x2  2x  xy) 

4x  y 1

(1)  4x 1 xy y 1  0  xy  4x  y 1  y  

dx dx 1 1 2 (b) 2x  2x  xy  xy  1 2x2  x  y   2x 1, so y    2. 2 x x (c) From 4xpart  y(a), 1

y  

 4 

y

 4 

11

(a)

d 

(b)

dx x

( x 



y) 

x 

 2





 2,



. x

y x



1 x [from part (b)]   1  1, which agrees with part (b). x x

cosx  y   d 5  sinx  1 y

1/2

dx y 5

y  1 x  y  (1 x )2  y  1 2 x  x, so

(b) cosx 



(1)  1 x1/2  1 y1/2 y  0  1 y   1  y   y 2 x 2 2 x 2 y dx

 y 1 



y  2 12 x1/2 1  1

(a)

 4 



3.5.3

3.5.4 d

1

x  x

which agrees with part (b).

 2x 1

 y  0 

 y  sinx  y  2 ysinx

2 y

y  5  cosx  y  (5  cosx)2 , so y  2(5  cosx)' sinx  2sinx 5  cosx

. 2 (c) From part (a), y  2 ysinx  2 (5  cosx)  2  5  cosxsinx[ since 5  cosx  0].

3.5.5

379

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y2  x  5  2 yy 1  0  y  

1 2y

, so the slope of the tangent line where y  2 is



1  . 4 2  2 

3.5.6 x2  y2  a2  2x  2 yy  0  y  



3.5.7 x2  y2  25 2x  2 yy  0  y 



x x2 y2  (25  y2 ) y(1)  xy y  x y y  y 25 y y       y2 y2 y2 y2 y3 





3.5.8 d

d (x2  4xy  y2 )  (4)  2x  4[xy  y(1)]  2yy  0  2 yy 2xy  4y  2x  dx dx 2y  x y( y  4x)  2 y  x  y  y  4x

3.5.9 d

(2x2  xy  y2 ) 

dx 4x  y (x  2 y) y  4x  y  y  x  2y

(2)  4x  xy y(1)  2yy  0  xy 2yy  4x  y 

3.5.10 d

(x4  x2 y2  y3) 

d dx

(5)  4x3  x2  2yy y2 (2x)  3y2 y  0  2x2 yy 3y2 y

4x  2xy 2x(2x  y )  4x  2xy  (2x 2 y  3y 2) y  4x 3  2xy 2  y  2 2 2x y  3y   y(2x2  3y) 3.5.11 3

380

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(x3  xy2  y3) 

(1)  3x2  x  2yy y2 (1)  3y2 y  0  3y2 y 2xyy  y2  3x2 

dx 2 2 2 2 (3y2 2xy) y y2  3x2  y y  3x   y  3x 3y2  2xy y(3y  2x)

3.5.12 d  x2  

(x  y)(2x)  x2 (1 y)





 (x  y)  xey y  ey 1  1 ex  xey y y  1 ey  y(xey 1)  1 e y 





 2 yy 

( y 1) (x  y)2   dx  x  y  dx 2x2  2xy  x2  x2 y  2 y(x  y)2 y x2  2xy  2 y(x  y)2 y  x2 y  x(x  2 y) x(x  2 y)  [2 y(x2  2xy  y2 )  x2 ]y y  2 2x y  4xy2  2 y3  x2 Or: Start by clearing fractions and then differentiate implicitly.

3.5.13 d

(xey ) 

y 

1 e xey 1

3.5.14 d

( y cos x) 

dx 2x  y sin x y(cos x  2 y)  2x  y sin x  y  cos x  2 y

(x2  y2 )  y(sin x)  cos x  y  2x  2 yy  cos xy  2 yy  2x  y sin x 

3.5.15 d

(cos xy) 

dx  y[x sin(xy)  cos y]  y sin(xy)  y 

(1 sin y)  (sin xy)(xy  y)  cos y  y   xysin(xy)  cos y  y  y sin(xy) y sin(xy) y sin(xy)  x sin(xy)  cos y x sin(xy)  cos y

3.5.16 d 4 4 1/2 3 3 d 1 x  y  (x  y )  2(x  y) (1 y)  4x  4 y y  dx dx

381

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 1   4x3  4 y3  y    2 2 2 2 x y x y x y x  y   1 8x3 x  y 8 y3 x  y 1 1 8x3 x  y y   y  3  2 x y 8 y x  y 1 2 x y 1

 



y  4x3  4 y3 y 



 

3.5.17 d

(ey sin x) 

(x  xy)  ey cos x sin xey y  1 xy y 1 

dx dx 1 y  ey cos x ey sin x        y   y     y    y xy 1 y e cos x y (e sin x x) 1 y e cos x y ey sin x  x

3.5.18 d x/ y d d x y 1 x  y (e )  (x  y)  ex/ y   1 y  ex/ y   1 y  dx

  2 dx dx  y  ex / y y   xex/ y   y  ex / y  xex/2 y      xex2/ y     x/ y 1   y 1 e  y  y y 1 y y  y  y 1 y y2   y     



y  ex/ y  x/ y y  y( y  e )  y  xex/ y y2  xex/ y 1 2 y

3.5.19 d d (xy)  x2  y2  xy y(1)  1 (x2  y2 )1/2 (2x  2yy)  2 dx dx x x y y y  xy  y  y  xy   y  x2  y2 x2  y2 x2  y2 x2  y2 

x x2  y2  y x2  y 2

 

y 

x  y x2  y2 x2  y2

 y

x  y x2  y2 x x2  y2  y

 3.5.20

382

 

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

d  ysinx2   xsin y2   ycosx2   2x  sin  x 2   y  xcos y2   2yysin  y2 1   dx   dx  sin  y2   2xycosx2  d

y sin  x 2   2xycos y2  sin  y2   2xycosx2   y 

sin  x 2   2xycos y2 

3.5.21 d d  x sin y  y sin x  (1)  x cos y  y   sin y 1 y cos x  sin x  y  0  dx dx x cos y  y  sin x  y  sin y  y cos x  y(x cos y  sin x)  sin y  y cos x  y 

sin y  y cos x x cos y  sin x

3.5.22

d sin(xy)  cos(x  y)  cos(xy) (xy y 1)  sin(x  y)  (1 y)  dx dx x cos(xy) y  y cos(xy)  sin(x  y)  ysin(x  y)  x cos(xy) y  ysin(x  y)   y cos(xy) sin(x  y)  y cos(xy)  sin(x  y) [x cos(xy)  sin(x  y)]y  [ y cos(xy)  sin(x  y)]  y   x cos(xy)  sin(x  y)

3.5.23



tan(x  y) 

y 2 2 2 2  (1 x ) tan(x  y)  y  (1 x )sec (x  y)  (1 y)  tan(x  y)  2x  y  1 x



(1 x2 ) sec2 (x  y)  (1 x2 ) sec2 (x  y) y  2x tan(x  y)  y  (1 x2 ) sec2 (x  y)  2x tan(x  y)  [1 (1 x2 ) sec2 (x  y)] y 

y 

(1 x2 ) sec2 (x  y)  2x tan(x  y) 1 (1 x2 ) sec2 (x  y)

3.5.24

cos(xy)  sin y  xy  sin(xy) [xy  y 1]  cos yy  xy y 1  sin(xy)  xy  y sin(xy)  cos yy  xy  y  sin(xy)  xy cos yy xy  y  y sin(xy) 

383

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 [cos y  x sin(xy)  x]y  y[1 sin(xy)]  y 

y[1 sin(xy)] cos y  x sin(xy)  x

3.5.25

2xex  2 yey  4  2x  ex  2  ex  (2 y  e y  y e y  2 y)  0  yey y  ey y  (xe x  ex ) ( yey  e y ) y (xe x  ex )  y  3.5.26

 f (x)  x [ f (x)]   2



(xex  ex ) yey  ey

 (10)  f (x)  x2 3[ f (x)]2  f (x) [ f (x)]3  2x  0. If x  1, we have

dx dx  2 2 3 f (1) 1 3[ f (1)]  f (1) [ f (1)]  2(1)  0  f (1) 12 3 22  f (1)  23  2  0  16 f (1) 12 f (1)  16  13 f (1)  16  f (1)   . 13

3.5.27

(x2 )  g(x)  x cos g(x) g(x) sin g(x) 1  2x. If x  0, we have dx dx g(0)  0  sin g(0)  2(0)  g(0)  sin 0  0  g(0)  0  0  g(0)  0. [g(x)  x sin g(x)] 

3.5.28

(x4 y2  x3 y  2xy3) 

(0)  x4  2y  y2  4x3x  (x3 1 y3x2  x)  2(x 3y 2  y3  x)  0

dy dy 3 2 2 3 4x y x  3x yx  2 y x  2x4 y  x3  6xy2  (4x3 y2  3x2 y  2y 3 )x  2x4 y  x3  6xy2  

x 

2x4 y  x3  6xy2 4x3 y2  3x2 y  2 y3

3.5.29

d ( y sec x)  (x tan y)  y sec x tan x  x  sec x 1  x sec2 y  tan y  x  dy dy y sec x tan x  x  tan y  x x sec2 y sec x  ( y sec x tan x  tan y)x x sec2 y sec x  dx x sec2 y sec x  x  dy y sec x tan x  tan y

384

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.30

x2 y  2x  15  x2 y  y2x  2  0  x2 y  2(xy 1)  y 

2(xy 1) x2

. When x  3 and y  1,

8 2(311) 2(4) 8    , so an equation of the tangent line is y   9 (x  3) 1 or 2 3 9 9 8 11 y  9 x 3 .

we have y 

3.5.31

y sin 2x  x cos 2 y  y cos 2x  2 sin 2x  y  x(2sin 2 y  2 y)  cos(2 y) 1  sin 2x  y  2x sin 2 y  y  2 y cos 2x  cos 2 y  y(sin 2x  2x sin 2 y)  2 y cos 2x  cos 2 y  y 

2 y cos 2x  cos 2 y

. When x   and y   , we have y 



 /2

0   1

sin 2x  2x sin 2 y equation of the tangent line is y  21 (x  2)  4 or y  12 x. 2

( / 2)(1)  0





, so an

3.5.32

sin(x  y)  2x  2 y  cos(x  y)  (1 y)  2  2 y  cos(x  y)  y 2 yy  2 cos(x  y)  2  cos(x  y) y[cos(x  y)  2]  2  cos(x  y)  y  . When x   and y   , we have cos(x  y)  2 2 1 1 y   , so an equation of the tangent line is y  1 (x   )   or y  1 x  2 . 3 3 3 1 2 3 





3.5.33

x2  xy  y2  1  2x  (xy  y 1)  2 yy  0  2x  xy y  2 yy  0  2x  y  xy 2yy  2x  y 4 1 3 2x  y  (x  2 y) y  y  . When x  2 and y  1, we have y   , so an equation of x  2y 22 4 the tangent line is y  3 (x  2) 1 or y  3 x  1 . 4

3.5.34

x2  2xy  4 y2  12  2x  2xy  2 y  8yy  0  2xy  8yy  2x  2 y  x  y . When x  2 and y  1, we have y   2 1   1 , so an (x  4 y) y  x  y  y   x  4y 24 2 equation of the tangent line is y   1 (x  2) 1 or y   1 x  2. 2

385

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.35

x2  y2  (2x2  2 y2  x)2  2x  2 yy  2(2x2  2 y2  x)(4x  4 yy 1). When x  0 and y  12 , we

have 0  y  2  12  (2 y 1)  y  2 y 1  y  1, so an equation of the tangent line is y  1(x  0)  1 or y  x  1 . 2

2/3

y

 



3.5.36 2/3

4

When x  3

2 3



1/3

2 3

1/3





3 y 1 y  0  y   . y  0   3  3 x 3y x



and y  1, we have y 

tangent line is y  13





  3 3



1/3

3 3 1 x  3 3 1 or y  3 x  4.



3

  3  1 , so an equation of the 2/3

3.5.37



2 x2  y2

  25(x  y )  4  x  y  (2x  2 yy)  25(2x  2 yy)  2

4(x  yy)(x2  y2 )  25(x  yy)  4 yy(x2  y2 )  25yy  25x  4x(x2  y2 )  25x  4x(x2  y2 ) y  . When x  3 and y  1, we have y  75120   4565   913, so an equation of the 2440 25y  4 y(x2  y2 ) tangent line is y   9 (x  3) 1or y   9 x  40 . 

3.5.38

y2 ( y2  4)  x2 (x2  5)  y4  4 y2  x4  5x2  4 y3 y  8yy  4x3  10x. When x  0 and y  2, we have 32 y 16 y  0  16 y  0  y  0, so an equation of the tangent line is y  0(x  0)  2 or y  2. 3.5.39

tan(xy)  x  sec2 (xy)[xy y y(1)]  1  x sec2 (xy) y  1 y sec2 (xy)  2 2 2 1 1 y sec (xy) cos (xy)  y cos2 ( xy) cos (xy)  y     , x   x sec2 (xy) 2 cos (xy)  cos ( xy) x 2

386



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.40

x2  3xy  y2  20  2x  3xy  3y  2 yy  0  y(3x  2 y)  3y  2x 

3(4)  2(6)  0 so the tangent line is horizontal and the equation . At the point6, 4  , y  3(6)  2(4) 3x  2 y of the tangent line is y  4. y 

3y  2x



3.5.41





x 12x y  y  15  2x 12x y  36x y  3yy  0  2x  36x2 y  12x3  3y 2 y  2

2x  36x y   y. At the point 1, 2  , the slope of the tangent line is undefined, and the tangent line is 12x3  3y2 vertical. Thus the equation of the tangent line is , x  1. 2

3.5.42

3x  xy  y2  9  3  (xy  y)  2 yy  0  xy 2 yy  3  y  (x  2 y) y  3  y  y 

3  y

x  2y

. When y  2, 3x  xy  y2  9  3x  x(2)  22  9 

5 5x  9  4  x  1. And when y  2, the slope of the tangent line is y  3  2    1. Then 1 2(2) 5  1  the slope of the normal line is    1,  1 



3.5.43

x  xy  y  7  2x  (xy  y)  2 yy  0  xy  2 yy  2x  y  2

(x  2 y) y  2x  y  y  5

2x  y

equal y. So   4 x  2y

2x  y

x  2y

. The slope at the point (2,1) is

2(2) 1



and this must

2  2(1) 4  5x 10 y  8x  4 y  6 y  3x  2 y  x. Substituting this for x in

x2  xy  y2  7 we have 4 y2  2 y2  y2  7  7 y2  7  y  1. Then x  2 y  2(1)  2, so the point on the curve with the same slope is 2, 1. 3.5.44

387



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

ey  sin x  ey

 cos x   dy  cos x  cos x  tan x, dx dx ey sin x 

3.5.45

y dy   d2y 1  dy  2 1/ 2 2 1/ 2  2  25  y 25  y2 2 2  (2 y)     (25  y )  (25  y )   y.  dx   dx dx 2   25  y d2y When y  3,   3, dx2 

388

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.46

3.5.47

3.5.48

x2  4 y2  4  2x  8yy  0  y  x / (4 y)  1 y 1 x  y 1 y  x[x / (4 y)] 1 4 y2  x2 1 4 since x and y must statisfy the  y"   4  4  4  4 4 y3  original equation x  y  4  y2 y2 y3   1 . Thus, y "   3 4y 2

389

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.49

x2  xy  y2  3  2x  xy y  2 yy  0  (x  2 y) y  2x  y  y 

2x  y

x  2y Differentiating 2x  xy  y  2yy  0 to find y gives 2  xy   y  y  2 yy   2 yy  0 

 2x  y  2x  y 2  (x  2 y) y   2  2 y  2( y)2  2 1      x  2 y  x  2 y   2 (x 2 2 y     2 y)  (2x  y)(x 2 2 y)  (2x  y)   x  2 y  (x  2 y)    2  (x2  4xy  4 y2  2x2  4xy  xy  2 y2  4x2  4xy  y2 ) 3 (x  2 y) 2 2  (3x2  3xy  3y2 )   (9) 3 (x  2 y)3 (x  2 y) since x and y must statisfy the   original equation x  xy  y  3   18 Thus, y   . (x  2 y)3 



3.5.50

sin y  cos x  1  cos y  y sin x  0  y 

sin x

 cos y cos y cos x sin x(sin y) y cos y cos x sin x sin y(sin x / cos y) y   (cos y)2 (cos y)2

cos2 y cos x  sin2 x sin y cos2 y cos x  sin2 x sin y  cos2 y cos y cos3 y Using sin y  cos x  1, the expression for y can be simplified to y  (cos2 x  sin y) / cos3 y. 

3.5.51

  

x3  y 3  7  3x2 3y2 y   0  y  

x 2  y2

2 2 2 2 3 2 3 3 y  y (2x)  x (2 yy)  2xy[ y  x(x / y )]  2x[ y  x / y )]  2x( y  x )  2x(7)  14x y3 y3 y5 y5 ( y2 )2 y4

390



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.52 If x  0 in xy  ey  e, then we get 0  e y  e, so y  1and the point where x  0 is 0,1. Differentiating implicitly with respect to x gives us xy  y 1 e y y  0. Substituting 0 for x and 1 for y gives us

0 1 ey  0  ey  1  y  1/ e. Differentiating xy  y 1 e y y  0 implicitly with respect to x gives us xy   y1 y  e y y   y e y y  0. Now substitute 0 for x, 1 for y and 1/ e for y. 2 1 1 1  1  1  1  1 0      ey    (e)   0    ey    0  ey    y   .  e  e  e   e  e e e e2         3.5.53



If x 1 in x2  xy  y3  1, then we get 1 y  y3  1  y3  y  0  y y2 1

  y  0 , so the

point where x 1 is 1, 0 . Differentiating implicitly with respect to x gives us

2x  xy  y 1 3y 2  y  0 . Substituting 1 for x and 0 for y gives us 2  y  0  0  0  y  2 . Differentiating 2x  xy  y  3y 2 y  0 implicitly with respect to x gives us





2  xy  y  1 y  3 y2 y  y 2 yy  0 . Now substitute 1 for x,0 for y , and 2 for y . 2 2  y  2  2  3  0  0  0  y   2 . Differentiating 2  xy  2 y  3y 2 y   6 y  y    0 implicitly with respect to x gives us 2 xy   y 1 2y   3  y 2 y   y  2yy   6  y  2yy    y  y  0 . Now substitute 1 for x,0 for   y, 2 for y , and 2 for y .y   2  4  3  0  0   6 0   8   0  y   2  4  48  42 .

3.5.54

391

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.5.55 (a)

(b) There are 9 points with horizontal tangents: 3 at x = 0, 3 at x = 1/2, and 3 at x = 1. The three horizontal tangents along the top of the wagon are hard to find, but by limiting the y-range of the graph (to [1.6, 1.7], for example) they are distinguishable.

3.5.56

x  . dx dx y dy dy x 2 2 0   . (b) x  y  25  2x  2 y dx dx y

(a) x2  y2  16  2x  2 y

0 

(c) The derivatives in parts (a) and (b) are the same, and would be the same for a circle of any positive radius. 3.5.57 From Exercise 37, a tangent to the lemniscate will be horizontal if

𝑦′ = 0 ⇒ 25𝑥 − 4𝑥(𝑥2 + 𝑦2) = 0 ⇒ 𝑥[25 − 4(𝑥2 + 𝑦2)] = 0 ⇒ 𝑥2 + 𝑦2 =

(1)

4 (Note that when x is 0, y is also 0, and there is no horizontal tangent at the origin.) Substituting 25/4 for 𝑥2 + 𝑦2 in the equation of the lemniscate

2(𝑥2 + 𝑦2)2 = 25(𝑥2 − 𝑦2) we get 𝑥2 − 𝑦2 =

25 8

(2)

Solving (1) and (2), we have 𝑥2 =

75 16

and 𝑦2 =

, so the four points are

5√3 (±

5 ,± ) . 4 4

392



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.5.58



b2 x       an equation of the tangent line at 1 0 y a2 y 2 22 a2 b2 a b y b x0 y0 y y02 x0 x x02 0 x , y y   (x  x )  y . gives  0 0  is    . Since 0 0 Multiplying both sides by 2 2 2 2 b a2 y0 b b a a2 2 2 xx y y x y  x0 , y0  lies on the ellipse, we have 0  0  0  0  1. a2 b2 a2 b2 



  2x

2 yy







3.5.59

2x 2 yy x2 y2    1  2  0  a2 b2 a2 b2 y  b x  an equation of the tangent line at  x0 , y0  is a2 y 2 y b x0 y y y2 x x x2 y (x  x0 )  y0. Multiplying both sides by 0 gives 0  0  0  0 . b2 a2 y0 b2 b2 a2 a2 Since  x , y  lies on the hyperbola, we have 0

x0 x



y0 y

 1.

3.5.60

x y  c 

y



2 x

y 2 y

 0  y  

y0 (x  x )  y . Now x  0  y  y  x0

 an equation of the tangent line at  x0 , y0  is

y0 (x )  y 

x0 y0 , so the y-intercept is

 y  x y . And y  0   y  y0 (x  x )  x  x  y0 x0  x  x  x y , so the x0 0 0 0 0 0 0 0 0 x0 y0

intercept is x0  x0



 

y0 . The sum of the intercepts is y0  x0 y0  x0  x0

 x0  2 x0 y0  y0 



 x  y    c   c. 2

3.5.61 If the circle has radius r, its equation is x2  y2  r2  2x  2 yy  0  y  

, so the slope of the y x0 1 y0 which is the tangent line at P  x , y  is  . The negative reciprocal of that slope is  0 0 x / y x , y 0

slope of OP, so the tangent line at P is perpendicular to the radius OP.

393

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 

3.5.62 



p1 p1 ( p/q) p1 p yq x p  qyq1 y px p1  y   px  px y  px x  x( p/ q)1 q1 q p qy qy qx q

3.5.63 x2  y2  r 2 is a circle with center O and ax  by  0 is a line through O [assume a and b are not both zero]. x2  y2  r 2  2x  2 yy  0  y  x / y, so the slope of the tangent line at P0  x0, y0  is x0 / y0 . The slope of the line OP0 is y0 / x0 which is the negative reciprocal of x0 / y0 . Hence, the curves are orthogonal, and the families of curves are orthogonal trajectories of each other. 3.5.64 The circles x2  y2  ax and x2  y2  by intersect at the origin where the tangents are vertical and horizontal [assume a and b are not both zero]. If P0  x0 , y0  is the other point of intersection, then

x02  y02  ax 0 (1) and x02  y 02  by 0 (2). Now x2  y2  ax  2x  2 yy  a  y  x2  y2  by  2x  2 yy  by  y 

a  2x

and

2x . Thus, the curves are orthogonal at  x0, y0   b  2y

a  2x0 b  2 y0  2ax  4x2  4y 2  2by  ax  by  2(x2  y2 ), which is true by (1) and   0 0 0 0 0 0 0 0 2y 2x 



(2). 3.5.65 y  cx2  y  2cx and x2  2 y2  k [assume k > 0]  2x  4 yy  0 

2 yy  x  y 

x



, so the curves are orthogonal if

2( y) 2(cx ) 2cx c  0, y  cx2  0 intersects x2  2 y2  k c  0. If then the horizontal line





2 2 orthogonally at  k , 0 , since the ellipse x  2 y  k has vertical tangents at

those two points. 3.5.66

394

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  ax3  y  3ax2 and x2  3y 2  b [assume b > 0]  2x  6 yy  0  3yy  x 

y  



 3

3( y)

3(ax )

3ax

, so the curves are orthogonal if a  0. If a  0, then the horizontal line





2 2 y  ax  0 intersects x  3y 2  b orthogonally at  b, 0 , since the ellipse x  3y  b has vertical 3

tangents at those two points. 3.5.67

3.5.68 −1

′

−2

′ ⟹ 𝑦 = −(𝑥 + 𝑐) and 𝑦 = 𝑎(𝑥 + 𝑘)3 ⇒ 𝑦 = 3 𝑎(𝑥 + 𝑘)3, so the curves are 1 orthogonal if the product of the slopes is -1, that is, − (𝑥+𝑐)2 · 𝑎 2 = −1 ⟹

𝑦 = (𝑥 + 𝑐)

3(𝑥+𝑘)3 1 2 𝑎 = 3(𝑥 + 𝑐)2(𝑥 + 𝑘)2/3 ⟹ 𝑎 = 3 ( ) 1

𝑦

𝑦 2

( )

[since 𝑦2 = (𝑥 + 𝑐)−2 and 𝑦2 = 𝑎^2(𝑥 + 𝑘)2/3]

𝑎

⇒ 𝑎 = 3 ( ) ⇒ 𝑎3 = 3 ⇒ 𝑎 = √3. 𝑎2

3.5.69



 n2a  n2a n3ab (a)  P  2 (V  nb)  nRT  PV  Pnb  V  V 2  nRT  V   d d (PV  Pnb  n2aV 1  n3abV 2 )  (nRT )  dP dP PV  V 1 nb  n2aV 2 V   2n3abV 3 V   0  V (P  n2aV 2  2n3abV 3 )  nb V  nb V dV V 3(nb V )  V  . 2 2 3 3 or 3 2 3 P  n aV  2n abV dP PV  n aV  2n ab (b) Using the last expression for dV/dP from part (a), we get

395



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(10 L)3[(1 mole)(0.04267 L/mole) 10 L] dV    dP (2.5)(10 L)3  (1 mole)2 (3.592 L2 -atm / mole2 )(10 L)   2(1 mole)3 (3.592 L2 -atm / mole2 )(0.04267 L/mole)   3 9957.33 L    4.04 L/atm. 2464.386541 L3-atm 

3.5.70 (a) x2  xy  y2 1  0  2x  xy  y 1 2 yy  0  0  y(x  2y)  2x  y 

y 

2x  y x  2y

(b) Plotting the curve in part (a) gives us an empty graph, that is, there are not points that satisfy the equation. If there were any points that satisfied the equation, then x and y would have opposite signs; otherwise, all the terms are positive and their sum cannot equal zero. x2  xy  y2 1  0  x2  2xy  y2  xy 1  0  (x  y)2  xy 1. The left side of the last equation is non-negative, but the right side is at most –1, so that proves that there are no points that satisfy the equation. Another solution: x2  xy  y2 1  12 x2  xy  12 y2  21 x2  21 y2 1

 21 (x2  2xy  y2 )  12 (x2  y2 ) 1  21 (x2  y2 )  21 (x2  y2 ) 1  1 Another solution: Viewing x2  xy  y2 1  0 as a quadratic in x, the discriminant is y2  4( y2 1) 1  3y2  4. This is negative, so there are no real solutions. (c) The expression for y in part (a) is meaningless; that is, because the equation in part (a) has no solution, it does not implicitly define a function y of x which makes it meaningless to consider y.

3.5.71 To find the points at which the ellipse x2  xy  y2  3 crosses the x-axis, let y  0 and solve for x. y  0  x2  x(0)  02  3  x   3. So the graph of the ellipse crosses the x-axis at the points

 3, 0. Using implicit differentiation to find ywe get 2x  xy  2 yy  0  y  2x 02 3 . So y at  3, 0 is y(2 y  x)  y  2x  y   2 and at  3, 0is 2y  x 2(0)  3 02 3 2(0)  3

 2. Thus the tangent lines at these points are parallel.

3.5.72

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.5.73

x2 y2  xy  2  x2  2 yy  y2  2x  x  y  y 1  0  y(2x2 y  x)  2xy2  y 

2xy2  y 2xy2  y 2 2 y   2 . So  2  1  2xy  y  2x y  x  y(2xy 1)  x(2xy 1)  2x y  x 2x y  x y(2xy 1)  x(2xy 1)  0  (2xy 1) ( y  x)  0  xy   1 or y  x. But xy   1  

x y  xy    2, so we must have x  y. Then x y  xy  2  x  x  2  2 2

1 4

2 2

1 2

x4  x2  2  0  (x2  2)(x2 1)  0. So x2  2, which is impossible, or x2  1  x 1. Since

x  y, the points on the curve where the tangent line has a slope of –1 are1, 1 and 1,1.



3.5.74

x2  4 y2  36  2x  8yy  0  y  

. Leta, b be a point on x2  4 y2  36 whose

tangent line passes through 12, 3. The tangent line is then

a a y   (x 12)  3 so b   (a 12)  3. Multiplying both sides by 4b 4b 4b gives 4b2 12b  a2 12a, so 4b2  a2  12(a  b). But 4b2  a2  36, so 36  12(a  b)  a  b  3 b  3  a. Substituting 2 2 3  a for b into a  4b  36 gives a2  4(3  a)2  36  a2  36  24a  4a2  36  5a2  24a  0  a(5a  24)  0, so a  0 or a  24 . If a  0, b  3 0  3, and if a  24 , b  3  24   9 . So the two points on the ellipse are 5

0, 3 and  245 ,  59 . Using y  4b(x 12)  3 witha, b  0, 3 gives us the tangent line y  3. Witha, b   24 ,  9  , we have y   24/5 (x 12)  3  y  (x 12)  3  y  x  5. A a

4(9/5)

graph of the ellipse and the tangent lines confirms our results. 3.5.75

397

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) y  J (x) and xy   y  xy  0  xJ (x)  J (x)  xJ (x)  0. If x  0, we have

0  J (x)  0  0, so J (x)  0. (b) Differentiating xy   y  xy  0 implicitly, we get xy   y 1 xy   xy  y 1  0 

xy   2 y   xy  y  0, so xJ (x)  2J (x)  xJ (x)  J (x)  0. If x  0, we have 0  2J (x)  0 1 [J(0) = 1 is given]  0  2J (0)  1  J (0)  12 . 3.5.76

x2  4 y2  5  2x  4(2 yy)  0  y  

. Now let h be the height of the lamp, and let a, b be

the point of tangency of the line passing through the points 3, h and 5, 0  . This line has slope (h  0) / [3  (5)]  18 h. But the slope of the tangent line through the point a, b can be expressed as

y  

a b   , or as b  0  b [since the line passes through 5, 0and a, b ], so  4b a  5 4b a  (5) a  5 a





4b2  a2  5a  a  4b  5a. But a  b  5 [sincea, b is on the ellipse], so 5  5a  2

a  1. Then 4b2  a2  5a  1 5(1)  4  b  1, since the point is on the top half of the ellipse. h b 1 1 So     h  2. Therefore the lamp is located 2 units above the x-axis. 8 a  5 1 5 4

398

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.6 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS



3.6.1  3   3     (a) sin1  sincesin  and is in  ,  . 3  2  3 3 2  2 2 



(b) cos

1





1   since cos  1 and  is in 0,  .

3.6.2  (a) tan1  1 

      since tan  1 and is in   ,  .  6 6 6 2 2  3 3            3  1 (b) sec 2  since sec is in 0, .  2 and  ,   3 3 3  2   2  



3.6.3       (a) arctan1  since tan is in  , .  1 and  2 2  4 4 4

    since sin  1 and is in  , . (b) sin1 1    4 4 2 2 4  2 2 







3.6.4 3    (a) tan1 tan  tan1 1     4  4  



1

 

2 

(b) cos arcsin

 cos

 



6  

3.6.5

399

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

2 .  3   

Let   sin1



Then tan sin

1



 2   tan  2 .  3  5   



3.6.6

Let   arccos .

5  5  3  Then cscarccos   csc  . 4 5    



3.6.7 1

Let   tan



2 . Then

1

sin 2tan



2  sin 2   2sin  2





 2  1   2 2   3 3 3   

3.6.8 Let x  tan12 and y  tan13 . Then





cos tan12  tan13  cos x  y   cosxcosy  sinxsiny

1 1 2 3  5 10 5 10 5 5 1    50 5 2 2



3.6.9   Let y  sin1x . Then   y  cosy  0 ,





so cos sin1x  cosy  1sin2 y  1 x2 . 3.6.10 1 Let y  sin x . Then siny  x , so from the triangle we see that





tan sin1x  tany 

x 1 x2

3.6.11 1 Let y  tan x . Then tany  x , so from the triangle we see that

400

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1





sin tan1x  siny 

x 1 x2

3.6.12 Let y  arccos x. Then cos y  x, so from the triangle we see that

sin arccos x  sin y  1 x2 .

3.6.13

The graph of sin1x is the reflection of the graph of sinx about the line y  x . 3.6.14

1

The graph of tan x is the reflection of the graph of tanx about the line y  x . 3.6.15 1 Let y  cos x . Then cosy  x and 0  y    siny

 1

1 dy  1 1  . [Note that siny  0 for0  y  .]    2 dx siny 1 x2 1 cos y

401

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.6.16 1

1

(a) Let a  sin x and b  cos x . Then cosa  1 sin a  1 x





a







since cosa  0 for

. Similarly, sinb  1 x2 . So



1

sin sin x  cos x

But 

  sin a  b  sinacosb  cosasinb  x  x  1 x2 1 x2  x2  1 x2   1

 sin1x  cos1x 

3

, and so1sin1x  cos1x 

2 1



2 1



(b) We differentiate sin x  cos x   with respect to x , and get

1 1 x2



cos1x  0  d cos1x  

1 1 x2



3.6.17

1 2 tan1 x y  (tan 1 x)2  y 2(tan 1 x)1  d (tan1 x)  2 tan1 x   dx 1 x2 1 x2 



3.6.18



d 1  (x2 )  1  2x  2x y  tan1(x2 )  y  1 x4 1 (x2 )2 dx 1 x4

3.6.19 y  sin1(2x 1) 

y 



1 (2x 1)

d dx

(2x 1) 

1 1 (4x  4x 1) 2

2 

2 4x  4x 2



1 x x 2

3.6.20 1

𝐹(𝜃) = arcsin √sin 𝜃 = arcsin(𝑠i𝑛 𝜃)2 ⇒ 1 𝑑 1 ′ 𝐹 (𝜃) =

(sin 𝜃)2 = · · (sin 𝜃) √1 − ( sin 𝜃 2 𝑑𝜃 √1 − sin 𝜃 2 ) √

cos 𝜃

−1 2 · cos 𝜃 =

2√1 − sin 𝜃 √sin 𝜃

3.6.21 𝐺(𝑥) = √1 − 𝑥2 arccos 𝑥 ⇒ −1 𝑥 arccos 𝑥 1 ′ 2 −1 + arccos 𝑥 · (1 − 𝑥 2) 2(−2𝑥) = −1 − 𝐺 (𝑥) = √1 − 𝑥 · √1 − 𝑥2 √1 − 𝑥2 2 3.6.22 𝑦 = cos−1(𝑒2𝑥) ⇒ © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

402

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

𝑦′ = −

2𝑥 · 𝑑 (𝑒2𝑥) = − 2𝑒 √1 − 𝑒4𝑥 √1 − (𝑒2𝑥)2 𝑑𝑥

3.6.23 𝑦 = tan−1 (𝑥 − √1 + 𝑥2) ⇒ 1 𝑦′ =

𝑥

√𝑥2 + 1 − 𝑥) 2 ( √𝑥 + 1

)= 1 + (𝑥 − √𝑥2 + 1) 2 (1 − 2 √𝑥 + 1 1 + 𝑥2 − 2𝑥√𝑥2 + 1 +2 𝑥2 + 1 √𝑥2 + 1 − 𝑥 √𝑥 + 1 − 𝑥 = = 2(1 + 𝑥2 − 𝑥√𝑥2 + 1)√𝑥2 + 1 2[√𝑥2 + 1(1 + 𝑥2) − 𝑥(𝑥2 + 1)] √𝑥2 + 1 − 𝑥 1 = = 2 2 2[(1 + 𝑥 )(√𝑥 + 1 − 𝑥)] 2(1 + 𝑥2) 3.6.24 𝑦 = arctan(cos 𝜃) ⇒ 𝑦′ =

1 (1 + cos 𝜃)2 (− sin 𝜃) = −

3.6.25

y  cos1(sin1 t)  y  



1 (sin1 t)2 dt

sin 𝜃 1 + cos2𝜃

(sin1 t)  

1 1 (sin1 t)2



1 1 t 2

3.6.26

R(t)  arcsin(1/ t)  1 1 1 1 1 R(t)   1    d1    t2   2 4  1 (1 / t) t t4  t2 1 (1 / t)2  1 (1 / t)2 dt t 1 1   t 2 (t 2 1) t t 2 1 3.6.27

y  x sin1 x  1 x2  1  (sin1 x)(1)  1 (1 x2 )1/2 (2x)  x x y  x   sin1 x   sin1 x 2 2 2 2 1 x 1 x 1 x 3.6.28

 1 x 1/2 1 x y  arctan  arctan    1 x  1 x 

403

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 

1/ 2

1/ 2

1  1  1 x   (1 x)(1)  (1 x)(1) d  1 x  1 x 2 1 x  y    (1 x)2  1 x  2 dx  1 x      1 1    1 x  1 x  1 1 1 x   2 1 x 1 (1 x)1/2  2         1/2 1 x 1 x 2  1 x (1 x)2 2 2 (1 x) (1 x)2  1 x 1 x  1  1  1

2(1 x)1/2 (1 x)1/2



2 1 x2

3.6.29

 b  a cos x  y  arccos   a    bcosx 1 (a  b cos x)(a sin x)  (b  a cos x)(b sin x) y   (a  b cos x)2  b  a cos x 2 1   a  b cos x  (a2  b2 ) sin x  2 2 a  b cos2 x  b2  a2 cos2 x a  b cos x 1





(a2  b2 ) sin x  a2  b2 sin x  a  b cos x sin x 1 cos2 x a  b cos x

1 a2  b2

But 0  x   , so sin x  sin x. Also a  b  0  b cos x  b  a, so a  b cos x  0.



Thus, y 

a2  b2 . a  b cos x

3.6.30 ƒ(𝑥) = arcsin(𝑒𝑥) ⇒ ƒ′(𝑥) =

· 𝑒𝑥 =

𝑒𝑥

√1 − (𝑒𝑥)2 √1 − (𝑒𝑥)2 Domain(ƒ) = {𝑥|−1 ≤ 𝑒𝑥 ≤ 1} = {𝑥|0 < 𝑒𝑥 ≤ 1} = (−∞, 0] Domain(ƒ′) = {𝑥|1 − 𝑒2𝑥 > 0} = {𝑥|𝑒2𝑥 < 1} = {𝑥|2𝑥 < 0} = (−∞, 0) 3.6.31 𝑔(𝑥) = cos−1(3 − 2𝑥) ⇒ 𝑔′(𝑥) = −

(−2) =

√1 − (3 − 2𝑥)2 √1 − (3 − 2𝑥)2 Domain(𝑔) = {𝑥|−1 ≤ 3 − 2𝑥 ≤ 1} = {𝑥|−4 ≤ −2𝑥 ≤ −2} = {𝑥|2 ≥ 𝑥 ≥ 1} = [1,2] Domain(𝑔′) = {𝑥|1 − (3 − 2𝑥)2 > 0} = {𝑥|(3 − 2𝑥)2 < 1} = {𝑥| |3 − 2𝑥| < 1} = {𝑥|−1 < 3 − 2𝑥 < 1}| = {𝑥|−4 < −2𝑥 < −2} = {𝑥|2 > 𝑥 > 1} = (1,2) 3.6.32

404

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

tan−1(𝑥𝑦) = 1 + 𝑥2𝑦 ⇒

(𝑥𝑦′ + 𝑦 · 1) = 0 + 𝑥2𝑦′ + 2𝑥𝑦 ⇒ 1 + 𝑥2𝑦2 𝑥 𝑦 𝑦′ ( − 𝑥 2) = 2𝑥𝑦 − ⇒ 2 2 1+𝑥 𝑦 1 + 𝑥2𝑦2 𝑦 2𝑥𝑦 − 1 + 𝑥2𝑦2 2𝑥𝑦(1 + 𝑥 2𝑦2 ) − 𝑦 𝑦(−1 − 2𝑥 − 2𝑥 3𝑦2) ′ 𝑦 = = = 𝑥(1 − 𝑥 − 𝑥3𝑦2) 𝑥 2 𝑥 − 𝑥2(1 + 𝑥2𝑦2) − 𝑥 1 + 𝑥2𝑦2 3.6.33 𝑔(𝑥) = 𝑥 sin−1(𝑥/4) + √16 − 𝑥2 ⇒ 𝑥 𝑥 𝑔′(𝑥) = sin−1(𝑥/4) + − = sin−1(𝑥/4) ⇒ 4√1 − (𝑥/4)2 √16 − 𝑥2 𝜋 𝑔′(2) = sin−1(1/2) = 6 3.6.34 𝑦 = 3 arccos(𝑥/2) ⇒ 𝑦′ = 3 [−

1 √1−(𝑥/2)2

] ( 1), so at (1, 𝜋), 𝑦 ′ = − 2

3 = −√3. An equation of the 2√1−1/4

tangent line is 𝑦 − 𝜋 = −√3(𝑥 − 1), or 𝑦 = −√3𝑥 + 𝜋 + √3. 3.6.35



f (x)  1 x2 arcsin x  1 1  arcsin x   1 (2x)  1 arcsin x . f (x)  1 x2  2 1 x2 1 x2 1 x2 Note that f   0 where the graph of f has a horizontal tangent. Also note that f  is negative when f is decreasing and f  is positive when f is increasing.

405

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.6.36



f (x)  arctan(x  x)  f (x)  2



(x2  x) 

2x 1 1 (x2  x)2

1 (x  x) dx  Note that f  0 where the graph of f has a horizontal tangent. Also note that f  is negative when f is decreasing and f  is positive when f is increasing. 2

3.6.37

lim sin−1 𝑥 = sin−1(−1) = −

𝑥→−1+

𝜋 2

3.6.38

Let 𝑡 =

1+𝑥2

1+2𝑥2

lim arccos (

𝑥→∞

1+𝑥2

. As 𝑥 → ∞, 𝑡 =

1+2𝑥

= 2

1/1+𝑥2 1/1+2𝑥2

1 + 𝑥2 1 + 2𝑥

→ . 2

) = lim arccos 𝑡 = arccos 2 𝑡→1/2

1 2

𝜋 3

3.6.39

Let 𝑡 = 𝑒 𝑥 . As 𝑥 → ∞, 𝑡 = ∞. 𝜋 lim arctan(𝑒𝑥) = lim arctan 𝑡 = . 𝑥→∞ 𝑡→∞ 2

3.6.40

Let 𝑡 = ln 𝑥. As 𝑥 → 0+, 𝑡 → −∞. 𝜋 lim tan−1(ln 𝑥) = lim tan−1 𝑡 = − . 𝑥→0+ 𝑡→−∞ 2

406

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.6.41 (a) If 𝑦 = ƒ−1(𝑥), then ƒ(𝑦) = 𝑥. Differentiating implicitly with respect to x and remembering that y is a function of x, we get 𝑑𝑦 𝑑𝑦 1 1 ƒ′(𝑦) = 1, so = ⇒ (ƒ−1)′(𝑥) = . ′ −1 𝑑𝑥 𝑑𝑥 ƒ′(𝑥) (𝑥)) ƒ (ƒ (b) ƒ(4) = 5 ⇒ ƒ−1(5) = 4. By part (a), (ƒ−1)′(5) =

ƒ′(ƒ −1(5))

= 1/ ( 3 ƒ′(4)

) = 3/2.

3.6.42 (a) Since ƒ(𝑥) = 2𝑥 + cos 𝑥 is one-to-one, ƒ−1(1) = 𝑘 - ƒ(𝑘) = 1. By inspection, we see that ƒ(0) = 2(0) + cos 0 = 1, so 𝑘 = ƒ−1(1) = 0. (b) 1 1 1 1 (ƒ−1)′(1) = =2 = = ′ − sin 0 2 ƒ′(ƒ−1(1)) ƒ (0)

3.6.43 (a) Let ƒ(𝑥) = sin 𝑥, so ƒ−1(𝑥) = sin−1 𝑥 and ƒ′(𝑥) = cos 𝑥. 1 1 𝑑 1 1 (sin−1 𝑥) = (ƒ−1)′(𝑥) = ⇒ = = −1 𝑑𝑥 cos(sin 𝑥) ƒ′(ƒ−1(𝑥)) √1 − 𝑥2/1 √1 − 𝑥2 which is Formula 1.

(b) Let ƒ(𝑥) = tan 𝑥, so ƒ−1(𝑥) = tan−1 𝑥 and ƒ′(𝑥) = sec2 𝑥. 1 (ƒ−1)′(𝑥) = ⇒ ƒ′(ƒ−1(𝑥)) 1 1 1 1 𝑑 (tan−1 𝑥) = = = = 2 2 −1 2 −1 [sec(tan 𝑥)] 1 + 𝑥2 sec (tan 𝑥) 𝑑𝑥 (√1 + 𝑥2) which is Formula 4.

407

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.6.44 (a) ƒ(𝑥) = sin(𝑠i𝑛−1 𝑥)

(b) 𝑔(𝑥) = sin−1(sin 𝑥)

cos 𝑥

cos 𝑥 |cos 𝑥| 𝜋 𝜋 Notice that ℎ(𝑥) = − 𝑔(𝑥) because sin−1 𝑡 + cos−1 𝑡 = for all t. ℎ′(𝑥) = −

√1 − sin2 𝑥

=−

408

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.7 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS

 

3.7.1 The differentiation formula for logarithmic functions,

d dx

because ln e  1.

(loga x) 

1 , is simplest when a  e x ln a



3.7.2



f (x)  x ln x  x  f (x)  x 

1  (ln x) 11  1 ln x 1  ln x x

3.7.3

f (x) 

ln x 2

x2   f (x) 

1  ln x(2x) x

x  2



x  2x ln x x(1 2 ln x) 1 2 ln x   x4 x4 x3

3.7.4

f (x)  sin(ln x)  f (x)  cos(ln x) 

1 cos(ln x) ln x  cos(ln x)   dx x x

3.7.5

f (x)  ln(sin2 x)  ln(sin x)2  2ln sin x  f (x)  2 

1 cos x  2cot x sin x

409

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.7.6

1 1 d 1 1 f (x)  ln  f (x)   1  x 1/ x dx  x   x   x2    x     1 1 Another solution: f (x)  ln  ln1 ln x  ln x  f (x)   x x 



3.7.7

y

 (ln x)1  y  1(ln x)2 

ln x



x(ln x)2

3.7.8

f (x)  log10 (1 cos x)  f (x) 

sin x 1 d  (1 cos x)   (1 cos x) ln10 dx (1 cos x) ln10

3.7.9

f (x)  log 10

d 1 1 1  x  1   x ln10 2 x 2(ln10)x x ln10 dx 1 1 1 1/2  x  log x  1 log x  f (x)   10 10 2 2 x ln10 2(ln10)x

(x)  x f

Or: f (x)  log 10

3.7.10

g(x)  ln(xe2x )  ln x  ln e2x  ln x  2x  g(x) 

2

x 3.7.11

g(t)  1 ln t  g(t)  12 (1 ln t)1/2

(1 ln t) 

1 1 1   2 1 ln t t 2t 1 ln t

3.7.12

1 2sin t   F (t)  (ln t)2 sin t  F (t)  (ln t)2 cos t  sin t  2 ln t   ln t ln t cos t    

3.7.13







 1 h(x)  ln x  x 2 1  h(x)  1 x  x2 1  

3.7.14

G( y)  ln

(2 y 1)5 y2 1

1/ 2

 ln(2 y 1)  ln( y 1)

 x

 



1  2 x   x  x2 1  1

x2 1  x  x 1 2

1 x 1 2

 5ln(2 y 1)  12 ln( y 1) 

410

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 



G( y)  5

2 y 1

 2   2  2y   2 y 1 2 y 1 y2 1

8y 2  y 10



(2 y 1)( y2 1)

3.7.15

P(v) 

ln v 1 v

 P(v) 

(1 v)(1/ v)  (ln v)(1) v 1 v  v ln v   (1 v)2 v v(1 v)2

3.7.16

F(s)  ln ln s  F(s) 

1 d

ln s  1  1  1 ln s ds ln s s s ln s

3.7.17

d 1 3t2 3 y  ln 1 t  t3  y  1 t1 3  (1 t  t ) 1 t  t3 t dt 3.7.18

T (z)  2z log 2 z  T (z)  2z Note that log2 z(ln 2)  z

ln z

1 ln 2

 1  1  log2 z  2z ln 2  2z   log 2 z(ln 2)  z ln 2 z ln 2   

(ln 2)  ln z by the Change of Base Theorem. 

Thus, T (z)  2  z ln 2  ln z 





3.7.19

y  ln(csc x  cot x)  1 csc x(csc x cot x) 1 d (csc x cot x  csc2 x)   csc x y   (csc x  cot x)  csc x cot x dx csc x cot x csc x cot x 3.7.20



y  ln e x  xex

  ln(ex (1 x))  ln(ex )  ln(1 x)  x  ln(1 x) 

1 1 x 1  x   y  1 1 x 1 x 1 x 3.7.21 1/ 2

2 2  a2  z2  1  2  2  H (z)  ln a  z  ln    ln  a z   1 ln(a 2 z 2)  1 ln(a 2 z 2)   2 2 2 a2  z2 2  a2  z2  2 a z  





411



 

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

H (z)  

  (2z)   2 2  (2z)  2 2  2  2 a2  z2 2 a z z a z  a2

z(z 2  a2 )  z(z 2  a2 ) (z 2  a2 )(z2  a2 )

z3  za2  z3  za2 )  2a2 z  4 (z 2  a2 )(z2  a2 ) z  a4

3.7.22

y  tan[ln(ax  b)]  y  sec2[ln(ax  b)]

 a  sec2[ln(ax  b)]

ax  b

a ax  b

3.7.23

y  log2 (x log5 x)  1 d  1 1  log 5 x   y  (x log 5 x)  1 1   (x log x)(ln 2) dx (x log x)(ln 2) x  x ln 5 (x log x)(ln 2)(ln 5)  x(ln 2)   5 5 5 Note that log5 x(ln 5) 

ln x

(ln 5)  ln x by the Change of Base theorem. ln 5 1  1  1 ln x   Thus, y  x ln x ln 2 x(ln 2) x ln x ln 2 3.7.24





f (x)  2sin 2x  ln( f (x))  ln 2sin 2x  sin 2x  ln 2  f (x)  ln 2  2cos 2x  f (x)  f (x) ln 2  2cos 2x  2sin2x ln 2  2cos 2x  f (x)



3.7.25

f (x)  3x x3  ln( f (x))  ln 3x x3   ln 3x  ln x3  x ln 3  3ln x  f (x) 1 3 x 3 3 x 2   ln 3  3  f (x)  f (x) ln 3   3 x ln 3   3 x  x ln 3  3     f (x) x x x       

3.7.26 𝑦 = [ln(1 + 𝑒𝑥)]2 ⇒ 𝑦′ = 2[ln(1 + 𝑒𝑥)] ·

1 2𝑒𝑥 ln(1 + 𝑒 𝑥 ) 𝑥 = · 𝑒 1 + 𝑒𝑥 1 + 𝑒𝑥

412

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 

3.7.27

x2  4  e2x  12 ln(x2  4)  g(x)  e2x  12 ln(x2  4) 2x  x 2x  1 ln(x 2  4)e2x  2  xe 1  ln(x 2  4)e2x  e 2 x  2  ln(x2  4)  2  2 e2 x 2  x 4 2 x 4  x 4  g(x)  e2x ln

3.7.28 

y  (x2  2)2 (x4  4)4  ln y  ln[(x2  2)2 (x4  4)4 ]  ln y  2 ln(x2  2)  4 ln(x4  4)   4x 16x3  1 1 1 3  y  2    2x  4 4 4x  y  y  2  4  y x2  2 x 4 x  4  x  2 3 4x 2 2 4 4 16x    y  (x  2) (x  4)  4   2 x  2  x  4  3.7.29 



ex cos2 x ex cos2 x  ln y  ln 2  y 2 x  x 1 x  x 1 ln y  ln ex  ln cos x 2  ln(x2  x 1)  x  2 ln cos x  ln(x2  x 1)  1 1 1   y  1 2 (sin x)  (2x 1)  y  y 1 2 tan x  2x 1   y cos x x2  x 1   x2  x 1  x 2 e cos x    2x 1   y  2 1 2 tan x  2  x  x 1  x  x 1  



3.7.30

1 1  x 1 1/2 x 1 y 4  ln y  ln  4   ln y  ln(x 1)  ln(x4 1)  x 1 2  x 1   1 2 3 3   x 1  1  2x 2x 1 1 1 1 41 4x 3  y  y   4 4 4    y    y 2 x 1 2 x 1 2(x 1) x 1 x 1 2x  2 x 1      





3.7.31

y  xex 2 x (x 1)2/3  ln y  ln x1/2ex 2 x (x 1)2/3     1 1 1 2 1 2 1 2 ln y  ln x  ln(x  x)  ln(x 1)  y    2x 1   2 3 y 2 x 3 x 1 2  2   1 x2  x 2/3  1 y  y   2x 1  y  xe (x 1)   2x 1   2x 3x  3 2x 3x  3      





3.7.32

413



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  xx  ln y  ln xx  ln y  x ln x  y / y  x(1/ x)  (ln x) 1  y  y(1 ln x)  y  xx (1 ln x) 3.7.33

1 y  cos x   ln x  (sin x)  y x  cos x   cos x  cos x y  y  ln x sin x  x  ln x sin x  y  xcos x  ln y  ln xcos x  ln y  cos x ln x 

 

 

 

 

3.7.34

y 1 y  xsin x  ln y  ln xsin x  ln y  sin x ln x   (sin x)   (ln x)(cos x)  y x  sin x   sin x  y  y  ln x cos x  xsin x  ln x cos x   x 

 

 x 

 

3.7.35 x

1/2 y  x x  ln y  ln x  ln y  x ln x  ln y  1 x ln x  2

y  



1 1 x   ln x   2 x 2 

y  y  12  12 ln x  12 x (1 ln x) x

3.7.36

y  (cos x)x  ln y  ln(cos x)x  ln y  x ln cos x 

y  x  1  (sin x)  ln cos x 1  cos x y

x sin x   y  y ln cos x   (cos x)x ln cos x  x tan x  

cos x  

3.7.37

y  (sin x)ln x  ln y  ln(sin x)ln x  ln y  ln x  lnsin x 

y y

 ln x 

cos x  lnsin x 

sin x



cos x ln sin x  ln sin x    y  y ln x   (sin x)ln x ln x cot x    

sin x

 

 

 

3.7.38

y  (tan x)1/x  ln y  ln(tan x)1/x  ln y 

tan x 

414

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

ln tan x  y   2  y  x x tan x  2   x  ln tan ln tan x   1/ x 1  1/ x  sec x     csc x sec x    2 x or y  (tan x) x y  (tan x)   x x tan x      y  

 sec2 x

 1   x  ln tan x x tan x sec   x2   y 2



3.7.39

y  (ln x)cos x  ln y  cos x ln(ln x)  y  (ln x)

cos x  cos x

 sin x ln ln x  x ln x   

y y

 cos x 



 (ln ln x)(sin x) 

ln x x

3.7.40 𝑦 = 𝑥2 ln(2𝑥) ⇒ 𝑦′ = 𝑥2 ·

1 2𝑥

𝑦′′ = 1 + 2𝑥 ·

· 2 + ln(2𝑥) · (2𝑥) = 𝑥 + 2𝑥 ln(2𝑥) ⇒

1 2𝑥

· 2 + ln(2𝑥) · 2 = 1 + 2 + 2 ln(2𝑥) = 3 + 2 ln(2𝑥)

3.7.41 𝑦=

ln 𝑥 𝑥2

𝑦′′ =

3.7.42



⇒ 𝑦′ =

1 𝑥2 ( ) − (ln 𝑥 )(2𝑥) 𝑥(1 − 2 ln 𝑥) 1 − 2 ln 𝑥 𝑥 = = ⇒ 𝑥4 𝑥3 (𝑥2)2

𝑥3(−2/𝑥) − (1 − 2 ln 𝑥)(3𝑥2) 𝑥2(−2 − 3 + 6 ln 𝑥) 6 ln 𝑥 − 5 = = (𝑥3)2 𝑥4 𝑥6

 1

1 2  ln x  y  x ln x  y  x   (ln x)   x 2 x 2 x 2 x (1 / x)  (2  ln x)(1 / y  2 x (1 / x)  (2  ln x)(1 / x ) 

2 x 

 2  (2  ln x)   ln x x (4x)

4x x

3.7.43

y

ln x (1 ln x)(1/ x)  (ln x)(1/ x)  1   y    2 2 (1 ln x) 1 ln x x(1 ln x)

415



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

d [x(1 ln x)2 ] x  2(1 ln x)  (1/ x)  (1 ln x)2 y  dx   [Reciprocal Rule] x2 (1 ln x)4 [x(1 ln x)2 ]2 (1 ln x)[2  (1 ln x)]  3  ln x   2 2 4 x (1 ln x) x (1 ln x)3 3.7.44

y  ln sec x  y 

(sec x) 

sec x dx

sec x tan x  tan x  y   sec2 x

sec x

3.7.45

y  ln(1 ln x)  y 

1 1 1     1 ln x x x(1 ln x) 



d [x(1 ln x)] [Reciprocal Rule] y  dx [x(1 ln x)]2 x(1/ x)  (1 ln x)(1)  11 ln x 2  ln x   2  2 2 2 2 x (1 ln x)2 x (1 ln x) x (1 ln x)



3.7.46



x f (x)   1 ln(x 1) x

f (x) 

1 ln(x 1) 

 

1 (x 1)[1 ln(x 1)]  x x 1 (x 1) ln(x 1)  x x 1  x 1  2 2 [1 ln(x 1)] [1 ln(x 1)] [1 ln(x 1)]2

[1 ln(x 1)]1 x 



2x 1 (x 1) ln(x 1) (x 1)[1 ln(x 1)]2

Dom( f )  x | x 1  0 and 1 ln(x 1)  0  x | x  1 and ln(x 1)  1





 x | x  1 and x 1  e1  x | x  1 and x  1 e  1,1 e  (1 e, ) 3.7.47

 



f (x)  2  ln x  (2  ln x)1/2  f (x) 

(2  ln x)1/2 



1 2x 2  ln x 2 Dom( f )   x | 2  ln x  0   x | ln x  2    x | ln x  e   e 2 ,   2

3.7.48

416

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (x)  ln(x2  2x)  f (x) 

(2x  2) 

x2  2x Dom( f )  x | x(x  2)  0  , 0   2,  



x(x  2)



3.7.49

2(x 1)



f (x)  ln ln ln x  f (x) 

1 1 1   ln ln x ln x x Dom( f )  x | ln ln x  0  x | ln x  0  x | x  e  e,  3.7.50

1 d (x  ln x)  1 1 1  x  ln x dx x  ln x x    1 (11)  1 2  2. Substitute 1 for x to get f (1)  1 1 1   1 ln1 1 1 0   f (x)  ln(x  ln x)  f (x) 







3.7.51

f (x)  cos(ln x2 )  f (x)  sin(ln x2 )

ln x 2  sin(ln x2 ) 1 2x   2sin(ln x ) dx x2 x

3.7.52 Substitute 1 for x to get f (1)  

2sin(ln12)  2sin 0  0. 1



3.7.53



1 y  ln(x2  3x 1)  y  2  (2x  3)  y(3)  113  3, so an equation of the tangent x  3x 1 line at (3, 0) is y  3(x  3)  0 or y  3x  9. 3.7.54 𝑦 = ln(𝑥𝑒

𝑥2

) = ln 𝑥 + ln 𝑒 𝑥 = ln 𝑥 + 𝑥2 ⇒ 𝑦′ = 𝑥 + 2𝑥. At (1, 1), the slope of the tangent line is

𝑦′(1) = 1 + 2 = 3, and an equation of the tangent line is 𝑦 − 1 = 3(𝑥 − 1), or 𝑦 = 3𝑥 − 2.

3.7.55

417

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.7.56

3.7.57 2 f  x  2ln x  f  x  . An equation of the tangent line at a, f  a    a, 2 ln a  is x 2 2 y   x  a  2ln a or y  x  2ln a  2. We know that the y-intercept is 4, so a a 4  2ln a  2  6  2ln a  3  ln a  e3  a. Thus the slope of the tangent line is 2  2 . a e3

418

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.7.58 f  x 

 1  x ln x   0  1 x   ln x 1    f  x   x

x ln x horizontal when f  x  

1 ln x

 x ln x 



1 ln x . The tangent line is

 x ln x 

1  0  1 ln x  0  ln x  1  x  . 2 e  x ln x 

e 1 1  e, so the point on the graph at which the tangent line is horizontal is f    1 1 e ln 1   e e 1   e , e .   

 3.7.59 1 x   ln x 1 ln x g  x   g  x  x  . The tangent line is horizontal when x2 x2 x ln e 1  , so the point on the graph at which the tangent 1ln x  0 1  ln x  x  e. g e  e e  1 line is horizontal is  e, . This is the only such point. e   ln x



 3.7.60

1 2  x 2 ln x    ln x  1 2   2 ln x  ln x  ln x  2  ln x. The tangent x  g  x ln x  g  x    x2 x2 x x2 line to the graph of g is horizontal when ln x 2  ln x 4 g x  0   0  ln x 2  ln x  0  x  1 or x  e2. g 1  0 and g e2  ,   2 x2  2 4 e so the points on the graph of g at which the tangent line is horizontal are 1, 0 and  e , 2 . e   2

419



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.7.61 f (x)  cx  ln(cos x)  f (x)  c  

6  f (4 )  c  tan 4  c 1  7  c.



3.7.62



sin x

 c  tan x. Then

cos x

2



2 2 . Then 3  f (1)   ln b   b  e2/3. f (x)  log (3x )  f (x)  6x  1  b 3x2 ln b x ln b ln b 3 2



420

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.7.63 s t   ln 1 t2   v t   st  

2t . The acceleration is 1 t2 2 2 a t   vt   1 t   2   22t 2t  2  2t 22 2 1 t 2. The acceleration is zero when  1 t 2  1 t 2  1 t 2  



2 1 t 2 

1 t 

2 2



 0  1 t 2  0  t  1since t  0. At this time, the particle’s position is





s 1  ln 112  ln 2.

3.7.64 1 1 1 2 1 s t    ln t  v t   st     . The acceleration is a t   vt    . The t3 t 2 t t2 t 2 1 2t acceleration is zero when   0   0  2  t  0  t  2. 3 2 3 t t t 3.7.65 y  ln(x2  y2 )  y 

(x2  y2 )  y  

2x  2 yy

 x2 y  y2 y  2x  2yy 



x2  y2 dx x2  y2 2x x2 y  y2 y  2 yy  2x  (x2  y2  2y) y  2x  y  x2  y2  2y 3.7.66 x y  yx  y ln x  x ln y  y 

1 x

y 

 (ln x)  y  x 

 y ln y  yln x 

y  ln y 



ln y  y / x ln x  x / y

3.7.67 ƒ(𝑥) = ln(𝑥 − 1) ⇒ ƒ′(𝑥) =

= (𝑥 − 1)−1 ⇒ ƒ′′(𝑥) = −(𝑥 − 1)−2 ⇒ ƒ′′′(𝑥) = 2(𝑥 − 1)−3 𝑥−1 ⇒ ƒ(4)(𝑥) = −2 · 3(𝑥 − 1)−4 ⇒ ⋯ ⇒

ƒ(𝑛)(𝑥) = (−1)𝑛−1 · 2 · 3 · 4 · … · (𝑛 − 1)(𝑥 − 1)−𝑛 = (−1)𝑛−1

(𝑛 − 1)! (𝑥 − 1)𝑛

421

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.7.68

y  x8 ln x, s0 D9 y  D8 y  D8 (8x7 ln x  x7 ). But the eighth derivative of x7 is 0, so now we have D8 (8x7 ln x)  D7 (8 7x6 ln x  8x6 )  D7 (8 7x6 ln x)  D6 (8 7  6x5 ln x)  D(8!x0 ln x)  8!/ x 3.7.69 1 , so f (0)  1. 1 x ln(1 x) f (x) f (x)  f (0) Thus, lim  lim  lim  f (0)  1. x0 x0 x0 x x x0 If f (x)  ln(1  x), then f (x) 

3.7.70 Let m  n / x. Then n  xm, and as n  , m  . n

1   x Therefore, lim1   lim 1  n  m  m  n 

m   1   lim 1    ex.  m  m  

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.8 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.8.1 (a) s  f (t)  t3  8t2  24t ft  v(t)  3t 2 16t  24 ft/s (b) v(1)  3(1)2 16(1)  24  11 ft/s

422

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



t

  16  

 16   4(3)(24) 16  32 2

2(3)

. The negative discriminant indicates that v is

never 0 and that the particle never rests. (d) From parts (b) and (c), we see that v(t)  0 for all t, so the particle is always moving in the positive direction. (e) The total distance traveled during the first 6 seconds (since the particle doesn’t change direction) is f (6)  f (0)  72  0  72 ft. (f) (f)

v(t)  3t2 16t  24  a(t)  v(t)  6t 16 ft/s2

(g)(g)

a(1)  6(1) 16  10 ft/s2 (h) (h)

(i) The particle is speeding up when v and a have the same sign. The velocity v is always positive and a is positive when 6t 16  0  t  8 , so the particle is speeding up when t  8 . It is slowing 3

down when v and a have opposite signs; that is when 0  t  . 8 3



3.8.2 (a) s  f (t) 

 t 2  9(9)  9t 2t  9t2  81 9 t 2  9  ft  v(t)  f (t)    ft/s

9t 

t2  9 (b) v(1) 



 t  9 2

91 9 72  0.72 ft/s

1 9 



 t  9 

t 2  9

100

 9t 2  9 (c) The particle is at rest when v(t)   0  t2  9  0  t  3 s [since t  0]. 2 2 t  9   (d) The particle is moving in the positive direction when v(t)  0.

9  t 2  9

t 2  9

2 2  0   9  t 2  9  0  t  9  0  t  9  0  t  3.

(e) Since the particle is moving in both the positive and negative directions, we need to calculate the distance traveled in the intervals0, 3 and3, 6, respectively. f (3)  f (0)  27  0  3 ; f (6)  f (3)  54  27  3 . 18

423

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 3 The total distance traveled is 32  10  9 5or 1.8 ft.

(f) (f)

(g)(g) v(t)  9

t  9  a(t)  v(t)  9  t  9 (2t)   t  9  2 t  9 (2t)  t  9  t  9     2



 9



   t  9   









2 2



2t t  9  t  9  2 t  9  18t t 2  27 ft/s2    2

2 2

a(1) 

 18t 26



t  9 2



 0.468 ft/s2

10 (h)

(i) The particle is speeding up when v and a have the same sign. The acceleration a is always negative for 0  t 

27  5.2 s, so from the figure in part (h), we see that v and a are both negative

for 3  t  3 3. The particle is slowing down when v and a have opposite signs. This occurs when

0  t  3 and when t  3 3. 3.8.3

  t   t  ft  v(t)  f (t) cos ft/s     2    2  2   t   (b) v(1)  cos   (0)  0 ft/s 2 2 2     t   t   0  cos    0 (c) The particle is at rest when v(t)  cos t   2  2  n   2  2 2      t    n  t  1 2n, where n is a non-negative integer since t ≥ 0. 2 2 (d) The particle is moving in the positive direction when v(t)  0. From part (c), we see that v changes sign at every positive odd integer. The velocity is positive when 0  t  1, 3  t  5, 7  t  9, and so on. (a) s  f (t)  sin



(e) The velocity changes sign at t = 1, 3, and 5 in the interval0, 6. The total distance traveled during the first 6 seconds is

424

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (1)  f (0)  f (3)  f (1)  f (5)  f (3)  f (6)  f (5)

 1 0  11  1 (1)  0 1

 1 2  2 1  6 ft. s (f)

(g)



 2   t    t     t  2  v(t)  cos   a(t)  v (t)   sin       sin  ft/s   2 2 2 2 2 4 2 2 2   2 a(1)   sin     ft/s 4 4 2

 

(h)

(i) The particle is speeding up when v and a have the same sign. From the figure in part (h), we see that v and a are both positive when 3  t  4 and both negative when1  t  2 and 5  t  6. Thus, the particle is speeding up1  t  2, 3  t  4, and 5  t  6. The particle is slowing down when v and a have opposite signs. This occurs when 0  t  1, 2  t  3, and 4  t  5. 3.8.4

 (a) s  f (t0  t 2et in feet  v(t)  f (t)  t 2 (et )  et (2t)  tet 2  t  (in ft/s) (b) v(1)  (1)e1(2 1)  e1 ft/s. (c) The particle is at rest when v(t)  0  t  0 or 2 s. (d) The particle is moving in the positive directions when v(t)  0  tet 2  t   0 

t(2  t)  0  0  t  2. (e) The velocity changes sign at t = 2 in the interval0, 6. The total distance traveled during the first 6 seconds is f (2)  f (0)  f (6)  f (2)  4e2  0  36e6  4e2  4e2  4e2  36e6

 8e2  36e6  0.99 ft. (f)(f)

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



 et  a(t)  v(t)   2t  t 2  e t   et  2  2t   et   2t  t 2    2  2t   e t t 2  4t  2  ft/s .  

(g) v(t)  2t  t

a(1)  e1 1 4  2  e1 ft/s2. (h)

a(t)  0  t 2  4t  2  0 e t  0 

(i)

t

4  8  2  2  0.6 and 3.4. The particle is speeding 2

up when v and a have the same sign. Using the previous information and the figure in part (h), we see that v and a are both positive when 0  t  2  2 and 2  t  2  2. both negative when The particle is slowing down when v and a have opposite signs. This occurs when 2  2  t  2 and

t  2  2.



3.8.5

v  0 on 0, 4. a  0 on 0,1 and 3, 4  , and a  0 on 1, 3. The particle is speeding

(b)

up on 0,1 and 3, 4 v  0, a  0. The particle is slowing down on 1, 3

v  0, a  0. 3.8.6 (a) The velocity v is positive when s is increasing, that is, on the intervals 0,1 and3, 4; and it is negative when s is decreasing, that is, on the interval 1, 3. The acceleration a is positive when the graph of s is concave upward (v is increasing), that is, on the interval 2, 4; and it is negative when the graph of s is concave downward (v is decreasing), that is, on the interval 0, 2. The particle is speeding up on the interval 1, 2 and on3, 4  . The particle is slowing down on the interval 0,1

v  0, a  0 and on 2, 3 v  0, a  0. (b) The velocity v is positive on 3, 4and negative on 0, 3. The acceleration a is positive on 0,1 and 2, 4 and negative on 1, 2  . The particle is speeding up on the interval 1, 2 v  0, a  0 and on 3, 4 v  0, a  0 The particle is slowing down on the interval 0,1 v  0, a  0 and on 2, 3 v  0, a  0. 3.8.7

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) h(t)  2  24.5t  4.9t 2  v(t)  h(t)  24.5  9.8t. The velocity two seconds is v(2)  24.5  9.8(2)  4.9 m/s and after four seconds is v(4)  24.5  9.8(4)  14.7 m/s. (b) The projectile reaches its maximum height when the velocity is zero.

v(t)  0  24.5  9.8t  0  t 

24.5

 2.5 s.

9.8

(c) The maximum height occurs when t = 2.5 s. h(2.5)  2  24.5(2.5)  4.9(2.5)2  32.625 m. (d) The projectile hits the ground when h  0  2  24.5t  4.9t2  0 

t



24.5  24.52  4(4.9)(2)  t  t f  5.080 s [ since t ≥ 0]. 24.9

   24.5  9.8t  25.3 m/s

(e) The projectile hits the ground when t  t f . Its velocity is v t f

[downward].

3.8.8 (a) At maximum height the velocity of the ball is 0 ft/s. v(t)  s(t)  80  32t  0  32t  80  t  52 . So the maximum height is

s  52   80  52  16  52   200 100  100 ft. 2





(b) s(t)  80t 16t 2  96  16t 2  80t  96  0  16 t 2  5t  6  0  16(t  3)(t  2)  0. So the ball has a height of 96 ft on the way up at t = 2 and on the way down at t = 3. At these times the velocities are v(2)  80  32(2)  16 ft/s and v(3)  80  32(3)  16 ft/s, respectively.

3.8.9 (a) h(t)  15t 1.86t2  v(t)  h(t)  15  3.72t. The velocity after 2 seconds is v(2)  15  3.72(2)  7.56 m/s. (b) 25  h  1.86t 2 15t  25  0  t 

15  152  4(1.86)(25)  t  t1  2.353 or 2(1.86)

t  t2  5.711s. The velocities are v t1   15  3.27t1  6.24 m/s [upward] and v t2   15  3.27t2  6.245 m/s or 6.245 m/s downward.

3.8.10 (a) s(t)  t 4  4t3  20t 2  20t s  v(t)  s(t)  4t3 12t 2  40t  20





v  20  4t3 12t 2  40t  20  20  4t3 12t 2  40t  0  4t t 2  3t 10  0  4t(t  5)(t  2)  0  t  0 s or t  5 s for t  0. (b) a(t)  v(t)  12t2  24t  40.

427

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 





a  0  12t 2  24t  40  0  4 3t 2  6t 10  0  t 

 1

1 3

6  62  4(3)(10)  2(3)

39; so t  3.082 s for t  0. At this time, the acceleration changes from negative to

positive and the velocity attains its minimum value.

3.8.11 (a) A(x)  x2  A(x)  2x. A(15)  30 mm2/mm is the rate at which the area is increasing with respect to the side length as x reaches 15 mm. (b) The perimeter is P(x)  4x, so A(x)  2x  1 2 4x   12 P(x).The figure suggests that if x is small, then the change in the area of the square is approximately half of its perimeter (2 of the 4 sides) times

x. From the figure, A  2x  x    x  . If Δx is small, then A A  2x x and so  2x. x 2



3.8.12 (a) V (x)  x3 

 3x2, and dV  3(3)2  27 mm3/mm is dx x3 dx

the rate at which the volume is increasing as x increases past 3 mm. (b) The surface area is S(x)  6x2 , so





V (x)  3x2  12 6x 2 21 S (x). The figure suggests that if x is small, then the change in the volume of the cube is approximately half of its surface area (the area of 3 of the 6 faces) times x. From the figure V  3x 2  x   3x  x    x  . If x is small, then 2

V  3x2 x and so

V x

 3x2.

3.8.13 (a) Using A(r)   r 2 , we find that the average rate of change is

A(3)  A(2) 9  4 A(2.5)  A(2) 6.25  4   5   4.5 (ii) 3 2 1 2.5  2 0.5 A(2.1)  A(2) 4.41  4   4.1 (iii) 2.1 2 0.1 (b) A(r)   r 2  A(r)  2r, so A(2)  4 . (c) The circumference is C(r)  2r  A(r). The figure suggests that if r is (i)

small, then the change in the area of the circle (a ring around the outside) is

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

approximately equal to its circumference times r. Straightening out this rings gives us a shape that is approximately rectangular with length 2r and width r, so A  2r r . Algebraically,

A  A  r  r   A(r)    r  r    r 2  2 r  r     r  . So we see that if r is small, A then A  2r r  and therefore  2r. r 2

3.8.14 After t seconds the radius is r = 60t, so the area is A(t)   (60t)2  3600t2  A(t)  7200t  (b) A(3)  21, 600 cm2 /s (c) A(5)  36, 000 cm2 /s (a) A(1)  7200 cm2 /s As time goes by, the area grows at an increasing rate. In fact, the rate of change is linear with respect to time.

3.8.15 S(r)  4r2  S(r)  8r  (b) S(1)  8 ft2 /ft (b) S(2)  16 ft2 /ft (c) S(3)  24 ft2 /ft As time goes by, the area grows at an increasing rate. In fact, the rate of change is linear with respect to the radius.

3.8.16 (a) UsingV (r) 

 r 3 , we find that the average rate of change is

3 V (8) V (5)

 (512)  4  (125)

  172 m / m 85 3 4 4  (216)   (125) V (6) V (5) 3 3 (ii) 3  121.3 m / m6 5 1 3 4  (5.1)  43 (125) (iii) V (5.1) V (5)  3  102.013 m3 / m 5.1 5 0.1 3 4  (5.1)  43 (125) (iv) V (5.1) V (5)  3  102.013 m3 / m 5.1 5 0.1 2 (b)V (r)  4r , so V (5)  100 m3 / m. 3

(i)

4 3

r3  V (r)  4r2  S(r). By analogy with Exercise 13(c), we can say that the

change in the volume of the spherical shell, V , is approximately equal to its thickness, r, times the surface area of the inner sphere. Thus V  4r 2 r  and so V / r  4r2.

3.8.17

429

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

The mass is f (x)  3x2 , so the linear density at x is (x)  f (x)  6x. (a) (1)  6 kg/m (b) (2)  12kg/m (c) (3)  18kg/m Since ρ is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end of the rod.

3.8.18



V (t)  5000 1 1 t 2 40

 V (t)  5000  2 1 t     250 1 t  1

5 (a) V (5)  250 1   218.75 gal/min

(b) V (10)  2501  10 40   187.5 gal/min (c) V (20)  2 50 1  40   125 gal/min 20

(d) V (40)  2 501  40   0 gal/min 40

The water is flowing out the fastest at the beginning – when t  0, V (t)  250 gal/min The water is flowing out the slowest at the end – when t  50, V (t)  0 gal/min As the tank empties, the water flows out more slowly.

3.8.19 The quantity of charge is Q(t)  t3  2t 2  6t  2, so the current is Q(t)  3t2  4t  6. (a) Q(0.5)  3(0.5)2  4(0.5)  6  4.75 A

(b) Q(1)  3(1)2  4(1)  6  5 A

The current is lowest when Q has a minimum. Q(t)  6t  4  0 when t  23 . So the current decreases when t  2 and increases when t  2 . Thus the current is lowest at t  2 s. 3

3.8.20 (a) F 

GmM r2

 GmM r2 

 2GmM  r 3  

2GmM r

, which is the rate of change of

the force with respect to the distance between the bodies. The minus sign indicates that as the distance r between the bodies increases, the magnitude of the force F exerted by the body of mass m on the body of mass M is decreasing. (b)Given F(20, 000)  2, find F(10, 000). 2  



2 20, 0003 F(10, 000)  

10, 000



2GmM  GmM  20, 0003. 20, 0003

 2  23  16 N/km

3.8.21 1/ 2

With m  m0 1 v2  , c  2   1  1/ 2 3/ 2 d d d   2v  d  2  v F  (mv)  m (v)  v (m)  m0 1   a  v  m0  1 v2      (v) c2    c2  dt dt dt dt  c2   2  

430

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 3/ 2

 m0 1 cv22   

 v  v  m0a  a0 1 c22   c22   2  1 v / c2  



Note that we factored out 1 v 2 / c 2



3/ 2



3/ 2

since 3 / 2 was the lesser exponent. Also note that

d (v)  a. dt 3.8.22 (a) D(t)  7  5cos0.503(t  6.75)  D(t)  5sin0.503(t  6.75)(0.503)  2.515sin0.503(t  6.75). At 3:00 AM, t  3, and D(3)  2.515sin0.503(3.75)  2.39 m/h (rising) (b) At 6:00 AM, t  6, and D(6)  2.515sin0.503(0.75)  0.93 m/h (rising) (c) At 9:00 AM, t  9, and D(9)  2.515sin0.503(2.25)  2.28 m/h (falling) (d) At noon, t  12, and D(12)  2.515sin0.503(5.25)  1.21 m/h (falling)

3.8.23 (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of P:

PV  C  V 





. 2

dP P (b) From the formula for dV dP in part(a), we see that as P increases, the absolute value of dV dP decreases.1Thus, dV the volume is decreasing more rapidly at the beginning. 1  C2  C (c)    C 1

V dP   V  P 



(PV )P 





3.8.24 (a) C 

a2kt akt 1

 rate of reaction

d C  (akt 1)  a k   a kt (ak ) a2k(akt 1 akt) a2k    2 2 2 dt  akt 1  akt 1  akt 1 2



(b) If x   C, then a  x  a 

a2kt



a2kt  a  a2kt



. akt 1 akt 1 d C  a 2 dx 2   So k(a  x)  k  [from part (a)]  .  2 dt dt  akt 1  akt 1 akt 1 a2k

431

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

a2kt

a kt  / t 2



a 2k

   a moles/L. (akt 1) / t ak  (1 / t) ak a2k d C (d) As t  ,  0. (akt 1)2 dt (e) As t increases, nearly all of the reactants A and B are converted into product C. In practical terms, the reaction virtually stops. akt 1

3.8.25 In Example 6, the population function was n  2t n0 . Since we are tripling instead of doubling and the initial population is 400, the population function is n(t)  400 3t. The rate of growth is

n(t)  400 3t  ln 3, so the rate of growth after 2.5 hours is n(2.5)  400 32.5  ln 3  6850 bacteria/hour.

3.8.26 n



0.7t f (t) 1 be



 n

a be0.7t (0.7) .

When t  0, n  20 and n  12.

1 be 

0.7t 2

a  a  20(1 b). 1 b 12 b b  f (0)  12  12  0.7ab  12  0.7(20)(1 b)b  14 1 b 1 b2  1 b2 

f (0)  20  20 







6(1 b)  7b  6  6b  7b  b  6 and a  20(1 6)  140. For the long run, we let t increase without bound. Then lim f (t)  lim t

t

140 140  140, indicating that the yeast  0.7t 1 6e 1 6 0

population stabilizes at 140 cells.

3.8.27 P 100  P 80 6080  4450   81.5 million/year 100  80 20 P 120  P 100 7755  6080   83.75 million/year 2010: P110  120 100 20

(a)

1990: P90 

(b)

P t   at3  bt 2  ct  d (in millions of people), where a  0.0008143939394, b  0.5875999001, c  8.037691475, and d  1731.167582.

(c)

P t   at3  bt2  ct  d  Pt   3at2  2bt  c (in millions of people per year)

432

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(d)

1990: P90  77.94 million people per year 2010: P110  91.67 million people per year The 1990 estimate is smaller than the estimate in part (a), and the 2010 estimate is larger.

(e)

f t   pqt (where p  1368.59 and q  1.01471)  f   t   ln q  pqt (in millions of

people per year) (f)

f 90  74.39 million people per year [smaller than the estimates in parts (a) and (d)] f 110  99.61 million people per year [larger than the estimates in parts (a) and (d)]

(g)

P115  94.80 million people per year, and f 115  107.16 million people per year

433

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.8.28 A  t   at4  bt3  ct 2  dt  e inches, where a  1.76484788106, b  0.0140195683,

(a)

c  41.76016548, d  55, 281.13001, and e  27, 440,551.53. At   at4  bt3  ct2  dt  e  At   4at3  3bt2  2ct  d. The units are inches per

(b)

year. (c)

Part (b) gives A1990  0.115 inches per year.

(d)

Part (b) gives A2015  0.092 inches per year. This is higher than the data suggest near 2015 because the function At  takes into account a larger time range over which there was, on average, a significant increase in rainfall.

(e)

3.8.29 (a)

Using v  of r: v(r) 

R  r with R = 0.01, l = 3, P= 3000, and η = 0.027, we have v as a function 4l 2

3000

2 0.01  r2 ;  4(0.027)3

v(0)  0.925 cm/s, v(0.005)  0.694 cm/s, v(0.01)  0. P (b) v(r)  R2  r2   v(r)  P (2r)   Pr . When l = 3, P = 3000, and η = 0.027, we 4l 4l 2l 3000r have v(r)  ; v(0)  0, v(0.005)  92.592 (cm/s) / cm, and 2(0.027)3 v(0.01)  185.185 (cm/s) / cm. (c) The velocity is greatest where r = 0 (at the center) and the velocity is changing most where r = R = 0.01 cm (at the edge).

434

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.8.30   1 T  L2   1 T 1 T  1 T  1  df 2    L  dL   2L   2   2L2      1 (ii) f  1 T   1 T1/ 2  df  1  1 T 1/ 2  dT 2 2L   2L  2L   4L T       1 T   T   1/ 2  df   1 T  3/ 2   T  (iii) f  2L   2L  d 2  2L  4L3/ 2 

(a) (i) f 







(b) Note: Illustrating tangent lines on the generic figures may help to explain the results.

 0 and L is decreasing  f is increasing  higher note dL df  (ii)  0 and T is increasing  f is increasing  higher note dT df (iii)  0 and ρ is decreasing  f is decreasing  lower note d  (i)

3.8.31 (a) C(x)  2000  3x  0.01x2  0.0002x3  C(x)  3  0.02x  0.0006x2 (b) C(100)  3  0.02(100)  0.0006(100)2  $11 /pair. C(100) is the rate at which the cost is increasing as the 100th pair of jeans is produced. It predicts the approximate cost of the 101st pair. (c) The cost of manufacturing the 101st pair of jeans is C(101)  C(100)  2611.0702  2600  11.0702  $11.07. This is close to the marginal cost from part (b).

3.8.32 (a) C(q)  84  0.16q  0.0006q2  0.000003q3  C(q)  0.16  0.0012q  0.000009q2 , and

C(100)  0.16  0.0012(100)  0.000009(100)2  0.13. This is the rate at which the cost is increasing as the 100th item is produced. st (c) The actual cost of producing the 101 item is C(101)  C(100)  97.13030299  97  $0.13.

435

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.8.33 (a) A(x) 

p(x) xp(x)  p(x)  A(x)  . A(x)  0  A(x) is increasing; that is, the average x x2

productivity increases as the size of the workforce increases. (b) p(x) is greater than the average productivity

 p(x)  A(x)  p(x) 

p(x) 

 A(x)  0.

 xp(x)  p(x)  xp(x)  p(x)  0  xp(x)  p(x)  0 x2

3.8.34 (a) S 

 1 4x 

9.6x0.6   40  24x0.4 1.6x0.6  1 4x  

0.4 2

dx 

0.4

9.6x0.6  38.4x0.2  64x0.6  38.4x0.2



1 4x 

0.4 2

54.4x0.6

1 4x 

0.4 2

(b) At low levels of brightness, R is quite large [R(0) = 40] and is quickly decreasing, that is, S is negative with large absolute value. This is to be expected: at low levels of brightness, the eye is more sensitive to slight changes than it is at higher levels of brightness.

3.8.35

 3c  9c2  8c 

t  ln

 





 ln 3c  9c  8c  ln 2   





2 dt 1 d 3c  9c 2  8c  0  3  12  9c  8c  (18c  8)   dc 3c  9c2  8c dc 3c  9c2  8c

3



1/ 2

9c  4

3 9c2  8c  9c  4 9c2  8c  . 2 2 3c  9c2  8c 9c  8c 3c  9c  8c





This derivative represents the rate of change of duration of dialysis with respect to the initial urea concentration.

3.8.36 f (r)  2 Dr  f (r)  2  12 (Dr)1/2  D 

D  Dr

D . This derivative, f (r), is the rate of r

change of the wave speed with respect to the reproductive rate.

436

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.8.37 1  PV . Using the Product Rule, we have nR (10)(0.0821) 0.821 1 dT  P(t)V (t) V (t)P(t)  1 8(0.15) 10(0.10)  0.2436 K/min.  dt 0.821 0.821

PV  nRT  T 



3.8.38 (a) If dP / dt  0, the population is stable (it is constant).  P P  dP   1   1   P     (b)  0   P  r0 1 P   P P Pc 1 . dt r P P r c c 0  c  0  r0  4 If P  10, 000, r  5%  0.05, and   4%  0.04, then P  10, 0001   2000. 



3.8.39 (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, dC/dt = 0 and dW/dt = 0. (b) ―The caribou go extinct‖ means that the population is zero, or mathematically, C = 0. (c) We have the equations dC/dt = aC – bCW and dW/dt = -cW + dCW. Let dC/dt = dW/dt = 0, a = 0.05, b = 0.001, c = 0.05, and d = 0.0001 to obtain 0.05𝐶 − 0.001𝐶W = 0 (1) and − 0.05W + 0.0001𝐶W = 0 (2) Adding 10 times (2) to (1) eliminates the CW-terms and gives us 0.05𝐶 − 0.5W = 0 ⇒ 𝐶 = 10W Substituting C = 10W into (1) results in 0.05(10W) − 0.001(10W)W = 0 - 0.5W − 0.01W2 = 0 - 50W − W2 = 0 - W(50 − W) = 0 So W = 0 or 50. Since C = 10W, C = 0 or 500. Thus, the population pairs (C, W) that leads to stable populations are (0, 0) and (500, 50). So it is possible for the two species to live in harmony.

Solution and Answer Guide

437

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION 3.9 TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.9.1

f (x)  x3  x2  3  f (x)  3x2  2x, so f (2)  9 and f (2)  16.

Thus, L(x)  f (2)  f (2)  x  (2)  9 16(x  2)  16x  23. 3.9.2

f (x)  sin x  f (x)  cos x, so f     1 and f     1 3. 6

Thus, L(x)  f 



  f   x      x   



  3 . 1

3.9.3

f (x)  x  f (x)  12 x1/2

1 2 x

, so f  4  2 and f   4   14 .

Thus, L(x)  f  4   f   4  x  4  2  41  x  4  2  41 x 1  41 x 1. 3.9.4

f (x)  2x  f (x)  2x ln 2, so f  0   1and f   0   ln 2.

Thus, L(x)  f 0  f   0  x  0  1 ln 2x.

3.9.5 ƒ(𝑥) = √1 − 𝑥 ⇒ ƒ′(𝑥) =

−1

, so ƒ(0) = 1 and ƒ′(0) = − . Therefore,

2√1−𝑥

√1 − 𝑥 = ƒ(𝑥) ≈ ƒ(0) + ƒ′(0)(𝑥 − 0) = 1 + (− ) (𝑥 − 0) = 1 − 𝑥. 2

438

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 1

So √0.9 = √1 − 0.1 ≈ 1 − (0.1) = 0.95 and √0.99 = √1 − 0.01 ≈ 1 − (0.01) = 0.995.

3.9.6 1

3 𝑔(𝑥) = √1 + 𝑥 = (1 + 𝑥)1/3 ⇒ 𝑔′ (𝑥) = (1 + 𝑥)−2/3, so 𝑔(0) = 1 and 𝑔′ (0) = .

Therefore, 3√1 + 𝑥 = 𝑔(𝑥) ≈ 𝑔(0) + 𝑔′(0)(𝑥 − 0) = 1 + 𝑥. 1

̅, So 3√0.95 = 3√1 + (−0.05) ≈ 1 + (−0.05) = 0.983 1

3 ̅ and 3√1.1 = √1 + 0.1 ≈ 1 + (0.1) = 1.03

3.9.7 1

3 ƒ(𝑥) = √1 − 𝑥 = (1 − 𝑥)1/3 ⇒ ƒ ′ (𝑥) = − (1 − 𝑥)−2/3 , so ƒ(0) = 1 and ƒ ′ (0) = − . Thus, ƒ(𝑥) ≈

ƒ(0) + ƒ ′ (0)(𝑥 − 0) = 1 − 𝑥. We need 3√1 − 𝑥 − 0.1 < 1 − 𝑥 < √1 − 𝑥 + 0.1, which is true 3

when −1.204 < 𝑥 < 0.706.

3.9.8 ƒ(𝑥) = tan 𝑥 ⇒ ƒ′(𝑥) = sec2 𝑥, so ƒ(0) = 0 and ƒ′(0) = 1. Thus, ƒ(𝑥) ≈ ƒ(0) + ƒ′(0)(𝑥 − 0) = 0 + 1(𝑥 − 0) = 𝑥. We need tan 𝑥 − 0.1 < 𝑥 < tan 𝑥 + 0.1, which is true when −0.63 < 𝑥 < 0.63.

3.9.9

439

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

ƒ(𝑥) =

1 (1+2𝑥)4

= (1 + 2𝑥)−4 ⇒ ƒ′(𝑥) = −4(1 + 2𝑥)−5(2) =

−8 (1+2𝑥)5

, so ƒ(0) = 1 and ƒ′(0) = −8.

Thus, ƒ(𝑥) ≈ ƒ(0) + ƒ′(0)(𝑥 − 0) = 1 + (−8)(𝑥 − 0) = 1 − 8𝑥. We need 1 1 − 0.1 < 1 − 8𝑥 < + 0.1, which is true when −0.045 < 𝑥 < 0.055. (1+2𝑥)4

(1+2𝑥)4

3.9.10 ƒ(𝑥) = 𝑒𝑥 ⇒ ƒ′(𝑥) = 𝑒𝑥, so ƒ(0) = 1 and ƒ′(0) = 1. Thus, ƒ(𝑥) ≈ ƒ(0) + ƒ′(0)(𝑥 − 0) = 1 + 1(𝑥 − 0) = 1 + 𝑥. We need 𝑒𝑥 − 0.1 < 1 + 𝑥 < 𝑒𝑥 + 0.1, which is true when −0.483 < 𝑥 < 0.416.

3.9.11



For y  f  x  x 1 , f  x   x 11   x 11

 x 1

x 1



2

 x 1

, so dy 

2 2 dx.

 x 1

3.9.12 For 𝑦 = ƒ(𝑟) = (1 + 𝑟3)−2, ƒ′(𝑟) = −2(1 + 𝑟3)−3(3𝑟2) =

−6𝑟2 (1+𝑟3)3

, so 𝑑𝑦 =

−6𝑟2 (1+𝑟3)3

𝑑𝑟.

3.9.13 For y  f  x  etan x , f   x  etan x sec2  x    , so dy   sec2  xetan xdx. 3.9.14 1 For y  f  x  1 ln x, f   x   1 ln x  2

1/ 2

1 1  , so dy  dx. x 2x 1 ln x 

3.9.15 To estimate (1.999)

we’ll find the linearization of f (x)  x at a  2. Since f (x)  4x , 4

f (2)  16,

4 and f (x)  32, we have L(x)  16  32(x  2). Thus, x  16  32(x  2). when x is near 2, so

(1.999)4  16  32(1.999  2)  160  0.032  15.986. 3.9.16

440

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  f (x) 

 dy 

1

dx. When x  4 and dx  0.002, dy   

x 1



 f (4.002)  f (4)  dy  1  4

4.002

1 (0.002)   , so 16 8000

 1999  0.249875.

8000

3.9.17

y  f (x)  3 x  dy 

x2/3dx. When x  1000 and dx  1, dy 

3 3

(1000)2/3(1) 

1001  f (1001)  f (1000)  dy  10 



1

, so 300

1  10.003  10.003. 300

3.9.18

 1 1 y  f (x)  x  dy  1 dx. When x  100 and dx  0.5, dy   1  , so 2 2 100 40 2 x 100.5  f (100.5)  f (100)  dy  10 

1 40

 10.025.

3.9.19

y  f (x)  ex  dy  exdx. When x  0 and dx  0.1, dy  e0(0.1)  0.1, so e0.1  f (0.1)  f (0)  dy  1 0.1  1.1.

3.9.20

30   29  To estimate cos , we’ll find the linearization of f  x  cos x at a   . Since   180 6  180  3 1  3 1       x   . Thus,  , and f  f   x  sin x, f   , we have L  x  6 2 2 2 6  6 2       3 1    x  when x is near , so cos x  2 2 6 6

   29    3 1 29   3 1   cos 180   2  2  180  6   2  2   180  

















3   0.875.  2 360

3.9.21

441

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

𝑦 = ƒ(𝑥) = sec 𝑥 ⇒ ƒ′(𝑥) = sec 𝑥 tan 𝑥, so ƒ(0) = 1 and ƒ′(0) = 1 · 0 = 0. The linear approximation of f at 0 is ƒ(0) + ƒ′(0)(𝑥 − 0) = 1 + 0(𝑥) = 1. Since 0.08 is close to 0, approximating sec 0.08 with 1 is reasonable.

3.9.22 If 𝑦 = 𝑥6, 𝑦′ = 6𝑥5 and the tangent line approximation at (1, 1) has slope 6. If the change in x is 0.01, the change in y on the tangent line is 0.06, and approximating (1.01)6 with 1.06 is reasonable.

3.9.23 𝑦 = ƒ(𝑥) = ln 𝑥 ⇒ ƒ′(𝑥) = 1/𝑥, so ƒ(1) = 0 and ƒ′(1) = 1. The linear approximation of f at 1 is ƒ(1) + ƒ′(1)(𝑥 − 1) = 0 + 1(𝑥 − 1) = 𝑥 − 1. Now ƒ(1.05) = ln 1.05 ≈ 1.05 − 1 = 0.05, so the approximation is reasonable.

3.9.24 (a) 𝑦 = 𝑒𝑥/10 ⇒ 𝑑𝑦 = 𝑒𝑥/10 ·

𝑑𝑥 =

𝑒𝑥/10𝑑𝑥

10 0

(b) 𝑥 = 0 and 𝑑𝑥 = 0.1 ⇒ 𝑑𝑦 = 1 𝑒10(0.1) = 0.01 10

∆𝑦 = ƒ(𝑥 + ∆𝑥) − ƒ(𝑥) = 𝑒0+0.1 − 𝑒0 = 𝑒0.1 − 1 ≈ 0.0101

3.9.25 1 −1/2 1 (a) 𝑑𝑥 = 𝑑𝑥 𝑦 = √𝑥 ⇒ 𝑑𝑦 = 2 𝑥 2√𝑥 (b) 𝑥 = 1 and 𝑑𝑥 = 1 ⇒ 𝑑𝑦 =

1 2(1)

(1) = . 2

∆𝑦 = ƒ(𝑥 + ∆𝑥) − ƒ(𝑥) = √1 + 1 − √1 = √2 − 1 ≈ 0.414. (c) Remember, ∆𝑦 represents the amount that the curve 𝑦 = ƒ(𝑥) rises or falls when x changes by an amount dx, whereas dy represents the amount that the tangent line rises or falls (the change in the linearization).

442

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.9.26 (a) f (x)  1 x   f (x)  2 1 x  , so f (0)  1and f (0)  2. 2

Thus, f (x)  Lf (x)  f (0)  f (0)  x  0  1 2x.

g(x)  e2x  g(x)  2e2 x , so g 0  1and g   0   2. Thus, g(x)  Lg (x)  g 0  g   0  x  0  1 2x.

 h(x)  1 ln(1 2x)  h(x) 

2 , so g 0  1and g   0   2. 1 2x

Thus, h(x)  Lh (x)  h 0  h   0  x  0  1 2x. Notice that Lf  Lg  Lh . This happens because f, g, and h have the same function values and the same derivative values at a  0. (b) The linear approximation appears to be best for the function f since it is closer to f for a larger domain than it is to either g or h. The approximation looks worst for h since h moves away from L faster than does either f or g. 3.9.27 As in Example 1, 𝑇(0) = 1.85, 𝑇(10) = 172, 𝑇(20) = 160 and 𝑇(10)−𝑇(20) 172−160 𝑇′(20) ≈ = = −1.2°F/min. 10−20

−10

𝑇(30) ≈ 𝑇(20) + 𝑇′(20)(30 − 20) ≈ 160 − 1.2(10) = 148°F. We would expect the temperature of the turkey to get closer to 75°F as time increases. Since the temperature decreased 13°F in the first 10 minutes and 12°F in the second 10 minutes, we can assume that the slopes of the tangent line are increasing through negative values: -1.3, -1.2, ..... Hence, the tangent lines are under the curve and 148°F is an underestimate. From the figure, we estimate the slope of the tangent line at t = 20 to be 37 184−147 = − . Then the linear approximation becomes 37 2 0−30 30 𝑇(30) ≈ 𝑇(20) + 𝑇′(20) · 10 = 160 − ( 10) ≈ 147 ≅ 147.7. 30

3.9.28 𝑃′(2) ≈

𝑃(1)−𝑃(2)

87.1−74.9

1−2

= −12.2 kilopascals/km.

−1

𝑃(3) ≈ 𝑃(2) + 𝑃′(2)(3 − 2) ≈ 74.9 − 12.2(1) = 62.7 kPa. From the figure, we estimate the slope of 35

the tangent line at h = 2 to be 98−63 = − . Then the linear approximation becomes 𝑃(3) ≈ 𝑃(2) + 𝑃′(2) · 1 ≈ 74.9 −

35 3

0−3

≈ 63.23 kPa.

443



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.9.29 Extend the tangent line at the point 2020,17 to the t-axis. Answers will vary based on this approximation—we’ll use t 1980 as our t-intercept. The linearization is then

P t   P 2020  P2020t  2020  17  1740t  2020 P 2030  17 

2030  2020  21.25%

40 P 2040  17 

2040  2020  25.5%

40 These predictions may be too low since the tangent line lies below the graph at t  2020.

3.9.30

444

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

N 2005  N 2010 32.16  34.15 N 2015  N 2010   0.398 and B   2005  2010 5 2015  2010  36.03  34.15  0.376. Then N2010  lim N t   N 2010  A  B  0.387 million/year. t  2010 2 x2010 5 So N 2009  N 2010  N20102009  2010  34.15  0.387 1  33.763 million.

Let A 





N2020 

N 2015  N 2020



36.03  37.74

2015  2020

 0.342 million/year.

5

N 2025  N 2020  N20202025  2020  37.74  0.342 5  39.45 million.

3.9.31 (a) If x is the edge length, then 𝑉 = 𝑥3 ⇒ 𝑑𝑉 = 3𝑥2𝑑𝑥. When x = 30 and dx = 0.1, dV = 3(30)2(0.1) = 270, so the maximum possible error in computing the volume of the cube is about 270 cm3. The relative error is calculated by dividing the change in V, ∆V, by V. We approximate ∆V with dV. 𝑑𝑉 3𝑥2𝑑𝑥 𝑑𝑥 Relative error = ∆𝑉 ≈ = =3 0.1 = 3 ( ) = 0.01. 𝑉

𝑥3

𝑉

𝑥

Percentage error = relative error × 100% = 0.01 × 100% = 1%. (b) 𝑆 = 6𝑥2 ⇒ 𝑑𝑆 = 12𝑥𝑑𝑥. When x = 30 and dx = 0.1, dS = 12(30)(0.1) = 36, so the maximum possible error in computing the𝑑𝑆surface area of 𝑑𝑥 the cube is about 36 cm2. 12𝑥𝑑𝑥 Relative error = ∆𝑆 ≈ = =2 0.1 = 2 ( ) = 0.006̅ 𝑆

6𝑥2

𝑆

𝑥

̅ × 100% = 0. 6 ̅% Percentage error = relative error × 100% = 0.006

3.9.32 (a) 𝐴 = 𝜋𝑟2 ⇒ 𝑑𝐴 = 2𝜋𝑟𝑑𝑟. When r = 24 and dr = 0.2, 𝑑𝐴 = 2𝜋(24)(0.2) = 9.6𝜋, so the maximum possible error in the calculated area of the disk is about 9.6𝜋 ≈ 30 cm2. (b) Relative error = ∆𝐴 ≈ 𝐴

𝑑𝐴 𝐴

2𝜋𝑟𝑑𝑟 𝜋𝑟2

2𝑑𝑟 𝑟

2(0.2) 24

0.2 12

̅. = 0.016

̅ × 100% = 1. 6 ̅% Percentage error = relative error × 100% = 0.016

3.9.33 (a) For a sphere of radius r, the circumference is 𝐶 = 2𝜋𝑟 and the surface area is 𝑆 = 4𝜋𝑟2,so 𝐶 𝐶 2 𝐶2 2 ⇒ 𝑑𝑆 = 𝐶𝑑𝐶 𝑟= ⇒ 𝑆 = 4𝜋 ( ) = 2𝜋 2𝜋 𝜋 𝜋

445

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 2

When C = 84 and dC = 0.5, 𝑑𝑆 = (84)(0.5) = 84/𝜋, so the maximum possible error is about 2

84/𝜋 ≈ 27 cm . Relative error ≈ (b)

𝐶

𝜋 𝑑𝑆 𝑆

𝐶3

84 𝜋 842 𝜋

1 84

≈ 0.012.

𝑉 = 𝜋𝑟 = 𝜋 ( ) = ⇒ 𝑑𝑉 = 2𝜋2 𝐶 𝑑𝐶 3 3 2𝜋 6𝜋2 1 1764 (84)2(0.5) = When C = 84 and dC = 0.5, 𝑑𝑉 = , so the maximum error is about 1764 ≈ 179 2𝜋2

𝜋2 1764/𝜋2

cm . The relative error is approximately 𝑑𝑉 = 𝑉

(84)3/(6𝜋2)

𝜋2

≈ 0.018.

3.9.34 2

For a hemispherical dome, 𝑉 = 𝜋𝑟3 ⇒ 𝑑𝑉 = 2𝜋𝑟2𝑑𝑟. When 𝑟 = (50) = 25 m and dr = 0.05 cm = 3

0.0005 m, 𝑑𝑉 = 2𝜋(25)2(0.0005) =

5𝜋

, so the amount of paint needed is about 5𝜋 ≈ 2 m3.

3.9.35 (a) 𝑉 = 𝜋𝑟2ℎ ⇒ ∆𝑉 ≈ 𝑑𝑉 = 2𝜋𝑟ℎ 𝑑𝑟 = 2𝜋𝑟ℎ ∆𝑟 (b) The error is ∆𝑉 − 𝑑𝑉 = [𝜋(𝑟 + ∆𝑟)2ℎ − 𝜋𝑟2ℎ] − 2𝜋𝑟ℎ ∆𝑟 = 𝜋𝑟2ℎ + 2𝜋𝑟ℎ ∆𝑟 + 𝜋(∆𝑟)2ℎ − 𝜋𝑟2ℎ − 2𝜋𝑟ℎ ∆𝑟 = 𝜋(∆𝑟)2ℎ

3.9.36 (a) sin 𝜃 =

20 𝑥

⇒ 𝑥 = 20 csc 𝜃 ⇒ 𝑑𝑥 = 20(− csc 𝜃 cot 𝜃)𝑑𝜃 = −20 csc 30° cot 30° (±1°)

2√3 𝜋 𝜋 ) = ± = −20(2)(√3) (± 180 9 2 So the maximum error is about ± √3𝜋 ≈ ±1.21 cm. 9

± √3𝜋 𝑑𝑥 ∆𝑥 (b) The relative error is ≈ 𝑥 = 9 = ± √3 𝜋 ≈ ±0.03, so the percentage error is approximately 𝑥 20(2) 180

±3%.

3.9.37 𝐹 = 𝑘𝑅4 ⇒ 𝑑𝐹 = 4𝑘𝑅3𝑑𝑅 ⇒

𝑑𝑅 𝑑𝐹 4𝑘𝑅3𝑑𝑅 ) = = 4 ( 𝑘𝑅4 𝑅 𝐹

446

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Thus, the relative change in F is about 4 times the relative change in R. So a 5% increase in the radius corresponds to a 20% increase in blood flow.

3.9.38 (a) ƒ(𝑥) = sin 𝑥 ⇒ ƒ′(𝑥) = cos 𝑥, so ƒ(0) = 0 and ƒ′(0) = 1. Thus, ƒ(𝑥) ≈ ƒ(0) + ƒ′(0)(𝑥 − 0) = 0 + 1(𝑥 − 0) = 𝑥. (b) We want to know the values of x for which y = x approximates 𝑦 = sin 𝑥 with less than a 2% difference; that is, the values of x for which 𝑥 − sin 𝑥 𝑥 − sin 𝑥 < 0.02 | | < 0.02 - −0.02 < sin 𝑥 sin 𝑥 −0.02 sin 𝑥 < 𝑥 − sin 𝑥 < 0.02 sin 𝑥 , if sin 𝑥 > 0 { −0.02 sin 𝑥 > 𝑥 − sin 𝑥 > 0.02 sin 𝑥 , if sin 𝑥 < 0 0.98 sin 𝑥 < 𝑥 < 1.02 sin 𝑥 , if sin 𝑥 > 0 -{ 1.02 sin 𝑥 > 𝑥 > 0.98 sin 𝑥 , if sin 𝑥 < 0 In the first figure, we see that the graphs are very close to each other near x = 0. Changing the viewing rectangle and using an intersect feature (see the second figure) we find that 𝑦 = 𝑥 intersects 𝑦 = 1.02 sin 𝑥 at x≈ 0.344. By symmetry, they also intersect at 𝑥 ≈ −0.344 (see the third figure). 180° Converting 0.344 radians to degrees, we get , which verifies the 0.344 ( ) ≈ 19.7° ≈ 20° 𝜋 statement.

3.9.39 (a) The graph shows that ƒ′(1) = 2, so 𝐿(𝑥) = ƒ(1) + ƒ′(1)(𝑥 − 1) = 5 + 2(𝑥 − 1) = 2𝑥 + 3. ƒ(0.9) ≈ 𝐿(0.9) = 4.8 and ƒ(1.1) ≈ 𝐿(1.1) = 5.2.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(b) From the graph, we see that ƒ′(𝑥) is positive and decreasing. This means that the slopes of the tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too large.

3.9.40 (a) 𝑔′(𝑥) = √𝑥2 + 5 ⇒ 𝑔′(2) = √9 = 3, 𝑔(1.95) ≈ 𝑔(22) + 𝑔′(2)1.95 − 2) = −4 + 3(−0.05) = −4.15. (b) The formula 𝑔′(𝑥) = √𝑥2 + 5 shows that 𝑔′(𝑥) is positive and increasing. This means that the slopes of the tangent lines are positive and the tangents are getting steeper. So the tangent lines lie below the graph of g. Hence, the estimate in part (a) is too small.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION CONCEPT CHECK TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3. CV.1

(a) The Power Rule: If n is any real number, then

x  nx . dx n

n1

To find the derivative of a variable raised to a constant power, we multiply the expression by the exponent and then subtract one from the exponent. (b) The Constant Multiple Rule: If c is a constant and f is a differentiable function, then d d [c  f (x)]  c f (x) dx dx The derivative of a constant times a function is the constant times the derivative of a function.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(c) The Sum Rule: If f and g are both differentiable, then d d d [ f (x)  g(x)]  f (x)  g(x) dx dx dx The derivative of a sum of functions is the sum of the derivatives. (d) The Difference Rule: If f and g are both differentiable, then d d d [ f (x)  g(x)]  f (x)  g(x) dx dx dx The derivative of a difference of functions is the difference of the derivatives. (e) The Product Rule: If f and g are both differentiable, then d d d [ f (x)g(x)]  f (x)  g(x)   g(x)  f (x) dx dx dx The derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. (f) The Quotient Rule: If f and g are both differentiable, then

d d g(x) f (x)  f (x)   g(x)    d  f (x)  dx dx  dx  g(x)   g(x)2 

The derivative of the quotient of functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. (g) The Chain Rule: If g is differentiable at x and f is differentiable at g(x) then the composite function defined by

F (x)  f (g(x)) is differentiable at x and F  is given by the product F(x)  f (g(x))g(x) The derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

3.CV.2 n1 (a) y  nx

(b) y  e

(d) y  1/ x

(e) y  1/ (x ln b)

(f) y  cos x

(g) y  sin x

2 (h) y  sec x

(i) y  csc x cot x

(j) y  sec x tan x

2 (k) y  csc x

(l) y  1/ 1 x2

(m) y  1/ 1 x

(n) y  1/ (1 x ) 2

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.CV.3

eh 1  1. h0 h

(a) e is the number such that lim (b) e  lim(1 x)

1/x

x0

x (c) The differentiation formula for y  b  dy / dx  b x ln b  is simplest when b  e because ln e  1.

(d) The differentiation formula for y  logb x dy / dx  1/ (x ln b) is simplest when b  e because

ln e  1. 3.CV.4 (a) Implicit differentiation consists of differentiating both sides of an equation involving x and y with respect to x, and then solving the resulting equation for 𝑦′. Use implicit differentiation when it is difficult to solve an equation for y in terms of x. (b) Logarithmic differentiation consists of taking natural logarithms of both sides of an equation 𝑦 = ƒ(𝑥), simplifying, differentiating implicitly with respect to x, and then solving the resulting equation for 𝑦′. Use logarithmic differentiation when the calculation of derivatives of complicated functions involving products, quotients, or powers can be simplified by taking logarithms. 3.CV.5 𝑑𝑦/𝑑𝑡 g𝘍(𝑡) You can find 𝑑𝑦 as a function of t by calculating 𝑑𝑦 = = [if 𝑑𝑥/𝑑𝑡 = ƒ′(𝑡) G 0]. 𝑑𝑥

𝑑𝑥

𝑑𝑥/𝑑𝑡

ƒ𝘍 (𝑡)

3.CV.6 In physics, interpretations of the derivative include velocity, linear density, electrical current, power (the rate of change of work), and the rate of radioactive decay. Chemists can use derivatives to measure reaction rate and the compressibility of a substance under pressure. In biology the derivative measures rates of population growth and blood flow. In economics, the derivative measures marginal cost (the rate of change of cost a more items are produced) and marginal profit. Other examples include the rate of heat flow in geology, the rate of performance improvement in psychology, and the rate at which a rumor spreads in sociology.

3.CV.7 The linearization L of f at x = a is 𝐿(𝑥) = ƒ(𝑎) + ƒ′(𝑎)(𝑥 − 𝑎).

3.CV.8 If y  f  x  , then the differential is dy  f   x dx. The differential dy represents the amount that the tangent line rises or falls when x changes from x0 by an amount dx  x. However, y represents the amount that the curve y  f  x  rises or falls when x changes from x0 by an

450

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

amount dx  x. As these amounts may differ, dy could be an overestimate or an underestimate for y.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION PRINCIPLES OF PROBLEM SOLVING TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.PofPS.1 We must find a value 𝑥0 such that the normal lines to the parabola 𝑦 = 𝑥2 at 𝑥 = ±𝑥0 intersect at a point 1 one unit from the points (±𝑥 , 𝑥2). The normals to 𝑦 = 𝑥2 at 𝑥 = ±𝑥 have slopes − and pass 0

through (±𝑥 , 𝑥2) respectively, so the normals have the equations 𝑦 − 𝑥2 = − 0

±2𝑥0

(𝑥 − 𝑥 ) and 𝑦 − 𝑥2 =

2𝑥0

(𝑥 + 𝑥 ). The common 𝑦-intercept is 𝑥2 + . We want to find the value of 𝑥 for which the distance

2𝑥0

from (0, 𝑥2 + ) to (𝑥 , 𝑥2) equals 1 . The square of the distance is

 x  0

2  x2  x2  1   x2  1  1  x  

 0

 0

2 

1 5    5 x02   ,so the center ofthe circle is at 0, . 2

 

4  

3 . For these values of x , the y -intercept is 2

Another solution: Let the center of the circle be 0, a  . Then the equation of the circle is

x2  ( y  a)2  1 .

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Solving with the equation of the parabola, y  x , we get 2



x2  x2  a

  1  x  x  2ax  a  1  x4  1 2a x2  a2 1  0 . The parabola and the 2

circle will be tangent to each other when this quadratic equation in x2 has equal roots; that is, when the





discriminant is 0 . Thus, (1 2a)2  4 a 2 1  0  1 4a  4a  4a  4  0  4a  5 , so a 

  

The center of the circle is 0,

5 . 4  



3.PofPS.2

y  x3  3x  4  y  3x2  3 , and y  3  x 2  x  y  6x  3 .

The slopes of the tangents of the two curves are equal when 3x2  3  6x  3 ; that is, when x  0 or 2 . At x  0 , both tangents have slope 3 , but the curves do not intersect. At x  2 , both tangents have slope 9 and the curves intersect at 2, 6 . So there is a common tangent line at 2, 6  , y  9x 12 . 3.PofPS.3 Similarly, an equation of the tangent line at x  q is We can eliminate y and solve for x by subtracting equation (1) from equation (2).

 2aq  b    2ap  b   x  aq 2  ap 2 2aq  2ap x

0  aq2  ap2

 a  q 2  p2 

2a q  p  x x



a  q  p  q  p   p  q  2a q  p  2

Thus, the x -coordinate of the point of intersection of the two tangent lines, namely r , is  p  q  / 2 . 3.PofPS.4 We could differentiate and then simplify or we can simplify and then differentiate. The latter seems to be the simpler method.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

sin2x  cos x sin2x sinx  cos2x cosx sin3x      1 tanx sinx cosx sinx  cosx cosx sinx 1 cotx 1 1 cosx sinx  o s3 2

x c o s x  s i n x

sinx  cosx sin x  sinx cosx  cos x  sin3x  cos3x  [factor sum of cubes]  sinx  cosx sinx  cosx 1 1  sin2x  sinx cosx  cos2x  1 sinx cosx  1 2sinx cosx   1 sin2x 2 2 2

d  sin2 x cos2 x  d  1 1  Thus, 1 sin2x    cos2x  2  cos2x .   dx 1cotx  1 tanx   dx  2 2  3.PofPS.5 Using f   a   lim xa

f  x  f a

, we recognize the given expression,

xa

sect secx , as tx      g   x with g  x  secx . Now f     g   , so we will find g x  . 4   4 g   x  secxtanx  g x   secxsec2 x  tanxsecxtanx  secx sec2x  tan 2x  , so f  x  limtx

   g    2  4 

 2 1   2 2 1  3 2. 2

3.PofPS.6

 f  x   f  a  x  a   f  x   f  a     lim xa  limxa f  x  f a lim xa   x  a x a x a  x  a     

  a 





453

x Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 x  a   f a   a  a   2 a f  a

f  x  f a  lim  lim xa xa xa

 

3.PofPS.7 We can assume without loss of generality that   0 at time t  0 , so that   12t rad. [The angular velocity of the wheel is 360rpm  360 2 rad / 60 s  12 rad / s .] Then the position of A as a

454



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

function of time is A  40cos , 40sin   40cos12t, 40sin12t  , so

sin 



1.2 m

40sin



sin

1  sin12t . 3 3

120

d

1  12 cos12t  4 cos . When dt 3

(a) Differentiating the expression for sin , we get cos 





, we have

d

 3  11 and sin  sin  , so cos  1     12 3 6 dt  6 

 4 cos  3  cos

2 4 3  6.56rad / s.  11/ 12 11



 2 2 2 (b) By the Law of Cosines, AP |  OA |  | OP | 2 OA OP cos 



1202  402  OP |2 2  40 OP cos  OP |2 80cos  OP 12,800  0  



OP 

80cos  6400cos   51, 200   40cos  40 cos   8  40 cos  8  cos  cm 2

[since OP  0] . As a check, note that OP  160 cm when   0 and OP  80 2 cm when  







   dx d  40 sin  2cossin 12  480sin 1 cos cm / s        2  2 8  cos2  dt d dt 8  cos 

In particular, dx / dt  0 cm / s when   0 and dx / dt  480 cm / s when  



2 3.PofPS.8 The equation of T1 is y  x1  2x 1  x  x1   2x1 x  2x1 or y  2x1x  x12 . 2

The equation of T2 is y  2x x  x 2. Solving for the point of intersection, we get 1  2 1 2  . Therefore, the coordinates of P are  x  x , x x . So 2 2

2x  x  x   x  x  x  1

x  x

2 

1 2

 

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 



if the point of contact of T is

a, a2  , then Q is  1  a  x , ax  and Q is  1 a  x , ax . 1

1 1 2 2 2 2 2     1 2 2 2 2  PQ  a  x 1 2  x a  x  a  x        Therefore,  x 2 and 1 2 1 2 2  4 4 1    2 2 2 2 2 1 2 1 PQ 2 2  a  x2  , and similarly 1 PP1   x1  x2   x1  x1  x2    x1  x 2    x1  . So  2 2 4 4 x  x PP   1 2   1 PQ 2  x  a 2 PQ1 PQ2 a  x2 x a 2 1    1 1. 2  2 . Finally, PP x  x  PP PP x x x x 

3.PofPS.9

Consider the statement that



n ax

e sinbx  r e sin bx  n  . For n 1 ,

e sinbx  ae sinbx  be cosbx , and dx ax

b a  reaxsin bx     reax sinbxcos  cosbxsin   reax  sinbx  cosbx  aeaxsinbx  beaxcosbx r r   b b a since tan   sin  and cos  . So the statement is true for n 1 . a r r Assume it is true for n  k . Then ax d k 1 e sinbx  

dx k 1

 k ax

k ax  k ax  r e sin bx  k   r ae sin bx  k   r e bcosbx  k  dx  r k e ax asin  bx  k    bcos  bx  k  





But

sin bx  k 1 

 sin  bx  k      sin  bx  k  cos  sin cos  bx  k  

a b sin bx  k   cosbx  k . r r

Hence, asin  bx  k   bcos  bx  k   rsin bx   k 1  . So ax k ax      d k 1 e sinbx  r e asin bx  k   bcos bx k   dxk 1  r k eax rsin  bx   k 1     r k 1eax sin  bx   k 1    . 

Therefore, the statement is true for all n by mathematical induction.

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Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.PofPS.10 Using f   0  lim x0

f  x  f 0 x0

, we see that for the given equation,

5 ax  b  2  5 , we have f  x  3 ax  b , f 0  2 , and f   0   . Now x 12 12 1 2/3 f 0  2  3 b  2  b  8 . Also f  x  (ax  b)  a , so 3 5 1 5 f   0    (8)2/3  a  11  5  a 5.  a   12 3 12 3  4  12 3

limx0

3.PofPS.11 It seems from the figure that as P approaches the point 0, 2 from the right, xT   and

yT  2 . As P approaches the point 3, 0 from the left, it appears that xT  3 and yT  

. So we guess that xT 3,  and yT 2,  . It is more difficult to estimate the range of

values for xN and yN . We might perhaps guess that xN   0, 3 , and yN  ,0  or 2, 0 . In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to find the 2

equation of the tangent line: x  y  1  2x  2y y  0 , so y  

. So at the point  x0 , y0  on

9 4 9y 4 x0  x  x  or 4x x  9y y  4x2  9y2 . This the ellipse, an equation of the tangent line is y  y   0 0 0 0 0 0 9 y 0 2 2 xx y y x y can be written as 0  0  0  0  1 , because  x0 , y0  lies on the ellipse. So an equation of the 9 4 9 4 x0 x y0 y  1. tangent line is 9 4 Therefore, the x -intercept xT for the tangent line is given by intercept yT is given by

9 x0 xT  1  xT  , and the y 9 x0

4 y0 yT  1  yT  . 4 y0

So as x0 takes on all values in 0, 3  , xT takes on all values in 3,  , and as y0 takes on all values in

0, 2  , yT takes on all values in 2,  . At the point  x0 , y0  on the ellipse, the slope of the normal line © 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

457

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

is 

1 y x , y  0



9 y0

, and its equation is y  y 

normal line is given by 0  y  0

given by y  y  N

9 y0

 x  x  . So the x -intercept x 0

4 x0

x  x   x   N

4 x0

0  x   y   0

4 x0

9 y0

So as x takes on all values in 0, 3  , x 0

9 y0

y  0

4x0

9 5y0

x 

5x0

, and the y -intercept y is N

4   

takes on all values in 0, N

0, 2  , y N takes on all values in   2 ,0  5



5 , and as y takes on all values in  0 3 





3.PofPS.12

f  x



lim x0

for the N

g  x

 lim x0

f  x  0

g  x  0

 lim x0

f  x   f 0

g  x   g 0

f  x  f 0 lim

 lim x0

f  x   f 0

x0 f   0  x0 x0   g  x   g 0 g   0  g  x  g 0 limx0 x0 x0

3.PofPS.13

 x

y

y 

1 a2 1 1

a 1 2



a 1 2

arctan

sinx

. Let k  a  a 1 . Then 2

a  a 1  cosx 2

1  cosx  k  cosx   sin x   2 2 2 (k  cosx )2 a 1 1 sin x / (k  cosx ) 2

kcosx  cos2x  sin2x 1 2 kcosx 1    2 2 2 2 2 2 a 1 a 1 k  2kcosx 1 a 1 a2 1 (k  cosx )  sin x k 2  2kcosx 1 2kcosx  2  k 2 1   a2 1  k 2  2kcosx 1 a2 1 k 2  2kcosx 1 









But k 2  2a2  2a a2 1 1  2a a  a 2 1 1  2ak 1 , so k 2 1  2ak , and

k 2 1  2ak 1 . 2ak 1

So y 

a 12ak  2kcosx 2

ak 1



. But ak 1  a  a

a 1k a  cosx 2

a2 1 1  k a2 1 , so

y  1/  a  cosx.

458

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.PofPS.14 We see that at x  0, f  x  ax  1 x  1, so if y  ax is to lie above y  1 x , the two curves must just touch at 0,1 , that is, we must have f   0   1 . [To see this analytically, note that x

ax 1

a  1 x  a 1  x  x  0, a x 1  x 

ax 1



 1 for x  0 , so f 0  lim x0



 1. Similarly, for

ax 1 a x 1  1, so f   0  lim x0  1 x x 

Since 1  f   0   1, we must have f   0   1 .] But f  x  a lna  f   0   lna , so we have x

lna 1  a  e . Another method: The inequality certainly holds for x  1, so consider x  1, x  0 . Then ax  1 x  a  (1 x)1/x for x  0  a  lim x 0 (1 x)

1/x

 e , by Equation 3.7.5. Also,

ax  1 x  a  (1 x)1/x for x  0  a  lim x 0 (1 x)1/x  e . So since e  a  e , we must have 

a  e. 3.PofPS.15 Let f  x  e

and g  x  k x [k  0]. From the graphs of f and g , we see that f will intersect g

exactly once when f and g share a tangent line. Thus, we must have f  g and f   g at x  a .

f  a   g  a   e2a  k a and f   a   g   a   2e2a 

k a

k 2 a

 e2a 

k 4 a

. So we must have

k  ( a )2  k  a  1 . From  å , e21/4  k 1/ 4  k  2e1/2  2 e  3.297 4k 4 4 a

3.PofPS.16 3 2 2 (a) f  x  x  x  2  x  6  x  8x 12x  f   x  3x 16x 12 . The average of the first pair

of zeros is 0  2 / 2  1 . At x 1 , the slope of the tangent line is f 1  1, so an equation of the tangent line has the form y  1x  b . Since f 1  5 , we have 5  1 b  b  6 and the tangent has equation y  x  6 . Similarly, at x 

06 2

 3, y  9x 18 ; at x 

26

 4, y  4x . From

the graph, we see that each tangent line drawn at the average of two zeros intersects the graph of f at the third zero.

459



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(b) A CAS gives f  x   x  b  x  c   x  a  x  c   x  a  x  b or

f  x  3x2  2  a  b  c x  ab  ac  bc . Using the Simplify command, we get ab  (a  b)2  a  b  (a  b)2   f  and f       8 a  b  2c , so an equation of the tangent line at 4  2   2  2 (a  b)  a  b  (a  b)2 ab x  2 is y   x  a  b  2c . To find the x -intercept, let y  0 and 8  4  2  use the Solve command. The result is x  c . Using Derive, we can begin by authoring the expression  x  a  x  b  x  c . Now load the utility file DifferentiationApplications. Next we author tangent 

#1, x, a  b / 2 - this is the command to find an equation of the tangent line of the function in #1 whose independent variable is x at the x -value  a  b /2 . We then simplify that expression and obtain the equation y  #4 . The form in expression #4 makes it easy to see that the x -intercept is the third zero, namely c . In a similar fashion we see that b is the x -intercept for the tangent line at a  c /2 and a is the x -intercept for the tangent line at b  c / 2 .

# 1:  x  a    x  b    x  c # 2: LOAD (c : \ Program Files\M I Education\Derive6\Math\DifferentiationApplications.mth a  b  # 3: TANGENT x  a    x  b    x  c  , x ,   2  # 4: 0

a2  2  a  b  b2    c  x  4 

3.PofPS.17 (a) If the two lines L1 and L2 have slopes m1 and m2 and angles of inclination 1 and 2 , then

m1  tan1 and m2  tan2 . The triangle in the figure shows that 1    180  2   180 and so

  2 1 . Therefore, using the identity for tan  x  y  , we have tan  tan      tan2  tan1 and so tan  m2  m1 . 2 1 1 tan2tan1 1 m1m2 (b) (i) The parabolas intersect when x2  (x  2)2  x  1 . If y  x2 , then y  2x , so the slope of the tangent to y  x2 at 1,1 is m1  2 1  2 . If y  (x  2)2 , then y  2  x  2 , so the slope of the tangent to y  (x  2)2 at 1,1 is m  21 2  2 . Therefore, tan  m2  m1 2

and so   tan1



1 m1m2

2  2 1 22



4 3

4  3  53  or 127  .  

460

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 2 2 2 2 (ii) x  y  3 and x  4x  y  3  0 intersect when





x2  4x  x 2  3  3  0  2x  x  2  0  x  0 or 2 , but 0 is extraneous. If x  2 , then y  1. If x2  y2  3 then 2x  2 yy  0  y  x / y and

x2  4x  y2  3  0  2x  4  2 yy  0  y 



2 x

. At 2,1 the slopes are m  2 and m  0 , 1

02  2    117 . At 2, 1 the slopes are m1  2 and m2  0 , so 1 2 0 0  2  tan   2    63 [or 117 ]. 1  2  0 

so tan 

3.PofPS.18 2

y  4 px  2 yy  4 p  y  2 p / y  slope of tangent at P  x , y  is m  2 p / y . The slope of 1

FP is m2 

y1 , so by the formula from Problem 17  a  , x1  p y1 2p x 1  p  y1

y  x  p y12  2 p x 1 p  1 1   2 p  y1 p  y1 x1  p y1 x1  p  2 py1 1  y  x  1  1  2 2 p  p  x1  2 p 4 px1  2 px1  2 p    x1y1  py1  2 py1 y1  p  x1  y1

tan 

 slope of tangent at P  tan  Since 0   ,  



, this proves that   

2 3.PofPS.19 Since  ROQ  OQP   , the triangle QOR is isosceles, so

QR  RO  x . By the Law of Cosines, x2  x2  r2  2rxcos . Hence, 2rxcos  r2 , so x 

 r 2  r  . Note that as y  0 ,  0(since 2rcos 2cos

sin  y / r) , and hence x 

r 2cos0



. Thus, as P is taken closer and closer

to the x -axis, the point R approaches the midpoint of the radius AO .

461

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.PofPS.20 Suppose that y  mx  c is a tangent line to the2 ellipse. Then it intersects the ellipse at only one point, so 2

(mx  c) 2 2 2 2 2 2 2 2 2 the discriminant of the equation 2   1  b  a m x  2mca x  a c  a b  0 a b2 x





must be 0; that is,

  2mca 2   4  b 2  a 2 m2  a 2 c 2  a 2b 2   4a 4 c 2 m2  4a 2b 2c 2  4a 2b 4  4a 4 c 2 m2  4a 4b 2 m 2 2

 4a2b2  a 2 m 2  b2  c2 



Therefore, a2m2  b2  c2  0 . Now if a point  ,   lies on the line y  mx  c , then c    m , so from above, 2



0  a m  b  (   m )  a  

 m  2m  b  

b2   2 m   0.  m  2 a  2 a2  2 2

2

(a) Suppose that the two tangent lines from the point  ,   to the ellipse have slopes m and

. Then m and

m 2

2

are roots of the equation

b2   2





1 

z 2  0 . This implies that  z  m   z   0  m a2  2 a  2 1 1     z2   m   z  m   0 , so equating the constant terms in the two m m      2 2  b  1 m  1 , and hence b2   2  a2   2 . So  ,   lies on the quadratic equations, we get   a2   2  m  2 2 2 2 hyperbola x  y  a  b . (b) If the two tangent lines from the point  ,   to the ellipse have slopes 1 1 1  m and  m , then m and  m are roots of the quadratic equation, and so  z  m  z  m  0 , and z







b2   2 equating the constant terms as in part (a), we get

a  2

 2

 2

 1 , and hence b      a . So the

point  ,   lies on the circle x  y  a  b . 2

3.PofPS.21

462

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  x4  2x2  x  y  4x3  4x 1. The equation of the tangent line at x  a is



 





 

y  a 4  2a2  a  4a3  4a 1  x  a  or y  4a3  4a 1 x  3a4  2a2

 and similarly for

x  b . So if at x  a and x  b we have the same tangent line, then 4a  4a 1  4b3  4b 1 and 3





3a4  2a2  3b4  2b2 . The first equation gives a3  b3  a  b  a  b  a 2  ab  b2  a  b . Assuming a  b , we have 1  a  ab  b . The second equation gives 2



 





3 a 4  b4  2 a 2  b2  3 a 2  b2

a 2  b2   2 a 2  b2  which is true if a  b . Substituting

into 1  a2  ab  b2 gives 1  a2  a2  a2  a  1 so that a 1 and b  1 or vice versa. Thus, the points 1, 2 and 1, 0 have a common tangent line.

As long as there are only two such points, we are done. So we show that these are in fact the only two



such points. Suppose that a2  b2  0 . Then 3 a 2  b2



 



1 1 1 2  , so b  . Hence, a2   , 3 3 3 3a 9a2 3 2 1 1 1 2 2 2 so 9a 4 1  6a 2  0  9a 4  6a 2 1   3a 2 1 . So 3a 1  0  a   b    a2 , 3 9a2 3 contradicting our assumption that a2  b2 . or a2  b2 

 a 2  b2   2 a 2  b2  gives 3a 2  b2   2

. Thus, ab  a2  ab  b2  a 2  b2  1

3.PofPS.22







Suppose that the normal lines at the three points a ,1a2 1 , a 2, a22 , and

a , a2  intersect at a common 3

point. Now if one of the ai is 0 (suppose a1  0 ) then by symmetry a2  a3 , so a1  a2  a3  0 . So we can assume that none of the ai is 0 . The slope of the tangent line at

a , a2  is 2a , so the slope of the normal line is  1 and its equation i

2ai

1  x  ai  . We solve for the x -coordinate of the intersection of the normal lines from 2ai  1 1  1 1 a , a2  and a , a2  : y  a21   x  a   a2   x  a 2   x  2a    a22  a12  1 1 2 2 1 2 2a 2a 2a 1 2  2 1   a1  a2   x  a  a  a  a  x  2a a a  a  1 2   1 2 (1). Similarly, solving for the x -coordinate  1 2 1 2 2a a  1 2 

is y  a i   2







of the intersections of the normal lines from a ,1a2 1 and a 3, a23

 gives x  2a a a  a  1 3



Equating (1) and (2) gives

a a  a   a a  a   a a  a   a2  a2    a  a  a  a   2

a1   a2  a3   a1  a2  a3  0 . 3.PofPS.23

463

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Because of the periodic nature of the lattice points, it suffices to consider the points in the 5 2 grid shown. We can see that the minimum value of r occurs when there is a line with slope

2 5

the circle centered at 3,1 and the circles centered at 0, 0 and 5, 2

which touches

To find P , the point at which the line is tangent to the circle at 0, 0 , we simultaneously solve

5 25 2 4 2 2 x  r2  x2  r x x2  y2  r 2 and y   x  x  2 4 29 2 2 2 we either use symmetry or solve (x  3)  ( y 1)  r and y 1  



r, y   5 r . To find Q , 29 29

 x  3 . As above, we get

2 2 5 r . Now the slope of the line PQ is 2 , so x  3  r, y  1 5 29 29 10 5  5  1 r 1 r   r  2 29 29 m   29 10r    29   PQ 2 2 4 r 3 r 3 29  4r 5 3 r  29 29 29 29 . So the minimum value of r for which any line 5 29  50r  6 29  8r  58r  29  r  58 2 with slope intersects circles with radius r centered at the lattice points on the plane is 5 29 r  0.093 . 58 

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION REVIEW TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

464

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

END OF SECTION EXERCISE SOLUTIONS 3.R.1

y  (x2  x3)4  y  4(x2  x3)3(2x  3x2 )  4(x2 )3(1 x)3 x(2x  3)  4x7 (x 1)3(3x  2) 3.R.2

y





 x1/ 2  x3/5  y   12 x3/ 2  53 x8/5  



5x 5 x3



 1 x8/5 (5x1/10  6)

2x x

3.R.3

x2  x  2 y

3/2

x

1/2

1/2

x

 2x



 y 2x

1/2

3/2

1/2

2x

x

1 1 3 x   2 2 x x3

3.R.4

tan x

y

 y  

1 cos x

(1 cos x) sec2 x  tan x(sin x)

(1 cos x) sec2 x  tan x(sin x) 

(1 cos x)2

3.R.5

(1 cos x)2

y  x2 sin  x  y  x2 (cos x)  (sin  x)(2x)  x( x cos x  2sin  x) 3.R.6  1  x y  x cos1 x  y  x   (cos1 x)(1)  cos1 x    1 x2  1 x2  

3.R.7

t 4 1



y 4  y  t 1

(t4 1)4t3  (t4 1)4t3 (t4 1)2

3.R.8

(xey ) 

4t3[(t4 1)  (t4 1)] 

(t4 1)2

8t3 

(t4 1)2

( y sin x)  xey y  ey 1  y cos x sin x  y xey ysin xy  y cos x  e y

 (xey sin x) y y cos x  ey  y 

y cos x  ey xey sin x

3.R.9

465

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

1 1  1  1 ln x (x ln x)   x   ln x 1  x ln x x ln x x x ln x   1 1 1 ln x 1 Another method: y  ln(x ln x)  ln x  ln(ln x)  y     x ln x x x ln x y  ln(x ln x)  y 

3.R.10

y  emx cos nx  y  emx (cos nx)  cos nx(emx )  emx (sin nx  n)  cos nx(emx  m)  emx (m cos nx  n sin nx) 3.R.11

y  x cos x

 y  x sin x( 1 x1/2 )  cos x( 1 x1/2 )  cos x  x sin x 2 2   2 x

3.R.12

y  (arcsin 2x)2  y  2(arcsin 2x)  (arcsin 2x)  2 arcsin 2x  

1 (2x)



3.R.13

x 2  e1/ x   e1/ x (x 2 )

1/ x

y  e  y  x2 



(x2 )2

 2 1/x 

x e (1/ x2 )  e1/x (2x) x4





 2 

4 arcsin 2x 1 4x2

e1/x (1 2x)



3.R.14

y  ln(sec x)  y 

sec x dx

(sec x) 

(sec x tan x)  tan x

sec x

3.R.15

( y  x cos y) 

x y  y  x(sin y  y)  cos y  x y y  2x   dy 2

y  x sin y  y  x2 y  2xy  cos y  (1 x sin y  x2 ) y  2xy  cos y  y 



2xy  cos y 1 x sin y  x2

3.R.16 4

2  u 1   u 1 3 (u  u 1)(1)  (u 1)(2u 1)  y 2 y  4  2    u  u 1  u  u 1  (u2  u 1)2     3  u 1 u2  u 1 2u2  u 1  4  u 13 u 2  2u  2  4 2 (u2  u 1)2 (u2  u 1)5 (u  u 1)3







3.R.17

466



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  arctan x  y 

(arctan x)1/2 





1 x

1 2 arctan (1 x2 )

3.R.18

y  cot(csc x)  y  csc2 (csc x)  (csc x cot x)  csc2 (csc x) csc x cot x



3.R.19







2 t  1 t (1)  t(2t)  t   t  1 t2 2 y  sec   y  tan   sec2      2 2 2 2 2  1 t   1 t   1 t  1 t 2 1 t 2 3.R.20 

















y  exsecx  y  exsecx (x sec x tan x  sec x 1)  sec xexsecx (x tan x 1) 3.R.21  1  y  3xln x  y  3xln x (ln 3)   x   ln x 1  3xln x (ln 3)(1 ln x) x   3.R.22

y  sec(1 x2 )  y  2x sec(1 x2 ) tan(1 x2 ) 3.R.23

y  (1 x 1 ) 1  y  1(1 x 1 )2 [(x2 )]  (11/ x) 2 x2    (x 1) / x 2 x 2  (x 1)2 3.R.24

y 3





 x  x

 x x x



1/3

 y  



1 3

 x  x 

4/3

 1  1  2  

3.R.25

sin(xy)  x2  y  cos(xy)(xy  y)  2x  y  x cos(xy) y  y  2x  y cos(xy)  2x  y cos(xy) y[x cos(xy) 1]  2x  y cos(xy)  y  x cos(xy) 1 3.R.26

467

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



y  sin x  y  12 sin x



1/ 2

cos

 x  x  2 1 x  4 cos x sin x

3.R.27

y  log5 (1 2x)  y 

2 (1 2x) ln 5

3.R.28

y  (cos x)x ln y  ln(cos x)x  x ln(cos x) 

1 y  x  (sin x)  ln(cos x) 1  y cos x

y  (cos x)x (ln(cos x)  x tan x) 3.R.29 



y  lnsin x  12 sin2 x  y 

1 cos x  12  2sin x cos x  cot x sin x cos x sin x

3.R.30

(x2 1)4

y

 ln y  4 ln(x 1)  3ln(2x 1)  5ln(3x 1)  (2x 41)3(3x 1)53 5 6 15  y  (x2 31)4 5  8x 2    2  3  y  2 2x    y 1 x 2x 1 3x 1 (2x 1) (3x 1) 1 x  2x 1  3x 1     



3.R.31

y  x tan1(4x)  y  x 



4x 1 1  tan1(4x) 2  4  tan (4x) 1  1 (4x) 116x2

3.R.32

y  ecos x  cos(ex )  y  ecos x (sin x) [sin(ex )  ex ]  sin x  ecos x  ex sin(ex ) 3.R.33

y  ln sec 5x  tan 5x  1 5sec5x(tan 5x sec5x) y  sec5x tan 5x sec2 5x 5   5sec5x sec5x  tan 5x sec5x  tan 5x





3.R.34

y  10tan  y  10tan  ln10 sec2       (ln10)10tan sec2   

3.R.35

468



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y  cot(3x2  5)  y  csc2 (3x2  5)(6x)  6x csc2 (3x2  5) 3.R.36

y

 y 

 

t ln t



4 1    ln  t4   t   4t3   ln t  4   t4 2 t ln  t 4   2 t ln  t 4 



3.R.37







y  sin tan 1 x3  y  cos tan 1 x3

sec 1 x [3x / (2 1 x )] 2

3.R.38





y  arctan arcsin x  y 



1 arcsin x







1 x 2 x

3.R.39

y  tan2 (sin )  [tan(sin )]2  y  2[tan(sin )]sec2 (sin ) cos 3.R.40

xey  y 1  xey y  ey  y  ey  y  xey y  y  e y / (1 xey ) 3.R.41

y

x 1(2  x)5  ln y  1 ln(x 1)  5 ln(2  x)  7 ln(x  3) 

y  

(x  3)7 1 5 7  



5    x 1(2  x) 

2(x 1) 2  x x  3 (2  x)4 3x2  55x  52  2 x 1(x  3)8 3.R.42 y



(x  )4 y

x4  4

 y  

(x  3)7



1 

  2(x 1)

(x4   4 )(4)(x  )3  (x  )4 (4x3) (x4   4 )2

3.R.43



5 2 x



7   x  3 

4(x  )3(3  x3) 

(x4   4 )2

y  (sin mx) / x  y  (mx cos mx sin mx) / x2

469

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.44

2 2(x 1)(x  4) 2x 2 2   y  ln x  4  ln x  4  ln 2x  5  y   2 x  4 2x  5 (x  2)(x  2)(2x  5) 2x  5



3.R.45



y  cos e tan 3x





  



y  sin e tan 3x  e





3sin e tan 3x e tan3x sec2 (3x)  tan3x

 (tan 3x) 1 2

1/2

sec (3x) 3  2

2 tan 3x

3.R.46

  1 1 cos x  y  2sincos sin  x coscos sin  x    sin sin  x   2 sin  x  sin cos sin  x coscos sin  x sin sin  x cos x 2

y  sin2 cos sin  x  sin(cos sin  x ) 



sin  x



3.R.47   f (t)  4t 1  f (t)  12 (4t 1)1/2  4  2(4t 1)1/2  f (t)  2   1 (4t 1)3/2  4  4 / (4t 1)3/2, so f (2)  4 / 93/2   4 . 2

3.R.48

g( )   sin  g( )   cos sin  g ( )   (sin )  cos  cos  2 cos    sin  , so g  6   2cos  6   6 sin  6  2

 3 / 2    (1/ 2)  3   /12  6

3.R.49

x6  y6  1  6x5  6 y5 y  0  y  x5 / y5  5 4 5 4 4 4 5 5 4 6 6 5 4 y   y (5x )  x (5y y)   5x y [ y  x(x / y )]   5x [( y  x ) / y )]   5x y10 y10 y10 y11

470

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.R.50



f (x)  (2  x)1  f (x)  (2  x)2  f (x)  2(2  x)3  f (x)  2 3(2  x)4  n! f (4) (x)  2 3 4(2  x)5. In general, f (n) (x)  2 3 4   n(2  x)(n1)  (2  x)n1.



3.R.51



x x x We first show it is true for n = 1: f (x)  xe  f (x)  xe  e  (x 1)e . We now assume it is true x

(x)  (x  k)e x . With this assumption, we must show it is true for n  k 1:

for n = k: f

(k)

f (k1) (x) 

d  f (k) (x)  (x  k)ex  (x  k)ex  ex  (x  k) 1ex   x  (k 1)ex.    dx dx

Therefore, f

(n)

(x)  (x  n)ex by mathematical induction.

3.R.52

y  4sin2 x  y  4(2sin x cos x). 3 At   ,1, y  8 1   2 3 so an equation of the tangent line is y  2 3  x    1, or 6

y  2 3x 1  3 / 3. 3.R.53

(x2 1)(2x)  (x2 1)(2x) 4x x2 1  y  2 1  y   2 2 . 2 2 x   x 1  x 1

At 0, 1, y  0 so an equation of the tangent line is y  0(x  0) 1, or y  1. 3.R.54

 



1 1 2cos x 2 y  1 4sin x  y  . At 0,1, y   2, so an 4 cos x  2 1 4sin x 1 4sin x 1 equation of the tangent line is y  2(x  0) 1, or y  2x 1. 3.R.55 𝑥 = ln 𝑡 , 𝑦 = 𝑡2 + 1, (0, 2). If (𝑥, 𝑦) = (0, 2), then 𝑥 = ln 𝑡 = 0 ⇒ 𝑡 = 1. 𝑑𝑦 𝑑𝑥 1 𝑑𝑦 𝑑𝑦/𝑑𝑡 2𝑡 = 2𝑡2 = 2𝑡, = , and = = 𝑑𝑡 𝑑𝑡 𝑡 𝑑𝑥 𝑑𝑥/𝑑𝑡 1/𝑡 So 𝑑𝑦 = 2 when t = 1. An equation of the tangent to the curve at (0, 2) is 𝑑𝑥

𝑦 − 2 = 2(𝑥 − 0), or 𝑦 = 2𝑥 + 2.

471

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.56 𝑥 = 𝑡3 − 2𝑡2 + 𝑡 + 1, 𝑦 = 𝑡2 + 𝑡, (1, 0). If (𝑥, 𝑦) = (1, 0), then 𝑦 = 𝑡2 + 𝑡 = 0 ⇒ 𝑡(𝑡 + 1) = 0 ⇒ 𝑡 = 0 or 𝑡 = 1, but we see that t = 0 satisfies the equation 𝑥 = 𝑡3 − 2𝑡2 + 𝑡 + 1 = 1. 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑦/𝑑𝑡 2𝑡 + 1 = 2𝑡 + 1, = 3𝑡2 − 4𝑡 + 1, and = = 𝑑𝑡 𝑑𝑡 𝑑𝑥 𝑑𝑥/𝑑𝑡 3𝑡2 − 4𝑡 + 1 𝑑𝑦 0+1 = 1 when t = 0. An equation of the tangent to the curve at (1, 0) is So = 𝑑𝑥

0−0+1

𝑦 − 0 = 1(𝑥 − 1), or 𝑦 = 𝑥 − 1.

3.R.57

x2  4xy  y2  13  2x  4(xy  y)  2 yy  0  x  2xy  2 y  yy  0  x  2 y 2xy  yy  x  2 y  y(2x  y)  x  2y  y . 2x  y 2  2 4 At 2,1, y    , so an equation of the tangent line is y   4 (x  2) 1, or y   4 x  13 . 4 1 5 5 5 3 The slope of the normal line is 5 , so an equation of the normal line is y  (x  2) 1, or y  x  . 5

3.R.58





y  (2  x)ex  y  (2  x) e x  ex (1)  ex[(2  x) 1]  ex (x 1). At 0, 2  ,

y  1(1)  1, so an equation of the tangent line is y  1(x  0)  2, or y  x  2. The slope of the normal line is1, so an equation of the normal line is y  1(x  0)  2, or y  x  2. 3.R.59





 





 x 11  x 1 e x/ x1 

y  e x/ x 1  y  e x/ x 1 

 x 1



 x 1



. At 4, e

4/5

, y 

e4/5

4 1

 2

e4/5

4/5

4/5 4/5 x 29e . The so an equation of the tangent line is y  e 4/5   e  x  4, or y  e 25 25 25 4/5 slope of the normal line is 25e , so an equation of the normal line is

4/5 4/5 4/5 y  e4/5  25e4/5  x  4  , or y  25e x 100e  e .

3.R.60

f (x)  xesin x  f (x)  x esin x (cos x)   esin x (1)  esin x (x cos x 1). As a check of our work, we notice from the graphs that f (x)  0 when f in increasing. Also, we see in the larger viewing

rectangle a certain similarity in the graphs of f and f  :the sizes of the oscillations of f and f  are linked.

471

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.61

3.R.62



3.R.63



y  sin x  cos x  y  cos x sin x  0  cos x  sin x and 0  x  2  x   or 5 , so the points are

 , 2  and  ,  2 .  4

5 4

3.R.64

x2  2 y2  1  2x  4 yy  0  y 

x

 1  x  2 y. Since the points lie on the ellipse, we have

 2 y   2 y 2  1  6 y 2  1  y   16 . The points are   26 , 16  and  26 ,  16  . 2

3.R.65

f (x)  (x  a)(x  b)(x  c)  f (x)  (x  b)(x  c)  (x  a)(x  c)  (x  a)(x  b) f (x) (x  b)(x  c)  (x  a)(x  c)  (x  a)(x  b) 1  1  1 . So   f (x) (x  a)(x  b)(x  c) x a x b x c Or: f (x)  (x  a)(x  b)(x  c)  ln f (x)  ln x  a  ln x  b  ln x  c  

1 1 f (x)  1   .  f (x) x  a x  b x  c 

472

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.66 2 2 (a) cos 2x  cos x sin x  2sin 2x  2 cos x sin x  2sin x cos x  sin 2x  2sin x cos x (b) sin(x  a)  sin x cos a  cos x sin a  cos(x  a)  cos x cos a sin x sin a. 3.R.67 (a) S(x)  f (x)  g(x)  S(x)  f (x)  g(x)  S(1)  f (1)  g(1)  3 1  4 (b) P(x)  f (x)g(x)  P(2)  f (2)g(2)  g(2) f (2) 1 4 1 2  6 (c) Q(x)  f (x)  Q(1)  g(1) f (1)  f (1)g(1)  33  2 1  7

g(x) [g(1)]2 32 9 (d) C(x)  f (g(x))  C(2)  f (g(2))g(2)  f (1)  4  3 4 12 3.R.68 (a) P(x)  f (x)g(x)  P(x)  f (x)g(x)  g(x) f (x) 

P(2)  f (2)g(2)  g(2) f (2)  (1)  60  (4)  03  1 2  4 (1)  2  4  2 30 30 

f (x) g(x) f (x)  f (x)g(x)   Q(x)   g(x) [g(x)]2 g(2) f (2)  f (2)g(2) 4 (1) 1 2 6 3 Q(2)     2 2 [g(2)] 4 16 8 (c) C(x)  f (g(x))  C(x)  f (g(x))g(x)  (b) Q(x) 

C(2)  f (g(2))g(2)  f (4)g(2)   60 53(2)  3 2  6



3.R.69

f (x)  x g(x)  f (x)  x2 g(x)  g(x)2x  x xg(x)  2g(x)  2



3.R.70



f (x)  g(x 2 )  f (x)  g(x2 )2x  2xg(x2 ) 3.R.71



f (x)   g(x)   f (x)  2g(x)  g(x) 3.R.72



f (x)  g(g(x))  f (x)  g(g(x))  g(x) 3.R.73

473



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (x)  g(ex )  f (x)  g(ex )  ex

 

3.R.74

f (x)  e g(x)  f (x)  e g(x) g(x)

 

3.R.75

f (x)  ln (x)  f (x) 

1 g(x)  g(x) g(x) g(x)

3.R.76

f (x)  g(ln x)  f (x)  g(ln x) 



g(ln x)

3.R.77







 11x2

f  x  g tan1 x  f  x  g tan1 x 

3.R.78 f  x  tan1 g  x  f  x 

1 1  g  x 

 g x  2

g x 1  g  x 

3.R.79

[ f (x)  g(x)][ f (x)g(x)  f (x)g(x)]  f (x)g(x)[ f (x)  g(x)] f (x)g(x)   h(x)  f (x)  g(x) [ f (x)  g(x)]2 [ f (x)]2 g(x)  f (x)g(x) f (x)  f (x)g(x)g(x) [g(x)]2 f (x)  f (x)g(x) f (x)  f (x)g(x)g (x)  [ f (x)  g(x)]2 f (x)[g(x)]2  g(x)[ f (x)]2  [ f (x)  g(x)]2 h(x) 

3.R.80

h(x) 



f (x) f (x)g(x)  f (x)g(x) f (x)g(x)  f (x)g(x)  h(x)   g(x) 2 f (x) / g(x)[g(x)] 2 2[g(x)]3/2 f (x)

3.R.81 Using the Chain Rule repeatedly, h(x)  f (g(4sin x))  d d h(x)  f (g(sin 4x))  (g(sin 4x))  f (g(sin 4x))  g(sin 4x) (sin 4x) dx dx

474

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 4 f (g(sin 4x))  g(sin 4x) cos 4x

3.R.82





h  x   f g  f  x   g  x  h x  f  g  f  x   g  x   

 dxg  f  x   g  x 



 f  g  f  x   g  x    g  f  x   g  x   

 f  x   g  x    dx  f   g  f  x   g  x     g  f  x   g  x     f  x  g  x   g  x  f   x   d

3.R.83



3.R.84



1  2(ln(x  4) and y  0  ln(x  4)  0  x4 x4

y  ln(x  4)  y  2 ln(x  4)  2

x  4  e0  x  4  1  x  3, so the tangent line is horizontal at the point 3, 0. 3.R.85 (a) The line x  4 y  1 has slope 1 . A tangent to y  e has slope 1 when y  ex  1  x

and the point of tangency is  ln 4, 14 . Thus, an equation of the tangent line is y  (x  ln 4)  , or y  1 x  1 ln 4 1.

x  ln 4  ln 4. Since y  e , x

the y-coordinate is 14 1 1 4



 dx e

(b) The slope of the tangent at the point a, ea is

 ea. Thus, an equation of the tangent line is xa

y  e (x  a)  e . We substitute x  0, y  0 into this equation, since we want the line to pass a

475

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 a a through the origin: 0  e (0  a)  e  0  ea  ea  a  1. So an equation of the tangent line



at the point a, e

  1, e is y  e(x 1)  e or y  ex.

3.R.86 Any point on the curve f (x)  e

x





has coordinates P  x 1, y 1  x 1, e  x1 . So a line through

y 0

4, 0 that is tangent to the curve must go through P, and therefore has slope m  1

x1  4



e x1 x1  4

. But a

line tangent to the curve at P also has slope f (x1)  ex1 . So e 1x 

 e x1  e x1 (x1  4)   x1  1  x1  1  4  3 and y1  e 3. e x1  4

Therefore, the slope of the given line is m 

y1  e3   1 , x1  4 3  4 e3

3.R.87

y  f (x)  ax2  bx  c  f (x)  2ax  b. We know that f (1)  6 and f (5)  2, so 2a  b  6 2 and10a  b  2. Subtracting the first equation from the second gives12a  8  a   3 . 2 14 Substituting  for a in the first equation gives b  . Now f (1)  4  4  a  b  c, so 3

c  4  3  3  0 and hence, f (x)   3x  2

14 3

3.R.88 The graph of C increases, though more slowly as time increases, from t  0 to t  ln 2, corresponding to a positive (though decreasing) graph of C. The concentration of the injected drug is increasing during this time, but more slowly as time passes. The concentration is its largest at t  ln 2, corresponding to a maximum of the graph of C and a zero of the graph of C. Beyond t  ln 2, the concentration decreases toward zero. The graph of C is negative here. The concentration decreases more and more slowly beyond a point that corresponds to a minimum of the graph of C.

476

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



3.R.89

s(t)  Ae

ct

cos(t   )  v(t)  s(t)  A ect[ sin(t   )]  cos(t   )(cect ) 

ct

  Ae





 sin(t   )  c cos(t   ) 





a(t)  v(t)  A ect  2 cos(t   )  c sin(t   )    sin(t   )  c cos(t   )  (cect )  Ae ct (c 2   2 ) cos(t   )  2c sin(t   ) 

 3.R.90





3 24  20  23  48 22  48 2  0 (a) average velocity = s(2)  s(0)   208 ft/s. 20 2 3 2 (b) The particle is at rest when its velocity is zero. So v(t)  s(t)  12t  60t  96t  48  0

 t  t1  0.4055, or t  t2  6.1883 seconds t  0. (c) v(t)  0  t1  x  t2 , and v(t)  0  t  t1and t  t2. Thus the particle changes direction at

t  t1  0.4055 s and at t  t2  6.188 seconds (d) Distance traveled to the right is s t1   s(0)  10.319 feet. Distance traveled to the left is s t1   s 4  1098.319 feet. Total distance traveled is 10.319 1098.319 1108.638 feet 2 2 (e) a(t)  v(t)  36t 120t  96  a(0)  96 ft/s .

(f) When t  0.5, v(t)  13.5  0, and a(0.5)  147  0. Since the velocity and acceleration have the same sign at t  0.5, the speed of the particle is increasing.

3.R.91 s(50)  s(0) 10  50e10 10  10    e  0.0000454. 50  0 50 (b) v(t)  s(t)  tet/5( 1 5)  et/5   15et/5(t  5)  0  (t  5)  0  t  5. Thus, the particle is moving to the left when t  5. (a) Average velocity 

477

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

The particle starts at position 10 when t  0, and moves to the right until t  5. When t  5 the particle is

5 at position s(5)  10  . The particle then moves to the left until t  50, when its position is e 50 s(50)  10   10. So the particle starts at its leftmost position, 10. e10  t/5 (c) The particle is at rest when v(t)  0  t  5. a(t)  v(t)   1 e (t  5) 

 5  1   e (1)  (t  5)  e    e (t 10)  a(5)    0.0736. 5 5 5 25 5e (d) v(t)  0 for 0  t  5 and v(t)  0 for 5  t  50. 5 5 Distance traveled to the right is s(5)  s(0)  10  10   1.839. e e 5 50 Distance traveled to the left is s(5)  s(50)    1.837. e e10 5 5 50 Total distance traveled is10  10    3.677. e e e10 4 4 (e) v(20)  3e  0 and a(20)  52 e  0. Because the velocity and acceleration have different signs 1

t/5

at t  20, the speed of the particle is decreasing.

3.R.92 (a) The particle is moving to the right when its velocity is positive. v(t)  etsint 1  0  etsint  1  t sin t  ln1  0  sin t  0 t 2k , (2k 1) , k a nonnegative integer. (b) speed = v(t) . v(t)  0.5  v(t)  0.5  t  3.006 or v(t)  0.5  t  3.350.

v(3)  v(1) (c) Average acceleration =

3 1



e

3sin 3

 

 e

1  esin1 1 2

3sin 3

 esin1

 0.396.

v(3)  v(1) e3sin3  esin1  (using technology)  t  2.05235. 3 1 2 (e) v(6)  0.813  0, a(6)  1.025  0. Because the velocity and acceleration have opposite signs when t  6, the speed of the particle is decreasing. (d) a(t) v(t)  etsint   t cos t  sin t  

3.R.93

s(t)  t3  2t2  4t  5  0  v(t)  s(t)  3t 2  4t  4  (3t  2)(t  2)  0  t  2. Therefore, the distance traveled to the right is s(5)  s(2)  60  (3)  63 and the distance traveled to the left is s(2)  s(0)  3  5  8. Therefore the total distance traveled is 63 + 8 = 71

3.R.94

478

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

dy dy 1 dx dx dx dx dy  12, at  x  ln x 1  1 ln x   / 1 ln x. Given that dt xdxdt dy dt dt dt dt dt 12 12 12  / 1 ln x     4, the point e2 , 2e2 ,

y  x ln x 





  1 2

1 ln e2

479

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.95 2

s(t)  sin t  e cos t  v(t)  s(t)  sin t  e cos t  2 cos t  ( sin t)  e cos t (cos t)





 2 cos3 t  cos t ecos t  0  t  4 , 2 , 34 . Since v(t) changes sign at each of these values in the 2

interval 0  t   , the particle changes direction 3 times

3.R.96 1 1  1   (t ln t)1  v(t)  s(t)  (1)(t ln t)2 t   ln t 1   1 ln t .  t 2 2 3  3t ln t 3 3t (ln t) 1 1 ln t v(t)  0   2  0  1 ln t  0  ln t  1  t  . The particle changes from moving to 3t (ln t)2 e 1 the right to moving to the right at t  . e s(t) 







3.R.97 (a) 𝑠(𝑡) = √𝑏2 + 𝑐2𝑡 ⇒ 𝑣(𝑡) = 𝑠′(𝑡) = [1/(2√𝑏2 + 𝑐2𝑡)]2𝑐2𝑡 = 𝑐2𝑡/√𝑏2 + 𝑐2𝑡 ⇒ 𝑏 2𝑐 2 𝑐2√𝑏2 + 𝑐2𝑡 − 𝑐2𝑡(𝑐2𝑡/√𝑏2 + 𝑐2𝑡) ′ (𝑡) 𝑎(𝑡) = 𝑣 = = 2 𝑏 2 + 𝑐 2𝑡 (𝑏 + 𝑐2𝑡2)3/2 (b) 𝑣(𝑡) > 0 for t > 0, so the particle always moves in the positive direction.

3.R.98 3 2 (a) y  t 12t  3  v(t)  y  3t 12  a(t)  v(t)  6t

(b) v(t)  3(t  4)  0 when t  2, so it moves upward when t  2 and downward when 0  t  2. 2

(c) Distance upward  y(3)  y(2)  6  (13)  7, Distance downward  y(0)  y(2)  3  (13)  16. Total distance  7 16  23. (d)

(e) The particle is speeding up when v and a have the same sign, that is, when t  2. The particle is slowing down when v and a have opposite signs; that is, when 0  t  2.

3.R.99

480

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

The linear density ρ is the rate of change of mass m with respect to length x.





m  x 1 x  x  x3/2    dm / dx  1 32 x, so the linear density when x  4 is 1 32 4 = 4 kg/m.

3.R.100

r is a constant h is a constant

(a) V  13  r h  dV / dh  13  r 2

(b) V  13  r h  dV / dr  23 rh 2

3.R.101 

(a) C(x)  920  2x  0.02x  0.00007x  C '(x)  2  0.04x  0.00021x (b) C '(100)  2  4  2.1  $0.10 / unit. This value represents the rate at which costs are increasing as the hundredth unit is produced and is the approximate cost of producing the 101st unit. (c) The cost of producing the 101st item is C(101)  C(100)  990.10107  990  $0.10107, 2

slightly larger than C '(100).

3.R.102 1 3

′

−2

(a) ƒ(𝑥) = √1 + 3𝑥 = (1 + 3𝑥) ⇒ ƒ (𝑥) = (1 + 3𝑥) 3, so the linearization of f at a = 0 is 𝐿(𝑥) = 1

−2

′ ƒ(0) + ƒ (0)(𝑥 − 0) = 13 + 1 3𝑥 = 1 + 𝑥. Thus √1 + 3𝑥 ≈ 1 + 𝑥 ⇒ 3√1.03 = 3√1 + 3(0.1) ≈ 1 + (0.01) = 1.01.

(b) The linear approximation is 3√1 + 3𝑥 ≈ 1 + 𝑥, so for the required accuracy we want 3√1 + 3𝑥 − 3 0.1 < 1 + 𝑥 < √1 + 3𝑥 + 0.1. From the graph, it appears that this is true when -0.23 < x < 0.40.

𝜋) 𝑥𝑑𝑥. When x = 60 and dx = 0.1, 4

3.R.103 2

𝐴 = 𝑥2 + 𝜋 (1 𝑥) = (1 + 𝜋) 𝑥2 ⇒ 𝑑𝐴 = (2 + 2

𝑑𝐴 = (2 + 12 +

3𝜋 2

𝜋) 60(0.1)

= 12 + 3𝜋, so the maximum error is approximately 2

≈ 16.7 cm2.

481

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.104 x17 1

d



 17(1)  17   x  x1 x 1  dx x1

lim

3.R.105 4 lim 16  h  2

h0

1  d 4    x 3/ 4   x   x16 4  dx x16 4

 16  4



1 32

3.R.106

 3   sin   lim cos  0.5   d cos    /3 3 2 d   /3   / 3  

3.R.107 lim h0

e33h  e9   d

 

 e dx



  x3

33

 3e



 3e 9

482

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.R.108 lim

√1 + tan 𝑥 − √1 + sin 𝑥 𝑥

𝑥→0

= lim

(√1 + tan 𝑥 − √1 + sin 𝑥)(√1 + tan 𝑥 + √1 + sin 𝑥) 3

𝑥 (√1 + tan 𝑥 + √1 + sin 𝑥) (1 + tan 𝑥) − (1 + sin 𝑥) 𝑥→0

= lim

𝑥→0 𝑥3(√1 + tan 𝑥 + √1 + sin 𝑥)

sin 𝑥 (1/ cos 𝑥 − 1)

cos 𝑥 𝑥→0 𝑥3(√1 + tan 𝑥 + √1 + sin 𝑥) cos 𝑥 sin 𝑥 (1 − cos 𝑥) · 1 + cos 𝑥 = lim 𝑥→0 𝑥3(√1 + tan 𝑥 + √1 + sin 𝑥) cos 𝑥 1 + cos 𝑥 sin 𝑥 · sin2 𝑥 = lim 𝑥→0 𝑥3(√1 + tan 𝑥 + √1 + sin 𝑥) cos 𝑥 (1 + cos 𝑥) sin 𝑥 3 1 ) lim = (lim 𝑥→0 𝑥3(√1 + tan 𝑥 + √1 + sin 𝑥) cos 𝑥 (1 + cos 𝑥) 𝑥→0 𝑥 1 1 = 13 · = (√1 + √1) · 1 · (1 + 1) 4 = lim

3.R.109 Differentiating the first given equation implicitly with respect to x and using the Chain Rule, we 1 obtain ƒ(𝑔(𝑥)) = 𝑥 ⇒ ƒ′(𝑔(𝑥))𝑔′(𝑥) = 1 ⇒ 𝑔′(𝑥) = . Using the second given equation to expand the denominator of this expression give 𝑔′(𝑥) =

ƒ′(𝑔(𝑥)) 1

1+[ƒ(g(𝑥))]

states that ƒ(𝑔(𝑥)) = 𝑥, so 𝑔′(𝑥) =

. But the first given equation

1 1 + 𝑥2

3.R.110 𝑑 [ƒ(2𝑥)] = 𝑥2

𝑑𝑥

1 1 ⇒ ƒ′( 2𝑥) · 2 = 𝑥2 ⇒ ƒ′( 2𝑥) = 1 𝑥2. Let 𝑡 = 2𝑥. Then ƒ′(𝑡) = 2 (2 𝑡) = 𝑡 2, so 2

ƒ′(𝑥) = 1 𝑥2. 8

3.R.111 Let (b, c) be on the curve, that is, 𝑏2/3 + 𝑐2/3 = 𝑎2/3. Now 2 2 −1/3 𝑑𝑦 = 0 , so at (b, c), the slope of the tangent line is −(𝑐/ 𝑥2/3 + 𝑦2/3 = 𝑎2/3 ⇒ 𝑥−1/3 + 𝑦 3

𝑑𝑥

483

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 𝑐

𝑏)1/3 and an equation of the tangent line is 𝑦 − 𝑐 = − ( )3 (𝑥 − 𝑏) or 𝑏

𝑦 = −(𝑐/𝑏)1/3𝑥 + (𝑐 + 𝑏2/3𝑐1/3). Setting y = 0, we find that the x-intercept is 𝑏1/3𝑐2/3 + 𝑏 = 𝑏1/3(𝑐2/3 + 𝑏2/3) = 𝑏1/3𝑎2/3 and setting x = 0, we find the y-intercept is 𝑐 + 𝑏2/3𝑐1/3 = 𝑐1/3(𝑐2/3 + 𝑏2/3) = 𝑐1/3𝑎2/3. So the length of the tangent line between these two points is √(𝑏1/2𝑎2/3)2 + (𝑐1/3𝑎2/3)2 = √𝑏2/3𝑎4/3 + 𝑐2/3𝑎4/3 = √(𝑏2/3 + 𝑐2/3)𝑎4/3 = √𝑎2/3𝑎4/3 = √𝑎2 = 𝑎 = constant.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 3: SECTION TRUE/FALSE TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 3.TF.1 True

This is the Sum Rule

3.TF.2 False. See the discussion before the Product Rule.

3.TF.3 True

This is the Chain Rule

484

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.TF.4 d dx

True

f (x) 

[ f (x)]1/2 

[ f (x)]1/2 f (x)  f (x) 2 2 f (x) 1

 

3.TF.5



x) f ( x )  f ( x )  1 x1/ 2  f ( , which is not f (x) . 2 dx 2 x 2 x d

False

3.TF.6 y  e2 is a constant, so y  0, not 2e.

False

3.TF.7 d

10  10 ln x, which is not equal to x10 . dx

False

x1

3.TF.8 ln10 is a constant, so its derivative,

False

d dx

(ln10) is 0, not 1 . 10

3.TF.9 d d (tan2 x)  2 tan x sec2 x, and (sec2 x)  2sec x(sec x tan x)  2 tan x sec2 x. dx dx d d d 2 2 (sec x)  (1 tan x)  (tan2 x). Or: dx dx dx

True



3.TF.10





f (x)  x2  x  x2  x for x  0 or x  1 and x2  x   x 2  x for 1  x  0. So

False

f (x)  2x 1 for x  0 or x  1 and f (x)  (2x 1) for 1  x  0. But 2x 1  2x 1for x   1 and 2x 1  2x 1 for x   1 . 2

3.TF.11 If p(x)  a n xn  a n1xn1 

True

p(x)  na nx

n1

 (n 1)a n1x

n2



 a1 x  a0 , then

 a1, which is a polynomial

485

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3.TF.12 f (x)  (x6  x4 )5 is a polynomial of degree 30, so its 31st derivative,

True

f (31) (x), is 0. 3.TF.13 If r(x) 

True

p(x)

, then r(x) 

q(x) - that is, a rational function.

3.TF.14

q(x) p(x)  p(x)q(x) [q(x)]

, which is a quotient of polynomials   



A tangent line to the parabola y  x2 has slope dy / dx  2x, so at 2, 4 the slope of the

False

tangent line is 2(2)  4 and an equation of the tangent line is y  4  4(x  2). [The given equation, y  4  2x(x  2), is not even linear!]

3.TF.15 g(x)  x5  g(x)  5x4  g(2)  5(2)4  80, and by the definition of the g(x)  g(2) derivative, lim  g(2)  5(2)4  80. x2 x2 True

3.TF.16 False. y represents the amount that the curve y  f  x  rises or falls, but dy represents the amount the tangent line rises or falls. These amounts may be different.

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 4: SECTION 4.1

TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

486

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

END OF SECTION EXERCISE SOLUTIONS 4.1.1

V  x3 

dV dx dx  3x2 dx dt dt



dt 4.1.2 (a) A   r 2 



dA dr

 2r

dt dr dt dt dA dr  2r  2 (30 m)(1 m/s)  60 m2 /s (b) dt dt 4.1.3 Let s denote the side of a square. The square's area A is given by A  s2 . Differentiating with

dA respect to t gives us

 2s

. When A  16, s  4 . Substituting 4 for s and 6 for

ds dt

gives us

 2  4  6   48 cm2 / s

dt 4.1.4

A  w 



dA dw d    w  203 108  140 cm2 / s dt dt dt

4.1.5

V   r 2 h   (5)2 h  25 h 

 25

 3  25 









m / min .

25

4.1.6

1 2 3 V  r3     3r2   4  80 4  25, 600 mm / s . 3 dt 3 dt dt 2  4

S  4r2 



4.1.7

dt 4.1.8

 4  2r

dr dt



 4  28 2  128 cm2 / min

  dA  1 55  2 5   2s ds  1 55  2 5   s ds 4 dt 4 dt 2 dt 1 1  55  2 5   2 0.5  55  2 5  cm /min

(a) A 

5 5  2 5 s2 

487

  

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(b) dA  1



 ds  5  1 55  2 5  3 ds  2  5  2 5  ds 

5 5 2 5 s

dt 2 dt 2 ds   cm / min. dt 3 5  2 5

4.1.9 (a) A 

(b) A 

absin 



d 1  1  abcos   2  3  cos 0.2  3 0.2  0.3 cm2 / min .   2  2 dt 2 3    

absin 

dA 1  d db  1        a  bcos  sin   2  3 cos 0.2  sin 1.5    dt 2  dt dt 2   3 3        1 3 3  1 3 0.2  3    0.3  3 cm2 / min  1.6 2 2 2 4     1 (c) A  absin  2 dA 1  da db d   bsin  a sin  abcos  dt 2  dt dt dt  





by (1) 1 1  2.53

  1  1 3   2  3  3  21.5 0.2  



    2  2     21  2  2   3 2  15 3   3  0.3  3  0.3 cm / min  4.85     8 4   8      Note how this answer relates to the answer in part (a) [ changing] and part (b) [ b and  changing].

Note that  fgh     fg  h     fg  h   fg  h   f g  fg  h   fg  h  f gh  fgh  fgh (1) 4.1.10



(a) y  2x 1 and When x  4,

3

dy dt



dy dx dx dt



1 2

(2x 1)1/2  23 

3 . 2x 1

 3  1. dt 9

488



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1 2 2 1 2 1 (b) y  2x 1  y  2x 1  2x  y 1  x  y  and 2 2



dx dy dy dt

 y 5  5y. When x  12, y  25  5, so

 5  

 5(5)  25.

4.1.11

(4x2  9y 2 ) 

(36)  8x

18y

 0  4x

 0  dt dt dt dt dt dx dx dx 1 2  1  4(2)  9  5 0  8  2   5 dt 3  3  dt dt 4 5    dx dy dy dy 2  dy (b) 4x  9y  0  4(2)(3)  9  5 0 6 5  24   4 dt dt 3 dt dt dt   5 (a)

9y



4.1.12

xy  20  x

y

0



x dy





. When x  4, y 

dt dt dt y dt dx 5 10 5   (2)    2.5, option (D). dt 4 4 2

 5, and when

 2,

4.1.13

If x2  y2  45 and x  2 y, then  2 y   y 2  45  5 y 2  45  y 2  9  y  3  y  0 . dx dy dy x dx 2y dx dx 2 2 0    2 . Then x  y  45  2x  2 y dt dt dt y dt y dt dt dy dx  22  4.  2, When dt dt 2

4.1.14

0

x  y  z  dt9  2x dt  2y dt  2z dt  0  x dt y dt z dt  0 . dt If

 5,

 4 and  x, y, z   2, 2,1 , then 25  2  4  1

dz dt

 18 .

4.1.15

 3 cm / s and  x, y   4, 2 , then dt dt dt dt dt dx dx 43  2  0   6 . Thus, the x -coordinate is increasing at a rate of 6 cm / s . dt dt

 xy 

8  x

y

 0. If

4.1.16

489

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

(a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station. If we let t be time (in hours) and x be the horizontal distance traveled by the plane (in miles), then we are given that dx/dt = 500 mi/h. (b) Unknown: the rate at which the distance from the plane to the station is increasing when it is 2 mi from the station. If we let y be the distance from the plane to the station, then we want to find dy/dt when y = 2 mi. (c)(c)

(d) By the Pythagorean Theorem, y2  x2 1  2 y(dy / dt)  2x(dx / dt).

dy (e)(e)

x  (500). Since y2  x2 1, when y  2, x  3, so dt y dt y dy  3 (500)  250  3  433 mi/h. dt 2 

x dx

4.1.17 (a) Given: the rate of decrease of the surface are is 1 cm2/min. If we let t be time (in minutes) and S be the surface area (in cm2), then we are given that dS/dt = –1 cm2/min. (b) Unknown: the rate of decrease of the diameter is 10 cm. If we let x be the diameter, then we want to find dx/dt when x = 10 cm.

(d) If the radius is r and the diameter x  2r, then r  12 x and

S  4 r 2  4  1 x    x2  2

(e) 1 

 2 x



dt 1



dS dx

 2 x

dx dt

  . 2x dt dt dt 1 dx When x  10,   1 . So the rate of decrease is cm/min. 20 20 dt

4.1.18 If we let t be time (in s) and x be the distance from the pole to the person (in ft), then we are given that dx / dt  5 ft / s. If we let y be the distance from the person to the tip of their shadow (in ft), then we want to find

d dt

 x  y when x  40 ft. See the

490

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

diagram. By similar triangles,



x y

15y  6x  6 y  9 y  6x  y  2 x. The tip of their

3 6 y d d  2  5 dx 5 25  5  3 ft / s.  x  y   x  x   shadow moves at a rate of dt dt 3 3 dt 3  

4.1.19 (a) Given: at noon, ship A is 150 km west of ship B; ship A is sailing east at 35 km/h, and ship B is sailing north at 25 km/h. If we let t be the time (in hours), x be the distance traveled by ship A (in km), and y be the distance traveled by ship B (in km), then we are given that dx/dt = 35 km/h and dy/dt = 25 km/h. (b) Unknown: the rate at which the distance between the ships is changing at 4:00 PM. If we let z be the distance between the ships, then we want to find dz/dt when t = 4 h. (c)(c)

(d) z2  (150  x)2  y2  2z

dz dy  dx   2(150  x)     2 y dt dt dt   

(e) At 4:00 PM, x  4(35)  140 and y  4(25)  100  z  (150 140)2 1002  10,100. dz 1  dx dy  So  (x 150) y  10(35) 100(25  215  21.393 km/h. 4.1.20

z 



dx We are given that

dt 

 60mi / h and

10,100 dy

101

 25mi / h.z2  x2  y2 

dt dt dz dx dy dz dx dy dz 1  dx dy  dt dt dt dt dt 2z  2x  2y z x  y dt  dt  z x dt  y dt .  

After 2 hours, x  2 60  120 and

y  225  50  z  1202  502  130, so dz 1  dx dy  12060  5025  x  y    65mi / h . dt z  dt dt   130 4.1.21

 1.6 m / s . By similar triangles, dt y 2 24 dy 24 dx 24  y     1.6. When 12 x x dt x2 dt x2

We are given that

491

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

x  8,

dy dt



24 1.6

 0.6 m / s , so the shadow is decreasing at a rate of 0.6 m / s .

4.1.22

dx We are given that

 4ft / s and

 5ft / s.z2  (x  y)2  5002 

dt dz  dx dy  2z  2 x  y   15 minutes after the second person  dt

 dt 

starts, we have x  4ft / s  20 min60 s / min  4800ft

and

y  515 60  4500  z  (4800  4500)2  5002  86, 740, 000 , so dz x  y  dx dy   4800  4500  4  5  837  8.99ft / s .      8674 dt z  dt dt  86, 740, 000 

4.1.23 Note that 9 gallons of water has volume approximately 1.203 cubic feet, so the rate at which the pool is filled is 9 gal/min  1.203 ft3/min. (a) Let V be the volume of the pool, h be the height, and r be the radius. Then

V  r 2 h 

 r2

. We have d  14 ft  r  7 ft.When

 1.203, we have dt dt dh 1.203 1.203    7    0.0078 ft / min. dt dt 49 dV dh , which is given as constant. Note that the (b) The radius is constant, so depends only on dt dt dt 2 dh

rates of height increase in (a) did not depend on knowing that the height was 2 ft or 3 ft. 4.1.24

3 2 s l, that the 4 length of the trough is l  6 ft, and the sides are s  2 ft. Therefore, when the trough is full, it 3 2  2  6  6 3 ft3 of water. holdsV  4

(a) We are told the trough is in the shape of a rectangular prism with volume V 

2 1 2 (b) The depth of the water is the height of the equilateral triangle, h  s  ( 2 s ) 

3 s2  4

3 2

When the trough is half full, it contains 3 3 ft3 of water, so

492

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

3 3  43  s2  6  3(4)  s2  6 full is 23 s  23

2  s. Therefore the depth of the water when the trough is half

2  62  1.225 ft.

(c) Note that the volume of the trough isV  43 s2 (6)  6 12 h   s, where h is the depth of the water, dh 3 ds h  3 s. This means  . So, to find the rate of change of the depth of the water, we 2 dt 2 dt must find the rate of change of the triangle side, s. We know ds dV 3 3  ds  V 6 3 2   3 3s . When the trough is half full, we also know

4 s  dt  2 



2s dt





s  2 from part (b). dV 3 2 1 1  0.5 ft3  ds  1 dV  ft/h. We are told (0.5)     dt 3 3s dt 3 3 2 36 dt 6 6

 

Then

3 2  3  ds =  3 2   ft/h. dt 2 dt 2 36 24

So the depth of the water is decreasing at a rate of 242  0.0589 ft/h. 4.1.25 We are given that V  34  r 3 . 3 (a) V  288 in3  43  r 3  288 34  r3  216  r  r  6 in.

(b) We are given that

 0.15 in/s, and r  6. We need to find

 4  3r2 

 4r2

dt dt 3 dt dt 2 3  4 (6) (0.15)  21.6 in /s. dV  3.6 in3 /s, withV  288 in3 and r  6. (c) Now we are given that dt dV dr  3.6  0.025 in/s 0.00797 in/s. dr dr dV 1 Now    4r2   dt dt dt dt 4r2 dt 4  62  

4.1.26 Let t be the time in hours, x be the distance traveled by Train A, y be the distance traveled by Train B, and z be the distance between the trains. We are given that

dx dt

 30 mph, and

 40 mph.We also

know that at noon (time 0), x  0, and y  z  10 m. (a) At 12:30 PM, t  12 , so x  0  12 (40)  20, and y  10  12 (30)  25. Therefore the distance between the trains at this time is z 

x2  y2  202  252  1025  5 41 miles.

493



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

dz dx dy dz 1  dx dy  (b) z2  x2  y2  2z dt  2x dt  2 y dt  dt  z  x dt  y dt .At 12:30 PM,

  dz  1 1550 310   mph. 20(40)  25(30)  dt 5 41 5 41 41 (c) Train A is 40a miles from the intersection after a hours and train B is10  30a miles from the intersection. Their distances are equal when 40a  10  30a  a  1 hour, or 1 PM. At this time, dz  1  x dx  y dy   1 (40  40  30  40)  70  49.498 mph. dt z dt dt  2   40 2 





4.1.27



We are given that

 24ft / s .

(a)

dy  dx  y2  (90  x)2  902  2 y dt  290  x   dt . When x  45 ,

 





y

452  902

 45







24  24 , so the distance from , so dy 90  x  dx   45    dt y  dt  5   45 5 



second base is decreasing at a rate of





24  10.7ft / s . 5

(b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer-and we do.

z2  x2  902  2z

dz dt

 2x

dx dt

. When x  45, z  45 5, so

 45 24  24  10.7ft / s. dt 45 5 5

4.1.28

dh 1  1 cm / min and A  bh , where b is the base and h is the altitude. We are given that dt  dA 2 dA 1  dh db 2 dt  2 cm / min Using the Product Rule, we have dt  2  b dt  h dt . When h 10 and   1 1 A  100 , we have 100  b 10  b  10  b  20 , so 2 2 1 db  db db 4  20 2  2  20 110 dt  4  20 10 dt  dt  10  1.6 cm / min .   4.1.29

494

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Given

 1 m/s, find

dt dx

when x = 8 m. y2  x2 1 2 y

 2x



y dy

y dx 65   . When x  8, y  65, so  . x dt x dt 8



65 Thus, the boat approaches the dockat

 1.01 m/s.

4.1.30 We are given that

dt z

720

 2(x  y)

 35 km/h and

dt  dx



dy 

 dt (140 100) 100 2

dt . At 4:00 PM, x  4(350  140 and y  4(25)  100 

 dt  

 325 km/h. z2  (x  y)2 1002 

 260, so 67, 600

x  y  dx



 55.385 km/h.

 dt



dy   dt 



140 100

(35  25)

260

13 4.1.31 The distance z of the particle to the origin is given by z  x2  y2 , so

z2  x2 [2sin  x / 2]2  dz dx dz dx        dx       dx 2z dt  2x dt  4  2sin  x cos  x  2 dt  z dt  x dt  2 sin  x cos  x  dt . 2 2 2 2          2 1 1  1 dz 1 When  x, y   ,1 , z   1  12  10  10 , so 10  10  2 sin   3   9 3 dt 3 3  3      1 dz 1  1  1 dz 3 3 cm / s . cos     2 3   1 10  2  2  3 dt 3 dt 2 6 6    



4.1.32 If C  the rate at which water is pumped in, then

 C 10, 000 ,

where V  r h is the volume at time t . By similar triangles, 2

2 1 1  r  h  V   1 h  h   h3  dV   h2 dh . dt 9 dt 2 6 3 3 3  27 dh  20 cm / min , so When h  200 cm , dt



495

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

C 10, 000 



(200)2 20  C  10, 000 

800, 000

  289, 253 cm3 / min.

4.1.33

When h 



, so b  3h . The trough has volume 1 h 1 dV dh dh 2 V  bh 10  53hh  15h2 12   30h   . 2 dt dt dt 5h

By similar triangles,

1 dh 2 4 ,   ft / min . 2 dt 5 1 5 2

4.1.34 The figure is labeled in meters. The area A of a trapezoid is

1 2

(base1

+ base2) (height), and the volume V of the 10 -meter-long trough is 10A .

1 2 a 0.25 1  , so 2a  h  V  50.6  h h  3h  5h2 . Now triangles,  h 0.5 2 dV dV dh dh dh   0.2  3 10h   0.2 . When h  0.3, dt dh dt dt dt 3 10h 10 1 cm / min . dh  0.2 0.2 m / min  m / min or   dt 3 100.3 6 30 3

Thus, the volume of the trapezoid with height h is V  10  0.3   0.3  2a  h . By similar





4.1.35 The figure is drawn without the top 3 feet.

V 

b 12h 20  10b 12h and, from

8  , so h 6 h 6 3  11h  8h 11h  110h2 . Thus, V  10 24  h  240h  and so b  x 12  y  h 12  3  12  3 3  3  dh dV  220  dh 3 0.8    240  h  . When h  5,  0.8   0.00132ft / min . dt 240  5220 / 3  2275 dt 3 dt

similar triangles,



and











4.1.36

496

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

1 2 1  h 2 h 3 3  30ft / min.V   r h   We are given that    h  3 3  2  dt 12 h 2 dh dh 120 dV dV dh . When   30    2 dt dh dt 4 dt dt h 6 dh 120   0.38ft / min . h  10ft,  2 dt 10  5 dV



4.1.37

x  x  100cot  100 dx d sin2 2 d  100csc    8 . When dt dt dt 100 100 1 (1/ 2)2 1 y  200,sin    d   8   rad / s . The angle is decreasing at a rate of 200 2 dt 100 50 1 rad / s . 50

We are given dx / dt  8ft / s cot 

4.1.38 The area A of an equilateral triangle with side s is given by A 

3s2 .

2 dA 1 3  2s ds  1 3  23010  150  3 cm / min . dt 4 dt 4

4.1.39

x d 1 dx dx  . From Example 2,  4 and cos   sin 10 dt 10 dt dt 8 when x  6, y  8 , so sin  . Thus, 10 8 d 1 d 1   4    rad / s . 10 dt 10 dt 2 4.1.40 From the figure and given information, we have x2  y2  L2 ,

 0.15 m/s, and

 0.2 m/s when x  3 m. Differentiating

2 2 2 implicitly with respect to t, we get x  y  L  2x

dy dt

 x

dx dt

 2y

 0 

.Substituting the given information gives us

497

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y(0.15)  3(0.2)  y  4 m. Thus, 32  42  L2  L  25 L  5 m. 2

4.1.41 According to the model in Example 2,

dy dt



x dx y dt

  as y  0 , which doesn't make physical

sense. For example, the model predicts that for sufficiently small y , the tip of the ladder moves at a speed greater than the speed of light. Therefore the model is not appropriate for small values of y . What actually happens is that the tip of the ladder leaves the wall at some point in its descent. For a discussion of the true situation see the article "The Falling Ladder Paradox" by Paul Scholten and Andrew Simoson in The College Mathematics Journal, 27, (1), January 1996, pages 49-54. Also see "On Mathematical and Physical Ladders" by M. Freeman and P. Palffy-Muhoray in the American Journal of Physics, 53 (3), March 1985, pages 276–277. 4.1.42 The area A of a sector of a circle with radius r and angle  is given by A 

1 2 r  . Here r is 2

1 d   r2 . The minute hand rotates through 360  2 radians each dt 2 dt d dA 1 2  2 and  r 2    r 2 cm2 / h . This answer makes sense because the hour, so dt dt 2 minute hand sweeps through the full area of a circle,  r 2 , each hour. constant and  varies, so

4.1.43

2 The volume of a hemisphere is

r3 , so the volume of a hemispherical basin of radius

3 30 cmis  (30)  18, 000 cm3. If the basin is half full, then 1  1  3 1   V    rh2  h3   9000   30h2  3 h3   3 h3  30h2  9000  0  2









h  H  19.58 [from a graph or numerical rootfinder; the other two solutions are less than0 and greater than 30]. 1  dV dh dh   L  cm3  dh   2 1000   60h  h    V    30h2  h3      60h  h2  2     3  dt dt dt   min     L  dt 2000 dh    0.804 cm / min. dt  60H  H 2  4.1.44

498

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Differentiating both sides of PV  C with respect to t and using the Product Rule gives us

dV dP dV   V dP . When V  600, P  150 and dP  20 , so we have V 0 dt dt dt P dt dt 600 dV   20  80 . Thus, the volume is decreasing at a rate of 80 cm3 / min . dt 150

4.1.45

1.4

 C  P 1.4V

0.4

V

1.4

0

dV dt



V 1.4 P 1.4V

dP 0.4



V dP

1.4P dt

250  10 , so we have dV   400 10  . Thus, the 1.480 dt dt 7 250  36 cm3 / min . volume is increasing at a rate of 7 When V  400, P  80 and

4.1.46 With R  80 and R  100, 1





1 R2



400 1 1 180  9 , so R .    80 100 8000 400 9

Differentiating 1  1  1 with respect to t , we have

R R1 R2 dR2  dR 1 . When R1  80 and 12 dR 12 dR1 12 dR2 dR 1 2 1  R dt   R dt  R dt  dt  R  R2 dt  R2 dt  1 2  1 2  dR 4002  1 1  107 R2  100,  2  2  0.3   0.132Ω / s .  0.2   dt 9  80 1002  810 4.1.47 We want to find

when L  18 using B  0.007W 2/3 and W  0.12L2.53 .

dt dB dB dW dL  2  20 15     0.007  W 1/3 0.12  2.53 L1.53    dt dW dL dt  3 10, 000, 000      2  2.53 1/3  1.53  5   0.007   0.12 18   1.045108 g / yr  0.12  2.5318   107  3     

4.1.48



We are given d / dt  2 min 



rad / min . By the Law of 90 Cosines, x2  122 152  21215cos  369  360cos 

499



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 360sin

d





180sin d

. When   60 ,

x  369  360cos60  189  3 21 , so dx 180sin60    3  7  0.396 m / min .  dt 21 3 21 90 3 21 4.1.49

dx Using Q for the origin, we are given

dy dt

 2ft / s and need to find

when x  5 . Using the Pythagorean Theorem twice, we have

x2 122  y2 122  39 , the total length of the rope. Differentiating with respect to t , we get

dx y dy  2  0 , so 2 x 12 dt y 12 dt 2

x y2 122 dx dy . Now when  dt y x2 122 dt x  5, 39  (5)2 122  y2 122  13  y2 122  y2 122  26 , and   5 26  2 dy 10 y 2 2  532 . So when x  5,    0.87ft / s . 26 12     dt 133 532 13 So cart B is moving towards Q at about 0.87ft / s .



4.1.50

(a) By the Pythagorean Theorem, 4000  y  2 . Differentiating with 2

respect to t , we obtain 2 y so when y  3000ft ,

y dy



3000

dy  2 d . We know that  600ft / s , dt dt dt

 40002  30002  25, 000, 000  5000ft

600 

1800

 360ft / s . 5000 5 y d d  y  1 dy d cos2 dy 2 d    .  tan     sec  (b) Here tan  4000 dt dt  4000  dt 4000 dt dt 4000 dt dy 4  600ft / s,  5000 and cos  4000  4000  , so When y  3000ft, 5000 5 dt 2 d  (4 / 5)  600  0.096rad / s . 4000 dt and



500

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

4.1.51

d We are given that

 42   8 rad / min.x  3tan 

1  1 2 10 dx d  3sec2 . When x  1, tan  , so sec2  1    and 3 dt dt 9 3   dx 80  10   3  8     83.8 km / min . dt 3  9 4.1.52

 2    1 dx  cot   csc    csc      3   6  5 dt 5 dt 5 dt  2 10 dx 5  2    km / min 130mi / h  .  dt  6 9 3   x

d

1 dx

4.1.53

d We are given that



2 rad

  rad / min . By the Pythagorean

2 min

Theorem, when h  6, x  8, so sin 

the figure, sin 

and cos 

8 10

. From

 h  10sin , so

dh d  10  8   8 m / min .  10cos  10  dt dt  



4.1.54

dx We are given that

 300 km / h . By the Law of Cosines,

 1 2 y  x 1  21 xcos120  x2 1 2x   x  x 1, so  2    dy dx dx dy 2x 1 dx 300 2y  2x    . After1 minute, x   5 km  dt dt dt dt 2 y dt 60 2 5 1   dy 1650 y  52  5 1  31 km   296 km / h 300   dt 2 31 31 2

4.1.55

dx We are given that

 3mi / h and

 2mi / h . By the Law of

Cosines,

501



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

z2  x2  y2  2xycos45  x2  y2  2xy  3 dz dx dy dy dx  1  2z  2x  2y  2x  2 y . After 15 minutes  h , we have x  and dt dt dt dt dt 4  4  2 2 2 1 3 2 3 2        y    z2        2     z  13  6 2 and 4 2 4 4 4  4  4  3 2  3 1 1  dz  2   3  2   2  2   2  2  3     4   dt 2 4  2   13  6 2    







13  6 2  13  6 2  2.125mi / h 2 13  6 2 2

4.1.56 Let the distance between the runner and the friend be of Cosines,

. Then by the Law

 2002 1002  2  200 100 cos  50, 000  40, 000cos  å  . Differentiating implicitly with respect to t , we obtain d d 2  40, 000sin  . Now if D is the distance run when the dt dt 2

angle is  radians, then by the formula for the length of an arc on a circle, s  r , we have

d 1 D  d  1 dD  7 . To substitute into the expression for  , we 100 dt dt 100 dt 100 must know sin at the time when  200 , which we find from 1 2 2 cos    sin  1  1   15 . Substituting, we  å  : 200  50, 000  40, 000cos   4  4 4 d 15   40, 000 get 2200  7  d / dt  7 15  6.78 m / s . Whether the distance   dt 4  100  4 D 100 , so  

between them is increasing or decreasing depends on the direction in which the runner is running. 4.1.57 The hour hand of a clock goes around once every 12 hours or, in radians

2





rad / h . The minute hand goes around once an hour, or 12 6 at the rate of 2 rad / h . So the angle  between them (measuring per hour,

clockwise from the minute hand to the hour hand) is changing at the rate of

 11  2   rad / h . Now, to relate  to , we use the Law 6 6 2 2 of Cosines: 2  4  8  2  4 8cos  80  64cos  å  . d / dt 

502

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

d Differentiating implicitly with respect to t , we get 2

 64sin 

d

. At 1: 00 , the angle

dt  radians. We use å    to find between the two hands is one-twelfth of the circle, that is,  12 6  1: 00 :  80  64cos   80  32 3 . Substituting, we get 2

1 11  64     88   11  d d  2  6    2  64sin     18.6 .  dt 6  6  dt 2 80  32 3 3 80  32 3 So at 1:00, the distance between the tips of the hands is decreasing at a rate of 18.6 mm / h  0.005 mm / s .

Solution and Answer Guide STEWART KOKOSKA, CALCULUS: CONCEPTS AND CONTEXTS, 5E, 2024, 9780357632499, CHAPTER 4: SECTION 4.2

TABLE OF CONTENTS End of Section Exercise Solutions ............................................................................................................. 1

END OF SECTION EXERCISE SOLUTIONS 4.2.1 A function f has an absolute minimum at x  c if f c  is the smallest function value on the entire domain of f , whereas f has a local minimum at c if

f c  is the smallest function value

when x is near c . 4.2.2 (a) The Extreme Value Theorem (b) See the Closed Interval, or Table of Values, Method. 4.2.3 Absolute maximum at s , absolute minimum at r , local maximum at c , local minima at b and r , neither a maximum nor a minimum at a and d . 4.2.4

503

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Absolute maximum at r ; absolute minimum at a ; local maxima at b and r ; local minimum at d ; neither a maximum nor a minimum at c and s . 4.2.5 Absolute maximum value is f  4   5 ; there is no absolute minimum value; local maximum values are f  4   5 and f 6  4; local minimum values are f  2   2 and f 1  f 5  3 4.2.6 There is no absolute maximum value; absolute minimum value is g  4   1 ; local maximum values are g 3  4 and g 6  3 ; local minimum values are g  2   2 and g  4   1 .

4.2.7 Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima at 2 and 4

4.2.8 Absolute maximum at 4 , absolute minimum at 5, local maximum at 2, local minimum at 3

4.2.9 Absolute minimum at 3, absolute maximum at 4, local maximum at 2

4.2.10

504

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

Absolute maximum at 2, absolute minimum at 5, 4 is a critial number but there is no local maximum or minimum there.

4.2.11

4.2.12 (a) Note that a local maximum cannot occur at an endpoint.

Note: By the Extreme Value Theorem, f must not be continuous. 4.2.13 (a) Note: By the Extreme Value Theorem, f must not be continuous; because if it were, it would attain an absolute minimum.

4.2.14

505

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

4.2.15

f (x)  12 (3x 1), x  3. Absolute maximum f (3)  4; no local maximum. No absolute or local minimum.

4.2.16

f (x)  2  1 x, x  2. Absolute maximum f (2)  8 ; no local maximum; No absolute or local 3

minimum.

4.2.17

f  x  1 / x, x  1.. Absolute maximum f 1  1; no local maximum. No absolute or local

minimum.

4.2.18

f  x  x2 , 0  x  2. No absolute or local maximum or minimum value.

506

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

4.2.19

f  x  ex . No absolute or local maximum or minimum value.

4.2.20

f  x  sinx, 0  x   / 2 . No absolute or local maximum. Absolute minimum f  0   0 ; no local

minimum.

4.2.21

f t   cost, 

3

f  , 1

t 

3

. Absolute and local maximum f  0   1; absolute and local minima



4.2.22

f  x  lnx, 0  x  2 . Absolute maximum f  2   ln2  0.69 ; no local maximum. No absolute or

local minimum.

507

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

4.2.23



f  x  x . No absolute or local maximum. Absolute and local minimum f  0   0 .

4.2.24

f  x  1

x . Absolute maximum f  0   1; no local maximum. No absolute or local minimum.

4.2.25

x 2 f  x   2  3x

if 1  x  0 if 0  x  1

No absolute or local maximum. Absolute minimum f 1  1. Local minimum f  0   0 .

4.2.26

f  x 

2x 1  4  2x

if 0  x  1 if 1  x  3

508

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

No absolute or local maximum. Absolute minimum f 3  2; no local minimum.

4.2.27

f (x)  4  1 x  1 x2  f (x)  1  x. f (x)  0  x  1 . This is the only critical number. 3

4.2.28

f (x)  x3  6x2 15x  f (x)  3x2 12x 15  3(x2  4x  5)  3(x  5)(x 1). f (x)  0  x  5,1. These are the only critical numbers. 4.2.29

f (x)  2x3  3x2  36x  f (x)  6x2  6x  36  6(x2  x  6)  6(x  2)(x  3). f (x)  0  x  2,3. These are the only critical numbers. 4.2.30

f (x)  2x3  x2  2x  f (x)  6x2  2x  2  2(3x2  x 1). Using the quadratic formula, 1 11 f (x)  0  x  . Since the discriminant, –11, is negative, there are no real solutions, 6 and therefore, there are no critical numbers. 4.2.31

g(t)  t 4  t3  t 2 1  g(t)  4t3  3t2  2t  t(4t2  3t  2). Using the quadratic formula, we see that 4t  3t  2 has no real solutions (its discriminant is negative), so g(t)  0 only if t  0. 2

Hence, the only critical number is 0. 4.2.32

g(t)  3t  4   3t  4    4 if t  4 . (3t  4) if if 3t 3t  4 4  00 3t 4  3t 3 4 if t   3   3 if t  43 4 4 and g(t) does not exist at t  , so t  , is a critical number. g(t)   3 3 4 3 if t  3

4.2.33

g  y 

y 1



y2  y 1

509

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 2 2 2 2 y  y 1 1  y 1 2 y 1 y  y 1 2 y  3y 1       g  y      y  2 y 2  y 2  y 2 2 2



  y2  y 1









 y  y 1



 y2  y 1  y2  y 1

g   y   0  y  0, 2 . The expression y2  y 1 is never equal to 0, so g   y  exists for all real numbers. The critical numbers are 0 and 2.



    2 2 h  p  p 1  h p   p  41   p 12 p  p2  4  2 p2  2 p  p  2 p  4 .  2 2 p 4  p  4  p  4  p2  4 

4.2.34



h p  0  p 

2 

4 16 2



 1 5 . The critical numbers are 1 5. h  p  exists for all real 

numbers.] 4.2.35

h t   t3/4  2t1/4  ht  

3 1/4 2 3/4 1 3/4 1/2 3 t 2 t  t  t 3t  2  . 4 4 4 4 4 t3

ht   0  3 t  2  t  and





4  t   ht  does not exist at t  0 , so the critical numbers are 0 3 9

9 4.2.36





g  x  3 4  x2  4  x

  g  x   1  4  x2   2x  

2x

2/3

2 1/3



3 4 x



 g  x   0 x  0 .

2 2/3

g   2  do not exist. Thus, the three critical numbers are 2, 0 , and 2. 4.2.37

F  x  x4/5 (x  4)2  4 1 F   x   x4/5  2  x  4  (x  4)2  x 1/5  x 1/5  x  4  5  x  2   x  4   4 5 5 

 x  414x 16  2 x  4 7x  8 5x1/5

5x1/5

510

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

8 8 F  x  0  x  4,  F 0  does not exist. Thus, the three critical numbers are 0, , and 4. 7 7 4.2.38

g( )  4  tan  g( )  4 sec2 . g( )  0  sec2   4  sec  2  cos   12 

    2n , 5  2n , 2  2n , and 4  2n are critical numbers. Note: The values of θ that 3

make g( ) undefined are not in the domain of g. 4.2.39

f    2cos  sin2  f    2sin  2sin cos. f    0  2sin cos 1  0  sin  0 or cos  1   n[n an integer ] or   2n . The solutions   n include the solutions   2n , so the critical numbers are   n . 4.2.40

h(t)  3t  arcsin t  h(t)  3  1

h(t)  0  3 



1 t

1 1 t 2

 1 t 2  31  1 t 2  91 

t 2  89  t   23 2  0.943, both in the domain of h, which is [–1, 1]. 4.2.41 2 3x

f (x)  x e



 f (x)  x2 (3e3x )  e3x (2x)  xe3x (3x  2). f (x)  0  x  0, 23

[e3x is never equal to 0]. f (x) always exists, so the critical numbers are 0 and 23 . 4.2.42 2 3 3 3 3 f (x)  x2 ln x  f (x)  x (1/ x)  (ln x)(2x )  x  2x ln x  x (1 2ln x) 

1 2ln x x3

f (x)  0 1 2ln x  0  ln x  12  x  e1/2 1.649. f (0) does not exist, but 0 is not in the domain of f, so the only critical number is e. 4.2.43 The graph of f  x  5e0.1 x sinx 1 has 10 zeros and exists everywhere, so f has 10 critical numbers. 4.2.44

511

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

100cos2 x 1 is shown. There are 7 zeros 10  x2 between 0 and 10, and 7 more zeros since f  is an even function. f  exists everywhere, so f has 14 critical numbers. A graph of f  x 



4.2.45

f  x  12  4x  x ,0, 5. f  x  4  2x  0  x  2. f 0  12, f  2   16 , and f 5  7 . 2

So f  2   16 is the absolute maximum value and f 5  7 is the absolute minimum value. 4.2.46



f  x  5  54x  2x3,0, 4. f   x  54  6x2  6 9  x2

  6 3  x  3  x  0  x  3, 3 .

f  0   5 , f 3  113 , and f  4   93 . So f 3  113 is the absolute maximum value and f  0   5 is the absolute minimum value. 4.2.47





f  x  2x3  3x2 12x 1,2, 3. f   x   6x2  6x 12  6 x 2  x  2  6  x  2  x 1  0 

x  2, 1. f 2  3, f 1  8, f  2   19 , and f 3  8 . So f 1  8 is the absolute maximum value and f  2   19 is the absolute minimum value. 4.2.48

f  x  x3  6x2  5 , 3, 5. f  x  3x2 12x  3x  x  4  0  x  0, 4 . f 3  76, f 0  5 , f  4   27 , and f 5  20 . So f  0   5 is the absolute maximum value and f 3  76 is the absolute minimum value. 4.2.49 f  x  3x4  4x3 12x2 1,2, 3. f  x  12x3 12x2  24x  12x  x 2  x  2  12x  x 1  x  2  0 

x  1, 0, 2. f 2  33, f 1  4, f 0  1, f  2   31, and f 3  28 . So f 2  33 is the absolute maximum value and f  2   31 is the absolute minimum value. 4.2.50

512



Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1







f  t   t  4 ,  2,3. f   t   3 t 2  4 2



  2t   6t(t  2) (t  2)  0  t  2, 0, 2 . 2

f 2  0 , f 0  64 , and f 3  53  125 . So f 3  125 is the absolute maximum value and f 0  64 is the absolute minimum value.

1

4.2.51

1 x2 1  x 1 x 1  0  x  1, but x  1 is not in f  x  x  ,0.2, 4. f  x  1 2  2  x x2 x x the given interval, 0.2, 4. f  x does not exist when x  0 , but 0 is not in the given interval, so 1 is the only critical nuumber. f 0.2  5.2, f 1  2 , and f  4   4.25 . So f 0.2  5.2 is the absolute maximum value and f 1  2 is the absolute minimum value. 4.2.52



x f  x   2 ,0,3 x  x 1



2 1 x2  x 1 2x2  x f  x   x  x 1  x  2x  2 2 

 x 2  x 1





1 x2

x 2  x 1



1 x1 x2  0 

 x 2  x 1

x  1 , but x  1 is not in the given interval, [0,3]. f 0  0, f 1  1, and f 3 

. So

f 1  1 is the absolute maximum value and f  0   0 is the absolute minimum value.



4.2.53

1 1 1 1 . f t   0  1  f t   t  t ,1, 4. f t   1 t2/3  1  t2/3   3 3t2/3 3t2/3 3 3 3/ 2 1 1 1 t          . f   t  does not exist when t  0. f 1  0, f 0  0 27 3 3 3 9    2 3  1  2 3  1  1  1  1 3   0.3849, f  1  1  , and f     9 9 3 33 3  3 3  3 3  3 3  3 3 2 3 f 4  4  3 4  2.413 . So f 4  4  3 4  is the absolute maximum value and  3  f  9   9   3



is the absolute minimum value. 4.2.54

t

f (t)  2 1 t

 (1 t 2 )(1/ 2 t )  t (2t) (1 t 2 )  2 t t (2t)   , 0, 2. f (t)  (1 t 2 )2

2 t (1 t 2 )2

1 3t 2 . 2 t (1 t 2 )2

513

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f (t)  0  1 3t2  0  t 2  13  t  

1 1 , but t   is not in the given interval, 0, 2. 3 3

f (t) does not exist when t = 0, which is an endpoint. f 0  0, 2  1  33/ 4 f  1  1/ 4 3  31/ 4  33/ 4  11/ 3 4 / 3 4  0.570, and f  2   3 4 is the 5  0.283. So f  3     absolute maximum value and f  0   0 is the absolute minimum value. 4.2.55

f t   2cost  sin2t,0, / 2.









f   t   2sint  cos2t  2  2sint  2 1 2sin2t  2 2sin2t  sint 1  2 2sint 1sint 1 .

f t   0  sint 

or sint  1  t 



. f 0  2, f

 

   6





 2.60 , and 3

6 2 2   f  0.  2      3   So f    3 is the absolute maximum value and f   0 is the absolute minimum value. 6 2  2  

4.2.56

f (t)  t  cos(t / 2), 4 , 74  . f (t)  1 sin  12t   12 . f (t)  0  1  12 sin  12 t   2  sin  21 t   there are no critical numbers. f       cos     2 and f  7   7  cos 7  7  2 . So f       4

is the absolute

minimum value and f  7   7  2 is the absolute maximum value. 4

4.2.57

1 1 2lnx 1  f  x  x2lnx, , 4 . f  x  x2   lnx 2x3  x3  2x3lnx  x3 1 2lnx  . 2  x x3 1 f  x  0  1 2lnx  0  2lnx  1  lnx   x  e1/2  1.65. f x does not exist when ln1 ln2  1  2 1   4ln2  2.773 , ,4 . f  ln1/ 2  x  0 , which is not in the given interval,



2 



 (1/ 2)2  2 

1/ 4

514

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

f e

1/ 2

ln4

  lne    0.184 , and f 4  2   0.087 . So f e1/2   is the e 2e 4 16 2e e  1 1/ 2

1/ 2

 

absolute maximum value and f    4ln2 is the absolute minimum value. 2

 



4.2.58

f  x  xex/2 ,3,1. f  x  xex/2

1

 e x/2 1  e x/2

1

1  x 1 . f  x   0  x 1  0  x  2

2 2  2 3/2  1  0.736 , and f 1  e1/2  1.649 . So f 1  e1/2 is  0.669, f 2  2e . f 3  3e the absolute maximum value and f 2  2 / e is the absolute minimum value. 4.2.59

f  x  xex/8, 1, 4. f  x  xex/8  18   ex/8 1  ex/8  18 x 1. f  x  0   81 x 1  0  x  8, but 8 is not in 1, 4. f 1  e1/8  1.133 and f  4   4e4/8  4e1/2  2.426. So f  4   4e1/2 is the absolute maximum value and f 1  e1/8 is the absolute minimum value. 4.2.60

f  x  ln  x 2  x 1,1,1. f   x  

1   2x 1  0  x   . Since x2  x 1  0 x2  x 1 32  1  ln  0.29 , and for all x , the domain of f and f  is R. f 1  ln1  0, f   2  

4 f 1  ln3  1.10 . So f 1  ln3  1.10 is the absolute maximum value and 3  1 f   ln  0.29 is the absolute minimum value.  2  

4.2.61



 

1 2  1 x2  2  x2 1  f  x  x  2tan x,0, 4. f   x  1 2  2  0  1  1 x 1 1 x2  1 x  1. f 0  0, f 1  1  0.57 , and f  4   4  2tan 4  1.35 . So f  4   4  2tan 4    2  is the absolute maximum value and f 1  1  is the absolute minimum value. 2 1

4.2.62

515

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1



1 f (t)  t  cot(t / 2), [ / 4, 7 / 4]. f (t)  1 csc2 (t / 2)  . 2 1 2 1  1 2 f (t)  0  csc (t / 2)  1  csc (t / 2)  2  csc(t / 2)   2  t   or t  3 2 2 4 2 4  3  7 and csc(t /2)   in the last interval  t   or t =   t  7   1 .  t 2  4  4 8 2 8 2 2       3 3      3 3 f     cot  3.20, f     cot  1  2.57, f    2  cot 2  2 1  3.71, and 8 4 4  4  23  43 2  2     7  7  7    f  1 is the absolute maximum value and f 1 f   cot   3.08. So  2  2 4 4 8        2 2 is the absolute minimum value. 



4.2.63

f (x) 

 f (x)  ex

ex (2x)  x2 (ex )



ex (2x  x2 )



x(2  x)

. e2 x ex 4 f (x)  0  x(2  x)  0  x  0, 2. f (1)  e  2.718, f (0)  0, f (2)  2  0.541, and e 9 f (3)   0.448. So the maximum value on the interval [–1, 3] is e. e3 e2 x

4.2.64

g  x  3  2x  x2  g  x  2  2x  2 1 x  g  x  0  x  1. g  4   6  0, and f   2   6  0. Since 1 is the only critical number in the interval, and g changes sign at x  1, we see that statements I, II (g has an absolute maximum value and a relative maximum on the interval) and III (g has no absolute minimum value on the interval) are all true. 4.2.65

s t   t3  2t2  4t  8  v t   st   3t 2  4t  4  a t   vt   6t  4  2 3t  2 

a  t   0  t  2 . v  2   3 2   4  2   4   2

negative to positive at t  4.2.66



ft / s  5.333 Since a t  changes from

2 3 and has no other sign changes, this velocity is an absolute minimum.



 2 2 2 2 y  x2 1 , on 4, 4  y  (x 1)(1)  (x 1)(2x)  x 1 2x  2x  x  2x 1

x 1

(x 1)

516

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

y is defined for all real x, and y  0   x2  2x 1  0  x  1



f (4)  



17 and f (4) 

 0.294, f 1 2 

3 17





 1.207, f 1 2 

(1 2) 1 2

 0.176. Therefore, the function (D) y 

maximum, f 1 2 



x 1 x 1 2

 0.207, (1 2) 1 2

, on 4, 4 has both a relative





2 2 and a relative minimum, f 1 2  . 2 2 (1 2) 1 (1 2) 1

4.2.67

f  x  xa (1 x)b , 0  x  1, a  0, b  0 . f   x   x a b(1 x)b1  1  (1 x)b  ax a1  x a1 (1 x)b1 x  b  1  1 x   a  xa1(1 x)b1 a  ax  bx 

At the endpoints, we have f 0  f 1  0[ the minimum value of f ] . In the interval

0,1, f  x  0  x 

a ab

a a b b a a  a b f  a    1 a b  a a  a  b  a   a a  b b    b ab ab ab a b (a  b) a b (a    b) (a a b)  b) (a        

 a  aabb   ab is the absolute maximum value.  a  b  (a  b)

So f  4.2.68

The graph of f (x)  1 5x  x3 indicates that f (x)  0 at x  1.3 and that f (x) does not exist at  02.1, 0.2, and 2.3. Those five values of x are the critical numbers of f.

4.2.69

517

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

From the graph, it appears that the absolute maximum value is about f (0.77)  2.186, and the absolute minimum value is about f (0.77)  1.814.

f (x)  x5  x3  2  f (x)  5x4  3x2  x2 (5x2  3). So

        2

f (x)  0  x  0,  53 . f  35   35    53 

similarly, f

 

3 5

 2   35  925 35  2  256

    3 5

6 25

3 5

 2 (maximum) and

3 5

 2 (minimum).

4.2.70 From the graph, it appears that the absolute maximum value is about f (1)  2.85, and the absolute minimum value is about f (0.23)  1.89.

f (x)  ex  e2x  f (x)  ex  2e2x  e2x (e3x  2). So f (x)  0  e3x  23x  ln 2  x  13 ln2[  0.231] .

   e 

f  13 ln2   eln 2

1/3

ln 2

2/3

 21/3  22/3  1.890, the minimum value.

f 1  e1  e2  2.854, the maximum value. 4.2.71 From the graph, it appears that the absolute maximum value is about f (0.75)  0.32, and the absolute minimum value is f (0)  f (1)  0; that is, at both endpoints.

f (x)  x x  x  f (x)  x 2

1 2x 2 xx

2 

 xx

(x  2x2 )  (2x  2x2 ) 2 xx



3x  4x2 2 xx

So f (x)  0  3x  4x2  0  x(3  4x)  0  x  0 or 3 4 .

f (0)  f (1)  0 (minimum), and f  34  34

3 4

 ( 34 )2 34

3 16

3163

(maximum).

4.2.72

518

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

From the graph, it appears that the absolute maximum value is about f (2)  1.17, and the absolute minimum value is about f (0.52)  2.26.

f (x)  x  2cos x  f (x)  1 2sin x. So f (x)  0  sin x   12  x   6 on 2, 0. f (2)  2  2 cos(2) (maximum) and

     3 (minimum).

f ( 6 )   6  2cos( 6)    6 2



3 2



4.2.73

3 2 (a) f (x)  x  4x  2x 12x  5  f (x)  4x 12x  4x 12  4(x 1)(x  3)(x 1). 4

f (x)  0  x  1,3, 1. Therefore the critical numbers of f are –1, 1, and 3. (b) From the graph we can see that f has local minima at x = –1 and x =3. The local minima are f (1)  f (3)  14. (c) From the graph, we can see that the local maximum value of f occurs at x = 1, and it is f (1)  2. (d) The absolute minimum is the same as the local minima, f (1)  f (3)  14. 4.2.74 (a) R(t)  e  2t 1  R(2)  e  2(4) 1  e  7  0.3891000  389 ft /min. t

(b) R(t)  e  4t  R(t)  0  e  4t  e / t  4  t  ln t  ln 4  t  A  0.389056, and t

t  B  2.15329. There are two critical points, A and B, in the interval.

R(0)  e0 1  2 thousand ft3 /m, R( A)  0.340 thousand ft3 /m, R(B)  2.174 thousand ft3 /m, and R(3)  e 17  3.0855 thousand ft /m. Therefore,  B, 2.174 is a local maximum and 3

 A, 0.340 is a local minimum. (c) From the work in part (b) we see that the absolute minimum is R( A)  0.340 thousand ft /m, 3

and the absolute maximum is R(3)  e 17  3.0855 thousand ft /m. 3

4.2.75





Let a  0.135 and b  2.802 . Then C t   atebt  C   t   a t  ebt b  ebt 1  aebt bt 1 .

1 a a C   t   0  bt 1  0  t    0.36 h. C 0  0, C 1/ b   e1    0.0177 , b b be

519

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

and C 3  3ae

 0.00009 . The maximum average BAC during the first three hours is about

0.0177 g / dL and it occurs at approximately 0.36 h 21.4min 

4.2.76



C t   8 e0.4t  e0.6t

  Ct   80.4e



 0.6e0.6t . C  t   0  0.6e0.6t  0.4e0.4t  3 3 0.6 3  e0.4t 0.6t   e0.2t  0.2t  ln  t  5ln  2.027 h. C 0  811  0 , 0.4 2  3 22  3 3  2  4 8  32 3     1.185 , and C  5ln  8 e2ln3/ 2  e3ln3/ 2  8       8     2    2    9 27  27  2 



0.4t



C 12  8 e4.8  e7.2  0.060 . The maximum concentration of the antibiotic during the first 32 12hours is g / mL . 27 4.2.77



The density is defined as   mass

1000 (in g / cm3 ). But a critical point of  will also be volume V T  d dV and V is never 0 ] , and V is easier to a critical point of V [since  1000V 2 dT dT differentiate than  . 

V T   999.87  0.06426T  0.0085043T 2  0.0000679T 3  V T   0.06426  0.0170086T  0.0002037T 2 . Setting this equal to 0 and using the quadratic formula to find T , we get 2 T  0.0170086  0.0170086  4  0.0002037 0.06426  3.9665 C or 79.5318 C . Since we 2 0.0002037 

are only interested in the region 0 C  T  30 C , we check the density  at the endpoints and at

1000 1.00013 ; 999.87  30  1000  0.99625;  3.9665  1000  1.000255. So water has its maximum 1003.7628 999.7447 density at about 3.9665 C . 3.9665 C:  0 

4.2.78



W dF sin  cos  0   W cos sin  W cos sin  F    . 2 2 sin  cos d (sin  cos ) (sin  cos ) dF sin  So  0  cos sin  0     tan . Substituting cos d

520

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

 tan for  in F gives us tan W Wtan Wtan cos Wsin  Wsin . If tan   , then F    1 tan sin  cos sin2 cos sin2  cos2 cos   sin  (see the figure), so F  W. 2 2  1  1    W . Now because   2 

We compare this with the value of F at the endpoints: F 0  W and F 



 1 and

 2 1   F . Hence,  2   



W is less than or equal to each of F 0 and  2 1  2 1   W is the absolute minimum value of F   , and it occurs when tan    2 1  , we have that

. 4.2.79 Let

a  0.00003237, b  0.0009037, c  0.008956, d  0.03629, e  0.04458, and f = 0.4074. Then S(t)  at5  bl 4  cl3  dt2  et  f and S(t)  5at4  4bt3  3ct2  2dt  e. We now apply the Closed Interval Method to the continuous function S on the interval 0  t  10 . Since S exists for all t, the only critical numbers of S occur when S(t)  0. We use a rootfinder on a CAS (or a graphing device) to find that S(t)  0 when

t1  0.855, t2  4.618, t3  7.292, and t4  9.570. The values of S at these critical numbers are S(t1 )  0.39, S(t2 )  0.43645, S(t3 )  0.427, and S(t4 )  0.43641. The values of S at the endpoints of the interval are S(0)  0.41 and S(10)  0.435. Comparing the six numbers, we see that sugar was most expensive at t2  4.618 (corresponding roughly to March 1998) and cheapest at

t1  0.855 (June 1994). 4.2.80 (a) The equation of the graph in the figure is

v t   0.00146t3  0.11553t 2  24.98169t  21.26872. 2 (b) a t   vt   0.00438t  0.23106t  24.98169 

at   0.00876t  0.23106 .

521

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1

at   0  t  1

0.23106

 26.4.a 0  24.98, a t   21.93 , and a 125  64.54 . 1

0.00876

The maximum acceleration is about 64.5ft / s2 and the minimum acceleration is about 21.93ft / s2 . 4.2.81 2 2 3 2 (a) v r   k r  r  r  kr r  kr  v   r   2kr r  3kr . vr   0  kr 2r  3r   0 

r  0 or v

1

r (but 0 is not in the interval). Evaluating v at

, and r , we get

r, r  310 3  2  4 kr3 , and v r  0 . Since 24 0 31 0 attains0 its maximum value at  ,v  kr , v r   

0 0  2 0  8 0  3 0  27  2    r  r0 . This supports the statement in the text. 3 4 3 kr . (b) From part (a), the maximum value of v is 0 27

4.2.82 (a) f  x  ax  bx  cx  d , where a  0 . So f  x  3ax  2bx  c is a quadratic function 3

and hence, the quadratic equation f  x  0 has either 2,1 , or 0 real solutions. Thus, a cubic function can have two, one, or no critical number(s). Case (i) [2 critical numbers ] :

f  x  x3  3x  f  x  3x2  3  3(x2 1) so x= 1, 1 are critical numbers. Case (ii) [1 critical number]:

f  x  x3  f  x  3x2 , so x  0 is the only critical number. Case (iii) [no critical number]:

522

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 1: Section 1.1





f  x  x3  3x  f   x   3x2  3  3 x 2 1 , so there is no critical number. (b) Since there are at most two critical numbers, a cubic function can have at most two local extreme values, and by (a)(i), this can occur. By (a)(ii) and (a)(iii), it can have no local extreme value. Thus, a cubic function can have zero or two local extreme values.

523