Curve Sketching
Suppose f is defined at c and either f’(c)=0 or f’(c) does not exist.
Then the number c is called a critical number of f, and the point (c,f(c)) on the graph of f is called the critical point of f.
NOTE
(i) If f(c) is not defined, then c is not a critical number. (ii) If f’(x) =0, the point is called stationary point.
EXAMPLE 7
critical points (i) f (x ) = x − 12 x f ' (x ) = 3 x 2 − 12 f ' (x) = 0 3
Solution
3 x 2 − 12 = 0 3 x 2 = 12 x = 2
critical number
f (− 2) = (− 2)3 − 12(− 2) = 16 critical point (− 2,16 ) f (2) = (2)3 − 12(2) = −16 critical point (2,−16 )
critical value critical value
f (x ) = x 3 − 3 x + 1 (ii)
Solution
f ' (x ) = 3 x 2 − 3 f ' (x) = 0 3x 2 − 3 = 0 3x 2 = 3 x = 1 f (− 1) = (− 1)3 − 3(− 1) + 1 = 3 (− 1,3) f (1) = (1)3 − 3(1) + 1 = −1 (1,−1) Critical points are (-1, 3) and (1, -1)
f increases on I if f ( x) 0 for all x in I.
f decreases on I if f ( x) 0 for all x in I.
DEFINITION Increasing Function: A function f(x) is said to be increasing if x1 < x2 and f(x1) < f(x2) Decreasing Function: A function f(x) is said to be decreasing if x1 < x2 and f(x1) > f(x2) Constant Function: A function f(x) is said to be constant if f(x1) = f(x2) for every x1 and x2.
Theorem Let f be differentiable on the interval (a,b) (i) If f (x) > 0 for all x in (a,b), then f is increasing on [a,b]. (ii) If f (x) < 0 for all x in (a,b), then f is decreasing on
[a,b].
(iii) If f (x) =0 for all x in (a,b), then f is constant on
[a,b].
NOTE
Sign diagrams for derivative are useful for determining intervals where a function is increasing or decreasing
For the interval (-2,2)
-2
f â&#x20AC;&#x2122;(x) -
+
f(x)
2
Graph of f
Decreasin Falls in the g interval (2,0) increasing Rises in the interval (0,2)
Example s f(x)=x2
f(x)=x2
EXAMPLE 7 Find the intervals on which the following functions are increasing or decreasing. increasing: (-2,0) and
(1, ) a)
decreasing: (− ,−2) and (0,1)
EXAMPLE 7
increasing: (−,2) and
(2, ) decreasing: (−2,2)
EXAMPLE 9
Find the intervals on which the following functions
are increasing and the intervals on which they are decreasing (a) f (x ) = x 2 â&#x2C6;&#x2019; 6 x + 8
(b) f (x ) = x 3 + 3 x 2 â&#x2C6;&#x2019; 7
Solution f (x ) = x 2 + 4 x f ' (x ) = 2x + 4 = 0 x = −2
Critical number
(- , -2)
(-2, )
Test value, k
-5
5
Sign of f’(x)
-
+
decreasing
increasing
Interval
f(x)
f (x ) = x 2 + 4 x
Hence, (-2,
is increasing in the interval
) and decreasing in the interval (-
, -2).
Solution
3 2 (b) f (x ) = 4 x − 3 x − 6 x + 1
f ' (x ) = 12 x 2 − 6 x − 6 = 0
(
)
6 2x 2 − x − 1 = 0 6(2x + 1)(x − 1) = 0 1 x=− and x = 1. 2
Interval Test value, k Sign of f’(x) f(x)
1 − , − 2
1 − ,1 2
-4 + increasing
0 decreasing
(1, ) 4 + increasing
Hence, f (x ) = 4 x 3 − 3 x 2 − 6 x + 1 is increasing in the intervals 1 − ,− and 1, while decreasing in the 2
(
1 2
interval − ,1 .
)
Relative Maximum and Minimum Values (i) A number f(c1) is a relative maximum of a function f if f(x)ď&#x201A;Ł f(c1) for every x in some open interval that contains c1. (ii) A number f(c1) is a relative minimum of a function f if f(x) ď&#x201A;ł f(c1) for every x in some open interval that contains c1. Relative maxima and minima are called, in general, relative extrema.
Theorem: Critical value theorem If a continuous function f has a relative extremum at c, then c must be a critical value of f.
EXAMPLE 10
Locate the relative extrema of the following functions:
Solution: a) relative minimum at (0,0) b) neither relative maximum nor relative minimum c) relative maximum at (-1,5) and relative minimum at (1,1)
First Derivative Test (to find relative max and relative min) Let c be a critical number for f, and suppose f is continuous at c and differentiable on an open interval I containing c, except possibly at c itself If f’ changes from positive to negative at c then f(c) is a relative maximum of f. (ii) If f’ changes from negative to positive at c then f(c) is a relative minimum of f. (iii) If f’(x)>0 or if f’(x) <0 for every x in I except x=c, then f(c) is not a relative extrema of f. (i)
Example 10 Let f(x) = x3-3x +2. Find all the relative maximum and minimum of f. Solution:
f (x ) = x 3 − 3 x + 2 f ' (x ) = 3x 2 − 3 = 0 x = 1 Interval (- , -1) Test value, k -3 Sign of f’(x) + f(x) increasing
(-1,1) 0 decreasing
f (− 1) = ( −1)3 − 3( −1) + 2 = 4 (− 1,4 ) is a relative maximum point f (1) = (1)3 − 3(1) + 2 = 0 (1,0 ) is a relative minimum point
(1, ) 3 + increasing
The second Derivative Test for Relative Extrema (to find relative max and relative min) Second Derivative Test Let f be a function for which f’’ exists on (a,b) that contains the critical number c. (i) If f’(c)=0 and f’’ (c) > 0 , then f has a relative minimum at point (c,f(c)) (ii) If f’(c)=0 and f’’ (c) < 0 , then f has a relative maximum at point (c,f(c)) (iii) If f’(c)=0 and f’’(c)=0 , then f may have a relative minimum, relative maximum or neither at point (c,f(c))
EXAMPLE 10
Find the relative maximum point and relative minimum of the following function f (x) = 4 + 12x – 3x 2 – 2x 3 Solution:
f ( x ) = 4 + 12 x − 3 x 2 − 2x 3 f ' ( x ) = 12 − 6 x − 6 x 2
f ' ( x) = 0 12 − 6 x − 6 x 2 = 0 − 6( −2 + x + x 2 ) = 0 − 6( x + 2)( x − 1) = 0 x = −2 , x = 1
At x = −2 f ( −2) = 4 + 12( −2) − 3( −2) 2 − 2( −2)3 = −16 At x = 1
f (1) = 4 + 12(1) − 3(1) 2 − 2(1)3 = 11 Critical points are (-2, -16) and (1, 11)
f ' ' ( x ) = −6 − 12x Substitute
x = −2
and
x = 1 into f ' ' ( x )
f ' ' ( −2) = −6 − 12( −2) 0 f ' ' (1) = −6 − 12(1) 0
Therefore, f has a relative minimum at point (− 2,−16) and f has a relative maximum at point (1, 11)
Concavity and the second derivative Test Concavity: Increasing and Decreasing Derivatives The graphs of two functions are shown below. The graph of f is turning up and the graph of g is turning down.
f’(x) > 0 f’(x) < 0
From the graph of f, the slopes of the tangent lines are increasing, i.e f’ is increasing over the interval. This can be determined by noting that f” is positive.
f’(x) > 0 f’(x) < 0
• For the graph of g, the slopes of the tangent lines are decreasing, i.e g’ is decreasing over the interval. This can be determined by noting that g” is negative.
Definition Let f be differentiable on the interval (a,b) (i) If f’ is an increasing function on (a,b), then the graph of f is concave upward on the interval (ii) If f’ is an decreasing function on (a,b), then the graph of f is concave downward on the interval Test for concavity Let f be a function for which f’’ exists on (a,b) (i) If f ’’ (x) > 0 for all x on (a,b), then the graph of f is concave upward on (a,b). (ii) If f ’’ (x) < 0 for all x on (a,b), then the graph of f is concave downward on (a,b).
Note: For the interval (a,b) f ”(x)
f ’(x)
Graph of y=f(x)
+
increasing
Concave upward
-
decreasing
Concave downward
Examples
EXAMPLE 10
Find open intervals on which the following functions are concave up and open intervals on which they are concave down. (a) f(x) = x2 – 4x + 3 (b) f(x) = x3 +x-5
Solution:
f ( x ) = x 2 − 4x + 3 f ' ( x ) = 2x − 4 f ' ' ( x) = 2 0
Thus, the graph of f is concave up on the intervals (−∞,∞).
3 f ( x ) = x + x−5 (b)
f ' ( x ) = 3x 2 + 1
f ' ' ( x ) = 6x
f ' ' (x) = 0 6x = 0 x=0
Interval Test value Sign of f ’’(x) f(x)
(−∞,0) −5 − Concave down
(0,∞) 5 + Concave up
Thus, the graph of f is concave down on the interval (−∞,0) and concave up on the interval (0,∞)
Point of Inflection Suppose the graph of a function f has a tangent line (possibly vertical) at the point P(c, f(c) ) and that the graph is concave up on one side of P and concave down on the other side. Then P is called an inflection point of the graph. Definition The function f(x) have an inflection point at x = a if and only if f is a continuous function at a and f changes concavity at a. Note: In general , the concavity of the graph f will change or the point of inflection ( a, f(a)) will occur at a number a for which f’’(a) =0 or f’’ (x) does not exist. -A point of inflection at (a, f(a)) if f’’(a) =0 and f’’’(a) 0 - a is an inflection point if and only if f’’(a) changes sign at a, and a is in the domain of f.
EXAMPLE 10
Find any points of inflection of the functions f(x) = x3 -3x+1. Solution:
f ( x ) = x 3 − 3x + 1 f ' ( x ) = 3x 2 − 3 f ' ' ( x) = 6x
When f ' ' ( x ) = 0
6x = 0 x=0 At x = 0, f (0) = (0)3 − 3(0) + 1 = 1
Interval Test value Sign of f’’(x) f(x)
(−∞,0) −5 − Concave down
From the table above, since changes sign at x=0, so the curve has a point of inflection at (0,1)
(0,∞) 5 + Concave up
Sketching the graph of Polynomial function Step 1 : Analyse f(x) Find the intercepts, The x intercepts are the solutions to f(x)=0, if they exist. And y intercept is f(0) Step 2 : Analyse f’(x) Find the zeros of f’(x). Construct a sign chart for f’(x), determine the intervals where f(x) is increasing and decreasing, and find local maxima and minima.
Step 3 : Analyse f (x) Find the zeros of f(x).Construct a sign chart for f(x), determine where the graph f is concave upward and concave downward, and find any inflection points. Step 4 : Sketch the graph of f
Example 14 Consider the following functions: (i) f(x)= x3 - 6x2 + 16 f(x) = 0 x=0 (ii) f(x)=2x3-3x2-12x +5 f ’(x) a) Find the intercepts on x-axis and y-axis. b) Find the intervals on which f is increasing or decreasing. f ’(x) c) Find the relative maximum and relative minimum of f. f ’’(x) d) Find the intervals where the graph of f is concave up and f ’’(x) concave down. f ’’(x) e) Find the inflection point of f. f) Using the above information, sketch the graph of f.
Example 14 f(x) = 0 x=0 Consider the following function: f(x) = 2x3 + 12x2 + 18x a) Find the intercepts on the x- axis and y-axis. b) Find the intervals on which f is increasing or decreasing. f’(x) c) Find the relative maximum and relative minimum points of f. d) Find the intervals where the graph of f is concave up and concave down. e) Find the inflection point of f. f) Using the above information, sketch the graph of f.
Solution a) Find the intercepts x- intercept f(x) =0 2x3+12x2+18x=0 2x(x2+6x+9)=0 2x(x+3)2 0 x=0,-3 The graph cuts the x-axis at (0,0), (-3,0)
f’’(x)
y-intercept, x = 0, f(0)= 2(0)3+12(0)2+18(0)=0 The graph cuts the y-axis at the point (0,0).
b) Find the intervals where f(x) is increasing and decreasing f(x) = 2x3 + 12x2 + 18x f’(x) =6x2+24x+18 f’(x) =0 6x2+24x+18=0 (- ∞, -3) (-3,-1) x2+4x+3=0 Interval (x+1)(x+3)=0 Test value, k -5 -2 x=-1 , -3 Sign of f’(x) + f(x)
increasing
decreasing
(-1, ∞) 0 + increasing
So, f(x) is increasing in the interval (-∞, -3) and (-1, ∞), and decreasing in the interval (-3,-1).
c) Find the relative minimum and relative maximum points Using second derivatives f ”(x) = 12x + 24 f ' ' ( −1) = 12( −1) + 24 = 12 0
Relative minimum
f ' ' ( −3) = 12( −3) + 24 = −12 0
Relative maximum
When x = -1, f(-1) = 2(-1)3+12(-1)2+18(-1) = -8 When x = -3, f(-3) = 2(-3)3+12(-3)2+18(-3) = 0 Therefore, f has a relative minimum at point (-1, -8) and f has a relative maximum at point (-3, 0).
d)
Find the intervals where f is concave upward and concave downward, f ' ' ( x) = 0 12 x + 24 = 0 x = −2
Interval Test value Sign of f’’(x) f(x)
(−∞, -2) -5 − Concave down
(-2, ∞) 0 + Concave up
Thus, the graph of f is concave down on the interval (−∞,-2), and concave up on the interval (-2, ∞). e) Find the inflection points. when x = -2, f ( −2) = 2(− 2)3 + 12(− 2)2 + 18( −2) = −4
The concavity changes at x = -2, the point of inflection is (-2, -4).
f) Sketch the graph of f
f(x) = 2x3 + 12x2 + 18x y 4 2
.
(-3,0)
-5
-4
-3
-2
-1
.
(0,0)
-2
.
(-2,-4)
-4 -6
.
(-1,-8)
-8
x 1
2
Example 15 Sketch the graph of f(x)= - 2x3+3x2+12x -5.
Solution Find the intercepts x-intercept, f(x) = 0 -2x3+3x2+12x -5 = 0 x = -2.04, 0.39, 3.15 The graph cuts the x-axis at the points (-2.04, 0), (0.39, 0) and (3.15, 0)
y-intercept,
f (0) = −2(0) + 3(0) + 12(0) − 5 = −5 The graph cuts the y-axis at the point (0,-5). 3
2
Find the intervals where f(x) is increasing and decreasing, f(x)= - 2x3+3x2+12x -5. f’(x)= -6x2+6x+12 f’(x)=0 -6x2+6x+12=0 x2-x-2=0 (x-2)(x+1)=0 x=2,-1 Interval Test value, k Sign of f’(x) f(x)
(- ∞, -1) -5 decreasing
(-1,2) 0 + increasing
(2,∞) 5 decreasing
So, f(x) is decreasing in the interval (-∞, -1) and (2,∞), and increasing in the interval (−1,2).
Find the relative minimum and relative maximum points (using second derivative test) f’’(x)=-12x +6 Relative minimum f ' ' ( −1) = −12( −1) + 6 = 18 0 f ' ' (2) = −12(2) + 6 = −18 0
Relative maximum
When x=−1, 3 2 f ( −1) = −2(− 1) + 3(− 1) + 12( −1) − 5 = −12
When x=2, 3 2 f (2) = −2(2) + 3(2) + 12(2) − 5 = +15
Therefore, f has a relative minimum at point (-1,-12) and f has a relative maximum at point (2,15).
Find the intervals where f is concave upward and concave downward, f ' ' (x) = 0
Interval
− 12 x + 6 = 0 1 x= 2
Test value Sign of f’’(x) f(x)
1 − , 2 0 + Concave up
1 , 2 5 Concave down
Thus, the graph of f is concave up on the interval − , 1 , and concave down
1 on the interval , 2 Find the inflection points 1 when x = , 2
3
2
2
3 1 1 1 1 f = −2 + 3 + 12 − 5 = 2 2 2 2 2
1 3 The concavity changes at x = ½, the point of inflection is , 2 2
Sketch the graph of f f(x)= - 2x3+3x2+12x -5 y
.
15
(2,15)
10 5
.
(1/2,3/2)
-3
-2
-1
.
-5
(-1,-12)
.
-10
1 (0,-5)
x 2
3
4
Example 16 3 2 Consider the function f ( x) = x + 6x + 12x + 12 a) Find the intervals on which f is increasing or decreasing. b) Find the relative maximum and relative minimum points of f (if any). c) Find the intervals where the graph of f is concave up and concave down. d) Find the inflection point of f. e) Using the above information, sketch the graph of f.
Solution: a) f ( x) = x 3 + 6x 2 + 12x + 12 f ( x ) = 3 x 2 + 12 x + 12 = 0 3( x 2 + 4x + 4) = 0 3( x + 2)2 = 0 x = −2
Interval Test value, k Sign of f’(x) f(x)
(- ∞, -2) -5 + increasing
So, f(x) is increasing in the interval (- ∞, -2) and (-2, ∞).
(-2, ∞) 0 + increasing
b) Using second derivatives f’’(x)=6x+12 f’’(-2)= 6(-2)+12=0 The test is inconclusive, we have to use the first derivative test. Using first derivatives, f’(x) > 0 for all x. f has no relative maximum and minimum point. c) f ' ' ( x ) = 6x + 12 = 0 x = −2
Interval Test value Sign of f’’(x) f(x)
(- ∞, -2) -3 Concave down
(-2, ∞) 0 + Concave up
Thus, the graph of f is concave down on the interval (− ,−2) , and concave up on the interval (− 2, )
d) The inflection point is when x = -2, 3 2 f ( −2) = (− 2) + 6(− 2) + 12( −2) + 12 =4 The concavity changes at x= -2 , the point of inflection is (-2, 4). e) Sketch the graph y-intercept, f (0) = 12 f ( x ) = x 3 + 6x 2 + 12x + 12