Solution Manual For Finite Mathematics and Applied Calculus, 8th Edition by Stefan Waner,q Steven Co

Solutions Section 0.1 Section 0.1

1. 2(4 + (−1))(2 ⋅ −4) = 2(3)(−8) = (6)(−8) = −48

2. 3 + ([4 − 2] ⋅ 9) = 3 + (2 × 9) = 3 + 18 = 21

3. 20∕(3 * 4) − 1 = 20 − 1 = 53 − 1 = 23 12

4. 2 − (3 * 4)∕10 = 2 − 12 = 2 − 65 = 45 10

5.

3 + ([3 + (−5)]) 3−2×2 3 + (−2) 1 = = = −1 3−4 −1

7. (2 − 5 * (−1))∕1 − 2 * (−1) 2 − 5 ⋅ (−1) = − 2 ⋅ (−1) 1 2+5 = +2=7+2=9 1

6.

12 − (1 − 4) 2(5 − 1) ⋅ 2 − 1 12 − (−3) 15 = = =1 16 − 1 15

8. 2 − 5 * (−1)∕(1 − 2 * (−1)) 5 ⋅ (−1) =2− 1 − 2 ⋅ (−1) 5 11 −5 =2− +2=2+ = 1+2 3 5

2 × (−1) 2 2 2×1 2 = = =1 2 2

10. 2 + 4 ⋅ 3 2 = 2 + 4 × 9 = 2 + 36 = 38

11. 2 ⋅ 4 2 + 1 = 2 × 16 + 1 = 32 + 1 = 33

12. 1 − 3 ⋅ (−2) 2 × 2 = 1 − 3 × 4 × 2 = 1 − 24 = −23

13. 3^2+2^2+1 = 3 2 + 2 2 + 1 = 9 + 4 + 1 = 14

14. 2^(2^2-2)

9. 2 ⋅ (−1) 2∕2 =

15.

3 − 2(−3) 2 −6(4 − 1) 2 3 − 2 × 9 3 − 18 = = −6 × 9 −6(3) 2 5 −15 = = −54 18

17. 10*(1+1/10)^3 1 3 = 10(1 + ) = 10(1.1) 3 10 = 10 × 1.331 = 13.31

2

= 2 (2 −2) = 2 4−2 = 2 2 = 4

16.

1 − 2(1 − 4) 2 2(5 − 1) 2 ⋅ 2 1 − 2(−3) 2 1−2×9 = = 2 × 16 × 2 2(4) 2 ⋅ 2 1 − 18 17 = =− 64 64

18. 121/(1+1/10)^2 121 121 = = 2 1 1.1 2 !1 + " 10

121 = = 100 1.21

−2 ⋅ 3 2 [ −(4 − 1) 2 ] −2 × 9 ⎤ −18 = 3⎡ = 3[ 2 −9 ] ⎣⎢ −3 ⎦⎥ =3×2=6

19. 3

Solutions Section 0.1

1 2 2 21. 3⎡1 − (− ) ⎤ + 1 ⎣⎢ 2 ⎦⎥ 1 2 = 3[1 − ] + 1 4 3 2 9 = 3[ ] + 1 = 3[ ] + 1 4 16 27 43 = +1= 16 16 2 = # 12 $ − 12 = 14 − 14 = 0 2

8(1 − 4) 2 20. − [ −9(5 − 1) 2 ] 8×9 72 = −[ = −( ] −9 × 16 −144 ) 1 1 = −(− ) = 2 2 1 2 2 2 22. 3⎡ − ( ) ⎤ + 1 ⎣⎢ 9 3 ⎦⎥ 1 4 2 = 3[ − ] + 1 9 9 −3 2 −1 2 = 3[ ] + 1 = 3[ ] + 1 9 3 1 3 4 =3 +1= +1= 9 9 3 2

= 22 − # 21 $ = 21 − 41 = −2

23. (1/2)^2-1/2^2

24. 2/(1^2)-(2/1)^2

25. 3 × (2 − 5) = 3*(2-5)

26. 4 +

27.

3 = 3/(2-5) 2−5

3−1 = (3-1)/(8+6) 8+6 Note 3-1/8-6 is wrong, as it corresponds to 3 − 18 − 6. 29.

31. 3 −

33.

4+7 = 3-(4+7)/8 8

2 − 𝑥𝑦 2 = 2/(3+x)-x*y^2 3+𝑥

35. 3.1𝑥 3 − 4𝑥 −2 −

60

𝑥2 − 1

= 3.1x^3-4x^(-2)-60/(x^2-1)

1

28.

4−1 = (4-1)/3 3

30. 3 +

32.

5 = 4+5/9 or 4+(5/9) 9

3 = 3+3/(2-9) 2−9

4×2 2

!3"

34. 3 +

= 4*2/(2/3) or (4*2)/(2/3)

3+𝑥 = 3+(3+x)/(x*y) 𝑥𝑦

𝑥2 − 3 2 = 2.1x^(-3)-x^(-1)+(x^2-3)/2

36. 2.1𝑥 −3 − 𝑥 −1 +

37.

2

Solutions Section 0.1

2

!3"

5 = (2/3)/5

38.

3

!5"

= 2/(3/5)

39. 3 4−5 × 6 = 3^(4-5)*6 Note that the entire exponent is in parentheses.

40.

4 41. 3 1 + ( 100 )

42. 3

−3

= 2/(3+5^(7-9)) Note that the entire exponent is in parentheses. 1+4 ( 100 ) = 3((1+4)/100)^(-3)

= 3*(1+4/100)^(-3)

43. 3 2𝑥−1 + 4 𝑥 − 1 = 3^(2*x-1)+4^x-1 2

45. 2 2𝑥 −𝑥+1 = 2^(2x^2-x+1) Note that the entire exponent is in parentheses.

47.

4𝑒 −2𝑥 2 − 3𝑒 −2𝑥 = 4*e^(-2*x)/(2-3e^(-2*x)) or 4(*e^(-2*x))/(2-3e^(-2*x)) or (4*e^(-2*x))/(2-3e^(-2*x)) 2 2

49. 3(1 − !− 12 " ) + 1

= 3(1-(-1/2)^2)^2+1

2 3+5 7−9

−3

2

44. 2 𝑥 − (2 2𝑥) 2 = 2^(x^2)-(2^(2*x))^2 2

46. 2 2𝑥 −𝑥 + 1 = 2^(2x^2-x)+1 Note that the entire exponent is in parentheses. 𝑒 2𝑥 + 𝑒 −2𝑥 𝑒 2𝑥 − 𝑒 −2𝑥 = (e^(2*x)+e^(-2*x))/(e^(2*x)e^(-2*x)) 48.

1 2 2 2 50. 3⎛ − ( ) ⎞ + 1 ⎜⎝ 9 3 ⎟⎠ = 3(1/9-(2/3)^2)^2+1

Solutions Section 0.2 Section 0.2 1. 3 3 = 27

2. (−2) 3 = −8

3. −(2 ⋅ 3) 2 = −(2 2 ⋅ 3 2) = −(4 ⋅ 9) = −36 or −(2 ⋅ 3) 2 = −6 2 = −36

4. (4 ⋅ 2) 2 = 4 2 ⋅ 2 2 = 16 ⋅ 4 = 64 or (4 ⋅ 2) 2 = 8 2 = 64

5. (

−2 2 (−2) 2 4 = = 3 ) 9 32

7. (−2) −3 =

3 3 3 3 27 6. ( ) = 3 = 2 8 2

1 1 1 = =− 8 (−2) 3 −8

1 1 8. −2 −3 = − 3 = − 8 2

1 −2 1 1 9. ( ) = = = 16 2 4 1∕16 (1∕4)

10. (

11. 2 ⋅ 3 0 = 2 ⋅ 1 = 2

12. 3 ⋅ (−2) 0 = 3 ⋅ 1 = 3

13. 2 32 2 = 2 3+2 = 2 5 = 32 or 2 32 2 = 8 ⋅ 4 = 32

14. 3 23 = 3 23 1 = 3 2+1 = 3 3 = 27or 3 23 = 9 ⋅ 3 = 27

15. 2 22 −12 42 −4 = 2 2−1+4−4 = 2 1 = 2

16. 5 25 −35 25 −2 = 5 2−3+2−2 = 5 −1 =

17. 𝑥 3𝑥 2 = 𝑥 3+2 = 𝑥 5

18. 𝑥 4𝑥 −1 = 𝑥 4−1 = 𝑥 3

19. −𝑥 2𝑥 −3𝑦 = −𝑥 2−3𝑦 = −𝑥 −1𝑦 = −

𝑦 𝑥

1 1 9 −2 −2 = = = 2 3 ) 4∕9 4 (−2∕3)

1 5

20. −𝑥𝑦 −1𝑥 −1 = −𝑥 1−1𝑦 −1 = −𝑥 0𝑦 −1 1 =− 𝑦

21.

𝑥3 1 = 𝑥 3−4 = 𝑥 −1 = 𝑥 𝑥4

22.

𝑦5 = 𝑦 5−3 = 𝑦 2 3 𝑦

23.

𝑥 2𝑦 2 = 𝑥 2−(−1)𝑦 2−1 = 𝑥 3𝑦 𝑥 −1𝑦

24.

𝑥 −1𝑦 1 = 𝑥 −1−2𝑦 1−2 = 𝑥 −3𝑦 −1 = 3 𝑥 2𝑦 2 𝑥 𝑦

25.

(𝑥𝑦 −1𝑧 3) 2 𝑥 2(𝑦 −1) 2(𝑧 3) 2 = 𝑥 2𝑦𝑧 2 𝑥 2𝑦𝑧 2 2 −2 6 𝑥 𝑦 𝑧 = 2 2 = 𝑥 2−2𝑦 −2−1𝑧 6−2 𝑥 𝑦𝑧 𝑧4 = 𝑥 0𝑦 −3𝑧 4 = 3 𝑦

27. !

𝑥𝑦 −2𝑧 3 (𝑥𝑦 −2𝑧) 3 " = 𝑥 −1𝑧 (𝑥 −1𝑧) 3 3 −6 3 𝑥 𝑦 𝑧 = −3 3 = 𝑥 3−(−3)𝑦 −6𝑧 3−3 𝑥 𝑧 𝑥6 = 𝑥 6𝑦 −6𝑧 0 = 𝑦6 𝑥 −1𝑦 −2𝑧 2 −1−1 −2−1 2 −2 𝑦 𝑧 ) ) = (𝑥 𝑥𝑦 = (𝑥 −2𝑦 −3𝑧 2) −2 = 𝑥 4𝑦 6𝑧 −4 𝑥 4𝑦 6 = 4 𝑧 −2

29. (

31. √4 = 2

𝑥 2𝑦𝑧 2 𝑥 2𝑦𝑧 2 = (𝑥𝑦𝑧 −1) −1 𝑥 −1𝑦 −1(𝑧 −1) −1 𝑥 2𝑦𝑧 2 = −1 −1 = 𝑥 2−(−1)𝑦 1−(−1)𝑧 2−1 𝑥 𝑦 𝑧 = 𝑥 3𝑦 2𝑧 4

Solutions Section 0.2 26.

𝑥 2𝑦 −1𝑧 0 (𝑥 2𝑦 −1𝑧 0) 2 = 𝑥𝑦𝑧 ) (𝑥𝑦𝑧) 2 𝑥 4𝑦 −2 = 2 2 2 = 𝑥 4−2𝑦 −2−2𝑧 −2 𝑥 𝑦 𝑧 𝑥2 = 𝑥 2𝑦 −4𝑧 −2 = 4 2 𝑦 𝑧 2

28. (

30.

𝑥𝑦 −2 = (𝑥 1−2𝑦 −2+1𝑧 −1) −3 ( 𝑥 2𝑦 −1𝑧 ) = (𝑥 −1𝑦 −1𝑧 −1) −3 = 𝑥 3𝑦 3𝑧 3 −3

32. √5 ≈ 2.236

33.

1 √1 1 = = √ 4 √4 2

34.

1 √1 1 = = √ 9 √9 3

35.

16 √16 4 = = √9 3 √9

36.

9 √9 3 = = √ 4 √4 2

37.

√4

5

=

2 5

38.

6

√25

=

6 5

39. √9 + √16 = 3 + 4 = 7

40. √25 − √16 = 5 − 4 = 1

41. √9 + 16 = √25 = 5

42. √25 − 16 = √9 = 3

3

3

3

3 √27

43. √8 − 27 = √−19 ≈ −2.668 45. √27∕8 =

3 √8

=

3 2

4

4

44. √81 − 16 = √65 ≈ 2.839 3

3

3

46. √8 × 64 = √8 ⋅ √64 = 2 ⋅ 4 = 8

Solutions Section 0.2

47. √(−2) 2 = √4 = 2

49.

48. √(−1) 2 = √1 = 1

1 16 √16 4 (1 + 15) = = = =2 √4 √4 √4 2

50.

1 36 √36 6 (3 + 33) = = = =2 √9 √9 3 √9

51. √𝑎 2𝑏 2 = √𝑎 2√𝑏 2 = 𝑎𝑏

𝑎 2 √𝑎 2 𝑎 52. √ 2 = = 𝑏 𝑏 √𝑏 2

53. √(𝑥 + 9) 2 = 𝑥 + 9 (𝑥 + 9 > 0 because 𝑥 is positive.)

54. (√𝑥 + 9) = 𝑥 + 9

3

3

3

55. √𝑥 3(𝑎 3 + 𝑏 3) = √𝑥 3 √(𝑎 3 + 𝑏 3) 3

= 𝑥 √(𝑎 3 + 𝑏 3)

57.

(Not 𝑥(𝑎 + 𝑏))

4𝑥𝑦 3 4𝑦 2 √4√𝑦 2 2𝑦 = = = √ 2 √𝑥 √𝑥 𝑥 √𝑥 𝑦

2

4

56. √

58.

4

√𝑥 4 𝑥4 𝑥 = = 4 4 𝑎𝑏 𝑎 4𝑏 4 √𝑎 4 √𝑏 4

4(𝑥 2 + 𝑦 2) √4√𝑥 2 + 𝑦 2 = 𝑐2 √𝑐 2 √ =

2√𝑥 2 + 𝑦 2 𝑐

(Not 2(𝑥 + 𝑦)∕𝑐)

Solutions Section 0.3 Section 0.3

3 𝑥4

2.

1 −4 1 𝑥 = 4 2 2𝑥

3 −2∕3 3 𝑥 = 4 4𝑥 2∕3

4.

4 −3∕4 4 𝑦 = 5 5𝑦 3∕4

6.

1 0.1𝑥 −2 𝑥 4 0.1 + = + 2 3 3 3𝑥 −4 3𝑥

1. 3𝑥 −4 =

3.

6 −1 6 5. 1 − 0.3 = 1 − 0.3𝑥 2 − 5𝑥 −2 − 5 𝑥 𝑥

3

5

7. 2 2∕3 = √2 2

8. 3 4∕5 = √3 4

3

3

3

3

9. 𝑥 4∕3 = √𝑥 4 = √𝑥 3 √𝑥 = 𝑥 √𝑥 11. (𝑥 1∕2𝑦 1∕3)

1∕5

5

= √√ 𝑥 √ 𝑦 3

3 3 3 13. − 𝑥 −1∕4 = − =− 4 1∕4 2 2𝑥 2 √𝑥 15. 0.2𝑥 −2∕3 +

3

7𝑥 −1∕2 0.2 3√ 𝑥 = 3 + 7 √𝑥 2

17.

=

0.2 3𝑥 1∕2 + 7 𝑥 2∕3

3 3 = 5∕2 4(1 − 𝑥) 4√(1 − 𝑥) 5 3 = 4√(1 − 𝑥) 4√1 − 𝑥 3 = 4(1 − 𝑥) 2√1 − 𝑥

4

4

4

4

10. 𝑦 7∕4 = √𝑦 7 = √𝑦 4 √𝑦 3 = |𝑦| √𝑦 3 12. 𝑥 −1∕3𝑦 3∕2 =

14.

16.

18.

𝑦 3∕2 √𝑦 3 |𝑦|√𝑦 = 3 = 3 𝑥 1∕3 √𝑥 √𝑥

4 3∕2 4√𝑥 3 4|𝑥|√𝑥 𝑥 = = 5 5 5 3.1 11 −1∕7 11 𝑥 − = 3.1𝑥 4∕3 − −4∕3 7 𝑥 7𝑥 1∕7 3 3 11 11 = 3.1 √𝑥 4 − 7 = 3.1𝑥 √𝑥 − 7 7 √𝑥 7 √𝑥 9(1 − 𝑥) 7∕3 9 = 4 4(1 − 𝑥) −7∕3 3

3

9 √(1 − 𝑥) 7 9 √(1 − 𝑥) 6 √1 − 𝑥 = = 4 4 3 9(1 − 𝑥) 2 √1 − 𝑥 = 4

19. √3 = 3 1∕2

20. √8 = 8 1∕2

21. √𝑥 3 = 𝑥 3∕2

22. √𝑥 2 = 𝑥 2∕3

3

3

3

23. √𝑥𝑦 2 = (𝑥𝑦 2) 1∕3 25.

Solutions Section 0.3

𝑥2 𝑥2 = = 𝑥 2−1∕2 = 𝑥 3∕2 √𝑥 𝑥 1∕2

24. √𝑥 2𝑦 = (𝑥 2𝑦) 1∕2 26.

𝑥

√𝑥

=

𝑥

𝑥 1∕2

= 𝑥 1−1∕2 = 𝑥 1∕2

27.

3 3 = 𝑥 −2 2 5 5𝑥

28.

2 2 = 𝑥3 −3 5 5𝑥

29.

3𝑥 −1.2 1 3 1 − 2.1 = 𝑥 −1.2 − 𝑥 −2.1 2 2 3 3𝑥

30.

2 𝑥 2.1 2 1.2 1 2.1 − = 𝑥 − 𝑥 3 3 3 3𝑥 −1.2

31.

33.

2𝑥 𝑥 0.1 4 − + 1.1 3 2 3𝑥 2 1 4 = 𝑥 − 𝑥 0.1 + 𝑥 −1.1 3 2 3 3√𝑥 5 4 − + 4 3√𝑥 3𝑥√𝑥 3𝑥 1∕2 5 4 = − + 1∕2 4 3𝑥 3𝑥 ⋅ 𝑥 1∕2 3 5 4 = 𝑥 1∕2 − 𝑥 −1∕2 + 𝑥 −3∕2 4 3 3 5

3 √𝑥 2 7 3𝑥 2∕5 7 35. − = − 4 4 2𝑥 3∕2 2√𝑥 3 3 7 = 𝑥 2∕5 − 𝑥 −3∕2 4 2

34.

36.

38.

39. 4 −1∕24 7∕2 = 4 −1∕2+7∕2 = 4 6∕2 = 4 3 = 64

40.

(𝑥 2 + 1) 3

−

3

4 √𝑥 2 + 1 1 3 = − (𝑥 2 + 1) 3 (𝑥 2 + 1) 1∕3 3 = (𝑥 2 + 1) −3 − (𝑥 2 + 1) −1∕3 4

41. 3 2∕33 −1∕6 = 3 2∕3−1∕6 = 3 1∕2 = √3

4𝑥 2 𝑥 3∕2 2 + − 2 3 6 3𝑥 4 1 2 = 𝑥 2 + 𝑥 3∕2 − 𝑥 −2 3 6 3 3

5√𝑥

−

5√𝑥 7 + 3 8 2 √𝑥

3 5𝑥 1∕2 7 − + 1∕2 8 5𝑥 2𝑥 1∕3 3 −1∕2 5 1∕2 7 −1∕3 = 𝑥 − 𝑥 + 𝑥 5 8 2 =

3

37.

1

32.

1 2 1 2 − = − 5 3∕2 √ 3𝑥 3∕5 8𝑥 𝑥 3 √𝑥 3 8𝑥 1 2 = 𝑥 −3∕2 − 𝑥 −3∕5 8 3 3

3 √(𝑥 2 + 1) 7 2 − 4 3(𝑥 2 + 1) −3 2 2 3 = (𝑥 + 1) 3 − (𝑥 2 + 1) 7∕3 3 4 2 1∕𝑎 1 = 2 1∕𝑎−2∕𝑎 = 2 −1∕𝑎 = 2∕𝑎 1∕𝑎 2 2

42. 2 1∕32 −12 2∕32 −1∕3 = 2 1∕3−1+2∕3−1∕3 1 = 2 −1∕3 = 1∕3 2

43.

𝑥 3∕2 1 = 𝑥 3∕2−5∕2 = 𝑥 −1 = 5∕2 𝑥 𝑥

45.

𝑥 1∕2𝑦 2 = 𝑥 1∕2−(−1∕2)𝑦 2−1 = 𝑥𝑦 𝑥 −1∕2𝑦

47.

44.

𝑦 5∕4 = 𝑦 5∕4−3∕4 = 𝑦 1∕2 = √𝑦 𝑦 3∕4

46.

𝑥 −1∕2𝑦 = 𝑥 −1∕2−2𝑦 1−3∕2 = 𝑥 −5∕2𝑦 −1∕2 𝑥 2𝑦 3∕2

Solutions Section 0.3

𝑥 1∕3 𝑦 2∕3 𝑥 1∕3 𝑦 2∕3 ⋅ = ( 𝑦 ) (𝑥) 𝑦 1∕3 𝑥 2∕3 = 𝑥 1∕3−2∕3𝑦 −1∕3+2∕3 = 𝑥 −1∕3𝑦 1∕3 or ( 𝑥𝑦 ) 1∕3

48.

49. 𝑥 2 − 16 = 0 ⇒ 𝑥 2 = 16 ⇒ 𝑥 = ±√16 ⇒ 𝑥 = ±4

𝑥 −1∕3 𝑦 1∕3 𝑥 −1∕3 𝑦 1∕3 ⋅ = (𝑥) (𝑦) 𝑦 −1∕3 𝑥 1∕3 𝑥 −1∕3−1∕3𝑦 1∕3+1∕3 = 𝑥 −2∕3𝑦 2∕3 or ( 𝑥𝑦 ) 2∕3

50. 𝑥 2 − 1 = 0 ⇒ 𝑥 2 = 1 ⇒ 𝑥 = ±√1 ⇒ 𝑥 = ±1

4 4 = 0 ⇒ 𝑥2 = 9 9 4 2 ⇒𝑥=± ⇒𝑥=± √9 3

1 1 = 0 ⇒ 𝑥2 = 10 10 1 1 ⇒𝑥=± ⇒𝑥=± √ 10 √10

51. 𝑥 2 −

52. 𝑥 2 −

53. 𝑥 2 − (1 + 2𝑥) 2 = 0 ⇒ 𝑥 2 = (1 + 2𝑥) 2 ⇒ 𝑥 = ±1 + 2𝑥 If 𝑥 = 1 + 2𝑥, then −𝑥 = 1 ⇒ 𝑥 = −1. 1 If 𝑥 = −(1 + 2𝑥), then = −1 ⇒ 𝑥 = − . 3 1 So, 𝑥 = 1 or − . 3

54. 𝑥 2 − (2 − 3𝑥) 2 = 0 ⇒ 𝑥 2 = (2 − 3𝑥) 2 ⇒ 𝑥 = ±2 − 3𝑥 1 If 𝑥 = 2 − 3𝑥, then 4𝑥 = 2 ⇒ 𝑥 = . 2 If 𝑥 = −(2 − 3𝑥), then −2𝑥 = −2 ⇒ 𝑥 = 1. 1 So, 𝑥 = 1 or . 2

55. 𝑥 5 + 32 = 0 ⇒ 𝑥 5 = −32

56. 𝑥 4 − 81 = 0 ⇒ 𝑥 4 = 81 4 ⇒ 𝑥 = ± √81 = ±3

57. 𝑥 1∕2 − 4 = 0 ⇒ 𝑥 1∕2 = 4 ⇒ 𝑥 = 4 2 = 16

58. 𝑥 1∕3 − 2 = 0 ⇒ 𝑥 1∕3 = 2 ⇒ 𝑥 = 23 = 8

⇒𝑥=

5 √−32 = −2

1 1 =0⇒1= 2 2 𝑥 𝑥 ⇒ 𝑥 2 = 1 ⇒ 𝑥 = ±√1 = ±1

59. 1 −

61. (𝑥 − 4) −1∕3 = 2 ⇒ 𝑥 − 4 = 2 −3 = ⇒𝑥=4+

1 33 = 8 8

60.

1 8

2 6 2 6 − 4=0⇒ 3= 4 3 𝑥 𝑥 𝑥 𝑥 ⇒ 2𝑥 4 = 6𝑥 3 ⇒ 2𝑥 = 6 ⇒ 𝑥 = 3

62. (𝑥 − 4) 2∕3 + 1 = 5 ⇒ (𝑥 − 4) 2∕3 = 4 ⇒ 𝑥 − 4 = ±4 3∕2 = ±8 ⇒ 𝑥 = 4 ± 8 = −4 or 12

Solutions Section 0.4 Section 0.4

1. 𝑥(4𝑥 + 6) = 4𝑥 2 + 6𝑥

2. (4𝑦 − 2)𝑦 = 4𝑦 2 − 2𝑦

3. (2𝑥 − 𝑦)𝑦 = 2𝑥𝑦 − 𝑦 2

4. 𝑥(3𝑥 + 𝑦) = 3𝑥 2 + 𝑥𝑦

5. (𝑥 + 1)(𝑥 − 3) = 𝑥 2 + 𝑥 − 3𝑥 − 3 = 𝑥 2 − 2𝑥 − 3

6. (𝑦 + 3)(𝑦 + 4) = 𝑦 2 + 3𝑦 + 4𝑦 + 12 = 𝑦 2 + 7𝑦 + 12

7. (2𝑦 + 3)(𝑦 + 5) = 2𝑦 2 + 3𝑦 + 10𝑦 + 15 = 2𝑦 2 + 13𝑦 + 15

8. (2𝑥 − 2)(3𝑥 − 4) = 6𝑥 2 − 6𝑥 − 8𝑥 + 8 = 6𝑥 2 − 14𝑥 + 8

9. (2𝑥 − 3) 2 = 4𝑥 2 − 12𝑥 + 9

10. (3𝑥 + 1) 2 = 9𝑥 2 + 6𝑥 + 1

2

2

11. !𝑥 + 𝑥1 " = 𝑥 2 + 2 + 12

12. !𝑦 − 1𝑦 " = 𝑦 2 − 2 + 12

13. (2𝑥 − 3)(2𝑥 + 3) = (2𝑥) 2 − 3 2 = 4𝑥 2 − 9

14. (4 + 2𝑥)(4 − 2𝑥) = 4 2 − (2𝑥) 2 = 16 − 4𝑥 2

𝑥

𝑦

15. !𝑦 − 1𝑦 "!𝑦 + 1𝑦 " = 𝑦 2 − ! 1𝑦 " = 𝑦 2 − 12

16. (𝑥 − 𝑥 2)(𝑥 + 𝑥 2) = 𝑥 2 − (𝑥 2) 2 = 𝑥 2 − 𝑥 4

17. (𝑥 2 + 𝑥 − 1)(2𝑥 + 4) = 𝑥 2(2𝑥 + 4) + 𝑥(2𝑥 + 4) − 1(2𝑥 + 4) = 2𝑥 3 + 4𝑥 2 + 2𝑥 2 + 4𝑥 − 2𝑥 − 4 = 2𝑥 3 + 6𝑥 2 + 2𝑥 − 4

18. (3𝑥 + 1)(2𝑥 2 − 𝑥 + 1) = 3𝑥(2𝑥 2 − 𝑥 + 1) + 1(2𝑥 2 − 𝑥 + 1) = 6𝑥 3 − 3𝑥 2 + 3𝑥 + 2𝑥 2 − 𝑥 + 1 = 6𝑥 3 − 𝑥 2 + 2𝑥 + 1

2

𝑦

19. (𝑥 2 − 2𝑥 + 1) 2 = (𝑥 2 − 2𝑥 + 1)(𝑥 2 − 2𝑥 + 1) = 𝑥 2(𝑥 2 − 2𝑥 + 1) − 2𝑥(𝑥 2 − 2𝑥 + 1) + 1(𝑥 2 − 2𝑥 + 1) = 𝑥 4 − 2𝑥 3 + 𝑥 2 − 2𝑥 3 + 4𝑥 2 − 2𝑥 + 𝑥 2 − 2𝑥 + 1 = 𝑥 4 − 4𝑥 3 + 6𝑥 2 − 4𝑥 + 1 20. (𝑥 + 𝑦 − 𝑥𝑦) 2 = (𝑥 + 𝑦 − 𝑥𝑦)(𝑥 + 𝑦 − 𝑥𝑦) = 𝑥(𝑥 + 𝑦 − 𝑥𝑦) + 𝑦(𝑥 + 𝑦 − 𝑥𝑦) − 𝑥𝑦(𝑥 + 𝑦 − 𝑥𝑦) = 𝑥 2 + 𝑥𝑦 − 𝑥 2𝑦 + 𝑥𝑦 + 𝑦 2 − 𝑥𝑦 2 − 𝑥 2𝑦 − 𝑥𝑦 2 + 𝑥 2𝑦 2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2 − 2𝑥 2𝑦 − 2𝑥𝑦 2 + 𝑥 2𝑦 2

21. (𝑦 3 + 2𝑦 2 + 𝑦)(𝑦 2 + 2𝑦 − 1) = 𝑦 3(𝑦 2 + 2𝑦 − 1) + 2𝑦 2(𝑦 2 + 2𝑦 − 1) + 𝑦(𝑦 2 + 2𝑦 − 1) = 𝑦 5 + 2𝑦 4 − 𝑦 3 + 2𝑦 4 + 4𝑦 3 − 2𝑦 2 + 𝑦 3 + 2𝑦 2 − 𝑦 = 𝑦 5 + 4𝑦 4 + 4𝑦 3 − 𝑦

22. (𝑥 − 2𝑥 + 4)(3𝑥 − 𝑥 + 2) = 𝑥 3(3𝑥 2 − 𝑥 + 2) − 2𝑥 2(3𝑥 2 − 𝑥 + 2) + 4(3𝑥 2 − 𝑥 + 2) = 3𝑥 5 − 𝑥 4 + 2𝑥 3 − 6𝑥 4 + 2𝑥 3 − 4𝑥 2 + 12𝑥 2 − 4𝑥 + 8 = 3𝑥 5 − 7𝑥 4 + 4𝑥 3 + 8𝑥 2 − 4𝑥 + 8 3

2

2

Solutions Section 0.4

23. (𝑥 + 1)(𝑥 + 2) + (𝑥 + 1)(𝑥 + 3) = (𝑥 + 1)(𝑥 + 2 + 𝑥 + 3) = (𝑥 + 1)(2𝑥 + 5)

25. (𝑥 2 + 1) 5(𝑥 + 3) 4 + (𝑥 2 + 1) 6(𝑥 + 3) 3 = (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 + 3 + 𝑥 2 + 1) = (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 2 + 𝑥 + 4)

24. (𝑥 + 1)(𝑥 + 2) 2 + (𝑥 + 1) 2(𝑥 + 2) = (𝑥 + 1)(𝑥 + 2)(𝑥 + 2 + 𝑥 + 1) = (𝑥 + 1)(𝑥 + 2)(2𝑥 + 3)

26. 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 5 + 15𝑥 2(𝑥 2 + 1) 5(𝑥 3 + 1) 4 = 5𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4[2(𝑥 3 + 1) + 3𝑥(𝑥 2 + 1)] = 5𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4(5𝑥 3 + 3𝑥 + 2)

27. (𝑥 3 + 1)√𝑥 + 1 − (𝑥 3 + 1) 2√𝑥 + 1 = (𝑥 3 + 1)√𝑥 + 1 ⋅ [1 − (𝑥 3 + 1)] = −𝑥 3(𝑥 3 + 1)√𝑥 + 1

28. (𝑥 2 + 1)√𝑥 + 1 − √(𝑥 + 1) 3 = √𝑥 + 1 ⋅ [𝑥 2 + 1 − √(𝑥 + 1) 2] = √𝑥 + 1 ⋅ [𝑥 2 + 1 − (𝑥 + 1)] = (𝑥 2 − 𝑥)√𝑥 + 1 = 𝑥(𝑥 − 1)√𝑥 + 1

29. √(𝑥 + 1) 3 + √(𝑥 + 1) 5 = √(𝑥 + 1) 3 ⋅ [1 + √(𝑥 + 1) 2] = √(𝑥 + 1) 3 ⋅ (1 + 𝑥 + 1) = (𝑥 + 2)√(𝑥 + 1) 3

30. (𝑥 2 + 1) √(𝑥 + 1) 4 − √(𝑥 + 1) 7 3

3

3

3

= √(𝑥 + 1) 4 ⋅ [𝑥 2 + 1 − √(𝑥 + 1) 3] 3

= √(𝑥 + 1) 4 ⋅ [𝑥 2 + 1 − (𝑥 + 1)] 3

= (𝑥 2 − 𝑥) √(𝑥 + 1) 4 3

= 𝑥(𝑥 − 1) √(𝑥 + 1) 4

31. 𝑎 = 2, 𝑏 = 6, 𝑐 = 5, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 6 2 − 4(2)(5) = 36 − 40 = −4 As the discriminant is negative, the expression does not factor at all.

32. 𝑎 = 4, 𝑏 = −6, 𝑐 = 2, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−6) 2 − 4(4)(2) = 36 − 32 = 4 = 2 2 As the discriminant is a perfect square, the expression factors over the integers.

33. 𝑎 = −3, 𝑏 = −2, 𝑐 = 3, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−2) 2 − 4(−3)(3) = 4 + 36 = 40 As the discriminant is positive but not a perfect square, the expression factors, but not over the integers. 34. 𝑎 = 1, 𝑏 = −4, 𝑐 = −7, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−4) 2 − 4(1)(−7) = 16 + 28 = 44 As the discriminant is positive but not a perfect square, the expression factors, but not over the integers. 35. 𝑎 = 8, 𝑏 = 12, 𝑐 = 4, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 12 2 − 4(8)(4) = 144 − 128 = 16 = 4 2 As the discriminant is a perfect square, the expression factors over the integers.

Solutions Section 0.4 36. 𝑎 = 1, 𝑏 = 2, 𝑐 = 19, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 2 2 − 4(1)(19) = 4 − 76 = −72 As the discriminant is negative, the expression does not factor at all.

37. 𝑎 = 40, 𝑏 = −64, 𝑐 = 24, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−64) 2 − 4(40)(24) = 4,096 − 3,840 = 256 = 16 2 As the discriminant is a perfect square, the expression factors over the integers. 38. 𝑎 = −10, 𝑏 = −32, 𝑐 = −32, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−32) 2 − 4(−10)(−32) = 1,024 − 1,280 = −256 As the discriminant is negative, the expression does not factor at all.

39. 𝑎 = 6, 𝑏 = −22, 𝑐 = 16, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−22) 2 − 4(6)(16) = 484 − 384 = 100 = 10 2 As the discriminant is a perfect square, the expression factors over the integers. 40. 𝑎 = 48, 𝑏 = 32, 𝑐 = 4, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 32 2 − 4(48)(4) = 1,024 − 768 = 256 = 16 2 As the discriminant is a perfect square, the expression factors over the integers. 41. a. 2𝑥 + 3𝑥 2 = 𝑥(2 + 3𝑥) b. 𝑥(2 + 3𝑥) = 0 𝑥 = 0 or 2 + 3𝑥 = 0 𝑥 = 0 or −2∕3

42. a. 𝑦 2 − 4𝑦 = 𝑦(𝑦 − 4) b. 𝑦(𝑦 − 4) = 0 𝑦 = 0 or 𝑦 − 4 = 0 𝑦 = 0 or 4

43. a. 6𝑥 3 − 2𝑥 2 = 2𝑥 2(3𝑥 − 1) b. 2𝑥 2(3𝑥 − 1) = 0 𝑥 2 = 0 or 3𝑥 − 1 = 0 𝑥 = 0 or 1∕3

44. a. 3𝑦 3 − 9𝑦 2 = 3𝑦 2(𝑦 − 3) b. 3𝑦 2(𝑦 − 3) = 0 𝑦 2 = 0 or 𝑦 − 3 = 0 𝑦 = 0 or 3

45. a. 𝑥 2 − 8𝑥 + 7 = (𝑥 − 1)(𝑥 − 7) b. (𝑥 − 1)(𝑥 − 7) = 0 𝑥 − 1 = 0 or 𝑥 − 7 = 0 𝑥 = 1 or 7

46. a. 𝑦 2 + 6𝑦 + 8 = (𝑦 + 2)(𝑦 + 4) b. (𝑦 + 2)(𝑦 + 4) = 0 𝑦 + 2 = 0 or 𝑦 + 4 = 0 𝑦 = −2 or −4

47. a. 𝑥 2 + 𝑥 − 12 = (𝑥 − 3)(𝑥 + 4) b. (𝑥 − 3)(𝑥 + 4) = 0 𝑥 − 3 = 0 or 𝑥 + 4 = 0 𝑥 = 3 or −4

48. a. 𝑦 2 + 𝑦 − 6 = (𝑦 − 2)(𝑦 + 3) b. (𝑦 − 2)(𝑦 + 3) = 0 𝑦 − 2 = 0 or 𝑦 + 3 = 0 𝑦 = 2 or −3

49. a. 2𝑥 2 − 3𝑥 − 2 = (2𝑥 + 1)(𝑥 − 2) b. (2𝑥 + 1)(𝑥 − 2) = 0 2𝑥 + 1 = 0 or 𝑥 − 2 = 0 𝑥 = −1∕2 or 2

50. a. 3𝑦 2 − 8𝑦 − 3 = (3𝑦 + 1)(𝑦 − 3) b. (3𝑦 + 1)(𝑦 − 3) = 0 3𝑦 + 1 = 0 or 𝑦 − 3 = 0 𝑦 = −1∕3 or 3

51. a. 6𝑥 2 + 13𝑥 + 6 = (2𝑥 + 3)(3𝑥 + 2) b. (2𝑥 + 3)(3𝑥 + 2) = 0 2𝑥 + 3 = 0 or 3𝑥 + 2 = 0 𝑥 = −3∕2 or −2∕3

52. a. 6𝑦 2 + 17𝑦 + 12 = (3𝑦 + 4)(2𝑦 + 3) b. (3𝑦 + 4)(2𝑦 + 3) = 0 3𝑦 + 4 = 0 or 2𝑦 + 3 = 0 𝑦 = −4∕3 or −3∕2

53. a. 12𝑥 2 + 𝑥 − 6 = (3𝑥 − 2)(4𝑥 + 3) b. (3𝑥 − 2)(4𝑥 + 3) = 0 3𝑥 − 2 = 0 or 4𝑥 + 3 = 0 𝑥 = 2∕3 or −3∕4

54. a. 20𝑦 2 + 7𝑦 − 3 = (4𝑦 − 1)(5𝑦 + 3) b. (4𝑦 − 1)(5𝑦 + 3) = 0 4𝑦 − 1 = 0 or 5𝑦 + 3 = 0 𝑦 = 1∕4 or −3∕5

55. a. 𝑥 2 + 4𝑥𝑦 + 4𝑦 2 = (𝑥 + 2𝑦) 2 b. (𝑥 + 2𝑦) 2 = 0 𝑥 + 2𝑦 = 0 𝑥 = −2𝑦

56. a. 4𝑦 2 − 4𝑥𝑦 + 𝑥 2 = (2𝑦 − 𝑥) 2 b. (2𝑦 − 𝑥) 2 = 0 2𝑦 − 𝑥 = 0 𝑦 = 𝑥∕2

57. a. 𝑥 4 − 5𝑥 2 + 4 = (𝑥 2 − 1)(𝑥 2 − 4) = (𝑥 + 1)(𝑥 − 1)(𝑥 + 2)(𝑥 − 2) b. (𝑥 + 1)(𝑥 − 1)(𝑥 + 2)(𝑥 − 2) = 0 𝑥 + 1 = 0 or 𝑥 − 1 = 0 or 𝑥 + 2 = 0 or 𝑥 − 2 = 0 𝑥 = ±1 or ±2

58. a. 𝑦 4 + 2𝑦 2 − 3 = (𝑦 2 − 1)(𝑦 2 + 3) = (𝑦 + 1)(𝑦 − 1)(𝑦 2 + 3) b. (𝑦 + 1)(𝑦 − 1)(𝑦 2 + 3) = 0 𝑦 + 1 = 0 or 𝑦 − 1 = 0 or 𝑦 2 + 3 = 0 𝑦 = ±1 (Notice that 𝑦 2 + 3 = 0 has no real solutions.)

59. a. 𝑥 2 − 3 = (𝑥 − √3)(𝑥 + √3) b. (𝑥 − √3)(𝑥 + √3) = 0 𝑥 − √3 = 0 or 𝑥 + √3 = 0 𝑥 = ±√3

60. a. 𝑦 2 − 7 = (𝑦 − √7)(𝑦 + √7) b. (𝑦 − √7)(𝑦 + √7) = 0 𝑦 − √7 = 0 or 𝑦 + √7 = 0 𝑦 = ±√7

Solutions Section 0.4

Solutions Section 0.5 Section 0.5 1.

2.

3.

4.

5.

6.

7.

8.

9.

𝑥 − 4 2𝑥 + 1 (𝑥 − 4)(2𝑥 + 1) 2𝑥 2 − 7𝑥 − 4 ⋅ = = 𝑥+1 𝑥−1 (𝑥 + 1)(𝑥 − 1) 𝑥2 − 1 2𝑥 − 3 𝑥 + 3 (2𝑥 − 3)(𝑥 + 3) 2𝑥 2 + 3𝑥 − 9 ⋅ = = 2 𝑥−2 𝑥+1 (𝑥 − 2)(𝑥 + 1) 𝑥 −𝑥−2

𝑥 − 4 2𝑥 + 1 (𝑥 − 4)(𝑥 − 1) + (𝑥 + 1)(2𝑥 + 1) 3𝑥 2 − 2𝑥 + 5 + = = 𝑥+1 𝑥−1 (𝑥 + 1)(𝑥 − 1) 𝑥2 − 1 2𝑥 − 3 𝑥 + 3 (2𝑥 − 3)(𝑥 + 1) + (𝑥 − 2)(𝑥 + 3) 3𝑥 2 − 9 + = = 2 𝑥−2 𝑥+1 (𝑥 − 2)(𝑥 + 1) 𝑥 −𝑥−2 𝑥2 𝑥 − 1 𝑥 2 − (𝑥 − 1) 𝑥 2 − 𝑥 + 1 − = = 𝑥+1 𝑥+1 𝑥+1 𝑥+1

(𝑥 2 − 1)(𝑥 − 1) − (𝑥 − 2) 𝑥 3 − 𝑥 2 − 2𝑥 + 3 𝑥2 − 1 1 − = = 𝑥−2 𝑥−1 (𝑥 − 2)(𝑥 − 1) 𝑥 2 − 3𝑥 + 2 1

+𝑥−1= 𝑥 ( 𝑥−1 ) 2

𝑥−2

! 𝑥2 "

−

𝑥 − 1 + 𝑥(𝑥 − 1) 𝑥 2 − 1 𝑥−1 +𝑥−1= = 𝑥 𝑥 𝑥

1 2𝑥 2 1 2𝑥 2 − 1 = − = 𝑥−2 𝑥−2 𝑥−2 𝑥−2

1 ⎡ 𝑥 − 3 1 ⎤ 1 ⎡ 𝑥 − 3 + 𝑥 ⎤ 2𝑥 − 3 + = = 𝑥 ⎢⎣ 𝑥𝑦 𝑦 ⎥⎦ 𝑥 ⎢⎣ 𝑥𝑦 ⎥⎦ 𝑥 2𝑦

10.

11.

12.

13.

14.

𝑦 2 ⎡ 2𝑥 − 3 𝑥 ⎤ 𝑦 2 ⎡ 2𝑥 − 3 + 𝑥 ⎤ 𝑦 2(3𝑥 − 3) 𝑦(3𝑥 − 3) 3𝑥𝑦 − 3𝑦 + = = = = 𝑥 ⎢⎣ 𝑦 𝑦 ⎥⎦ 𝑥 ⎢⎣ 𝑥𝑦 𝑥𝑦 𝑥 𝑥 ⎥⎦

(𝑥 + 1) 2(𝑥 + 2) 3 − (𝑥 + 1) 3(𝑥 + 2) 2 (𝑥 + 1) 2(𝑥 + 2) 2[(𝑥 + 2) − (𝑥 + 1)] (𝑥 + 1) 2 = = (𝑥 + 2) 4 (𝑥 + 2) 6 (𝑥 + 2) 6 6𝑥(𝑥 2 + 1) 2(𝑥 3 + 2) 3 − 9𝑥 2(𝑥 2 + 1) 3(𝑥 3 + 2) 2 (𝑥 3 + 2) 6 2 2 3 3𝑥(𝑥 + 1) (𝑥 + 2) 2[2(𝑥 3 + 2) − 3𝑥(𝑥 2 + 1)] 3𝑥(𝑥 2 + 1) 2(−𝑥 3 − 3𝑥 + 4) = = (𝑥 3 + 2) 4 (𝑥 3 + 2) 6 (𝑥 2 − 1)√𝑥 2 + 1 − 𝑥2 + 1

𝑥√𝑥 3 − 1 −

𝑥3 − 1

3𝑥 4

√𝑥 3−1

𝑥4 √𝑥 2+1

=

=

(𝑥 2 − 1)(𝑥 2 + 1) − 𝑥 4 (𝑥 2 + 1)√𝑥 2 + 1

𝑥(𝑥 3 − 1) − 3𝑥 4

(𝑥 3 − 1)√𝑥 3 − 1

=

=

−2𝑥 4 − 𝑥

−1

√(𝑥 2 + 1) 3

√(𝑥 3 − 1) 3

=

−𝑥(2𝑥 3 + 1) √(𝑥 3 − 1) 3

15.

1 − 12 𝑥 (𝑥+𝑦) 2

16.

1 − 13 𝑥 (𝑥+𝑦) 3

𝑦

=

Solutions Section 0.5

𝑥 2 − (𝑥 + 𝑦) 2 𝑥 2 − 𝑥 2 − 2𝑥𝑦 − 𝑦 2 −𝑦(2𝑥 + 𝑦) −(2𝑥 + 𝑦) = = = 2 2 2 2 2 2 𝑦𝑥 (𝑥 + 𝑦) 𝑦𝑥 (𝑥 + 𝑦) 𝑦𝑥 (𝑥 + 𝑦) 𝑥 2(𝑥 + 𝑦) 2

𝑥 2 − (𝑥 + 𝑦) 2 𝑥 3 − 𝑥 3 − 3𝑥 2𝑦 − 3𝑥𝑦 2 − 𝑦 3 = 𝑦 𝑦𝑥 2(𝑥 + 𝑦) 2 𝑦𝑥 3(𝑥 + 𝑦) 3 2 2 2 −𝑦(3𝑥 + 3𝑥𝑦 + 𝑦 ) −(3𝑥 + 3𝑥𝑦 + 𝑦 2) = = 𝑦𝑥 3(𝑥 + 𝑦) 3 𝑥 3(𝑥 + 𝑦) 3 =

Solutions Section 0.6 Section 0.6

1. 𝑥 + 1 = 0 ⇒ 𝑥 = 0 − 1 ⇒ 𝑥 = −1

2. 𝑥 − 3 = 1 ⇒ 𝑥 = 1 + 3 ⇒ 𝑥 = 4

3. −𝑥 + 5 = 0 ⇒ −𝑥 = −5 ⇒ 𝑥 = 5

4. 2𝑥 + 4 = 1 ⇒ 2𝑥 = −3 ⇒ 𝑥 = −

5. 4𝑥 − 5 = 8 ⇒ 4𝑥 = 13 ⇒ 𝑥 =

13 4 43 7

8. 3𝑥 + 1 = 𝑥 ⇒ 2𝑥 = −1 ⇒ 𝑥 = −

9. 𝑥 + 1 = 2𝑥 + 2 ⇒ −𝑥 = 1 ⇒ 𝑥 = −1 11. 𝑎𝑥 + 𝑏 = 𝑐 ⇒ 𝑎𝑥 = 𝑐 − 𝑏 ⇒ 𝑥 =

3 3 4 𝑥 + 1 = 0 ⇒ 𝑥 = −1 ⇒ 𝑥 = − 4 4 3

6.

7. 7𝑥 + 55 = 98 ⇒ 7𝑥 = 43 ⇒ 𝑥 = −

𝑐−𝑏 𝑎

12. 𝑥 − 1 = 𝑐𝑥 + 𝑑 ⇒ (1 − 𝑐)𝑥 = 𝑑 + 1 ⇒ 𝑥 =

𝑑+1 1−𝑐

1 2

14. 𝑥 2 + 𝑥 + 1 = 0 ⇒ Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so this equation has no real solutions. 15. 𝑥 2 − 𝑥 + 1 = 0 ⇒ Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so this equation has no real solutions. 16. 2𝑥 2 − 4𝑥 + 3 ⇒ Δ = 𝑏 2 − 4𝑎𝑐 = −8 < 0, so this equation has no real solutions.

18. 3𝑥 2 − 1 = 0 ⇒ 𝑥 2 =

5 5 ⇒𝑥=± √2 2 1 1 ⇒𝑥=± √3 3

19. −𝑥 2 − 2𝑥 − 1 = 0 ⇒ −(𝑥 + 1) 2 = 0 ⇒ 𝑥 = −1

3 20. 2𝑥 2 − 𝑥 − 3 = 0 ⇒ (2𝑥 − 3)(𝑥 + 1) = 0 ⇒ 𝑥 = , −1 2 21.

1 2

10. 𝑥 + 1 = 3𝑥 + 1 ⇒ −2𝑥 = 0 ⇒ 𝑥 = 0

13. 2𝑥 2 + 7𝑥 − 4 = 0 ⇒ (2𝑥 − 1)(𝑥 + 4) = 0 ⇒ 𝑥 = −4,

17. 2𝑥 2 − 5 = 0 ⇒ 𝑥 2 =

3 2

1 2 3 𝑥 − 𝑥 − = 0 ⇒ 𝑥 2 − 2𝑥 − 3 = 0 ⇒ (𝑥 + 1)(𝑥 − 3) = 0 ⇒ 𝑥 = −1, 3 2 2

1 1 22. − 𝑥 2 − 𝑥 + 1 = 0 ⇒ 𝑥 2 + 𝑥 − 2 = 0 ⇒ (𝑥 + 2)(𝑥 − 1) = 0 ⇒ 𝑥 = −2, 1 2 2

−𝑏 ± √𝑏 2 − 4𝑎𝑐 1 ± √5 = 2𝑎 2

Solutions Section 0.6

23. 𝑥 2 − 𝑥 = 1 ⇒ 𝑥 2 − 𝑥 − 1 = 0 ⇒ 𝑥 =

24. 16𝑥 2 = −24𝑥 − 9 ⇒ 16𝑥 2 + 24𝑥 + 9 = 0(4𝑥 + 3) 2 = 0 ⇒ 𝑥 = − 25. 𝑥 = 2 −

3 4

1 ⇒ 𝑥 2 = 2𝑥 − 1 ⇒ 𝑥 2 − 2𝑥 + 1 = 0 ⇒ (𝑥 − 1) 2 = 0 ⇒ 𝑥 = 1 𝑥

1 ⇒ (𝑥 + 4)(𝑥 − 2) = 1 ⇒ 𝑥 2 + 2𝑥 − 8 = 1 ⇒ 𝑥 2 + 2𝑥 − 9 − 0 𝑥−2 −𝑏 ± √𝑏 2 − 4𝑎𝑐 −2 ± √40 ⇒𝑥= = = −1 ± √10 2𝑎 2

26. 𝑥 + 4 =

27. 𝑥 4 − 10𝑥 2 + 9 = 0 ⇒ (𝑥 2 − 1)(𝑥 2 − 9) = 0 ⇒ 𝑥 2 = 1 or 𝑥 2 = 0 ⇒ 𝑥 = ±1, ±3 28. 𝑥 4 − 2𝑥 2 + 1 = 0 ⇒ (𝑥 2 − 1) 2 = 0 ⇒ 𝑥 = ±1 29. 𝑥 4 + 𝑥 2 − 1 = 0 ⇒ 𝑥 2 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 −1 ± √5 −1 ± √5 ⇒ 𝑥2 = ⇒ 𝑥 = ±√ 2𝑎 2 2

30. 𝑥 3 + 2𝑥 2 + 𝑥 = 0 ⇒ 𝑥(𝑥 2 + 2𝑥 + 1) = 0 ⇒ 𝑥(𝑥 + 1) 2 = 0 ⇒ 𝑥 = 0, −1

31. 𝑥 3 + 16𝑥 2 + 11𝑥 + 6 = 0 ⇒ (𝑥 + 1)(𝑥 + 2)(𝑥 + 3) = 0 ⇒ 𝑥 = −1, −2, −3 32. 𝑥 3 − 6𝑥 2 + 12𝑥 − 8 = 0 ⇒ (𝑥 − 2) 3 = 0 ⇒ 𝑥 = 2

33. 𝑥 3 + 4𝑥 2 + 4𝑥 + 3 = 0 ⇒ (𝑥 + 3)(𝑥 2 + 𝑥 + 1) = 0 ⇒ 𝑥 = −3 (For 𝑥 2 + 𝑥 + 1 = 0, Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no real solutions to this quadratic equation.) 3

34. 𝑦 3 + 64 = 0 ⇒ 𝑦 3 = −64 ⇒ 𝑦 = √−64 = −4 3

35. 𝑥 3 − 1 = 0 ⇒ 𝑥 3 = 1 ⇒ 𝑥 = √1 = 1 36. 𝑥 3 − 27 = 0 3 ⇒ 𝑥 3 = 27 ⇒ 𝑥 = √27 = 3

37. 𝑦 3 + 3𝑦 2 + 3𝑦 + 2 = 0 ⇒ (𝑦 + 2)(𝑦 2 + 𝑦 + 1) = 0 ⇒ 𝑦 = −2 (For 𝑦 2 + 𝑦 + 1 = 0, Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no real solutions to this quadratic equation.) 38. 𝑦 3 − 2𝑦 2 − 2𝑦 − 3 = 0 ⇒ (𝑦 − 3)(𝑦 2 + 𝑦 + 1) = 0 ⇒ 𝑦 = 3 (For 𝑦 2 + 𝑦 + 1 = 0, Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no real solutions to this quadratic equation.) 39. 𝑥 3 − 𝑥 2 − 5𝑥 + 5 = 0 ⇒ (𝑥 − 1)(𝑥 2 − 5) = 0 ⇒ 𝑥 = 1, ± √5

40. 𝑥 3 − 𝑥 2 − 3𝑥 + 3 = 0 ⇒ (𝑥 − 1)(𝑥 2 − 3) = 0 ⇒ 𝑥 = 1, ± √3

Solutions Section 0.6 41. 2𝑥 − 𝑥 − 2𝑥 + 1 = 0 ⇒ (2𝑥 − 1)(𝑥 4 − 1) = 0 ⇒ (2𝑥 2 − 1)(𝑥 2 − 1)(𝑥 2 + 1) = 0 (Think of the cubic you get by substituting 𝑦 for 𝑥 2) 1 ⇒ 𝑥 = ±1, ± √2 6

4

2

2

42. 3𝑥 6 − 𝑥 4 − 12𝑥 2 + 4 = 0 ⇒ (3𝑥 2 − 1)(𝑥 4 − 4) = 0 ⇒ (3𝑥 2 − 1)(𝑥 2 − 2)(𝑥 2 + 2) = 0 (Think of the cubic you get by substituting 𝑦 for 𝑥 2) 1 ⇒ 𝑥 = ±√2, ± √3

43. (𝑥 2 + 3𝑥 + 2)(𝑥 2 − 5𝑥 + 6) = 0 ⇒ (𝑥 + 2)(𝑥 + 1)(𝑥 − 2)(𝑥 − 3) = 0 ⇒ 𝑥 = −2, −1, 2, 3 44. (𝑥 2 − 4𝑥 + 4) 2(𝑥 2 + 6𝑥 + 5) 3 = 0 ⇒ (𝑥 − 2) 4(𝑥 + 1) 3(𝑥 + 5) 3 = 0 ⇒ 𝑥 = −5, −1, 2

Solutions Section 0.7 Section 0.7

1. 𝑥 4 − 3𝑥 3 = 0, 𝑥 3(𝑥 − 3) = 0, 𝑥 = 0, 3

2. 𝑥 6 − 9𝑥 4 = 0, 𝑥 4(𝑥 2 − 9) = 0, 𝑥 = 0, ±3

3. 𝑥 4 − 4𝑥 2 = −4, 𝑥 4 − 4𝑥 2 + 4 = 0, (𝑥 2 − 2) 2 = 0, 𝑥 = ±√2

4. 𝑥 4 − 𝑥 2 = 6, 𝑥 4 − 𝑥 2 − 6 = 0, (𝑥 2 − 3)(𝑥 2 + 2) = 0, 𝑥 = ±√3

5. (𝑥 + 1)(𝑥 + 2) + (𝑥 + 1)(𝑥 + 3) = 0, (𝑥 + 1)(𝑥 + 2 + 𝑥 + 3) = 0, (𝑥 + 1)(2𝑥 + 5) = 0, 𝑥 = −1, −5∕2

6. (𝑥 + 1)(𝑥 + 2) 2 + (𝑥 + 1) 2(𝑥 + 2) = 0, (𝑥 + 1)(𝑥 + 2)(𝑥 + 2 + 𝑥 + 1) = 0, (𝑥 + 1)(𝑥 + 2)(2𝑥 + 3) = 0, 𝑥 = −1, −2, −3∕2 7. (𝑥 2 + 1) 5(𝑥 + 3) 4 + (𝑥 2 + 1) 6(𝑥 + 3) 3 = 0, (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 + 3 + 𝑥 2 + 1) = 0, (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 2 + 𝑥 + 4) = 0, 𝑥 = −3 (Neither 𝑥 2 + 1 = 0 nor 𝑥 2 + 𝑥 + 4 = 0 has a real solution.) 8. 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 5 − 10𝑥 2(𝑥 2 + 1) 5(𝑥 3 + 1) 4 = 0, 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4[𝑥 3 + 1 − 𝑥(𝑥 2 + 1)] = 0, 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4(1 − 𝑥) = 0, 𝑥 = −1, 0, 1 9. (𝑥 3 + 1)√𝑥 + 1 − (𝑥 3 + 1) 2√𝑥 + 1 = 0, (𝑥 3 + 1)√𝑥 + 1 [1 − (𝑥 3 + 1)] = 0, −𝑥 3(𝑥 3 + 1)√𝑥 + 1 = 0, 𝑥 = 0, −1

10. (𝑥 2 + 1)√𝑥 + 1 − √(𝑥 + 1) 3 = 0, √𝑥 + 1 [𝑥 2 + 1 − (𝑥 + 1)] = 0, (𝑥 2 − 𝑥)√𝑥 + 1 = 0, 𝑥(𝑥 − 1)√𝑥 + 1 = 0, 𝑥 = −1, 0, 1

11. √(𝑥 + 1) 3 + √(𝑥 + 1) 5 = 0, √(𝑥 + 1) 3 (1 + 𝑥 + 1) = 0, (𝑥 + 2)√(𝑥 + 1) 3 = 0, 𝑥 = −1 (𝑥 = −2 is not a solution because √(𝑥 + 1) 3 is not defined for 𝑥 = −2.) 3

3

12. (𝑥 2 + 1) √(𝑥 + 1) 4 − √(𝑥 + 1) 7 = 0, 3

√(𝑥 + 1) 4 [𝑥 2 + 1 − (𝑥 + 1)] = 0, 3

3

(𝑥 2 − 𝑥) √(𝑥 + 1) 4 = 0, 𝑥(𝑥 − 1) √(𝑥 + 1) 4 = 0, 𝑥 = −1, 0, 1

13. (𝑥 + 1) 2(2𝑥 + 3) − (𝑥 + 1)(2𝑥 + 3) 2 = 0, (𝑥 + 1)(2𝑥 + 3)(𝑥 + 1 − 2𝑥 − 3) = 0, (𝑥 + 1)(2𝑥 + 3)(−𝑥 − 2) = 0, 𝑥 = −2, −3∕2, −1

Solutions Section 0.7 14. (𝑥 − 1) (𝑥 + 2) − (𝑥 − 1) (𝑥 + 2) 2 = 0, (𝑥 2 − 1) 2(𝑥 + 2) 2(𝑥 + 2 − 𝑥 2 + 1) = 0, −(𝑥 2 − 1) 2(𝑥 + 2) 2(𝑥 2 − 𝑥 − 3) = 0, 𝑥 = −2, −1, 1, (1 ± √13)∕2 2

15.

16.

17.

2

3

6𝑥(𝑥 2 + 1) 2(𝑥 2 + 2) 4 − 8𝑥(𝑥 2 + 1) 3(𝑥 2 + 2) 3 = 0, (𝑥 2 + 2) 8 2𝑥(𝑥 2 + 1) 2(𝑥 2 + 2) 3[3(𝑥 2 + 2) − 4(𝑥 2 + 1)] = 0, (𝑥 2 + 2) 8 −2𝑥(𝑥 2 + 1) 2(𝑥 2 − 2) = 0, (𝑥 2 + 2) 5 −2𝑥(𝑥 2 + 1) 2(𝑥 2 − 2) = 0, 𝑥 = 0, ±√2 2(𝑥 2 − 1)√𝑥 2 + 1 − 𝑥2 + 1

𝑥4 − 2

4𝑥√𝑥 3 − 1 − 𝑥3 − 1

𝑥 4 − 4𝑥

19. 𝑥 − 20. 1 −

22.

23.

𝑥4 √𝑥 2+1

= 0,

2(𝑥 2 − 1)(𝑥 2 + 1) − 𝑥 4 (𝑥 2 + 1)√𝑥 2 + 1 4

= 0, 𝑥 4 − 2 = 0, 𝑥 = ± √2

3𝑥 4 √𝑥 3−1

(𝑥 3 − 1)√𝑥 3 − 1

21.

3

(𝑥 + 1) 2(𝑥 + 2) 3 − (𝑥 + 1) 3(𝑥 + 2) 2 = 0, (𝑥 + 2) 6 (𝑥 + 1) 2(𝑥 + 2) 2[(𝑥 + 2) − (𝑥 + 1)] = 0, (𝑥 + 2) 6 (𝑥 + 1) 2 = 0, (𝑥 + 1) 2 = 0, 𝑥 = −1 (𝑥 + 2) 4

(𝑥 2 + 1)√𝑥 2 + 1

18.

2

= 0,

4𝑥(𝑥 3 − 1) − 3𝑥 4 (𝑥 3 − 1)√𝑥 3 − 1

= 0,

= 0,

3

= 0, 𝑥 4 − 4𝑥 = 0, 𝑥(𝑥 3 − 4) = 0, 𝑥 = 0, ± √4

1 = 0, 𝑥 2 − 1 = 0, 𝑥 = ±1 𝑥

4 = 0, 𝑥 2 − 4 = 0, 𝑥 = ±2 𝑥2

1 9 − 3 = 0, 𝑥 2 − 9 = 0, 𝑥 = ±3 𝑥 𝑥

1 1 − = 0, 𝑥 + 1 − 𝑥 2 = 0, 𝑥 2 − 𝑥 − 1 = 0, 𝑥 = (1 ± √5)∕2 2 𝑥+1 𝑥 (𝑥 − 4)(𝑥 − 1) − 𝑥(𝑥 + 1) 𝑥−4 𝑥 − = 0, = 0, 𝑥+1 𝑥−1 (𝑥 + 1)(𝑥 − 1) −6𝑥 + 4 = 0, −6𝑥 + 4 = 0, 𝑥 = 2∕3 (𝑥 + 1)(𝑥 − 1)

Solutions Section 0.7 (2𝑥 − 3)(𝑥 + 1) − (2𝑥 + 3)(𝑥 − 1) 2𝑥 − 3 2𝑥 + 3 24. − = 0, = 0, 𝑥−1 𝑥+1 (𝑥 − 1)(𝑥 + 1) −2𝑥 = 0, −2𝑥 = 0, 𝑥 = 0 (𝑥 − 1)(𝑥 + 1) 25.

26.

3𝑥(𝑥 + 4) + (𝑥 + 1)(𝑥 + 4) 𝑥+4 𝑥+4 + = 0, = 0, 𝑥+1 3𝑥 3𝑥(𝑥 + 1) (𝑥 + 4)(3𝑥 + 𝑥 + 1) (𝑥 + 4)(4𝑥 + 1) = 0, = 0, 3𝑥(𝑥 + 1) 3𝑥(𝑥 + 1) (𝑥 + 4)(4𝑥 + 1) = 0, 𝑥 = −4, −1∕4

(2𝑥 − 3)(𝑥 + 1) − 𝑥(2𝑥 − 3) 2𝑥 − 3 2𝑥 − 3 − = 0, = 0, 𝑥 𝑥+1 𝑥(𝑥 + 1) (2𝑥 − 3)(𝑥 + 1 − 𝑥) (2𝑥 − 3) = 0, = 0, 2𝑥 − 3 = 0, 𝑥 = 3∕2 𝑥(𝑥 + 1) 𝑥(𝑥 + 1)

Solutions Section 0.8 Section 0.8

1. 𝑃 (0, 2), 𝑄(4, −2), 𝑅(−2, 3), 𝑆(−3.5, −1.5), 𝑇 (−2.5, 0), 𝑈(2, 2.5)

2. 𝑃 (−2, 2), 𝑄(3.5, 2), 𝑅(0, −3), 𝑆(−3.5, −1.5), 𝑇 (2.5, 0), 𝑈(−2, 2.5) 3.

4.

5. Solve the equation 𝑥 + 𝑦 = 1 for 𝑦 to get 𝑦 = 1 − 𝑥. Then plot some points: 𝑥

−2

𝑦=1−𝑥

3

−1 2

0

1

1

2

0

−1

Graph:

6. Solve the equation 𝑦 − 𝑥 = −1 for 𝑦 to get 𝑦 = −1 + 𝑥. Then plot some points: 𝑥

𝑦 = −1 + 𝑥

−2 −3

−1 −2

0

−1

1 0

2 1

Solutions Section 0.8 Graph:

7. Solve the equation 2𝑦 − 𝑥 2 = 1 for 𝑦 to get 𝑦 = 𝑦=

𝑥

−2

1 + 𝑥2 2

2.5

−1 1

0

0.5

1

2

1

1 + 𝑥2 . Then plot some points: 2

2.5

Graph:

8. Solve the equation 2𝑦 + √𝑥 = 1 for 𝑦 to get 𝑦 = 𝑦=

𝑥

1 − √𝑥 2

Graph:

0

0.5

1 0

4

−0.5

9

−1

16

−1.5

1 − √𝑥 . Then plot some points: 2

Solutions Section 0.8 4 9. Solve the equation 𝑥𝑦 = 4 for 𝑦 to get 𝑦 = . Then plot some points: 𝑥 𝑦=

𝑥

−3

4 𝑥

−2

− 43

1

−1

−2

2

4

−4

3 4 3

2

Graph:

1 10. Solve the equation 𝑥 2𝑦 = −1 for 𝑦 to get 𝑦 = − 2 . Then plot some points: 𝑥 𝑥

1 𝑦=− 2 𝑥

−3

− 19

−2

− 14

−1 −1

1

−1

2

− 14

3

− 19

Graph:

11. Solve the equation 𝑥𝑦 = 𝑥 2 + 1 for 𝑦 to get 𝑦 = 𝑥 + 𝑦=𝑥+

𝑥

1 𝑥

−2

− 52

−1 −2

− 12 − 52

1 2 5 2

1 2

2 5 2

1 . Then plot some points: 𝑥

Solutions Section 0.8 Graph:

12. Solve the equation 𝑥𝑦 = 2𝑥 3 + 1 for 𝑦 to get 𝑦 = 2𝑥 2 + 𝑦 = 2𝑥 2 +

𝑥

1 𝑥

−2 15 2

−1 1

− 12 − 32

1 2 5 2

1 3

2

17 2

Graph:

13. √(2 − 1) 2 + (−2 + 1) 2 = √2 14. √(6 − 1) 2 + (1 − 0) 2 = √26

15. √(0 − 𝑎) 2 + (𝑏 − 0) 2 = √𝑎 2 + 𝑏 2

16. √(𝑏 − 𝑎) 2 + (𝑏 − 0) 2 = √2(𝑏 − 𝑎) 2 or √2|𝑏 − 𝑎| 17. Set the two distances equal and solve: √(1 − 0) * 2 + (𝑘 − 0) 2 = √(1 − 2) 2 + (𝑘 − 1) 2 ⇒ √1 + 𝑘 2 = √2 − 2𝑘 + 𝑘 2 ⇒ 1 + 𝑘 2 = 2 − 2𝑘 + 𝑘 2 1 ⇒ 2𝑘 = 1 ⇒ 𝑘 = 2 18. Set the two distances equal and solve: √(𝑘 + 1) * 2 + (𝑘 − 0) 2 = √(𝑘 − 0) 2 + (𝑘 − 2) 2 ⇒ √2𝑘 2 + 2𝑘 + 1 = √2𝑘 2 − 4𝑘 + 4 ⇒ 2𝑘 2 + 2𝑘 + 1 = 2𝑘 2 − 4𝑘 + 4 1 ⇒ 6𝑘 = 3 ⇒ 𝑘 = 2

1 . Then plot some points: 𝑥

Solutions Section 0.8 19. Circle with center (0, 0) and radius 3. 20. The single point (0, 0).

Solutions Section 0.9 Section 0.9 1. Exponential Form

10 2 = 100

4 3 = 64

4 4 = 256

0.45 0 = 1

8 1∕2 = 2√2

1 Logarithmic Form log10 100 = 2 log4 64 = 3 log4 256 = 4 log0.45 1 = 0 log8 2√2 = 2

2. Exponential Form Logarithmic Form

10 1 = 10

5 5 = 3,125

log10 10 = 1

Exponential Form

0.5 3 = 0.125

Exponential Form

0.3 2 = 0.09

Logarithmic Form log0.5 0.125 = 3 3.

Logarithmic Form log0.3 0.09 = 2 Exponential Form Logarithmic Form 4. Exponential Form Logarithmic Form Exponential Form

log6 216 = 3

log4 √2 = 14

1 log3 ! 81 " = −4

0

10 −3 = 0.001

1 !2" = 1

1 3 −4 = 81

log1∕2 1 = 0

log10 0.001 = −3

1 log9 81 = −2

log2 1,024 = 10

log64 14 = − 13

2 −1 = 12

10 5 = 100,000

10 −5 = 0.00001

16 −1∕4 = 12

0.25 2 = 0.0625

1 9 −2 = 81

log2 12 = −1 2 11 = 2,048

Logarithmic Form log2 2,048 = 11

2 10 = 1,024

1 log4 ! 64 " = −3

6 3 = 216

log5 3,125 = 5 4 1∕4 = √2

1 4 −3 = 64

64 −1∕3 = 14

log10 100,000 = 5 log10 0.00001 = −5 log16 12 = − 14

log0.25 0.0625 = 2

5. log4 16 = the power to which we need to raise 4 in order to get 16. Because 4 boxed 2 = 16, this power is 2, so log4 16 = 2.

6. log5 125 = the power to which we need to raise 5 in order to get 125. Because 5 boxed 3 = 125, this power is 3, so log5 125 = 3.

1 1 1 7. log5 25 = the power to which we need to raise 5 in order to get 25 . Because 5 boxed −2 = 25 , this power is −2, so 1 log5 25 = −2.

1 1 1 8. log4 64 = the power to which we need to raise 4 in order to get 64 . Because 4 boxed −3 = 64 , this power is −3, so 1 log4 64 = −3.

Solutions Section 0.9 9. log 100,000 = log10 100,000 = the power to which we need to raise 10 in order to get 100,000. Because 10 boxed 5 = 100,000, this power is 5, so log 100,000 = 5. 10. log 1,000 = log10 1,000 = the power to which we need to raise 10 in order to get 1,000. Because 10 boxed 3 = 1,000, this power is 3, so log 1,000 = 3.

11. log16 16 = the power to which we need to raise 16 in order to get 16. Because 16 boxed 1 = 16, this power is 1, so log16 16 = 1. 1 1 1 1 boxed 1 1 = the power to which we need to raise in order to get . Because ! " = , this power is 1, 2 2 2 2 2 1 so log1∕2 = 1. 2 12. log1∕2

1 1 1 = the power to which we need to raise 4 in order to get . Because 4 boxed −2 = , this power is −2, 16 16 16 1 so log4 = −2. 16 13. log4

14. log2 log2

1 1 1 = the power to which we need to raise 2 in order to get . Because 2 boxed −3 = , this power is −3, so 8 8 8

1 = −3. 8

15. log2 √2 = the power to which we need to raise 2 in order to get √2. Because 2 boxed 1∕2 = √2, this power is 1 1 , so log2 √2 = . 2 2 16. log4 √2 = the power to which we need to raise 4 in order to get √2. Because 4 boxed 1∕4 = √2, this power is 1 1 , so log4 √2 = . 4 4 17. By Identity (1), log𝑏 3 + log𝑏 4 = log𝑏 (3 × 4) = log𝑏 boxed 12. 3 18. By Identity (2), log𝑏 3 − log𝑏 4 = log𝑏 boxed . 4

19. log𝑏 2 − log𝑏 5 − log𝑏 4 = log𝑏 2 − (log𝑏 5 + log𝑏 4) = log𝑏 2 − (log𝑏 (5 × 4)) (Identity 1) 2 1 = log𝑏 ! " (Identity 2) = log𝑏 boxed 5×4 10 20. log𝑏 3 + log𝑏 2 − log𝑏 7 = log𝑏 (3 × 2) − log𝑏 7 (Identity 1) = log𝑏 !

3×2 6 " (Identity 2) = log𝑏 boxed 7 7

3 3 21. log𝑏 3 − 3 log𝑏 2 = log𝑏 3 − log𝑏 2 3 (Identity 3) = log𝑏 ! 3 " (Identity 2) = log𝑏 boxed 8 2

22. 3 log𝑏 2 + 2 log𝑏 3 = log𝑏 2 3 + log𝑏 3 2 (Identity 3) = log𝑏 (2 3 × 3 2) (Identity 1) = log𝑏 boxed 72 23. 4 log𝑏 𝑥 + 5 log𝑏 𝑦 = log𝑏 𝑥 4 + log𝑏 𝑦 5 (Identity 3) = log𝑏 boxed 𝑥 4𝑦 5 (Identity 1)

Solutions Section 0.9 𝑝3 24. 3 log𝑏 𝑝 − 2 log𝑏 𝑞 = log𝑏 𝑝 3 − log𝑏 𝑞 2 (Identity 3) = log𝑏 boxed 2 (Identity 2) 𝑞 25. 2 log𝑏 𝑥 + 3 log𝑏 𝑦 − 4 log𝑏 𝑧 = log𝑏 𝑥 2 + log𝑏 𝑦 3 − log𝑏 𝑧 4 (Identity 3) 𝑥 2𝑦 3 = log𝑏 (𝑥 2𝑦 3) − log𝑏 𝑧 4 (Identity 1) = log𝑏 boxed 4 (Identity 2) 𝑧 26. 4 log𝑏 𝑝 − 3 log𝑏 𝑞 − 2 log𝑏 𝑟 = log𝑏 𝑝 4 − log𝑏 𝑞 3 − log𝑏 𝑟 2 (Identity 3)

= log𝑏 𝑝 4 − (log𝑏 𝑞 3 + log𝑏 𝑟 2) = log𝑏 𝑝 4 − log𝑏 (𝑞 3𝑟 2) (Identity 1) = log𝑏 boxed

27. 𝑥 log𝑏 2 − 2 log𝑏 𝑥 = log𝑏 2 𝑥 − log𝑏 𝑥 2 (Identity 3) = log𝑏 boxed

2𝑥 (Identity 2) 𝑥2

28. 𝑝 log𝑏 𝑞 + 𝑞 log𝑏 𝑝 = log𝑏 𝑞 𝑝 + log𝑏 𝑝 𝑞 (Identity 3) = log𝑏 boxed 𝑞 𝑝𝑝 𝑞 (Identity 1) 29. log 21 = log(3 × 7) = log 3 + log 7 = 𝑏 + 𝑐 (Identity 1)

30. log 14 = log(2 × 7) = log 2 + log 7 = 𝑎 + 𝑐 (Identity 1)

31. log 42 = log(2 × 3 × 7) = log 2 + log 3 + log 7 = 𝑎 + 𝑏 + 𝑐 (Identity 1) 32. log 28 = log(2 × 2 × 7) = log 2 + log 2 + log 7 = 2𝑎 + 𝑐 (Identity 1) 1 33. log! " = − log 7 = −𝑐 (Identity 5) 7

1 34. log! " = − log 3 = −𝑏 (Identity 5) 3

2 35. log! " = log 2 − log 3 = 𝑎 − 𝑏 (Identity 2) 3 7 36. log! " = log 7 − log 2 = 𝑐 − 𝑎 (Identity 2) 2

4 37. log! " = log 4 − log 7 (Identity 2) = log 2 + log 2 − log 7 (Identity 1) = 2𝑎 − 𝑐 7

2 38. log! " = log 2 − log 9 (Identity 2) = log 2 − (log 3 + log 3) (Identity 1) = 𝑎 − 2𝑏 9 39. log 16 = log 2 4 = 4 log 2 = 4𝑎 (Identity 3) 40. log 81 = log 3 4 = 4 log 3 = 4𝑏 (Identity 3) 41. log 0.03 = log!

3 " = log 3 − log 10 2 = log 3 − 2 log 10 = 𝑏 − 2 10 2

42. log 7,000 = log(7 × 10 3) = log 7 + log 10 3 = log 7 + 3 log 10 = 𝑐 + 3

𝑝4 (Identity 2) 𝑞 3𝑟 2

43. log 5 = log!

10 " = log 10 − log 2 = 1 − 𝑎 (Identity 2) 2

44. log 25 = log!

Solutions Section 0.9

100 " = log 100 − log 4 = log 10 2 − log 2 2 = 2 log 10 − 2 log 2 = 2 − 2𝑎 4

45. log √7 = log(7 1∕2) = 46. log!

1 𝑐 log 7 = 2 2

2 1 𝑏 " = log(2) − log √3 = log(2) − log(3 1∕2) = log 2 − log 3 = 𝑎 − √3 2 2

47. 4 = 2 𝑥 is exactly the equation we solve in order to calculate log2 4; answer: 2. Alternatively, write the equation 4 = 2 𝑥 in logarithmic form to get 𝑥 = log2 4 = 2.

48. 81 = 3 𝑥 is exactly the equation we solve in order to calculate log3 81; answer: 4. Alternatively, write the equation 81 = 3 𝑥 in logarithmic form to get 𝑥 = log3 81 = 4. 49. Take the base 3 logarithm of both sides of the given equation 27 = 3 2𝑥−1 to get log3 27 = log3 3 2𝑥−1 3 = (2𝑥 − 1) log3 3 3 = 2𝑥 − 1 3+1 𝑥= =2 2

By Identity 3 Solve for 𝑥.

By Identity 4

50. Take the base 4 logarithm of both sides of the given equation 4 2−3𝑥 = 256 to get log4 4 2−3𝑥 = log4 256 (2 − 3𝑥) log4 4 = 4 2 − 3𝑥 = 4 2−4 2 𝑥= =− 3 3

By Identity 3 Solve for 𝑥.

By Identity 4

1 51. Take the base 5 logarithm of both sides of the given equation 5 −𝑥+1 = to get 125 1 −𝑥+1 log5 5 = log5 ! " 125 (−𝑥 + 1) log5 5 = − log5 125 By Identities 3 and 5 −𝑥 + 1 = −3 𝑥=1+3=4

By Identity 4

𝑥 2 − 12 = −3 𝑥 = ±√9 = ±3

By Identity 4

Solve for 𝑥.

2 1 52. Take the base 3 logarithm of both sides of the given equation 3 𝑥 −12 = to get 27 2 1 log3 3 𝑥 −12 = log3 ! " 27 2 (𝑥 − 12) log3 3 = − log3 27 By Identities 3 and 5

Solve for 𝑥.

53. First divide both sides of the given equation by 50 to get 2.4 = 2 3𝑡. Take the common logarithm of both sides: log 2.4 = log(2 3𝑡)

log 2.4 = 3𝑡 log 2 log 2.4 3𝑡 = log 2 log 2.4 𝑡= ≈ 0.4210 3 log 2

Solutions Section 0.9 By Identity 3 Divide.

Solve for 𝑡.

If instead you take the base-2 logarithm, the answer is represented as

log2 (2.4) . 3

54. First divide both sides of the given equation by 500 to get 2 = 1.1 2𝑡. Take the common logarithm of both sides: log 2 = log(1.1 2𝑡) log 2 = 2𝑡 log 1.1 By Identity 3 log 2 2𝑡 = Divide. log 1.1 log 2 𝑡= ≈ 3.6363 Solve for 𝑡. 2 log 1.1 55. First divide both sides of the given equation by 300 to get sides:

10 " = log(1.3 4𝑡−1) 3 log 10 − log 3 = (4𝑡 − 1) log 1.3 log 10 − log 3 4𝑡 − 1 = log 1.3 log 10 − log 3 1 𝑡= ! + 1" ≈ 1.3972 4 log 1.3 log!

By Identities 2 and 3 Divide.

Solve for 𝑡.

56. First divide both sides of the given equation by 700 to get both sides:100 log! " = log(1.04 3𝑡+1) 7 log 100 − log 7 = (3𝑡 + 1) log 1.04 log 100 − log 7 3𝑡 + 1 = log 1.04 1 log 100 − log 7 𝑡= ! − 1" ≈ 22.2675 3 log 1.04

10 = 1.3 4𝑡−1. Take the common logarithm of both 3

100 = 1.04 3𝑡+1. Take the common logarithm of 7

By Identities 2 and 3 Divide.

Solve for 𝑡.

Solutions Section 1.1 Section 1.1 1. Using the table: 2. Using the table:

a. 𝑓(0) = 2

a. 𝑓(−1) = 4

b. 𝑓(2) = −0.5 b. 𝑓(1) = −1

3. Using the table: a. 𝑓(2) − 𝑓(−2) = −0.5 − 2 = −2.5 c. −2𝑓(−1) = −2(4) = −8 4. Using the table: a. 𝑓(1) − 𝑓(−1) = −1 − 4 = −5 c. 3𝑓(−2) = 3(2) = 6 5. From the graph, we estimate:

a. 𝑓(1) = 20

In a similar way, we find: c. 𝑓(3) = 30 f. 𝑓(3 − 2) = 𝑓(1) = 20 6. From the graph, we estimate:

7. From the graph, we estimate:

b. 𝑓(1)𝑓(−2) = (−1)(2) = −2

b. 𝑓(2) = 30

d. 𝑓(5) = 20\\e. 𝑓(3) − 𝑓(2) = 30 − 30 = 0

a. 𝑓(1) = 20

In a similar way, we find: c. 𝑓(3) = 10 f. 𝑓(3 − 2) = 𝑓(1) = 20

b. 𝑓(−1)𝑓(−2) = (4)(2) = 8

b. 𝑓(2) = 10

d. 𝑓(5) = 20 \\e. 𝑓(3) − 𝑓(2) = 10 − 10 = 0

a. 𝑓(−1) = 0

b. 𝑓(1) = −3 since the solid dot is on (1, −3).

In a similar way, we estimate c. 𝑓(3) = 3 d. Since 𝑓(3) = 3 and 𝑓(1) = −3,

𝑓(3) − 𝑓(1) 3 − (−3) = = 3. 3−1 3−1

8. From the graph, we estimate:

Solutions Section 1.1 a. 𝑓(−3) = 3 b. 𝑓(−1) = −2 since the solid dot is on (−1, −2).

In a similar way, we estimate c. 𝑓(1) = 0 𝑓(3) − 𝑓(1) 2 − 0 d. Since 𝑓(3) = 2 and 𝑓(1) = 0, = = 1. 3−1 3−1

1 , with its natural domain. 𝑥2 The natural domain consists of all 𝑥 for which 𝑓(𝑥) makes sense: all real numbers other than 0. 1 1 63 a. Since 4 is in the natural domain, 𝑓(4) is defined, and 𝑓(4) = 4 − 2 = 4 − . = 16 16 4 b. Since 0 is not in the natural domain, 𝑓(0) is not defined. 1 1 c. Since −1 is in the natural domain, 𝑓(−1) = −1 − = −1 − = −2. 2 1 (−1) 9. 𝑓(𝑥) = 𝑥 −

10. 𝑓(𝑥) =

2 − 𝑥 2, with domain [2, ∞) 𝑥

a. Since 4 is in [2, ∞), 𝑓(4) is defined, and 𝑓(4) = b. Since 0 is not in [2, ∞), 𝑓(0) is not defined.

2 1 31 − 4 2 = − 16 = − . 4 2 2 c. Since 1 is not in [2, ∞), 𝑓(1) is not defined.

11. 𝑓(𝑥) = √𝑥 + 10, with domain [−10, 0) a. Since 0 is not in [−10, 0), 𝑓(0) is not defined. b. Since 9 is not in [−10, 0), 𝑓(9) is not defined. c. Since −10 is in [−10, 0), 𝑓(−10) is defined, and 𝑓(−10) = √−10 + 10 = √0 = 0

12. 𝑓(𝑥) = √9 − 𝑥 2, with domain (−3, 3) a. Since 0 is in (−3, 3), 𝑓(0) is defined, and 𝑓(0) = √9 − 0 = 3. b. Since 3 is not in (−3, 3), 𝑓(3) is not defined. c. Since −3 is not in (−3, 3), 𝑓(−3) is not defined. 13. 𝑓(𝑥) = 4𝑥 − 3 a. 𝑓(−1) = 4(−1) − 3 = −4 − 3 = −7 b. 𝑓(0) = 4(0) − 3 = 0 − 3 = −3 c. 𝑓(1) = 4(1) − 3 = 4 − 3 = 1 d. Substitute 𝑦 for 𝑥 to obtain 𝑓(𝑦) = 4𝑦 − 3 e. Substitute (𝑎 + 𝑏) for 𝑥 to obtain 𝑓(𝑎 + 𝑏) = 4(𝑎 + 𝑏) − 3.

14. 𝑓(𝑥) = −3𝑥 + 4 a. 𝑓(−1) = −3(−1) + 4 = 3 + 4 = 7 b. 𝑓(0) = −3(0) + 4 = 0 + 4 = 4 c. 𝑓(1) = −3(1) + 4 = −3 + 4 = 1 d. Substitute 𝑦 for 𝑥 to obtain 𝑓(𝑦) = −3𝑦 + 4 e. Substitute (𝑎 + 𝑏) for 𝑥 to obtain 𝑓(𝑎 + 𝑏) = −3(𝑎 + 𝑏) + 4. 15. 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 3 a. 𝑓(0) = (0) 2 + 2(0) + 3 = 0 + 0 + 3 = 3

b. 𝑓(1) = 1 2 + 2(1) + 3 = 1 + 2 + 3 = 6

Solutions Section 1.1 c. 𝑓(−1) = (−1) + 2(−1) + 3 = 1 − 2 + 3 = 2 d. 𝑓(−3) = (−3) 2 + 2(−3) + 3 = 9 − 6 + 3 = 6 e. Substitute 𝑎 for 𝑥 to obtain 𝑓(𝑎) = 𝑎 2 + 2𝑎 + 3. f. Substitute (𝑥 + ℎ) for 𝑥 to obtain 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 + 2(𝑥 + ℎ) + 3. 2

16. 𝑔(𝑥) = 2𝑥 2 − 𝑥 + 1 a. 𝑔(0) = 2(0) 2 − 0 + 1 = 0 − 0 + 1 = 1 b. 𝑔(−1) = 2(−1) 2 − (−1) + 1 = 2 + 1 + 1 = 4 c. Substitute 𝑟 for 𝑥 to obtain 𝑔(𝑟) = 2𝑟 2 − 𝑟 + 1. d. Substitute (𝑥 + ℎ) for 𝑥 to obtain 𝑔(𝑥 + ℎ) = 2(𝑥 + ℎ) 2 − (𝑥 + ℎ) + 1.

1 𝑠 1 1 a. 𝑔(1) = 1 2 + = 1 + 1 = 2 b. 𝑔(−1) = (−1) 2 + =1−1=0 1 −1 1 1 65 1 c. 𝑔(4) = 4 2 + = 16 + = or 16.25 d. Substitute 𝑥 for 𝑠 to obtain 𝑔(𝑥) = 𝑥 2 + 4 4 4 𝑥 1 2 e. Substitute (𝑠 + ℎ) for 𝑠 to obtain 𝑔(𝑠 + ℎ) = (𝑠 + ℎ) + 𝑠+ℎ 1 1 f. 𝑔(𝑠 + ℎ) − 𝑔(𝑠) = Answer to part (e) − Original function = !(𝑠 + ℎ) 2 + " − !𝑠 2 + " 𝑠+ℎ 𝑠 17. 𝑔(𝑠) = 𝑠 2 +

1 𝑟+4 1 1 1 1 a. ℎ(0) = b. ℎ(−3) = = = =1 0+4 4 (−3) + 4 1 1 1 1 c. ℎ(−5) = d. Substitute 𝑥 2 for 𝑟 to obtain ℎ(𝑥 2) = 2 . = = −1 (−5) + 4 (−1) 𝑥 +4 1 1 e. Substitute (𝑥 2 + 1) for 𝑟 to obtain ℎ(𝑥 2 + 1) = 2 . = 2 (𝑥 + 1) + 4 𝑥 + 5 1 f. ℎ(𝑥 2) + 1 = Answer to part (d) + 1 = 2 + 1\\ 𝑥 +4 18. ℎ(𝑟) =

19. 𝑓(𝑥) = −𝑥 3 (domain (−∞, ∞))

20. 𝑓(𝑥) = 𝑥 3 (domain [0, ∞))

Technology formula: -(x^3)\\

Technology formula: x^3

21. 𝑓(𝑥) = 𝑥 (domain (−∞, ∞)) 4

3

22. 𝑓(𝑥) = √𝑥 (domain (−∞, ∞))

23. 𝑓(𝑥) =

1 (𝑥 ≠ 0) 𝑥2

24. 𝑓(𝑥) = 𝑥 +

1 (𝑥 ≠ 0) 𝑥

Solutions Section 1.1 Technology formula: x^4

Technology formula: x^(1/3)

Technology formula: 1/(x^2)

Technology formula: x+1/x

25. a. 𝑓(𝑥) = 𝑥 (−1 ≤ 𝑥 ≤ 1) Since the graph of 𝑓(𝑥) = 𝑥 is a diagonal 45 ∘ line through the origin inclining up from left to right, the correct graph is (A). b. 𝑓(𝑥) = −𝑥 (−1 ≤ 𝑥 ≤ 1) Since the graph of 𝑓(𝑥) = −𝑥 is a diagonal 45 ∘ line through the origin inclining down from left to right, the correct graph is (D). c. 𝑓(𝑥) = √𝑥 (0 < 𝑥 < 4) Since the graph of 𝑓(𝑥) = √𝑥 is the top half of a sideways parabola, the correct graph is (E). 1 d. 𝑓(𝑥) = 𝑥 + − 2 (0 < 𝑥 < 4) 𝑥 If we plot a few points like 𝑥 = 1∕2, 1, 2, and 3, we find that the correct graph is (F). e. 𝑓(𝑥) = |𝑥| (−1 ≤ 𝑥 ≤ 1) Since the graph of 𝑓(𝑥) = |𝑥|is a "V"-shape with its vertex at the origin, the correct graph is (C). f. 𝑓(𝑥) = 𝑥 − 1 (−1 ≤ 𝑥 ≤ 1)

Solutions Section 1.1 Since the graph of 𝑓(𝑥) = 𝑥 − 1 is a straight line through (0, −1) and (1, 0), the correct graph is (B).

26. a. 𝑓(𝑥) = −𝑥 + 3 (0 < 𝑥 ≤ 3) Since the graph of 𝑓(𝑥) = −𝑥 + 3 is a straight line inclining down from left to right, the correct graph must be (D). b. 𝑓(𝑥) = 2 − |𝑥| (−2 < 𝑥 ≤ 2) Since 𝑓(𝑥) = 2 − |𝑥| is obtained from the graph of 𝑦 = |𝑥| by flipping it vertically (the minus sign in front of |𝑥|) and then moving it 2 units vertically up (adding 2 to all the values), the correct graph is (F). c. 𝑓(𝑥) = √𝑥 + 2 (−2 < 𝑥 ≤ 2) The graph of𝑓(𝑥) = √𝑥 + 2 is similar to that of 𝑦 = √𝑥, which is half a parabola on its side, and the correct graph is (A). d. 𝑓(𝑥) = −𝑥 2 + 2 (−2 < 𝑥 ≤ 2) The graph of 𝑓(𝑥) = −𝑥 2 + 2 is a parabola opening down, so the correct graph is (C). 1 e. 𝑓(𝑥) = − 1 𝑥 1 The graph of 𝑓(𝑥) = − 1 (0 < 𝑥 ≤ 3) is part of a hyperbola, and the correct graph is (E). 𝑥 f. 𝑓(𝑥) = 𝑥 2 − 1 (−2 < 𝑥 ≤ 2) The graph of 𝑓(𝑥) = 𝑥 2 − 1 is a parabola opening up, so the correct graph is (B). 27. Technology formula: 0.1*x^2 - 4*x+5 Table of values: 𝑥

𝑓(𝑥)

0 5

1

2

1.1

3

4

5

−5

𝑔(𝑥) 39.9

−4

−3

30.3

21.5

−2

13.5

−1

6.3

0

1

0.5

8

9

10

1.5

2.5

3.5

2

3

4

5

−0.1 −5.7 −10.5 −14.5 −17.7 −20.1

29. Technology formula: (x^2-1)/(x^2+1) Table of values: 𝑥

7

−2.6 −6.1 −9.4 −12.5 −15.4 −18.1 −20.6 −22.9 −25

28. Technology formula: 0.4*x^2-6*x-0.1 Table of values: 𝑥

6

4.5

5.5

6.5

7.5

8.5

9.5

10.5

4

5

6

7

8

9

ℎ(𝑥) −0.6000 0.3846 0.7241 0.8491 0.9059 0.9360 0.9538 0.9651 0.9727 0.9781 0.9820 30. Technology formula: (2*x^2+1)/(2*x^2-1) Table of values: 𝑥

−1

0

1

2

3

𝑟(𝑥) 3.0000 −1.0000 3.0000 1.2857 1.1176 1.0645 1.0408 1.0282 1.0206 1.0157 1.0124

𝑥 if − 4 ≤ 𝑥 < 0 {2 if 0 ≤ 𝑥 ≤ 4 Technology formula: x*(x\lt 0)+2*(x\gt =0) (For a graphing calculator, use ≥ instead of >=.) 31. 𝑓(𝑥) =

Solutions Section 1.1

a. 𝑓(−1) = −1. We used the first formula, since −1 is in [−4, 0). b. 𝑓(0) = 2. We used the second formula, since 0 is in [0, 4]. c. 𝑓(1) = 2. We used the second formula, since 1 is in [0, 4].

−1 if − 4 ≤ 𝑥 ≤ 0 {𝑥 if 0 < 𝑥 ≤ 4 Technology formula: (-1)*(x\lt =0)+x*(x\gt 0) (For a graphing calculator, use ≤ instead of <=.) 32. 𝑓(𝑥) =

a. 𝑓(−1) = −1. We used the first formula, since −1 is in [−4, 0]. b. 𝑓(0) = −1. We used the first formula, since 0 is in [−4, 0]. c. 𝑓(1) = 1. We used the second formula, since 1 is in (0, 4].

⎧𝑥 2 if − 2 < 𝑥 ≤ 0 33. 𝑓(𝑥) = ⎪ ⎨ 1 if 0 < 𝑥 ≤ 4 ⎪𝑥 ⎩ Technology formula: (x^2)*(x<=0)+(1/x)*(0<x) (For a graphing calculator, use ≤ instead of <=.)

a. 𝑓(−1) = 1 2 = 1. We used the first formula, since −1 is in (−2, 0]. b. 𝑓(0) = 0 2 = 0. We used the first formula, since 0 is in (−2, 0]. c. 𝑓(1) = 1∕1 = 1. We used the second formula, since 1 is in (0, 4].

if − 2 < 𝑥 ≤ 0 −𝑥 {√𝑥 if 0 < 𝑥 < 4 Technology formula: Excel: (-1\*x^2)\*(x\lt =0)+SQRT(ABS(x))\*(x\gt 0) TI-83/84 Plus: (-1\*x^2)\*(x≤0)+ √(x)\*(x\gt 0) 2

34. 𝑓(𝑥) =

Solutions Section 1.1

a. 𝑓(−1) = −(−1) 2 = −1. We used the first formula, since −1 is in (−2, 0]. b. 𝑓(0) = −0 2 = 0. We used the first formula, since 0 is in (−2, 0]. c. 𝑓(1) = √1 = 1. We used the second formula, since 1 is in (0, 4).

if − 1 < 𝑥 ≤ 0 ⎧𝑥 35. 𝑓(𝑥) = ⎪ 𝑥 1 if 0 < 𝑥 ≤ 2 + ⎨ ⎪ 𝑥 if 2 < 𝑥 ≤ 4 ⎩ Technology formula: x*(x<=0)+(x+1)*(0<x)*(x<=2)+x*(2<x) (For a graphing calculator, use ≤ instead of <=.)

a. 𝑓(0) = 0. We used the first formula, since 0 is in (−1, 0]. b. 𝑓(1) = 1 + 1 = 2. We used the second formula, since 1 is in (0, 2]. c. 𝑓(2) = 2 + 1 = 3. We used the second formula, since 2 is in (0, 2]. d. 𝑓(3) = 3. We used the third formula, since 3 is in (2, 4].

if − 1 < 𝑥 < 0 ⎧−𝑥 𝑥 2 if 0 ≤ 𝑥 ≤ 2 − 36. 𝑓(𝑥) = ⎪ ⎨ ⎪ if 2 < 𝑥 ≤ 4 ⎩−𝑥 Technology formula: x*(x\lt 0)+(x-2)*(0\lt =x)*(x\lt =2)+(-x)*(2\lt x) (For a graphing calculator, use ≤ instead of <=.) y

1 -1

2 -2 -4

4

x

Solutions Section 1.1

a. 𝑓(0) = 0 − 2 = −2. We used the second formula, since 0 is in [0, 2]. b. 𝑓(1) = 1 − 2 = −1. We used the second formula, since 1 is in [0, 2]. c. 𝑓(2) = 2 − 2 = 0. We used the second formula, since 2 is in [0, 2]. d. 𝑓(3) = −3. We used the third formula, since 3 is in (2, 4]. 37. 𝑓(𝑥) = 𝑥 2 a. 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 Therefore,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (𝑥 + ℎ) 2 − 𝑥 2 = 𝑥 2 + 2𝑥ℎ + ℎ 2 − 𝑥 2 = 2𝑥ℎ + ℎ 2 = ℎ(2𝑥 + ℎ)

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ(2𝑥 + ℎ) = = 2𝑥 + ℎ ℎ ℎ

38. 𝑓(𝑥) = 3𝑥 − 1 a. 𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ) − 1 = 3𝑥 + 3ℎ − 1 Therefore, 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 + 3ℎ − 1 − (3𝑥 − 1) = 3𝑥 + 3ℎ − 1 − 3𝑥 + 1 = 3ℎ

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 3ℎ = =3 ℎ ℎ

39. 𝑓(𝑥) = 2 − 𝑥 2 a. 𝑓(𝑥 + ℎ) = 2 − (𝑥 + ℎ) 2 Therefore,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 2 − (𝑥 + ℎ) 2 − (2 − 𝑥 2) = 2 − 𝑥 2 − 2𝑥ℎ − ℎ 2 − 2 + 𝑥 2 = −2𝑥ℎ − ℎ 2 = −ℎ(2𝑥 + ℎ)

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) −ℎ(2𝑥 + ℎ) = = −(2𝑥 + ℎ) ℎ ℎ 40. 𝑓(𝑥) = 𝑥 2 + 𝑥 a. 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 + (𝑥 + ℎ)Therefore,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (𝑥 + ℎ) 2 + (𝑥 + ℎ) − (𝑥 2 + 𝑥) = 𝑥 2 + 2𝑥ℎ + ℎ 2 + 𝑥 + ℎ − 𝑥 2 − 𝑥 = 2𝑥ℎ + ℎ 2 + ℎ = ℎ(2𝑥 + ℎ + 1)

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ(2𝑥 + ℎ + 1) = = 2𝑥 + ℎ + 1 ℎ ℎ

41. From the table, a. 𝑝(2) = 0.67; Pemex produced 0.67 billion barrels of crude oil in 2017 (𝑡 = 2). 𝑝(3) = 0.61; Pemex produced 0.61 billion barrels of crude oil in 2018 (𝑡 = 3). 𝑝(6) = 0.62; Pemex produced 0.62 billion barrels of crude oil in 2021 (𝑡 = 6).

Solutions Section 1.1 b. 𝑝(6) − 𝑝(3) = 0.62 − 0.61 = 0.01; Annual crude oil production by Pemex increased by 0.01 billion barrels from 2018 (𝑡 = 3) to 2021 (𝑡 = 6). 42. From the table, a. 𝑠(0) = 0.69; Pemex produced 0.69 billion barrels of offshore crude oil in 2015 (𝑡 = 0). 𝑠(2) = 0.56; Pemex produced 0.56 billion barrels of offshore crude oil in 2017 (𝑡 = 2). 𝑠(4) = 0.51; Pemex produced 0.51 billion barrels of offshore crude oil in 2019 (𝑡 = 4). b. 𝑠(4) − 𝑠(0) = 0.51 − 0.69 = −0.18; Annual offshore crude oil production by Pemex decreased by 0.18 billion barrels from 2015 (𝑡 = 0) to 2019 (𝑡 = 4). 43. a. Graph of 𝑝:

The graph suggests a curve, so we exclude the linear models (A) and (C). Model (B) gives a parabola that curves up whereas the curve suggested by the graph curves down, leaving us with Model (D). Also, model (D) gives almost the exact values shown in the chart. (Use the technology formula -0.25x^2+3.5x+57.) b. Using Model (D), 𝑝(5) = −0.25(5) 2 + 3.5(5) + 57 = 68.25 Interpretation: 𝑡 = 5 represents 5 years since the start of 2013, or the start of 2018. Thus, we interpret the answer as follows: Approximately 68.25% of U.S. adults used Facebook at the start of 2018 c. The graph becomes less steep as time increases, indicating decelerating Facebook membership over the period 2013–2021. 44. a. Graph of 𝑝:

The graph suggests a curve, so we exclude the linear model (B). Model (D) gives a parabola that curves up whereas the curve suggested by the graph curves down, leaving us with Models (A) and (C). Model (A) gives values that round to the values shown in the chart wheeras Model (C) is way off. (Use the technology formula -0.32x^2+3.6x+13. b. Using Model (A), 𝑝(7.5) = −0.32(7.5) 2 + 3.6(7.5) + 13 = 22 Interpretation: 𝑡 = 7.5 represents 7.5 years since the start of 2013, or midway through 2020. Thus, we interpret the answer as follows: Approximately 22% of U.S. adults used Twitter midway through 2020. c. The graph becomes less steep as time increases, indicating decelerating Twitter membership over the period 2013–2021.

45. From the graph, 𝑓(11) ≈ 800. Because 𝑓 is the number of thousands of housing starts in year 11, we interpret the result as follows: Approximately 800,000 homes were started in 2016. Similarly, 𝑓(15) ≈ 1,000. Approximately 1,000,000 homes were started in 2020. Also, we estimate 𝑓(2.5) ≈ 800. Because 𝑡 = 2.5 is midway between 2007 and 2008, we interpret the result as follows: approximately 800,000 homes were started in the year beginning midway through 2007.

Solutions Section 1.1 46. From the graph, 𝑓(3) ≈ 600. Because 𝑓 is the number of thousands of housing starts in year 3, we interpret the result as follows: Approximately 600,000 homes were started in 2008. 𝑓(12) ≈ 900: Approximately 900,000 homes were started in 2016. 𝑓(14.5) ≈ 950: Approximately 950,000 homes were started in the year beginning midway through 2019.

47. 𝑓(14 − 10) = 𝑓(4) ≈ 400. Interpretation: Approximately 400,000 homes were started in 2009 (𝑡 = 4). 𝑓(14) − 𝑓(10) ≈ 900 − 700 = 200. Interpretation: 𝑓(14) − 𝑓(10) is the change in the number of housing starts (in thousands) from 2015 to 2019; there were approximately 200,000 more housing starts in 2019 than in 2015. 48. 𝑓(15 − 1) = 𝑓(14) ≈ 900. Interpretation: Approximately 900,000 homes were started in 2019 (𝑡 = 14). 𝑓(15) − 𝑓(1) ≈ 1,000 − 1,500 = −500. Interpretation: 𝑓(15) − 𝑓(1) is the change in the number of housing starts (in thousands) from 2006 to 2020; there were approximately 500,000 fewer housing starts in 2020 than in 2006.

49. 𝑓(𝑡 + 5) − 𝑓(𝑡) measures the change from year 𝑡 to the year five years later. It is greatest when the line segment from year 𝑡 to year 𝑡 + 5 is steepest upward-sloping. From the graph, or by computing differences of values estimated from the graph, this occurs when 𝑡 = 6 or 7: 𝑓(11) − 𝑓(6) = 800 − 400 = 400; 𝑓(12) − 𝑓(7) = 900 − 500 = 400. Interpretation: The greatest five-year increase in the number of housing starts occurred in 2011–2016 and again in 2012–2017. 50. 𝑓(𝑡) − 𝑓(𝑡 − 1) measures the change from year 𝑡 − 1 to the following year. It is least when the the line segment from year 𝑡 − 1 to year 𝑡 is steepest downward-sloping. From the graph, this occurs when 𝑡 = 2, for a change of 𝑓(2) − 𝑓(1) = 1,000 − 1,500 = −500. Interpretation: The greatest annual decrease in the number of housing starts occurred in 2006–2007.

51. a. From the graph, 𝑏(3) ≈ 50 and 𝑏(5) ≈ 35. The (50-day average) price of Bitcoin was approximately $50,000 on June 1, 2021 (𝑡 = 3) and $35,000 on August 1, 2021 (𝑡 = 5). 𝑏(7) − 𝑏(5) ≈ 50 − 35 = 15. The Bitcoin price increased by around $15,000 from August 1 to October 1, 2021. b. Increasing most rapidly when the graph is steepest upward from left to right on the interal [1, 3], which occurs at 𝑡 = 1 (integer answer required). Thus, during the period April 1–June 1, 2021, the price of Bitcoin was increasing most rapidly around April 1. c. Decreasing most rapidly when the graph is steepest down from left to right, which occurs at 𝑡 = 3 (integer answer required). Thus, during the period April 1–June 1, 2021, the price of Butcoin was decreasing most rapidly around June 1. 52. a. From the graph, 𝑒(2) ≈ 2.5 and 𝑒(4) ≈ 2.5. The price of Etherium was approximately $2,500 on May 1, 2021 (𝑡 = 2) and again on July 1, 2021 (𝑡 = 4) . 𝑒(3) − 𝑒(1) ≈ 2 − 3 = −1. The price of Etherium decreased by around $1,000 from April 1 to June 1, 2021. b. Increasing most rapidly when the graph is steepest upward from left to right on the interal [1, 5], which occurs at 𝑡 = 4 (integer answer required). Thus, during the period April 1–August 1, 2021, the price of Etherium was increasing most rapidly around July 1. c. Decreasing most rapidly when the graph is steepest down from left to right, which occurs at 𝑡 = 2 (integer answer required). Thus, during the period April 1–August 1, 2021, the price of Etherium was decreasing most rapidly around May 1. 53. 𝑟(𝑥) = 0.4𝑥 2 + 𝑥 + 26.5 (0 ≤ 𝑥 ≤ 5) a. The domain of 𝑟 is [0, 5].

𝑟(0) = 0.4(0) + 0 + 26.5 = 26.5; 𝑟(1) = 0.4(1) 2 + 1 + 26.5 = 27.9 𝑟(3) = 0.4(3) 2 + 3 + 26.5 = 33.1 𝑟(5) = 0.4(5) 2 + 5 + 26.5 = 41.5 Food delivery app revenues in the U.S. were projected to be $26.5 billion in 2020, $27.9 billion in 2021, $33.1 billion in 2023, and $41.5 billion in 2025. b. Graph: 2

Solutions Section 1.1

(As the curve is concave up the projected revenue was accelerating.)

54. 𝑟(𝑥) = −0.2𝑥 2 + 3𝑥 + 18 (0 ≤ 𝑥 ≤ 5) a. The domain of 𝑟 is [0, 5]. 𝑟(0) = −0.2(0) 2 + 3(0) + 18 = 18; 𝑟(1) = −0.2(1) 2 + 3(1) + 18 = 20.8 𝑟(3) = −0.2(3) 2 + 3(3) + 18 = 25.2 𝑟(5) = −0.2(5) 2 + 3(5) + 18 = 28 Food delivery app revenues in Europe were projected to be $18 billion in 2020, $20.8 billion in 2021, $25.2 billion in 2023, and $28 billion in 2025. b. Graph

(As the graph is concave down the projected revenue was decelerating). 55. a. The model is valid for the range 1958 (𝑡 = 0) through 1966 (𝑡 = 8). Thus, an appropriate domain is [0, 8]. 𝑡 ≥ 0 is not an appropriate domain because it would predict federal funding of NASA beyond 1966, whereas the model is based only on data up to 1966. 4.5 b. 𝑝(𝑡) = 2 1.07 (𝑡−8) 4.5 ⇒ 𝑝(5) = ≈ 2.4 Technology formula: 4.5/(1.07^((t-8)^2)) 2 1.07 (5−8) 𝑡 = 5 represents 1958 + 5 = 1963, and therefore we interpret the result as follows: In 1963, 2.4% of the U.S. federal budget was allocated to NASA. c. 𝑝(𝑡) is increasing most rapidly when the graph is steepest upward-sloping from left to right, and, among the given values of 𝑡, this occurs when 𝑡 = 5. Thus, the percentage of the budget allocated to NASA was

Solutions Section 1.1 increasing most rapidly in 1963.

56. a. [1, 55]; [0, 55] is not an appropriate domain because 𝑝 is undefined at 0. 5 b. 𝑝(𝑡) = 0.03 + 0.6 𝑡 5 ⇒ 𝑝(40) = 0.03 + ≈ 0.58 Technology formula: 0.03+5/t^0.6 40 0.6 𝑡 = 40 represents 1965 + 40 = 2005, and therefore we interpret the result as follows: In 2005, 0.58% of the US federal budget was allocated to NASA. c. If we evaluate 𝑝(𝑡) for 𝑡 = 100, 1,000, 100,000, 1,000,000 . . . , we find values of 𝑝(𝑡) decreasing toward 0.03. Thus, in the (very) long term, the percentage of the budget allocated to NASA is predicted to approach 0.03% 12, 200 57. a. 𝑝(𝑡) = 100⎛1 − 4.48 ⎞ (𝑡 ≥ 8.5) ⎝⎜ ⎠⎟ 𝑡 Technology formula: 100*(1-12200/t^4.48) b. Graph:

c. Table of values: 𝑡

9

𝑝(𝑡) 35.2

10

11

12

13

14

15

16

17

18

19

20

59.6

73.6

82.2

87.5

91.1

93.4

95.1

96.3

97.1

97.7

98.2

d. From the table, 𝑝(12) = 82.2, so that 82.2% of children are able to speak in at least single words by the age of 12 months. e. We seek the first value of 𝑡 such that 𝑝(𝑡) is at least 90. Since 𝑡 = 14 has this property (𝑝(14) = 91.1) we conclude that, at 14 months, 90% or more children are able to speak in at least single words. 5.27 × 10 17 (𝑡 ≥ 30) " 𝑡 12 Technology formula: 100*(1-5.27*10^17/t^12) b.Graph: 58. a. 𝑝(𝑡) = 100!1 −

c. Table of values: 𝑡

𝑝(𝑡)

30

31

32

33

34

35

36

37

38

39

40

0.8

33.1

54.3

68.4

77.9

84.4

88.9

92.0

94.2

95.7

96.9

d. From the table, 𝑝(36) = 88.9, so that 88.9% of children are able to speak in sentences of five or more words by the age of 36 months.

Solutions Section 1.1 e. We seek the first value of 𝑡 such that 𝑝(𝑡) is at least 75. Since 𝑡 = 34 has this property (𝑝(34) = 77.9) we conclude that, at 34 months, 75% or more children are able to speak in sentences of five or more words. if 0 ≤ 𝑡 < 16 ⎧8(1.22) 𝑡 ⎪ 59. 𝑣(𝑡) = ⎨400𝑡 − 6,200 if 16 ≤ 𝑡 < 25 ⎪0.45(𝑡 − 25) 3 + 3,800 if 25 ≤ 𝑡 ≤ 40 ⎩

a. 𝑣(10) = 8(1.22) 10 ≈ 58. We used the first formula, since 10 is in [0, 16). 𝑣(16) = 400(16) − 6,200 = 200. We used the second formula, since 16 is in [16, 25). 𝑣(40) = 0.45(40 − 25) 3 + 3,800 ≈ 5,319 We used the third formula, since 40 is in [25, 40]. Interpretation: Processor speeds were about 58 MHz in 1990, 200 MHz in 1996, and 5,319 MHz in 2020. b. Technology formula (using 𝑥 as the independent variable): (8*(1.22)^x)*(x<16)+(400*x-6200)*(x>=16)*(x<25) +(0.45*(x-25)^3+3800)*(x>=25) (For a graphing calculator, use ≤ instead of <=.) c. Using the above technology formula (for instance, on the Function Evaluator and Grapher on the Web site) we obtain the graph and table of values. Graph:

Table of values: 𝑡

𝑣(𝑡)

0

4

8

12

16

20

24

28

32

36

40

8

18

39

87

200 1,800 3,400 3,800 4,000 4,400 5,300

d. From either the graph or the table, we see that the speed reached 3,000 MHz between 𝑡 = 20 and 𝑡 = 24. We can obtain a more precise answer algebraically by using the formula for the corresponding portion of the graph: 3,000 = 400𝑡 − 6,200 giving 𝑡=

9,200 = 23 400

Since 𝑡 is time since 1980, 𝑡 = 23 corresponds to 2003.

⎧0.12𝑡 2 + 0.04𝑡 + 0.2 if 0 ≤ 𝑡 < 12 60. 𝑣(𝑡) = ⎪ if 12 ≤ 𝑡 < 26 ⎨1.1(1.22) 𝑡 ⎪400𝑡 − 10,200 if 26 ≤ 𝑡 ≤ 30 ⎩ 2 a. 𝑣(2) = 0.12(2) + 0.04(2) + 0.2 = 0.76. We used the first formula, since 2 is in [0, 12). 𝑣(12) = 1.1(1.22) 12 ≈ 12. We used the second formula, since 12 is in [12, 26). 𝑣(28) = 400(28) − 10, 200 = 1, 000. We used the third formula, since 28 is in [26, 30]. Interpretation: Processor speeds were about 0.76 MHz in 1972, 12 MHz in 1982, and 1,000 MHz in 1998.

Solutions Section 1.1

b. Technology formula (using 𝑥 as the independent variable): (0.12*x^2+0.04*x+0.2)*(x<12)+(1.1*(1.22)^x)*(x>=12)*(x<26) +(400*x-10200)*(x>=26) (For a graphing calculator, use ≤ instead of <=.) c. Using the above technology formula (for instance, on the Function Evaluator and Grapher on the Web site) we obtain the graph and table of values. Graph:

Table of values: 𝑡

0

2

𝑣(𝑡) 0.20 0.76

4

6

8

10

12

14

16

18

20

22

24

26

28

30

2.3

4.8

8.2

13

12

18

26

39

59

87

130

200 1,000 1,800

d. From either the graph or the table, we see that the speed reached 500 MHz around 𝑡 = 27. We can obtain a more precise answer algebraically by using the formula for the corresponding portion of the graph: 500 = 400𝑡 − 10,200

giving 𝑡=

10,700 = 26.75 ≈ 27 to the nearest year 400

Since 𝑡 is time since 1970, 𝑡 = 27 corresponds to 1997. 61. a. Each row of the table gives us a formula with a condition: First row in words: 10% of the amount over $0 if your income is over $0 and not over $9,950. Translation to formula: 0.10𝑥 if 0 < 𝑥 ≤ 9,950. Second row in words: $995.00 + 12% of the amount over $9,950 if your income is over $9,950 and not over $40,525. Translation to formula: 995.00 + 0.12(𝑥 − 9, 950) if 9,950 < 𝑥 ≤ 40,525. Continuing in this way leads to the following piecewise-defined function: ⎧0.10𝑥 ⎪995.00 + 0.12(𝑥 − 9,950) ⎪ ⎪4,664 + 0.22(𝑥 − 40,525) 𝑇 (𝑥) = ⎪ ⎨14,751 + 0.24(𝑥 − 86,375) ⎪33,603 + 0.32(𝑥 − 164,925) ⎪ ⎪47,843 + 0.35(𝑥 − 209,425) ⎪157,804.25 + 0.37(𝑥 − 523,600) ⎩

if 0 < 𝑥 ≤ 9,950 if 9,950 < 𝑥 ≤ 40,525 if 40,525 < 𝑥 ≤ 86,375 if 86,375 < 𝑥 ≤ 164,925 if 164,925 < 𝑥 ≤ 209,425 if 209,425 < 𝑥 ≤ 523,600 if 523,600 < 𝑥

Solutions Section 1.1

b. A taxable income of $45,000 falls in the bracket 40,525 < 𝑥 ≤ 86,375 and so we use the formula 4,664 + 0.22(𝑥 − 40,525) : 4,664 + 0.22(45,000 − 40,525) = 4,664 + 0.22(4,475) = $5,648.50 62. a. Each row of the table gives us a formula with a condition: First row in words: 10% of the amount over $0 if your income is over $0 and not over $8,700. Translation to formula: 0.10𝑥 if 0 < 𝑥 ≤ 8,700. Second row in words: $870.00 + 15% of the amount over $8,700 if your income is over $8,700 and not over $35,350. Translation to formula: 870.00 + 0.15(𝑥 − 8,700) if 8,700 < 𝑥 ≤ 35,350. Continuing in this way leads to the following piecewise-defined function: ⎧0.10𝑥 ⎪870.00 + 0.15(𝑥 − 8,700) ⎪ ⎪4,867.50 + 0.22(𝑥 − 35,350) 𝑇 (𝑥) = ⎨ 17,442.50 + 0.28(𝑥 − 85,650) ⎪ ⎪43,482.50 + 0.33(𝑥 − 178,650) ⎪112,683.50 + 0.35(𝑥 − 388,350) ⎩

if 0 < 𝑥 ≤ 8,700 if 8,700 < 𝑥 ≤ 35,350 if 35,350 < 𝑥 ≤ 85,650 if 85,650 < 𝑥 ≤ 178,650 if 178,650 < 𝑥 ≤ 388,350 if 388,350 < 𝑥

b. A taxable income of $45,000 falls in the bracket 35,350 < 𝑥 ≤ 85,650 and so we use the formula 4,867.50 + 0.22(𝑥 − 35,350) : 4,867.50 + 0.22(45,000 − 35,350) = 4,867.50 + 0.22(9,650) = $9,990.50.

63. The dependent variable is a function of the independent variable. Here, the market price of gold 𝑚 is a function of time 𝑡. Thus, the independent variable is 𝑡 and the dependent variable is 𝑚. 64. The dependent variable is a function of the independent variable. Here, the weekly profit 𝑃 is a function of the selling price 𝑠. Thus, the independent variable is 𝑠 and the dependent variable is 𝑃 .

65. To obtain the function notation, write the dependent variable as a function of the independent variable. Thus 𝑦 = 4𝑥 2 − 2 can be written as 𝑓(𝑥) = 4𝑥 2 − 2 or 𝑦(𝑥) = 4𝑥 2 − 2 66. To obtain the equation notation, introduce a dependent variable instead of the function notation. Thus 𝐶(𝑡) = −0.34𝑡 2 + 0.1𝑡 can be written as 𝑐 = −0.34𝑡 2 + 0.1𝑡 or 𝑦 = −0.34𝑡 2 + 0.1𝑡 67. False. A graph usually gives infinitely many values of the function while a numerical table will give only a finite number of values.

68. True. An algebraically specified function 𝑓 is specified by algebraic formulas for 𝑓(𝑥). Given such formulas, we can construct the graph of 𝑓 by plotting the points (𝑥, 𝑓(𝑥)) for values of 𝑥 in the domain of 𝑓. 69. False. In a numerically specified function, only certain values of the function are specified so we cannot know its value on every real number in [0, 10], whereas an algebraically specified function would give values for every real number in [0, 10].

Solutions Section 1.1 70. False. A graphically specified function is specified by a graph. However, we cannot always expect to find an algebraic formula whose graph is exactly the graph that is given. 71. Functions with infinitely many points in their domain (such as 𝑓(𝑥) = 𝑥 2) cannot be specified numerically. So, the assertion is false.

72. A numerical model supplies only the values of a function at specific values of the independent variable, whereas an algebraic model supplies the value of a function at every point in its domain. Thus, an algebraic model supplies more information.

73. (Answers may vary.) Take 𝑓(𝑥) = 𝑥 2. Then a. 𝑓(3 + 2) = 𝑓(5) = 5 2 = 25, whereas 𝑓(3) + 𝑓(2) = 3 2 + 2 2 = 9 + 4 = 13. Thus, 𝑓(3 + 2) ≠ 𝑓(3) + 𝑓(2). b. 𝑓(3 − 2) = 𝑓(1) = 1 2 = 1, whereas 𝑓(3) − 𝑓(2) = 3 2 − 2 2 = 9 − 4 = 5. Thus, 𝑓(3 − 2) ≠ 𝑓(3) − 𝑓(2). 74. (Answers may vary.) Take 𝑓(𝑥) = 𝑥 + 1. Then a. 𝑓(3 × 2) = 𝑓(6) = 6 + 1 = 7, whereas 3𝑓(2) = 3(2 + 1) = 3 × 3 = 9. Thus, 𝑓(3 × 2) ≠ 3𝑓(2). b. 𝑓(3 × 2) = 𝑓(6) = 6 + 1 = 7, whereas 𝑓(3)𝑓(2) = (3 + 1)(2 + 1) = 4 × 3 = 12. Thus, 𝑓(3 × 2) ≠ 𝑓(3)𝑓(2).

75. As the text reminds us: to evaluate 𝑓 of a quantity (such as 𝑥 + ℎ) replace 𝑥 everywhere by the whole quantity 𝑥 + ℎ : 𝑓(𝑥) = 𝑥2 − 1 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 − 1.

76. Knowing 𝑓(𝑥) for two values of 𝑥 does not convey any information about 𝑓(𝑥) at any other value of 𝑥. Interpolation is only a way of estimating 𝑓(𝑥) at values of 𝑥 not given.

77. If two functions are specified by the same formula 𝑓(𝑥), say, their graphs must follow the same curve 𝑦 = 𝑓(𝑥). However, it is the domain of the function that specifies what portion of the curve appears on the graph. Thus, if the functions have different domains, their graphs will be different portions of the curve 𝑦 = 𝑓(𝑥). 78. If we plot points of the graphs 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥), we see that, since 𝑔(𝑥) = 𝑓(𝑥) + 10, we must add 10 to the 𝑦-coordinate of each point in the graph of 𝑓 to get a point on the graph of 𝑔. Thus, the graph of 𝑔 is 10 units higher up than the graph of 𝑓. 79. Suppose we already have the graph of 𝑓 and want to construct the graph of 𝑔. We can plot a point of the graph of 𝑔 as follows: Choose a value for 𝑥 (𝑥 = 7, say) and then "look back" 5 units to read off 𝑓(𝑥 − 5) (𝑓(2) in this instance). This value gives the 𝑦-coordinate we want. In other words, points on the graph of 𝑔 are obtained by "looking back 5 units" to the graph of 𝑓 and then copying that portion of the curve. Put another way, the graph of 𝑔 is the same as the graph of 𝑓, but shifted 5 units to the right:

80. Suppose we already have the graph of 𝑓 and want to construct the graph of 𝑔. We can plot a point of

Solutions Section 1.1 the graph of 𝑔 as follows: Choose a value for 𝑥 (𝑥 = 7, say) and then look on the other side of the 𝑦-axis to read off 𝑓(−𝑥) (𝑓(−7) in this instance). This value gives the 𝑦-coordinate we want. In other words, points on the graph of 𝑔 are obtained by "looking back" to the graph of 𝑓 on the opposite side of the 𝑦-axis and then copying that portion of the curve. Put another way, the graph of 𝑔(𝑥) is the mirror image of the graph of 𝑓(𝑥) in the 𝑦-axis:

Solutions Section 1.2 Section 1.2

1. 𝑓(𝑥) = 𝑥 2 + 1 with domain (−∞, ∞) 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) a. 𝑠(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) = (𝑥 2 + 1) + (𝑥 − 1) = 𝑥 2 + 𝑥 b. Since both functions are defined for every real number 𝑥, the domain of 𝑠 is the set of all real numbers: (−∞, ∞). c. 𝑠(−3) = (−3) 2 + (−3) = 9 − 3 = 6

2. 𝑓(𝑥) = 𝑥 2 + 1 with domain (−∞, ∞) 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) a. 𝑑(𝑥) = 𝑔(𝑥) − 𝑓(𝑥) = (𝑥 − 1) − (𝑥 2 + 1) = −𝑥 2 + 𝑥 − 2 b. Since both functions are defined for every real number 𝑥, the domain of 𝑑 is the set of all real numbers: (−∞, ∞). c. 𝑑(−1) = −(−1) 2 + (−1) − 2 = −4 3. 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) 𝑢(𝑥) = √𝑥 + 10 with domain [−10, 0) a. 𝑝(𝑥) = 𝑔(𝑥)𝑢(𝑥) = (𝑥 − 1)√𝑥 + 10 b. The domain of 𝑝 consists of all real numbers 𝑥 simultaneously in the domains of 𝑔 and 𝑢; that is, [−10, 0). c. 𝑝(−6) = (−6 − 1)√−6 + 10 = (−7)(2) = −14

4. ℎ(𝑥) = 𝑥 + 4 with domain [10, ∞) 𝑣(𝑥) = √10 − 𝑥 with domain [0, 10] a. 𝑝(𝑥) = ℎ(𝑥)𝑣(𝑥) = (𝑥 + 4)√10 − 𝑥 b. The domain of 𝑝 consists of all real numbers 𝑥 simultaneously in the domains of ℎ and 𝑣; that is, the single point 𝑥 = 10. c. As 1 is not in the domain of 𝑝, 𝑝(1) is not defined.

5. 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) 𝑣(𝑥) = √10 − 𝑥 with domain [0, 10] 𝑣(𝑥) √10 − 𝑥 a. 𝑞(𝑥) = = 𝑔(𝑥) 𝑥−1 b. The domain of 𝑞 consists of all real numbers 𝑥 simultaneously in the domains of 𝑣 and 𝑔 such that 𝑔(𝑥) ≠ 0. Since 𝑔(𝑥) = 0 when 𝑥 − 1 = 0, or 𝑥 = 1 we exclude 𝑥 = 1 from the domain of the quotient. Thus, the domain consists of all 𝑥 in [0, 10] excluding 𝑥 = 1 (since 𝑔(1) = 0), or 0 ≤ 𝑥 ≤ 10; 𝑥 ≠ 1. c. As 1 is not in the domain of 𝑞, 𝑞(1) is not defined.

6. 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) 𝑣(𝑥) = √10 − 𝑥 with domain [0, 10] 𝑔(𝑥) 𝑥−1 a. 𝑞(𝑥) = = 𝑣(𝑥) √10 − 𝑥 b. The domain of 𝑞 consists of all real numbers 𝑥 simultaneously in the domains of 𝑣 and 𝑔 such that 𝑣(𝑥) ≠ 0. Since 𝑣(𝑥) = 0 when √10 − 𝑥 = 0, or 𝑥 = 10 we exclude 𝑥 = 10 from the domain of the quotient. Thus, the domain consists of all 𝑥 in [0, 10] excluding 𝑥 = 10; that is, [0, 10).

Solutions Section 1.2 1−1 c. 1 is in the domain of 𝑞, and 𝑞(1) = =0 √10 − 1 7. 𝑓(𝑥) = 𝑥 2 + 1 with domain (−∞, ∞) a. 𝑚(𝑥) = 5𝑓(𝑥) = 5(𝑥 2 + 1) b. The domain of 𝑚 is the same as the domain of 𝑓 : (−∞, ∞). c. 𝑚(1) = 5𝑓(1) = 5(1 2 + 1) = 10 8. 𝑢(𝑥) = √𝑥 + 10 with domain [−10, 0) a. 𝑚(𝑥) = 3𝑢(𝑥) = 3√𝑥 + 10 b. The domain of 𝑚 is the same as the domain of 𝑢 : [−10, 0) c. 𝑚(−1) = 3𝑢(−1) = 3√−1 + 10 = 9

9. Number of music files = Starting number + New files = 200 + 10 × Number of days So, 𝑁(𝑡) = 200 + 10𝑡 (𝑁 = number of music files, 𝑡 = time in days) 10. Free space left = Current amount − Decrease = 50 − 5× Number of months So, 𝑆(𝑡) = 50 − 5𝑡 (𝑆 = space on your HD, 𝑡 = time in months)

11. The number of hours you study, ℎ(𝑛), equals 4 on Sunday through Thursday and equals 0 on the remaining days. Since Sunday corresponds to 𝑛 = 1 and Thursday to 𝑛 = 5, we get 4 if 1 ≤ 𝑛 ≤ 5 . {0 if 𝑛 > 5

12. The number of hours you watch movies, ℎ(𝑛), equals 5 on Saturday (𝑛 = 7) and Sunday (𝑛 = 1) and equals 2 on the remaining days. 5 if 𝑛 = 1or 𝑛 = 7 . {2 otherwise

13. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $1,500 per piano, 𝑏 = fixed cost = $1,000. Thus, the daily cost function is 𝐶(𝑥) = 1,500𝑥 + 1,000. a. The cost of manufacturing 3 pianos is 𝐶(3) = 1,500(3) + 1,000 = 4,500 + 1,000 = $5,500. b. The cost of manufacturing each additional piano (such as the third one or the 11th one) is the marginal cost, 𝑚 = $1,500. c. Same answer as (b). d. Variable cost = part of the cost function that depends on 𝑥 = $1,500𝑥 Fixed cost = constant summand of the cost function = $1,000 Marginal cost = slope of the cost function = $1,500 per piano

Solutions Section 1.2 e. Graph: C 8,000 7,000 6,000 5,000 4,000 3,000 2,000 x

1,000

0

1

2

3

4

14. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $88 per tuxedo, 𝑏 = fixed cost = $20. Thus, the cost function is 𝐶(𝑥) = 88𝑥 + 20. a. The cost of renting 2 tuxes is 𝐶(2) = 88(2) + 20 = $196 b. The cost of each additional tux is the marginal cost 𝑚 = $88. c. Same answer as (b). d. Variable cost = part of the cost function that depends on 𝑥 = $88𝑥 Fixed cost = constant summand of the cost function = $20 Marginal cost = slope of the cost function = $88 per tuxedo e. Graph: C 400 350 300 250 200 150 100 50 0

1

2

3

x

4

15. a. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $0.40 per copy, 𝑏 = fixed cost = $70. Thus, the cost function is 𝐶(𝑥) = 0.4𝑥 + 70. The revenue function is 𝑅(𝑥) = 0.50𝑥. (𝑥 copies at 50¢ per copy) The profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.5𝑥 − (0.4𝑥 + 70) = 0.5𝑥 − 0.4𝑥 − 70 = 0.1𝑥 − 70

b. 𝑃 (500) = 0.1(500) − 70 = 50 − 70 = −20 Since 𝑃 is negative, this represents a loss of $20. c. For breakeven, 𝑃 (𝑥) = 0 :

0.1𝑥 − 70 = 0 0.1𝑥 = 70 70 𝑥= = 700 copies 0.1

Solutions Section 1.2

16. a. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $0.15 per serving, 𝑏 = fixed cost = $350. Thus, the cost function is 𝐶(𝑥) = 0.15𝑥 + 350. The revenue function is 𝑅(𝑥) = 0.50𝑥. The profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.50𝑥 − (0.15𝑥 + 350) = 0.35𝑥 − 350

b. For break-even, 𝑃 (𝑥) = 0 : 0.35𝑥 − 350 = 0 0.35𝑥 = 350 𝑥 = 1,000 servings c. 𝑃 (1,500) = 0.35(1,500) − 350 = 525 − 350 = $175, representing a profit of $175. 17. The revenue per jersey is $100. Therefore, Revenue 𝑅(𝑥) = $100𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 100𝑥 − (2,000 + 10𝑥 + 0.2𝑥 2) = −2,000 + 90𝑥 − 0.2𝑥 2

To break even, 𝑃 (𝑥) = 0, so −2,000 + 90𝑥 − 0.2𝑥 2 = 0. This is a quadratic equation with 𝑎 = −0.2, 𝑏 = 90, 𝑐 = −2,000 and solution −𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 −90 ± √(90) 2 − 4(−2,000)(−0.2) = ≈ 23.44 or 426.56 jerseys. 2(−0.2)

𝑥=

Since the second value is outside the domain, we use the first: 𝑥 = 23.44 jerseys. To make a profit, 𝑥 should be larger than this value: at least 24 jerseys. 18. The revenue per pair is $120. Therefore, Revenue 𝑅(𝑥) = $120𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 120𝑥 − (3,000 + 8𝑥 + 0.1𝑥 2) = −3,000 + 112 − 0.1𝑥 2

To break even, 𝑃 (𝑥) = 0, so −3,000 + 112 − 0.1𝑥 2 = 0. This is a quadratic equation with 𝑎 = −0.1, 𝑏 = 112, 𝑐 = −3,000 and solution 𝑥=

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 −90 ± √(112) 2 − 4(−3,000)(−0.1) = ≈ 27.46 or 1092.54 jerseys. 2(−0.1)

Solutions Section 1.2 Since the second value is outside the domain, we use the first: 𝑥 = 27.46 jerseys. To make a profit, 𝑥 should be larger than this value: at least 28 pairs of cleats.

19. The revenue from one thousand square feet (𝑥 = 1) is $0.1 million. Therefore, Revenue 𝑅(𝑥) = $0.1𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.1𝑥 − (1.7 + 0.12𝑥 − 0.0001𝑥 2) = −1.7 − 0.02𝑥 + 0.0001𝑥 2

To break even, 𝑃 (𝑥) = 0, so −1.7 − 0.02𝑥 + 0.0001𝑥 2 = 0. This is a quadratic equation with 𝑎 = 0.0001, 𝑏 = −0.02, 𝑐 = −1.7 and solution\\

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 0.02 ± √(−0.02) 2 − 4(0.0001)(−1.7) 0.02 ± 0.03286 = ≈ 264 thousand square feet = 2(0.0001) 0.0002

𝑥=

20. The revenue from one thousand square feet (𝑥 = 1) is $0.2 million. Therefore, Revenue 𝑅(𝑥) = $0.2𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.2𝑥 − (1.7 + 0.14𝑥 − 0.0001𝑥 2) = −1.7 + 0.06𝑥 + 0.0001𝑥 2

To break even, 𝑃 (𝑥) = 0, so −1.7 + 0.06𝑥 + 0.0001𝑥 2 = 0. This is a quadratic equation with 𝑎 = 0.0001, 𝑏 = 0.06, 𝑐 = −1.7 and solution\\

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 −0.06 ± √(−0.06) 2 − 4(0.0001)(−1.7) −0.06 ± 0.0654 = ≈ 27 thousand square feet = 2(0.0001) 0.0002

𝑥=

21. The hourly profit function is given by Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) (Hourly) cost function: This is a fixed cost of $10,200 only: 𝐶(𝑥) = 10,200 (Hourly) revenue function: This is a variable of $200 per passenger cost only: 𝑅(𝑥) = 200𝑥 Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 200𝑥 − 10,200 For the domain of 𝑃 (𝑥), the number of passengers 𝑥 cannot exceed the capacity: 381. Also, 𝑥 cannot be negative. Thus, the domain is given by 0 ≤ 𝑥 ≤ 381, or [0, 381]. For breakeven, 𝑃 (𝑥) = 0 200𝑥 − 10,200 = 0 10,200 200𝑥 = 10,200, or 𝑥 = = 51 200 If 𝑥 is larger than this, then the profit function is positive, and so there should be at least 52 passengers; 𝑥 ≥ 52, for a profit.

Solutions Section 1.2 22. The hourly profit function is given by Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) (Hourly) cost function: This is a fixed cost of $6,500 only: 𝐶(𝑥) = 6,500 (Hourly) revenue function: This is a variable of $100 per passenger cost only: 𝑅(𝑥) = 200𝑥 Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 200𝑥 − 6,500 For the domain of 𝑃 (𝑥), the number of passengers 𝑥 cannot exceed the capacity: 150 Also, 𝑥 cannot be negative. Thus, the domain is given by 0 ≤ 𝑥 ≤ 150, or [0, 150]. For breakeven, 𝑃 (𝑥) = 0 200𝑥 − 5,600 = 0 5,600 200𝑥 = 5,600, or 𝑥 = = 28 200 If 𝑥 is larger than this, then the profit function is positive, and so there should be at least 29 passengers; 𝑥 ≥ 29, for a profit. 23. To compute the break-even point, we use the profit function: Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑅(𝑥) = 2𝑥 $2 per unit 𝐶(𝑥) = Variable Cost + Fixed Cost = 40% of Revenue + 6,000 = 0.4(2𝑥) + 6,000 = 0.8𝑥 + 6,000 Thus, 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 2𝑥 − (0.8𝑥 + 6,000) = 1.2𝑥 − 6,000 For breakeven, 𝑃 (𝑥) = 0 1.2𝑥 − 6,000 = 0 6,000 1.2𝑥 = 6,000, so 𝑥 = = 5,000 1.2𝑥 Therefore, 5,000 units should be made to break even. 24. To compute the break-even point, we use the profit function: Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑅(𝑥) = 5𝑥 $5 per unit 𝐶(𝑥) = Variable Cost + Fixed Cost = 30% of Revenue + 7,000 = 0.3(5𝑥) + 7,000 = 1.5𝑥 + 7,000 Thus, 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 5𝑥 − (1.5𝑥 + 7,000) = 3.5𝑥 − 7,000 For breakeven, 𝑃 (𝑥) = 0 3.5𝑥 = 7,000, so 𝑥 = 2,000 units. 25. To compute the break-even point, we use the revenue and cost functions: 𝑅(𝑥) = Selling price × Number of units = 𝑆𝑃 𝑥 𝐶(𝑥) = Variable Cost + Fixed Cost = 𝑉 𝐶𝑥 + 𝐹 𝐶 (Note that "variable cost per unit" is marginal cost.) For breakeven 𝑅(𝑥) = 𝐶(𝑥) 𝑆𝑃 𝑥 = 𝑉 𝐶𝑥 + 𝐹 𝐶 Solve for 𝑥 : 𝑆𝑃 𝑥 − 𝑉 𝐶𝑥 = 𝐹 𝐶 𝐹𝐶 𝑥(𝑆𝑃 − 𝑉 𝐶) = 𝐹 𝐶, so 𝑥 = . 𝑆𝑃 − 𝑉 𝐶

Solutions Section 1.2 26. To compute the break-even point, we use the revenue and cost functions: 𝑅(𝑥) = 𝑆𝑃 𝑥; 𝐶(𝑥) = 𝑉 𝐶𝑥 + 𝐹 𝐶 At breakeven 𝑅(𝐵𝐸) = 𝐶(𝐵𝐸) 𝑆𝑃 (𝐵𝐸) = 𝑉 𝐶(𝐵𝐸) + 𝐹 𝐶 Thus, 𝐹 𝐶 = 𝑆𝑃 (𝐵𝐸) − 𝑉 𝐶(𝐵𝐸) = 𝐵𝐸(𝑆𝑃 − 𝑉 𝐶).

27. Take 𝑥 to be the number of grams of perfume he buys and sells. The profit function is given by Profit = Revenue − Cost: that is, 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) Cost function 𝐶(𝑥) : Fixed costs: 20,000 Cheap perfume @ $20 per g: 20𝑥 Transportation @ $30 per 100 g: 0.3𝑥 Thus the cost function is 𝐶(𝑥) = 20𝑥 + 0.3𝑥 + 20,000 = 20.3𝑥 + 20,000 Revenue function R(x) 𝑅(𝑥) = 600𝑥 $600 per gram Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 600𝑥 − (20.3𝑥 + 20,000) = 579.7𝑥 − 20,000, with domain 𝑥 ≥ 0. For breakeven, 𝑃 (𝑥) = 0 579.7𝑥 − 20,000 = 0 20,000 579.7𝑥 = 20,000, so 𝑥 = ≈ 34.50 579.7 Thus, he should buy and sell 34.50 grams of perfume per day to break even.

28. Take 𝑥 to be the number of grams of perfume he buys and sells. The profit function is given by Profit = Revenue − Cost Cost function: 𝐶(𝑥) = 400𝑥 + 30𝑥 = 430𝑥 Revenue: 𝑅(𝑥) = 420𝑥 Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 420𝑥 − 430𝑥 = −10𝑥, with domain 𝑥 ≥ 0. For breakeven, −10𝑥 = 0, so 𝑥 = 0 grams per day; he should shut down the operation. 29. a. To graph the demand function we use technology with the formula 760/x-1 and with xMin = 400 and xMax = 700. Graph:

760 760 − 1 = 0.9, 𝑞(500) = − 1 = 0.52 400 500 So, the change in demand is 0.52 − 0.9 = −0.38 billion units, which means that demand decreases by about 380 million units sold per year. b. 𝑞(400) =

Solutions Section 1.2 c. The value of 𝑞 on the graph decreases by smaller and smaller amounts as we move to the right on the graph, indicating that the demand decreases at a smaller and smaller rate (Choice (D)). 30. a. To graph the demand function we use technology with the formula 350/x+0.5 and with xMin = 100 and xMax = 300. Graph:

350 350 + 0.5 = 4, 𝑞(200) = + 0.5 = 2.25 100 200 So, the change in demand is 2.25 − 4 = −1.75 billion units, which means that demand decreases by about 1.75 billion units sold per year. c. The value of 𝑞 on the graph increases by larger and larger amounts as we move to the left on the graph, indicating that the demand increases at a greater and greater rate (Choice (A)). b. 𝑞(100) =

31. The price at which there is neither a shortage nor surplus is the equilibrium price, which occurs when demand = supply: −3𝑝 + 700 = 2𝑝 − 500 5𝑝 = 1200 𝑝 = $240 per skateboard 32. The price at which there is neither a shortage nor surplus is the equilibrium price, which occurs when demand = supply: −5𝑝 + 50 = 3𝑝 − 30 8𝑝 = 80 𝑝 = $10 per skateboard 33. a. The equilibrium price occurs when demand = supply: −𝑝 + 57 = 4𝑝 − 136 5𝑝 = 193 𝑝 = 38.6, or $38,600 per vehicle. b. Since $38,400 is below the equilibrium price, there would be a shortage at that price. To calculate it, compute demand and supply: Demand: 𝑞 = −38.4 + 57 = 18.6 million vehicles Supply: 𝑞 = 4(38.4) − 136 = 17.6 million vehicles Shortage = Demand − Supply = 18.6 − 17.6 = 1 million vehicles 34. a. The equilibrium price occurs when demand = supply: −0.15𝑝 + 23 = 3.85𝑝 − 131 4𝑝 = 154 𝑝 = 38.5, or $38,500 per vehicle. b. Since $38,600 is above the equilibrium price, there would be a surplus at that price. To calculate it, compute demand and supply: Demand: 𝑞 = −0.15(38.6) + 23 = 17.21 million vehicles

Solutions Section 1.2 Supply: 𝑞 = 3.85(38.6) − 131 = 17.61 million vehicles Surplus = Supply − Demand = 17.61 − 17.21 = 0.4 million vehicles 35. a. For equilibrium, Demand = Supply: 760 − 1 = 0.00304𝑝 − 1 𝑝 760 = 0.0.00304𝑝 𝑝 Cross-multiply: 760 0.00304𝑝 2 = 760 ⇒ 𝑝 2 = = 250,000 0.00304 So 𝑝 = √250,000 = $500. Thus, the equilibrium price is $500, and the equilibrium demand (or supply) is 760∕500 − 1 = 0.52 billion phones b. Graph:

Technology formulas: Demand: y = 760/x-1 Supply: y = 0.00304x-1 The graphs cross at (500, 0.52) confirming the calclation in part (a). c. Since $400 is below the equilibrium price, there would be a shortage at that price. To calculate it, compute demand and supply: 760 Demand: 𝑞 = − 1 = 0.9 billion phones 400 Supply: 0.00304(400) − 1 = 0.216 billion phones Shortage = Demand − Supply ≈ 0.9 − 0.216 = 0.684 billion phones, or 684 million phones. 36. a. For equilibrium, Demand = Supply: 350 + 0.5 = 0.0056𝑝 + 0.5 𝑝 350 = 0.0056𝑝 𝑝 Cross-multiply: 350 0.0056𝑝 2 = 350 ⇒ 𝑝 2 = = 62,500 0.0056 So 𝑝 = √62,500 = $250. Thus, the equilibrium price is $250, and the equilibrium demand (or supply) is 350∕250 + 0.5 = 1.9 billion phones

Solutions Section 1.2 b. Graph:

Technology formulas: Demand: y = 350/x+0.5 Supply: y = 0.0056x+0.5 The graphs cross at (250, 1.9) confirming the calclation in part (a). c. Since $200 is below the equilibrium price, there would be a shortage at that price. To calculate it, compute demand and supply: 350 Demand: 𝑞 = + 0.5 = 2.25 billion phones 200 Supply: 0.0056(200) + 0.5 = 1.62 billion phones Shortage = Demand − Supply ≈ 2.25 − 1.62 = 0.63 billion phones, or 630 million phones. 37. 𝐶(𝑞) = 2,000 + 100𝑞 2 a. 𝐶(10) = 2,000 + 100(10) 2 = 2,000 + 10,000 = $12,000 b. 𝑁 = 𝐶 − 𝑆, so 𝑁(𝑞) = 𝐶(𝑞) − 𝑆(𝑞) = 2,000 + 100𝑞 2 − 500𝑞

This is the cost of removing 𝑞 lb of PCPs per day after the subsidy is taken into account. c. 𝑁(20) = 2,000 + 100(20) 2 − 500(20) = 2,000 + 40,000 − 10,000 = $32,000

38. 𝐶(𝑞) = 1,000 + 100√𝑞 a. 𝐶(100) = 1,000 + 100√100 = 1,000 + 100(10) = $2,000 b. 𝑁 = 𝐶 − 𝑆, so 𝑁(𝑞) = 𝐶(𝑞) − 𝑆(𝑞) = 1,000 + 100√𝑞 − 200𝑞 This is the cost for dental coverage to the company if it has 𝑞 employees, after the subsidy is taken into account. c. 𝑁(100) = 1,000 + 100√100 − 200(100) = 1000 + 100(10) − 20,000 = −$18,000 The company makes $18,000 from the government for dental coverage if it employs 100 people. 39. The technology formulas are: (A) -0.2*t^2+t+16 (B) 0.2*t^2+t+16 (C) t+16 The following table shows the values predicted by the three models:

𝑡

Solutions Section 1.2 0

2

4

6

7

16

18

22

28

30

(A)

16

17.2

16.8

14.8

13.2

(B)

16

18.8

23.2

29.2

32.8

(C)

16

18

20

22

23

𝑆(𝑡)

As shown in the table, the values predicted by model (B) are much closer to the observed values 𝑆(𝑡) than those predicted by the other models. b. Since 1998 corresponds to 𝑡 = 8, 𝑆(𝑡) = 0.2𝑡 2 + 𝑡 + 16 𝑆(8) = 0.2(8) 2 + 8 + 16 = 36.8 So the spending on corrections in 1998 was predicted to be approximately $37 billion. 40. The technology formulas are: (A) 16+2*t (B) 16+t+0.5*t^2 (C) 16+t-0.5*t^2 The following table shows the values predicted by the three models: 𝑡

0

2

4

6

7

16

18

22

28

30

(A)

16

20

24

28

30

(B)

16

20

28

40

47.5

(C)

16

16

12

4

-1.5

𝑆(𝑡)

As shown in the table, the values predicted by model (A) are much closer to the observed values 𝑆(𝑡) than those predicted by the other models. b. Since 1998 corresponds to 𝑡 = 8, 𝑆(𝑡) = 16 + 2𝑡 𝑆(8) = 16 + 2(8) = 32 So the spending on corrections in 1998 was predicted to be $32 billion. 41. The technology formulas are: (A) 0.005*x+20.75 (B) 0.01*x+20+25/x (C) 0.0005*x^2-0.07*x+23.25 (The stars are optional for some technologies like graphing calculators and the Evaluator and Grapher App) Here is result from the Evaluator and Grapher App: Functions box (Cartesian mode):

Evaluator box (Set to 2 decimal places):

Model (C) fits the data perfectly to two decimal places—more closely than any of the other models.

Solutions Section 1.2 b. Graph of model (C):

0.0005*x^2-0.07*x+23.25

The lowest point on the graph occurs at 𝑥 = 70 with a 𝑦-coordinate of 20.8. Thus, the lowest cost per shirt is $20.80, which the team can obtain by buying 70 shirts. 42. The technology formulas are: (A) 0.05*x+20.75 (B) 0.1*x+20+25/x (C) 0.0008*x^2-0.07*x+23.25 (The stars are optional for some technologies like graphing calculators and the Evaluator and Grapher App) Here is result from the Evaluator and Grapher App: Functions box (Cartesian mode):

Evaluator box (Set to 2 decimal places):

Model (B) fits the data perfectly to two decimal places—more closely than any of the other models. b. Graph of model (B):

0.1*x+20+25/x

The lowest point on the graph occurs at 𝑥 = 16 with a y-coordinate of 23.1625. Thus, the lowest cost per hat is $23.16, which the team can obtain by buying 16 hats. 43. Here are the technology formulas as entered in the online Function Evaluator and Grapher at the Web Site: (A) 0.075*0.83^x (B) 0.24/(x+3) (C) 0.00054*x^2-0.012*x+0.075 (D) -0.006*x+0.065 The following graph shows all four curves together with the plotted points (entered as shown in the margin technology note with Example 5).

Solutions Section 1.2

As shown in the graph, the values predicted by models (A) and (C) are much closer to the observed values than those predicted by the other models. b. Since 2030 corresponds to 𝑡 = 20, Model (A): 𝑐(20) = 0.075 * 0.83 20 ≈ $0.00181 Model (C): 𝑐(20) = 0.00054(20 2) − 0.012(20) + 0.075 ≈ $0.0510 So model (A) gives the lower price: approximately $0.0018 per GB. 44. Here are the technology formulas as entered in the online Function Evaluator and Grapher at the Web Site: (A) -0.038+0.8/(x+7) (B) 0.075*1.21^(-x) (C) 0.00034*(x-14.5)^2 (D) -0.022+0.8/(x+9.32) The following graph shows all three curves together with the plotted points (entered as shown in the margin technology note with Example 5).

As shown in the graph, the values predicted by models (A), (B), and (C) are much closer to the observed values than those predicted by (D). b. Since 2030 corresponds to 𝑡 = 20, 0.8 Model (A): 𝑐(20) = −0.038 + ≈ −$0.084 (unreasonable as prices cannot be negative) 20 + 7 Model (B): 𝑐(20) = 0.075(1.21 −20) ≈ $0.0017 (reasonable given the current trends) Model (C): 𝑐(20) = 0.00034(20 − 14.5) 2 ≈ $0.010 (unreasonable that the price would be the same again in 10 years) 45. A plot of the given points gives a straight line (Option (A)). Options (B) and (C) give curves, so (A) is the best choice.

Solutions Section 1.2 46. Plotting the three data points suggests a concave down curve, suggesting that a linear model may not be the best fit. An exponential model 𝑝(𝑡) = 𝐴𝑏 𝑡 would result in a concave up curve, regardless of whether 𝑏 is larger than 1 or less than 1. This leaves a quadratic model as the only possible choice. In fact, a quadratic can always be found that passes through any three points not on the same straight line with different 𝑥-coordinates. Therefore, a quadratic model would give an exact fit. 47. A plot of the given data suggests a concave-down curve that becomes steeper downward as the price 𝑝 increases, suggesting Model (D). Model (A) would predict increasing demand with increasing price, Model (B) would correspond to a descending curve that becomes less steep as 𝑝 increases (a concave up curve), and Model (C) would give a concave-up parabola. 48. Model (B) is the best choice; Model (A) would predict increasing demand with increasing price, Model (D) would correspond to a concave down parabola, and Model (C), would predict demand that, rather than flattening out as the price increases, would begin to climb again. 49. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = $1,000, 𝑟 = 6∕100 = 0.06, 𝑛 = 4, 𝑡 = 4. 0.06 4×4 4 ) = 1,000(1.015) 16 ≈ $1,268.99

𝐴(4) = 1,000(1 +

50. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = $10,000, 𝑟 = 2∕100 = 0.02, 𝑛 = 4, 𝑡 = 5 0.02 4×5 4 ) = 1,000(1.005) 20 ≈ $11,048.96

𝐴(5) = 1,000(1 +

51. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 5,000, 𝑟 = 0.15∕100 = 0.0015, and 𝑛 = 12. We get the model 𝐴(𝑡) = 5,000(1 + 0.0015∕12) 12𝑡 In June 2028 (𝑡 = 7), the deposit would be worth 5,000(1 + 0.0015∕12) 12(7) ≈ $5,053. 52. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 4,000, 𝑟 = 0.0120, and 𝑛 = 365. We get the model 𝐴(𝑡) = 4,000(1 + 0.0120∕365) 365𝑡 In March 2029 (𝑡 = 8), the deposit would be worth 4,000(1 + 0.0120∕365) 365(8) ≈ $4,403.

53. 𝑃 (0) = 200, 𝑃 (1) = 230, 𝑃 (2) = 260, ... and so on. Thus, the population is increasing by 30 per year. 54. 𝐵(0) = 5000, 𝐵(1) = 4800, 𝐵(2) = 4600, ... and so on. Thus, the balance is decreasing by $200 per day.

Solutions Section 1.2 55. Curve fitting. The model is based on fitting a curve to a given set of observed data. 56. Analytical. The model is obtained by analyzing the situation being modeled.

57. The given model is 𝑐(𝑡) = 4 − 0.2𝑡. This tells us that 𝑐 is $4 at time 𝑡 = 0 (January) and is decreasing by $0.20 per month. So, the cost of downloading a movie was $4 in January and is decreasing by 20¢ per month. 58. The given model is 𝑐(𝑡) = 4 − 0.2𝑡 and therefore passes through the points (𝑡, 𝑐) = (0, 4) and (1, 3.8). So, the cost of downloading a movie was $4 in January and $3.80 in February. 59. In a linear cost function, the variable cost is 𝑥 times the marginal cost.

60. In a linear cost function, the marginal cost is the additional (or incremental) cost per item. 61. Yes, as long as the supply is going up at a faster rate, as illustrated by the following graph:

62. There would be a shortage at any given price. Therefore, consumers would be willing to pay more for a scarce commodity and sellers would naturally oblige by charging more, resulting in an upward spiral of prices for the commodity.

63. Extrapolate both models and choose the one that gives the most reasonable predictions.

64. No; as long as 𝑎 is negative, the value of 𝑠(𝑡) for large 𝑡 will be negative, making the model unreasonable for large value of 𝑡.

65. The value of 𝑓 − 𝑔 at 𝑥 is 𝑓(𝑥) − 𝑔(𝑥). Since 𝑓(𝑥) ≥ 𝑔(𝑥) for every 𝑥, it follows that 𝑓(𝑥) − 𝑔(𝑥) ≥ 0 for every 𝑥. 66. The value of

𝑓 𝑓(𝑥) 𝑓(𝑥) at 𝑥 is . Since 𝑓(𝑥) > 𝑔(𝑥) > 0 for every 𝑥, it follows that > 1 for every 𝑥. 𝑔 𝑔(𝑥) 𝑔(𝑥)

𝑓 𝑓(𝑥) 𝑓 at 𝑥 are ratios , it follows that the units of measurement of are units of 𝑓 𝑔 𝑔(𝑥) 𝑔 per unit of 𝑔; that is, books per person. 67. Since the values of

68. Write 𝑓(𝑥) = 𝑚𝑥 + 𝑏 and 𝑔(𝑥) = 𝑛𝑥 + 𝑐. Then 𝑓(𝑥) − 𝑔(𝑥) = 𝑚𝑥 + 𝑏 − (𝑛𝑥 + 𝑐) = (𝑚 − 𝑛)𝑥 + (𝑏 − 𝑐), also a linear function.

Solutions Section 1.3 Section 1.3 1.

𝑥 𝑦

−1 5

0 8

1

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 0 − (−1) = 1 Δ𝑦 = 8 − 5 = 3 This gives a slope of Δ𝑦 3 𝑚= = = 3. Δ𝑥 1 Now look at the second and third points: The change in 𝑥 is again Δ𝑥 = 1 − 0 = 1 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 Δ𝑦 = 3(1) = 3 This means that the missing value of 𝑦 is 8 + Δ𝑦 = 8 + 3 = 11. 2.

𝑥 𝑦

−1

0

−1

−3

2

3

1

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 0 − (−1) = 1 Δ𝑦 = −3 − (−1) = −2 This gives a slope of Δ𝑦 −2 𝑚= = = −2. Δ𝑥 1 Now look at the second and third points: The change in 𝑥 is again Δ𝑥 = 1 − 0 = 1 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 Δ𝑦 = −2(1) = −2 This means that the missing value of 𝑦 is −3 + Δ𝑦 = −3 + (−2) = −5. 3.

𝑥 𝑦

−1

−2

5

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 3 − 2 = 1 Δ𝑦 = −2 − (−1) = −1 This gives a slope of Δ𝑦 −1 𝑚= = = −1. Δ𝑥 1 Now look at the second and third points: The change in 𝑥 is Δ𝑥 = 5 − 3 = 2 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 Δ𝑦 = (−1)(2) = −2 This means that the missing value of 𝑦 is −2 + Δ𝑦 = −2 + (−2) = −4.

4.

𝑥

2

4

𝑦

−1

−2

𝑥

−2

0

5

Solutions Section 1.3

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 4 − 2 = 2 Δ𝑦 = −2 − (−1) = −1 This gives a slope of Δ𝑦 −1 1 𝑚= = =− . Δ𝑥 2 2 Now look at the second and third points: The change in 𝑥 is Δ𝑥 = 5 − 4 = 1 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 1 1 Δ𝑦 = !− "(1) = − 2 2 This means that the missing value of 𝑦 is 1 5 −2 + Δ𝑦 = −2 + !− " = − or −2.5. 2 2 5.

𝑦

4

2

10

We calculate the slope 𝑚 first. The first and third points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 2 − (−2) = 4 Δ𝑦 = 10 − 4 = 6 This gives a slope of Δ𝑦 6 3 𝑚= = = . Δ𝑥 4 2 Now look at the first and second points: The change in 𝑥 is Δ𝑥 = 0 − (−2) = 2 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 3 Δ𝑦 = !− "(2) = 3 2 This means that the missing value of 𝑦 is 4 + Δ𝑦 = 4 + 3 = 7. 6.

𝑥 𝑦

0

−1

3

6

−5

We calculate the slope 𝑚 first. The first and third points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 6 − 0 = 6 Δ𝑦 = −5 − (−1) = −4 This gives a slope of Δ𝑦 −4 2 𝑚= = =− . Δ𝑥 6 3 Now look at the first and second points: The change in 𝑥 is Δ𝑥 = 3 − 0 = 3 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 2 Δ𝑦 = !− "(3) = −2 3

Solutions Section 1.3 This means that the missing value of 𝑦 is −1 + Δ𝑦 = −1 + (−2) = −3 7. From the table, 𝑏 = 𝑓(0) = −2. The slope (using the first two points) is 𝑦 − 𝑦1 −2 − (−1) −1 1 𝑚= 2 = = =− . 𝑥2 − 𝑥1 0 − (−2) 2 2 Thus, the linear equation is 1 𝑥 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = − 𝑥 − 2, or 𝑓(𝑥) = − − 2. 2 2 8. From the table, 𝑏 = 𝑓(0) = 3. The slope (using the first two points) is 𝑦 − 𝑦1 2−1 1 𝑚= 2 = = 𝑥2 − 𝑥1 (−3) − (−6) 3 Thus, the linear equation is 1 𝑥 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = 𝑥 + 3, or 𝑓(𝑥) = + 3. 3 3

9. The slope (using the first two points) is 𝑦 − 𝑦1 −2 − (−1) −1 𝑚= 2 = = = −1. 𝑥2 − 𝑥1 −3 − (−4) 1 To obtain 𝑓(0) = 𝑏, use the formula for 𝑏 : 𝑓(0) = 𝑏 = 𝑦1 − 𝑚𝑥1 = −1 − (−1)(−4) = −5 This gives 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = −𝑥 − 5. 10. The slope (using the first two points) is 𝑦 − 𝑦1 6−4 2 𝑚= 2 = = =2 𝑥2 − 𝑥1 2 − 1 1 To obtain 𝑓(0) = 𝑏, use the formula for 𝑏 : 𝑓(0) = 𝑏 = 𝑦1 − 𝑚𝑥1 = 4 − (2)(1) = 2 This gives 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = 2𝑥 + 2.

Using the point (𝑥1 , 𝑦1 ) = (−4, −1)

Using the point (𝑥1 , 𝑦1 ) = (1, 4)

11. In the table, 𝑥 increases in steps of 1 and 𝑓 increases in steps of 4, showing that 𝑓 is linear with slope Δ𝑦 4 𝑚= = =4 Δ𝑥 1 and intercept 𝑏 = 𝑓(0) = 6 giving 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = 4𝑥 + 6. The function 𝑔 does not increase in equal steps, so 𝑔 is not linear.

12. In the table, 𝑥 increases in steps of 10 and 𝑔 increases in steps of 5, showing that 𝑔 is linear with slope Δ𝑦 5 1 𝑚= = = Δ𝑥 10 2 and intercept 𝑏 = 𝑔(0) = −4 giving 1 𝑔(𝑥) = 𝑚𝑥 + 𝑏 = 𝑥 − 4. 2

Solutions Section 1.3 The function 𝑓 does not increase in equal steps, so 𝑓 is not linear.

13. In the first three points listed in the table, 𝑥 increases in steps of 3, but 𝑓 does not increase in equal steps, whereas 𝑔 increases in steps of 6. Thus, based on the first three points, only 𝑔 could possibly be linear, with slope Δ𝑦 6 𝑚= = =2 Δ𝑥 3 and intercept 𝑏 = 𝑔(0) = −1 giving 𝑔(𝑥) = 𝑚𝑥 + 𝑏 = 2𝑥 − 1. We can now check that the remaining points in the table fit the formula 𝑔(𝑥) = 2𝑥 − 1, showing that 𝑔 is indeed linear.

14. In the first and last pairs of points listed in the table, 𝑥 increases in steps of 3, but 𝑓 does not increase in equal steps, whereas 𝑔 increases in steps of 9. Thus, based on those points, only 𝑔 could possibly be linear, with slope Δ𝑦 9 𝑚= = =3 Δ𝑥 3 and intercept 𝑏 = 𝑔(0) = −1 giving 𝑔(𝑥) = 𝑚𝑥 + 𝑏 = 3𝑥 − 1. We can now check that the remaining points in the table fit the formula 𝑔(𝑥) = 3𝑥 − 1, showing that 𝑔 is indeed linear. 15. Slope = coefficient of 𝑥 = − 16. Slope = coefficient of 𝑥 = 17. Slope = coefficient of 𝑥 =

2 3 1 6

18. Write the equation as 𝑦 = − Slope = coefficient of 𝑥 = −

2 3

3 2

2𝑥 1 + 3 3

19. If we solve for 𝑥 we find that the given equation represents the vertical line 𝑥 = −1∕3, and so its slope is infinite (undefined). 20. 8𝑥 − 2𝑦 = 1. Solving for 𝑦 : 2𝑦 = 8𝑥 − 1 1 𝑦 = 4𝑥 − 2 Slope = coefficient of 𝑥 = 4 21. 3𝑦 + 1 = 0. Solving for 𝑦 : 3𝑦 = −1 1 𝑦=− 3

Slope = coefficient of 𝑥 = 0

Solutions Section 1.3

22. If we solve for 𝑥 we find that the given equation represents the vertical line 𝑥 = −3∕2, and so its slope is infinite (undefined). 23. 4𝑥 + 3𝑦 = 7. Solve for 𝑦 : 3𝑦 = −4𝑥 + 7 4 7 𝑦=− 𝑥+ 3 3 4 Slope = coefficient of 𝑥 = − 3 24. 2𝑦 + 3 = 0. Solve for 𝑦 : 2𝑦 = −3 3 𝑦=− 2 Slope = coefficient of 𝑥 = 0 25. 𝑦 = 2𝑥 − 1 𝑦-intercept = −1, slope = 2

26. 𝑦 = 𝑥 − 3 𝑦-intercept = −3, slope = 1

27. 𝑦-intercept = 2, slope = −

1 28. 𝑦 = − 𝑥 + 3 2

𝑦-intercept = 3, slope = −

29. 𝑦 +

2 3

1 2

1 1 𝑥 = −4. Solve for 𝑦 to obtain 𝑦 = − 𝑥 − 4 4 4

y-intercept = −4, slope = −

1 4

Solutions Section 1.3

1 1 𝑥 = −2. Solve for 𝑦 to obtain 𝑦 = 𝑥 − 2 4 4 1 𝑦-intercept = −2, slope = 4 30. 𝑦 −

31. 7𝑥 − 2𝑦 = 7. Solve for 𝑦 : 7 7 −2𝑦 = −7𝑥 + 7, so 𝑦 = 𝑥 − 2 2 7 7 𝑦-intercept = − = −3.5, slope = = 3.5 2 2

32. 2𝑥 − 3𝑦 = 1. Solve for 𝑦 : 2 1 −3𝑦 = −2𝑥 + 1, so 𝑦 = 𝑥 − 3 3 1 2 𝑦-intercept = − , slope = 3 3

33. 3𝑥 = 8. Solve for 𝑥 to obtain 𝑥 = The graph is a vertical line:

8 . 3

Solutions Section 1.3 7 34. 2𝑥 = −7. Solve for 𝑥 to obtain 𝑥 = − = −3.5. 2 The graph is a vertical line:

35. 6𝑦 = 9. Solve for 𝑦 to obtain 𝑦 =

9 3 = = 1.5 6 2

36. 3𝑦 = 4. Solve for 𝑦 to obtain 𝑦 =

4 3

𝑦-intercept =

3 = 1.5, slope = 0. The graph is a horizontal line: 2

4 𝑦-intercept = , slope = 0. The graph is a horizontal line: 3

37. 2𝑥 = 3𝑦. Solve for 𝑦 to obtain 𝑦 = 𝑦-intercept = 0, slope =

2 3

2 𝑥 3

3 38. 3𝑥 = −2𝑦. Solve for 𝑦 to obtain 𝑦 = − 𝑥 2 3 𝑦-intercept = 0, slope = − 2

39. (0, 0) and (1, 2) 𝑦 − 𝑦1 2−0 𝑚= 2 = =2 𝑥2 − 𝑥1 1 − 0

40. (0, 0) and (−1, 2) 𝑦 − 𝑦1 2−0 𝑚= 2 = = −2 𝑥2 − 𝑥1 −1 − 0

41. (−1, −2) and (0, 0) 𝑦 − 𝑦1 0 − (−2) 𝑚= 2 = =2 𝑥2 − 𝑥1 0 − (−1)

42. (2, 1) and (0, 0) 𝑦 − 𝑦1 0−1 1 𝑚= 2 = = 𝑥2 − 𝑥1 0 − 2 2

Solutions Section 1.3

43. (4, 3) and (5, 1) 𝑦 − 𝑦1 1 − 3 −2 𝑚= 2 = = = −2 𝑥2 − 𝑥1 5 − 4 1

44. (4, 3) and (4, 1) 𝑦 − 𝑦1 1−3 𝑚= 2 Undefined = 𝑥2 − 𝑥1 4 − 4

45. (1, −1) and (1, −2) 𝑦 − 𝑦1 −2 − (−1) 𝑚= 2 Undefined = 𝑥2 − 𝑥1 1−1

46. (−2, 2) and (−1, −1) 𝑦 − 𝑦1 −1 − 2 −3 𝑚= 2 = = = −3 𝑥2 − 𝑥1 −1 − (−2) 1

47. (2, 3.5) and (4, 6.5) 𝑦 − 𝑦1 6.5 − 3.5 3 𝑚= 2 = = = 1.5 𝑥2 − 𝑥1 4−2 2

48. (10, −3.5) and (0, −1.5) 𝑦 − 𝑦1 −1.5 − (−3.5) 2 𝑚= 2 = = = −0.2 𝑥2 − 𝑥1 0 − 10 −10

49. (300, 20.2) and (400, 11.2) 𝑦 − 𝑦1 11.2 − 20.2 −9 𝑚= 2 = = = −0.09 𝑥2 − 𝑥1 400 − 300 100

50. (1, −20.2) and (2, 3.2) 𝑦 − 𝑦1 3.2 − (−20.2) 23.4 𝑚= 2 = = = 23.4 𝑥2 − 𝑥1 2−1 1

1 3 51. (0, 1) and !− , " 2 4 𝑦2 − 𝑦1 3∕4 − 1 −1∕4 2 1 𝑚= = = = = 𝑥2 − 𝑥1 −1∕2 − 0 −1∕2 4 2

1 1 3 52. ! , 1" and !− , " 2 2 4 𝑦2 − 𝑦1 3∕4 − 1 −1∕4 1 𝑚= = = = 𝑥2 − 𝑥1 −1∕2 − 1∕2 4 −1

53. (𝑎, 𝑏) and (𝑐, 𝑑) (𝑎 ≠ 𝑐) 𝑦 − 𝑦1 𝑑−𝑏 𝑚= 2 = 𝑥2 − 𝑥1 𝑐 − 𝑎

54. (𝑎, 𝑏) and (𝑐, 𝑏) (𝑎 ≠ 𝑐) 𝑦 − 𝑦1 𝑏−𝑏 𝑚= 2 = =0 𝑥2 − 𝑥1 𝑐 − 𝑎

55. (𝑎, 𝑏) and (𝑎, 𝑑) (𝑏 ≠ 𝑑) 𝑦 − 𝑦1 𝑑−𝑏 𝑚= 2 \ \ Undefined = 𝑥2 − 𝑥1 𝑎 − 𝑎

56. (𝑎, 𝑏) and (−𝑎, −𝑏) (𝑎 ≠ 0) 𝑦 − 𝑦1 −𝑏 − 𝑏 −2𝑏 𝑏 𝑚= 2 = = = 𝑥2 − 𝑥1 −𝑎 − 𝑎 −2𝑎 𝑎

57. (−𝑎, 𝑏) and (𝑎, −𝑏) (𝑎 ≠ 0) 𝑦 − 𝑦1 𝑏 −𝑏 − 𝑏 −2𝑏 𝑚= 2 = = =− 𝑥2 − 𝑥1 𝑎 − (−𝑎) 2𝑎 𝑎

58. (𝑎, 𝑏) and (𝑏, 𝑎) (𝑎 ≠ 𝑏) 𝑦 − 𝑦1 𝑎−𝑏 𝑚= 2 = = −1 𝑥2 − 𝑥1 𝑏 − 𝑎

59. a. 𝑚 =

Δ𝑦 1 = =1 Δ𝑥 1

Solutions Section 1.3

60. a. 𝑚 =

Δ𝑦 −1 = = −1 Δ𝑥 1

b. 𝑚 =

Δ𝑦 1 = Δ𝑥 2

b. 𝑚 =

Δ𝑦 2 = =2 Δ𝑥 1

c. 𝑚 =

Δ𝑦 0 = =0 Δ𝑥 1

c. 𝑚 =

Δ𝑦 0 = =0 Δ𝑥 1

d. 𝑚 =

Δ𝑦 3 = =3 Δ𝑥 1

d. 𝑚 =

Δ𝑦 1 = =1 Δ𝑥 1

e. 𝑚 =

Δ𝑦 −2 1 = =− Δ𝑥 4 2

e. 𝑚 =

f. 𝑚 =

Δ𝑦 −1 1 = =− Δ𝑥 3 3 Δ𝑦 −1 = = −1 Δ𝑥 1

f. Vertical line; undefined slope

g. 𝑚 =

Δ𝑦 3 = =3 Δ𝑥 1

g. Vertical line; undefined slope

h. 𝑚 =

Δ𝑦 −1 1 = =− Δ𝑥 4 4

h. 𝑚 =

Δ𝑦 −1 1 = =− Δ𝑥 3 3

i. 𝑚 =

Δ𝑦 −2 = = −2 Δ𝑥 1

i. 𝑚 =

Δ𝑦 −2 = = −2 Δ𝑥 1

61. Through (1, 3) with slope 3 Point: (1, 3) Slope: 𝑚 = 3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 3 − 3(1) = 0 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 3𝑥 + 0, or 𝑦 = 3𝑥 Solutions Section 1.3

62. Through (2, 1) with slope 2 Point: (2, 1) Slope: 𝑚 = 2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 1 − 2(2) = −3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 2𝑥 − 3 3 1 63. Through !1, − " with slope 4 4 3 1 Point: !1, − " Slope: 𝑚 = 4 4

Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 1 1 64. Through !0, − " with slope 3 3 1 1 Point: !0, − " Slope: 𝑚 = 3 3

3 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = − − (1) = −1 4 4

1 𝑥−1 4

1 1 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = − − (0) = − 3 3 3 1 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 − 3 3 65. Through (20, −3.5) and increasing at a rate of 10 units of 𝑦 per unit of 𝑥 Δ𝑦 10 Point: (20, −3.5) Slope: 𝑚 = = = 10 Δ𝑥 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −3.5 − (10)(20) = −3.5 − 200 = −203.5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 10𝑥 − 203.5

66. Through (3.5, −10) and increasing at a rate of 1 unit of 𝑦 per 2 units of 𝑥 Δ𝑦 1 Point: (3.5, −10) Slope: 𝑚 = = = 0.5 Δ𝑥 2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −10 − (0.5)(3.5) = −11.75 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 0.5𝑥 − 11.75 67. Through (2, −4) and (1, 1)

𝑦2 − 𝑦1 5 = = −5 𝑥2 − 𝑥1 −1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −4 − (−5)(2) = 6 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −5𝑥 + 6 Point:(2, −4)

Slope: 𝑚 =

68. Through (1, −4) and (−1, −1) 𝑦 − 𝑦1 3 Point: (1, −4) Slope: 𝑚 = 2 = = −1.5 𝑥2 − 𝑥1 −2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −4 − (−1.5)(1) = −2.5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −1.5𝑥 − 2.5

69. Through (1, −0.75) and (0.5, 0.75) 𝑦 − 𝑦1 0.75 − (−0.75) 1.5 Point: (1, −0.75) Slope: 𝑚 = 2 = = = −3 𝑥2 − 𝑥1 0.5 − 1 −0.5 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −0.75 − (−3)(1) = −0.75 + 3 = 2.25 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −3𝑥 + 2.25

70. Through (0.5, −0.75) and (1, −3.75) 𝑦 − 𝑦1 −3.75 − (−0.75) −3 Point: (0.5, −0.75) Slope: 𝑚 = 2 = = = −6 𝑥2 − 𝑥1 1 − 0.5 0.5 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −0.75 − (−6)(0.5) = −0.75 + 3 = 2.25 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −6𝑥 + 2.25 Solutions Section 1.3

71. Through (6, 6) and parallel to the line 𝑥 + 𝑦 = 4 Point: (6, 6) Slope: Same as slope of 𝑥 + 𝑦 = 4. To find the slope, solve for 𝑦, getting 𝑦 = −𝑥 + 4. Thus, 𝑚 = −1. Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 6 − (−1)(6) = 6 + 6 = 12 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −𝑥 + 12 72. Through (1∕3, −1) and parallel to the line 3𝑥 − 4𝑦 = 8 Point: (1∕3, −1)

Slope: Same as slope of 3𝑥 − 4𝑦 = 8. To find the slope, solve for 𝑦, getting 𝑦 = 3 1 1 5 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = (−1) − ! "! " = −1 − = − 4 3 4 4 3 5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 − 4 4

3 3 𝑥 − 2. Thus, 𝑚 = . 4 4

73. Through (0.5, 5) and parallel to the line 4𝑥 − 2𝑦 = 11 Point: (0.5, 5)

Slope: Same as slope of 4𝑥 − 2𝑦 = 11. To find the slope, solve for 𝑦, getting 𝑦 = 2𝑥 − Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 5 − (2)(0.5) = 5 − 1 = 4 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 2𝑥 + 4

74. Through (1/3, 0) and parallel to the line 6𝑥 − 2𝑦 = 11 Point: (1∕3, 0)

Slope: Same as slope of 6𝑥 − 2𝑦 = 11. To find the slope, solve for 𝑦, getting 𝑦 = 3𝑥 − 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (3)! " = −1 3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 3𝑥 − 1 75. Through (0, 0) and (𝑝, 𝑞)

𝑦2 − 𝑦1 𝑞−0 𝑞 = = 𝑥2 − 𝑥1 𝑝 − 0 𝑝 𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (0) = 0 𝑝 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 𝑝 Point: (0, 0)

Slope: 𝑚 =

76. Through (𝑝, 𝑞) parallel to 𝑦 = 𝑟𝑥 + 𝑠 Point: (𝑝, 𝑞) Slope: Since the line has the same slope as 𝑦 = 𝑟𝑥 + 𝑠, 𝑚 = 𝑟 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − 𝑟𝑝 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑟𝑥 + 𝑞 − 𝑟𝑝 or 𝑦 = 𝑟(𝑥 − 𝑝) + 𝑞 77. Through (𝑝, 𝑞) and (𝑟, 𝑞) (𝑝 ≠ 𝑟) 𝑦 − 𝑦1 𝑞−𝑞 Point: (𝑝, 𝑞) Slope: 𝑚 = 2 = =0 𝑥2 − 𝑥1 𝑟 − 𝑝

11 . Thus, 𝑚 = 2. 2

11 . Thus, 𝑚 = 2. 2

Solutions Section 1.3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − (0)𝑝 = 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏; that is, 𝑦 = 𝑞 78. Through (𝑝, 𝑞) and (−𝑝, −𝑞) (𝑝 ≠ 0) 𝑦 − 𝑦1 −𝑞 − 𝑞 −2𝑞 𝑞 Point: (𝑝, 𝑞) Slope: 𝑚 = 2 = = = 𝑥2 − 𝑥1 −𝑝 − 𝑝 −2𝑝 𝑝 𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − 𝑝 = 𝑞 − 𝑞 = 0 𝑝 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 𝑝

79. Through (−𝑝, 𝑞) and (𝑝, −𝑞) (𝑝 ≠ 0) 𝑦 − 𝑦1 𝑞 −𝑞 − 𝑞 −2𝑞 Point: (−𝑝, 𝑞) Slope: 𝑚 = 2 = = =− 𝑥2 − 𝑥1 𝑝 − (−𝑝) 2𝑝 𝑝 𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − !− "(−𝑝) = 𝑞 − 𝑞 = 0 𝑝 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥 𝑝

80. Through (𝑝, 𝑞) and (𝑟, 𝑠) (𝑝 ≠ 𝑟) 𝑦 − 𝑦1 𝑠−𝑞 Point: (𝑝, 𝑞) Slope: 𝑚 = 2 = 𝑥2 − 𝑥1 𝑟 − 𝑝 𝑠−𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − ! "𝑝 𝑟−𝑝 𝑠−𝑞 𝑠−𝑞 𝑠−𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = ! "𝑥 + 𝑞 − ! "𝑝, or 𝑦 = ! "(𝑥 − 𝑝) + 𝑞 𝑟−𝑝 𝑟−𝑝 𝑟−𝑝 81. We are given two points on the graph of the linear cost function: (100, 10,500) and (120, 11,000) (𝑥 is the number of bicycles, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 11,000 − 10,500 500 𝑚= 2 = $25 per bicycle = = 𝑥2 − 𝑥1 120 − 100 20 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 10,500 − (25)(100) = 10,500 − 2,500 = $8,000 82. We are given two points on the graph of the linear cost function: (1,000, 6,000) and (1,500, 8,500) (𝑥 is the number of cases, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 8,500 − 6,000 2,500 𝑚= 2 = $5 per case. = = 𝑥2 − 𝑥1 1,500 − 1,000 500 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 6,000 − (5)(1,000) = $1,000 83. We are given two points on the graph of the linear cost function: (10, 5,920) and (20, 11,820) (𝑥 is the number of iPhones made in an hour, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 11,820 − 5,920 5,900 𝑚= 2 = $590 per iPhone = = 𝑥2 − 𝑥1 20 − 10 10 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 5,920 − (590)(10) = $20 Thus, the cost equation is 𝐶 = 𝑚𝑥 + 𝑏 = 590𝑥 + 20. The cost to manufacture each additional iPhone is the marginal cost: $590.

Solutions Section 1.3 The cost to manufacture 40 iPhones in an hour is obtained by setting 𝑥 = 40 in the cost equation: 𝐶(40) = 590(40) + 20 = $23,620 84. We are given two points on the graph of the linear cost function: (8, 4,110) and (16, 8,190) (𝑥 is the number of consoles made in an hour, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 8,190 − 4,110 4,080 𝑚= 2 = $510 per unit = = 𝑥2 − 𝑥1 16 − 8 8 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 4,110 − (510)(8) = $30 Thus, the cost equation is 𝐶 = 𝑚𝑥 + 𝑏 = 510𝑥 + 30. The cost to manufacture each additional Kinect is the marginal cost: $510. The cost to manufacture 30 Kinects in an hour is obtained by setting 𝑥 = 30 in the cost equation: 85. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏 (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (1, 1,960) and (5, 1,800). Slope: 𝑞 − 𝑞1 1,800 − 1,960 −160 𝑚= 2 = = = −40 𝑝2 − 𝑝1 5−1 4 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 1,960 − (−40)(1) = 1,960 + 40 = 2,000 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −40𝑝 + 2,000 86. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (5, 3,950) and (10, 3,700). Slope: 𝑞 − 𝑞1 3,700 − 3,950 −250 𝑚= 2 = = = −50 𝑝2 − 𝑝1 10 − 5 5 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 3,950 − (−50)5 = 4,200 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −50𝑝 + 4,200

87. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: 2020: (𝑝, 𝑞) = (650, 220) 2024: (𝑝, 𝑞) = (450, 700) 𝑞 − 𝑞1 700 − 220 480 Slope: 𝑚 = 2 = = = −2.4 𝑝2 − 𝑝1 450 − 650 −200 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 220 − (−2.4)650 = 1,780 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −2.4𝑝 + 1,780. If 𝑝 = $400, then 𝑞 = −2.4(400) + 1,780 = 820 million phones. b. Since the slope is −2.4 million phones per unit increase in price, we interpret the slope as follows: For every $1 increase in price, sales of smartphones decrease by 2.4 million units. 88. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: 2013: (335, 1,010) 2017: (265, 1,710)

Solutions Section 1.3 𝑞 − 𝑞1 1,710 − 1,010 700 Slope: 𝑚 = 2 = = = −10 𝑝2 − 𝑝1 265 − 335 −70 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 1,010 − (−10)335 = 4,360 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −10𝑝 + 4,360. If 𝑝 = $385, then 𝑞 = −10(385) + 4,360 = 510 million phones. b. Since the slope is −10 million phones per unit increase in price, we interpret the slope as follows: For every $1 increase in price, sales of smartphones decrease by 10 million units.

89. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (3, 28,000) and (5, 19,000). 𝑞 − 𝑞1 19,000 − 28,000 −9,000 Slope: 𝑚 = 2 = = = −4,500 𝑝2 − 𝑝1 5−3 2 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 28,000 − (−4,500)3 = 28,000 + 13,500 = 41,500 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −4,500𝑝 + 41,500. b. The units of measurement of the slope are generally units of 𝑦 per unit of 𝑥. In this case: Units of 𝑞 per unit of 𝑝. That is, Rides per day per $1 increase in the fare. Since the slope is −4,500 rides/day per $1 increase in the price, we interpret it as saying that ridership decreases by 4,500 rides per day for every $1 increase in the fare. c. From part (a), the demand equation is 𝑞 = −4,500𝑝 + 41,500 If the fare is $6, we have 𝑝 = 6, so 𝑞 = −4500(6) + 41,500 = −27,000 + 41,500 = 14,500 rides/day.

90. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (5, 14) and (3, 18). 𝑞 − 𝑞1 18 − 14 4 Slope: 𝑚 = 2 = = = −2 𝑝2 − 𝑝1 3−5 −2 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 14 − (−2)(5) = 24 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −2𝑝 + 24. b. The units of measurement of the slope are generally units of 𝑦 per unit of 𝑥. In this case: Units of 𝑞 per unit of 𝑝. That is, Millions of rides/day per Z1 increase in the fare. Since the slope is −2 million rides/day per Z1 increase in the price, we interpret it as saying that ridership decreases by 2 million rides per day for every Z1 increase in the fare. c. From part (a), the demand equation is 𝑞 = −2𝑝 + 24 If the fare is Z10, we have 𝑝 = 10, so 𝑞 = −2(10) + 24 = 4 million rides/day.

91. a. In a linear model of 𝑦 versus time 𝑡, the slope is the number of units of 𝑦 per unit time, and we are given this quantity: 40 million pounds/year. Thus, working in millions of pounds, we can take 𝑚 = 40. We are also given the 𝑦-intercept (the value of 𝑦 at 𝑡 = 0) as 𝑏 = 290. Thus, the model is 𝑦 = 40𝑡 + 290 million pounds of pasta b. In 2005, 𝑡 = 15, and so 𝑦(15) = 40(15) + 290 = 890 million pounds. 92. a. In a linear model of 𝑦 versus time 𝑡, the slope is the number of units of 𝑦 per unit time, and we are given this quantity: 60 million kg/year. Thus, working in millions of kilograms, we can take 𝑚 = 60. We are also given the 𝑦-intercept (the value of 𝑦 at 𝑡 = 0) as 𝑏 = 550. Thus, the model is 𝑦 = 60𝑡 + 550 million kg of mercury b. The year 2230 corresponds to 𝑡 = 20, and so 𝑦 = 60(20) + 550 = 1,750 million kg.

Solutions Section 1.3 93. a. The desired linear model has the form 𝑛 = 𝑚𝑡 + 𝑏, where 𝑡 is time in years since 2013. We are given two points on its graph: 2015 data: (2, 90); 2019 data: (6, 270) 𝑛 − 𝑛1 270 − 90 180 Slope: 𝑚 = 2 = = = 45 𝑡2 − 𝑡1 6−2 4 Intercept: 𝑏 = 𝑛1 − 𝑚𝑡1 = 90 − (45)(2) = 0 Thus, the linear model is 𝑛 = 𝑚𝑝 + 𝑏 = 45𝑡. b. The units of measurement of the slope are units of 𝑛 per unit of 𝑡; that is, thousands of (daily) transactions per year. The number of daily Bitcoin transactions increased at a rate of 45,000 per year. c. The year 2021 corresponds to 𝑡 = 8, so 𝑛 = 45(8) = 360 thousand, somewhat higher than the actual number of transactions. 94. a. The desired linear model has the form 𝐸 = 𝑚𝑡 + 𝑏, where 𝑡 is time in years since 2010. We are given two points on its graph: 2016 data: (6, 213); 2021 data: (11, 210) 𝐸 − 𝐸1 210 − 213 −3 Slope: 𝑚 = 2 = = = −0.6 𝑡2 − 𝑡1 11 − 6 5 Intercept: 𝑏 = 𝐸1 − 𝑚𝑡1 = 213 − (−0.6)(6) = 216.6 Thus, the linear model is 𝐸 = 𝑚𝑝 + 𝑏 = −0.6𝑡 + 216.6 b. The units of measurement of the slope are units of 𝐸 per unit of 𝑡; that is, billions of dollars per year. Exxon Mobil's operating expenses decreased at a rate of $0.6 billion per year. c. The year 2018 corresponds to 𝑡 = 8, and so 𝐸 = −0.6(8) + 216.6 = $211.8 billion, which differs quite significantly from the actual operating expenses. 95. 𝑠(𝑡) = 2.5𝑡 + 10 a. Velocity = slope = 2.5 feet/sec. b. After 4 seconds, 𝑡 = 4, so 𝑠(4) = 2.5(4) + 10 = 10 + 10 = 20 Thus the model train has moved 20 feet along the track. c. The train will be 25 feet along the track when 𝑠 = 25. Substituting gives 25 = 2.5𝑡 + 10 Solving for time t gives 2.5𝑡 = 25 − 10 = 15 15 𝑡= = 6 seconds 2.5 96. 𝑠(𝑡) = −1.8𝑡 + 9 a. Velocity = slope = −1.8 feet/sec. b. 𝑠(4) = −1.8(4) + 9 = 1.8 feet from the ground. c. 0 = −1.8𝑡 + 9, giving 𝑡 = 5 seconds

97. a. Take 𝑠 to be displacement from Jones Beach, and 𝑡 to be time in hours. We are given two points: (𝑡, 𝑠) = (10, 0) (𝑡, 𝑠) = (10.1, 13)

𝑠 = 0 for Jones Beach. 6 minutes = 0.1 hours

We are asked for the speed, which equals the magnitude of the slope. 𝑠 − 𝑠1 13 − 0 13 𝑚= 2 = = = 130 𝑡2 − 𝑡1 10.1 − 10 0.1 Units of slope = units of 𝑠 per unit of 𝑡 = miles per hour Thus, the police car was traveling at 130 mph. b. For the displacement from Jones Beach at time 𝑡, we want to express 𝑠 as a linear function of 𝑡; namely, 𝑠 = 𝑚𝑡 + 𝑏. We already know 𝑚 = 130 from part (a). For the intercept, use

Solutions Section 1.3 𝑏 = 𝑠1 − 𝑚𝑡1 = 0 − 130(10) = −1,300 Therefore, the displacement at time 𝑡 is 𝑠 = 𝑚𝑡 + 𝑏 = 130𝑡 − 1,300

98. a. Take 𝑠 to be displacement from Jones Beach, and 𝑡 to be time in hours. We are given two points: (𝑡, 𝑠) = (9.9, 0) and (10.1, 13) We are asked for the speed, which equals the magnitude of the slope. 𝑠 − 𝑠1 13 − 0 𝑚= 2 = = 65 miles per hour 𝑡2 − 𝑡1 10.1 − 9.9 Thus, the perp was traveling at 65 mph. b. 𝑠 = 𝑚𝑡 + 𝑏, where 𝑚 = 65 from part (a), and 𝑏 = 𝑠1 − 𝑚𝑡1 = 0 − 65(9.9) = −643.5 Therefore, 𝑠 = 𝑚𝑡 + 𝑏 = 65𝑡 − 643.5

99. a. The desired linear model has the form 𝐿 = 𝑚𝑛 + 𝑏. We are given two points on its graph: Second edition data: (2, 585); Seventh edition data: (7, 782) 𝐿 − 𝐿1 782 − 585 197 Slope: 𝑚 = 2 = = = 39.4 𝑛2 − 𝑛1 7−2 5 Intercept: 𝑏 = 𝐿1 − 𝑚𝑛1 = 585 − (39.4)(2) = 506.2 Thus, the linear model is 𝐿 = 𝑚𝑛 + 𝑏 = 39.4𝑛 + 506.2 b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Applied Calculus is growing at a rate of 39.4 pages per edition. c. The length 𝐿 will equal 1,500 when 39.4𝑛 + 506.2 = 1,500. Solving for 𝑛 gives 39.4𝑛 = 1,500 − 506.2 = 993.8 993.8 𝑛= ≈ 25.2 39.4 Thus, by the 26th edition, the book will be over 1,500 pages long.

100. a. The desired linear model has the form 𝐿 = 𝑚𝑛 + 𝑏. We are given two points on its graph: Second edition data: (2, 603); Fifth edition data: (5, 690) 𝐿 − 𝐿1 690 − 603 87 Slope: 𝑚 = 2 = = = 29 𝑛2 − 𝑛1 5−2 3 Intercept: 𝑏 = 𝐿1 − 𝑚𝑛1 = 603 − (29)(2) = 545 Thus, the linear model is 𝐿 = 𝑚𝑛 + 𝑏 = 29𝑛 + 545. b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Finite Mathematics is growing at a rate of 29 pages per edition. 1,000 − 545 c. 𝐿 = 29𝑛 + 545 = 1,000 when 𝑛 = ≈ 15.7. Thus, by the 16th edition, the book will be over 29 1,000 pages long. 101. 𝐹 = Fahrenheit temperature, 𝐶 = Celsius temperature, and we want 𝐹 as a linear function of 𝐶. That is, 𝐹 = 𝑚𝐶 + 𝑏. (𝐹 plays the role of 𝑦 and 𝐶 plays the role of 𝑥.) We are given two points: (𝐶, 𝐹 ) = (0, 32) (𝐶, 𝐹 ) = (100, 212)

Slope: 𝑚 =

Freezing point Boiling point

𝐹2 − 𝐹1 212 − 32 180 = = = 1.8 𝐶2 − 𝐶1 100 − 0 100

Solutions Section 1.3 Intercept: 𝑏 = 𝐹1 − 𝑚𝐶1 = 32 − 1.8(0) = 32 Thus, the linear relation is 𝐹 = 𝑚𝐶 + 𝑏 = 1.8𝐶 + 32 When 𝐶 = 30 ∘ 𝐹 = 1.8(30) + 32 = 54 + 32 = 86 ∘ When 𝐶 = 22 ∘ 𝐹 = 1.8(22) + 32 = 39.6 + 32 = 71.6 ∘. Rounding to the nearest degree gives 72 ∘F. When 𝐶 = −10 ∘, 𝐹 = 1.8(−10) + 32 = −18 + 32 = 14 ∘. When 𝐶 = −14 ∘, 𝐹 = 1.8(−14) + 32 = −25.2 + 32 = 6.8 ∘. Rounding to the nearest degree gives 7°F.

102. 𝐹 = Fahrenheit temperature, 𝐶 = Celsius temperature, and we want 𝐹 as a linear function of 𝐶. That is, 𝐹 = 𝑚𝐶 + 𝑏. (𝐹 plays the role of 𝑦 and 𝐶 plays the role of 𝑥.) We are given two points: (𝐹 , 𝐶) = (32, 0) and (212, 100) 𝐶 − 𝐶1 100 − 0 100 5 Slope: 𝑚 = 2 = = = 𝐹2 − 𝐹1 212 − 32 180 9 5 160 Intercept: 𝑏 = 𝐶1 − 𝑚𝐹1 = 0 − (32) = − 9 9 Thus, the linear relation is 5 160 𝐶 = 𝑚𝐹 + 𝑏 = 𝐹 − 9 9 5 160 360 𝐶(104) = (104) − = = 40 ∘ 9 9 9 5 160 225 𝐶(77) = (77) − = = 25 ∘ 9 9 9 5 160 −90 𝐶(14) = (14) − = = −10 ∘ 9 9 9 5 160 −360 𝐶(−40) = (−40) − = = −40 ∘ 9 9 9 103. a. 𝑆 = Southwest Airlines net income (in $ millions), 𝐽 = JetBlue Airways net income (in $ millions), and we want 𝐽 as a linear function of 𝑆. That is, 𝐽 = 𝑚𝑆 + 𝑏 𝐽 plays the role of 𝑦 and 𝑆 plays the role of 𝑥. We are given two points: (𝑆, 𝐽) = (2,000, 1,100) (𝑆, 𝐽) = (−3,000, −1,400)

2017 data 2020 data

𝐽2 − 𝐽1 −1,400 − 1,100 −2,500 = = = 0.5 𝑆2 − 𝑆1 −3,000 − 2,000 −5,000 Intercept: 𝑏 = 𝐽1 − 𝑚𝑆1 = 1,100 − (0.5)(2,000) = 100 Thus, the linear relation is 𝐽 = 𝑚𝑆 + 𝑏 = 0.5𝑆 + 100. b. In 2019, Southwest Airlines' net income was 𝑆 = 2,300, so 𝐽 = 0.5(2,300) + 100 = 1,250, predicting a $1,250 million net income for JetBlue, $650 million higher than the actual $600 million net income JetBlue earned in 2019. c. The units of measurement of the slope are units of 𝐽 per unit of 𝑆; that is, millions of dollars of JetBlue Airways net income per million dollars of Southwest Airlines net income; JetBlue Airways earned an additional net income of $0.50 per $1 additional net income earned by Southwest Airlines. Slope: 𝑚 =

Solutions Section 1.3 104. a. 𝐴 = Alaska Air Group net income (in $ millions), 𝐽 = JetBlue Airways net income (in $ millions), and we want 𝐽 as a linear function of 𝐴. That is, 𝐽 = 𝑚𝐴 + 𝑏 𝐽 plays the role of 𝑦 and 𝐴 plays the role of 𝑥. We are given two points: (𝐴, 𝐽) = (800, 600) (𝐴, 𝐽) = (−1,200, −1,400)

2016 data 2020 data

𝐽2 − 𝐽1 −1,400 − 600 −2,000 = = =1 𝐴2 − 𝐴1 −1,200 − 800 −2,000 Intercept: 𝑏 = 𝐽1 − 𝑚𝐴1 = 600 − (1)(800) = −200 Thus, the linear relation is 𝐽 = 𝑚𝐴 + 𝑏 = 𝐴 − 200. b. 2017: 𝐴 = 1,000; 𝐽 = 1,000 − 200 = $800 million 2018: 𝐴 = 400; 𝐽 = 400 − 200 = $200 million 2019: 𝐴 = 800; 𝐽 = 800 − 200 = $600 million comparing these values with those in the table shows that the model gives a perfect prediction for every year except 2017 ($300 million lower than the actual JetBlue income). c. The units of measurement of the slope are units of 𝐽 per unit of 𝐴; that is, millions of dollars of JetBlue Airways net income per million dollars of Alaska Air Group net income; JetBlue Airways earned an additional net income of $1 per $1 additional net income earned by Alaska Air Group. Slope: 𝑚 =

105. Income = royalties + screen rights

𝐼 = 5\% of net profits + 50,000 𝐼 = 0.05𝑁 + 50,000 Equation notation 𝐼(𝑁) = 0.05𝑁 + 50,000 Function notation

For an income of $100,000,

100,00 = 0.05𝑁 + 50,000 0.05𝑁 = 50,000 50,000 𝑁= = 1,000,000 0.05

Her marginal income is her increase in income per $1 increase in net profit. This is the slope, 𝑚 = 0.05 dollars of income per dollar of net profit, or 5\hbox{\rm\rlap/c} per dollar of net profit. 106. 𝐼 = 𝑚𝑁 + 𝑏 𝑏 = 100,000, 𝑚 = 0.08, and so 𝐼 = 0.08𝑁 + 100,000 For an income of $1,000,000, 1,000,000 = 0.08𝑁 + 100,000 0.08𝑁 = 900,000 𝑁 = 11,250,000 Marginal income is 𝑚 = 8 hbox rlap ∕c per dollar of net profit.

107. The year 2000 corresponds to 𝑡 = 10, which is in the range 6 ≤ 𝑡 < 15, so we use the first equation: 𝑣(𝑡) = 400𝑡 − 2,200. The slope is 400 MHz/year, telling us that the speed of a processor was increasing by 400 MHz/year. 108. The year 2000 corresponds to 𝑡 = 25, which is in the range 20 ≤ 𝑡 ≤ 30, so we use the second equation: 𝑣(𝑡) = 174𝑡 − 3,420. The slope is 174 MHz/year, telling us that the speed of a processor was increasing by 174 MHz/year.

Solutions Section 1.3 109. a. The data are 𝑡

𝑦

0

20

40

78

2,100 2,950

1970–1990 (first two data points): 𝑦 − 𝑦1 700 − 78 Slope: 𝑚 = 2 = = 31.1 𝑡2 − 𝑡1 20 − 0 Intercept: 𝑏 = 78, specified in first data point Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 31.1𝑡 + 78. b. 1990–2010 (second and third data points): 𝑦 − 𝑦1 2,950 − 700 Slope: 𝑚 = 2 = = 112.5 𝑡2 − 𝑡1 40 − 20 Intercept: 𝑏 = 𝑦1 − 𝑚𝑡1 = 700 − (112.5)20 = −1,550 Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 112.5𝑡 − 1,550. c. Since the first model is valid for 0 ≤ 𝑡 ≤ 20 and the second one for 20 ≤ 𝑡 ≤ 40, we put them together as 31.1𝑡 + 78 if 0 ≤ 𝑡 < 20 𝑦= {112.5𝑡 − 1,550 if 20 ≤ 𝑡 ≤ 40. Notice that, since both formulas agree at 𝑡 = 20, we can also say 31.1𝑡 + 78 if 0 ≤ 𝑡 ≤ 20 𝑦= {112.5𝑡 − 1,550 if 20 < 𝑡 ≤ 40. d. Since 2004 is represented by 𝑡 = 34, we use the second formula to obtain 𝑦 = 112.5(34) − 1,550 = 2,275, or $2,275,000, in good agreement with the actual value shown in the graph. 110. a. The data are 𝑡

0

20

40

𝑦 222 2,100 5,600

1980–2000 (first two data points): 𝑦 − 𝑦1 2,100 − 222 Slope: 𝑚 = 2 = = 93.9 𝑡2 − 𝑡1 20 − 0 Intercept: 𝑏 = 222, specified in first data point Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 93.9𝑡 + 222. b. 2000–2020 (second and third data points): 𝑦 − 𝑦1 5,600 − 2,100 Slope: 𝑚 = 2 = = 175 𝑡2 − 𝑡1 40 − 20 Intercept: 𝑏 = 𝑦1 − 𝑚𝑡1 = 2,100 − (175)20 = −1,400 Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 175𝑡 − 1,400. c. Since the first model is valid for 0 ≤ 𝑡 ≤ 20 and the second one for 20 ≤ 𝑡 ≤ 40, we put them together as 93.9𝑡 + 222 if 0 ≤ 𝑡 < 20 𝑦= {175𝑡 − 1,400 if 20 ≤ 𝑡 ≤ 40. Notice that, since both formulas agree at 𝑡 = 20, we can also say

93.9𝑡 + 222 if 0 ≤ 𝑡 ≤ 20 {175𝑡 − 1,400 if 20 < 𝑡 ≤ 40. d. Since 1992 is represented by 𝑡 = 12, we use the first formula to obtain 𝑦 = 93.9(12) + 222 = 1,348.8, or $1,348,800, considerably higher than the actual value shown on the graph. The actual cost is not linear in the range 1980–2000. Solutions Section 1.3

𝑦=

111. 2006–2012: Points: 2006 data: (𝑡, 𝑆) = (0, 531) 2012 data: (𝑡, 𝑆) = (6, 547) 𝑆 − 𝑆1 547 − 531 8 Slope: 𝑚 = 2 = = 𝑡2 − 𝑡1 6−0 3 Intercept: We are given the 𝑆-intercept as 531. 8 Thus, the linear model is 𝑆 = 𝑚𝑡 + 𝑏 = 𝑡 + 531. 3 2012–2018: Points: 2000 data: (𝑡, 𝑆) = (6, 547) 2004 data: (𝑡, 𝑆) = (12, 529) 𝑆 − 𝑆1 529 − 547 Slope: 𝑚 = 2 = = −3 𝑡2 − 𝑡1 12 − 6 Intercept: 𝑏 = 𝑆1 − 𝑚𝑡1 = 547 − (−3)6 = 565 Thus, the linear model is 𝑆 = 𝑚𝑡 + 𝑏 = −3𝑡 + 565 Putting them together gives 𝑆=

8 𝑡 + 531 3

{−3𝑡 + 565

if 0 ≤ 𝑡 < 6

if 6 ≤ 𝑡 ≤ 12. The score in 2009 is 𝑆(3), so we use the first formula to obtain 8 𝑆(3) = (3) + 531 = 539, in exact agreement with the actual score. 3 112. 2001–2004: Points: 2001 data: (𝑡, 𝑃 ) = (0, 9.7) 2004 data: (𝑡, 𝑃 ) = (3, 4.3) 𝑃 − 𝑃1 4.3 − 9.7 Slope: 𝑚 = 2 = = −1.8 𝑡2 − 𝑡1 3−0 Intercept: 𝑏 = 9.7, specified in first data point Thus, the linear model is 𝑁 = 𝑚𝑡 + 𝑏 = −1.8𝑡 + 9.7. 2004–2007: Points: 2004 data: (𝑡, 𝑃 ) = (3, 4.3) 2007 data: (𝑡, 𝑃 ) = (6, 10.3) 𝑃 − 𝑃1 10.3 − 4.3 Slope: 𝑚 = 2 = =2 𝑡2 − 𝑡1 6−3 Intercept: 𝑏 = 𝑃1 − 𝑚𝑡1 = 4.3 − (2)(3) = −1.7 Thus, the linear model is 𝑁 = 𝑚𝑡 + 𝑏 = 2𝑡 − 1.7. Putting them together gives −1.8𝑡 + 9.7 if 0 ≤ 𝑡 ≤ 3 𝑃= {2𝑡 − 1.7 if 3 < 𝑡 ≤ 6. The percentage of delinquent borrowers in 2006 is 𝑃 (5), so we use the second formula to obtain 𝑃 (5) = 2(5) − 1.7 = 8.3%.

Solutions Section 1.3 113. Compute the corresponding successive changes Δ𝑥 in 𝑥 and Δ𝑦 in 𝑦, and compute the ratios Δ𝑦∕Δ𝑥. If the answer is always the same number, then the values in the table come from a linear function. 114. The desired equation has the form 𝑦 = 𝑚𝑥 + 𝑏. The slope 𝑚 is given by 𝑚 = Δ𝑦∕Δ𝑥, where Δ𝑥 and Δ𝑦 are corresponding changes in 𝑥 and 𝑦. The intercept 𝑏 is given by the 𝑦-value corresponding to 𝑥 = 0, if supplied. If it is not supplied, choose any point (𝑥1 , 𝑦1 ) and use the formula 𝑏 = 𝑦1 − 𝑚𝑥1 . 115. To find the linear function, solve the equation 𝑎𝑥 + 𝑏𝑦 = 𝑐 for 𝑦 : 𝑏𝑦 = −𝑎𝑥 + 𝑐 𝑎 𝑐 𝑦 =− 𝑥+ 𝑏 𝑏

𝑎 𝑐 Thus, the desired function is 𝑓(𝑥) = − 𝑥 + . 𝑏 𝑏 𝑎 𝑐 If 𝑏 = 0, then and are undefined, and 𝑦 cannot be specified as a function of 𝑥. (The graph of the 𝑏 𝑏 resulting equation would be a vertical line.)

116. The slope of the line with equation 𝑦 = 𝑚𝑥 + 𝑏 is the number of units that 𝑦 increases per unit increase in 𝑥 . 117. The slope of the line is 𝑚 = as fast as 𝑥, then its slope is 3 .

Δ𝑦 3 = = 3. Therefore, if, in a straight line, 𝑦 is increasing three times Δ𝑥 1

118. The slope is 𝑚 = −4∕3 units of 𝑦 per unit of 𝑥. We do not have enough information to compute the intercept. 119. If 𝑚 is positive, then 𝑦 will increase as 𝑥 increases; if 𝑚 is negative, then 𝑦 will decrease as 𝑥 increases; if 𝑚 is zero, then 𝑦 will not change as 𝑥 changes. 120. Since Δ𝑦 = −Δ𝑥, the function is linear with slope Δ𝑦 −Δ𝑥 𝑚= = = −1. Δ𝑥 Δ𝑥 121. The slope computed in cell C2 is given by 𝑦 − 𝑦1 −1 − 2 𝑚= 2 = = −1.5 𝑥2 − 𝑥1 3−1 If we increase the 𝑦-coordinate in cell B3, this increases 𝑦2 , and thus increases the numerator Δ𝑦 = 𝑦2 − 𝑦1 without affecting the denominator Δ𝑥. Thus the slope will increase. 122. The slope increases: Δ𝑦 is negative, and stays the same, while Δ𝑥 becomes a larger positive number, so the (negative) slope decreases in absolute value, meaning that it actually increases. 123. The units of the slope 𝑚 are units of 𝑦 (bootlags) per unit of 𝑥 (zonars). The intercept 𝑏 is on the

Solutions Section 1.3 𝑦-axis, and is thus measured in units of 𝑦 (bootlags). Thus, 𝑚 is measured in bootlags per zonar and 𝑏 is measured in bootlags. 124. Units of slope = units of 𝑦 per unit of 𝑥 = miles per dollar. Thus, the independent variable is measured in dollars and the dependent variable is measured in miles.

125. If a quantity changes linearly with time, it must change by the same amount for every unit change in time. Thus, since it increases by 10 units in the first day, it must increase by 10 units each day, including the third. 126. Since 𝑄(0) is positive, 𝑏 is positive. Since 𝑄 decreases with increasing 𝑇 , 𝑚 is negative.

127. Since the slope is 0.1, the velocity is increasing at a rate of 0.1 m/sec every second. Since the velocity is increasing, the object is accelerating (choice B). 128. Velocity = slope = 0.2 units of position per unit time. Thus, the object is moving with constant speed (choice A). 129. Write 𝑓(𝑥) = 𝑚𝑥 + 𝑏 and 𝑔(𝑥) = 𝑛𝑥 + 𝑐. Then 𝑓(𝑥) + 𝑔(𝑥) = 𝑚𝑥 + 𝑏 + (𝑛𝑥 + 𝑐) = (𝑚 + 𝑛)𝑥 + (𝑏 + 𝑐), also a linear function with slope 𝑚 + 𝑛.

130. Not necessarily; for instance, if 𝑓(𝑥) = 2𝑥 + 1 and 𝑔(𝑥) = 𝑥, then their ratio has values 2𝑥 + 1 1 = 2 + , not a linear function of 𝑥. (Only if 𝑔 is a nonzero constant will the ratio will be linear.) 𝑥 𝑥 131. Answers may vary. For example, 𝑓(𝑥) = 𝑥 1∕3, 𝑔(𝑥) = 𝑥 2∕3 gives 𝑓(𝑥)𝑔(𝑥) = 𝑥 1∕3𝑥 2∕3 = 𝑥.

132. Answers may vary. For example, 𝑓(𝑥) = 𝑥 3 + 2𝑥 2, 𝑔(𝑥) = 𝑥 2 gives 𝑓(𝑥) 𝑥 3 + 2𝑥 2 = = 𝑥 + 2 (with domain all real numbers other than 0). 𝑔(𝑥) 𝑥2 133. Increasing the number of items from the break-even number results in a profit: Because the slope of the revenue graph is larger than the slope of the cost graph, it is higher than the cost graph to the right of the point of intersection, and hence corresponds to a profit.

134. You should solve your equation for 𝑞 to obtain 𝑞 as a function of 𝑝. Simply switching 𝑝 and 𝑞 will result in the wrong equation, and starting from scratch would be a less efficient way of obtaining the result you want.

Solutions Section 1.4 Section 1.4

1. (1, 1), (2, 2), (3, 4); 𝑦 = 𝑥 − 1

𝒚̂ =𝒙 − 𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

1

1

0

1

1

2

2

1

1

1

3

4

2

2

4

SSE = Sum of squares of residuals (last column) = 1 + 1 + 4 = 6 2. (0, 1), (1, 1), (2, 2); 𝑦 = 𝑥 + 1

𝒚̂ =𝒙 + 𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

0

1

1

1

1

2

2

2

3

0

0

−1

1

−1

1

SSE = Sum of squares of residuals (last column) = 0 + 1 + 1 = 2 3. (0, −1), (1, 3), (4, 6), (5, 0); 𝑦 = −𝑥 + 2 𝒙 0

𝒚

−1

1

3

4

6

5

0

𝒚̂ = − 𝒙 + 𝟐 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐 2

−3

9

1

2

4

−3

8

64

3

9

−2

SSE = Sum of squares of residuals (last column) = 9 + 4 + 64 + 9 = 86 4. (2, 4), (6, 8), (8, 12), (10, 0); 𝑦 = 2𝑥 − 8 𝒙

𝒚

𝒚̂ =𝟐𝒙 − 𝟖 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

2

4

8

64

6

8

4

4

16

8

12

8

4

16

10

0

12

−4

−12

144

SSE = Sum of squares of residuals (last column) = 240 5. (1, 1), (2, 2), (3, 4) a. 𝑦 = 1.5𝑥 − 1

Solutions Section 1.4

𝒚̂ =𝟏.𝟓𝒙 − 𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

1

1

0.5

0.5

0.25

2

2

2

0

0

3

4

3.5

0.5

0.25

SSE = Sum of squares of residuals = 0.5 b. 𝑦 = 2𝑥 − 1.5

𝒚̂ =𝟐𝒙 − 𝟏.𝟓 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

1

1

0.5

2

2

2.5

3

4

4.5

0.5

0.25

−0.5

0.25

−0.5

0.25

SSE = Sum of squares of residuals = 0.75 The model that gives the better fit is (a) because it gives the smaller value of SSE. 6. (0, 1), (1, 1), (2, 2) a. 𝑦 = 0.4𝑥 + 1.1

𝒚̂ =𝟎.𝟒𝒙 + 𝟏.𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

0

1

1.1

1

1

1.5

2

2

1.9

−0.1 −0.5

0.01

0.1

0.01

0.25

SSE = Sum of squares of residuals = 0.27 b. 𝑦 = 0.5𝑥 + 0.9

𝒚̂ =𝟎.𝟓𝒙 + 𝟎.𝟗 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

0

1

0.9

1

1

1.4

2

2

1.9

0.1

−0.4

0.01

0.1

0.01

0.16

SSE = Sum of squares of residuals = 0.18 The model that gives the better fit is (b) because it gives the smaller value of SSE. 7. (0, −1), (1, 3), (4, 6), (5, 0) a. 𝑦 = 0.3𝑥 + 1.1 𝒙 0

𝒚

−1

𝒚̂ =𝟎.𝟑𝒙 + 𝟏.𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐 1.1

−2.1

4.41

1

3

1.4

1.6

2.56

4

6

2.3

3.7

13.69

5

0

2.6

−2.6

6.76

SSE = Sum of squares of residuals = 27.42

b. 𝑦 = 0.4𝑥 + 0.9 𝒙

𝒚

−1

0

Solutions Section 1.4

𝒚̂ =𝟎.𝟒𝒙 + 𝟎.𝟗 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐 −1.9

0.9

3.61

1

3

1.3

1.7

2.89

4

6

2.5

3.5

12.25

5

0

2.9

−2.9

8.41

SSE = Sum of squares of residuals = 27.16 The model that gives the better fit is (b) because it gives the smaller value of SSE. 8. (2, 4), (6, 8), (8, 12), (10, 0) a. 𝑦 = −0.1𝑥 + 7

𝒚̂ = − 𝟎.𝟏𝒙 + 𝟕 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

2

4

6.8

6

8

6.4

1.6

2.56

8

12

6.2

5.8

33.64

10

0

6

−2.8 −6

7.84

36

SSE = Sum of squares of residuals = 80.04 b. 𝑦 = −0.2𝑥 + 6

𝒚̂ = − 𝟎.𝟐𝒙 + 𝟔 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

2

4

5.6

6

8

4.8

3.2

10.24

8

12

4.4

7.6

57.76

10

0

4

−1.6 −4

2.56

16

SSE = Sum of squares of residuals = 86.56 The model that gives the better fit is (a) because it gives the smaller value of SSE. 9. Data points (𝑥, 𝑦) : (1, 1), (2, 2), (3, 4) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

2

2

4

4

1 3 6

1 4 7

1

12 17

1 9

14

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(17) − (6)(7) 9 𝑚= = = = 1.5 6 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(14) − 6 2

Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 7 − 1.5(6) −2 ≈ −0.6667 = = 𝑛 3 3

Solutions Section 1.4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ 1.5𝑥 − 0.6667. Graph:

6 5 4 3 2 1

y

x -1

0

1

2

𝒙

𝒚

𝒙𝒚

𝒙𝟐

1

1

1

1

3

10. Data points (𝑥, 𝑦) : (0, 1), (1, 1), (2, 2) 0 2 3

1 2 4

0

4

0

4

4

5

5

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(5) − (3)(4) 3 𝑚= = = = 0.5 6 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(5) − 3 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 4 − 0.5(3) 2.5 Intercept: 𝑏 = ≈ 0.8333 = = 𝑛 3 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ 0.5𝑥 + 0.8333. Graph:

2.5

y

2 1.5 1 0.5 -1

0

1

2

x

3

Solutions Section 1.4 11. Data points (𝑥, 𝑦) : (0, −1), (1, 3), (3, 6), (4, 1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

3

3

1

0

−1

3

6

18

9

25

1 4

1

8

0 4

0 9

16 26

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(25) − (8)(9) 28 𝑚= = = = 0.7 40 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(26) − 8 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 9 − 0.7(8) 3.4 Intercept: 𝑏 = = = = 0.85 𝑛 4 4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = 0.7𝑥 + 0.85. Graph:

6 5 4 3 2 1

y

x -1

0

1

2

𝒙

𝒚

𝒙𝒚

𝒙𝟐

4

8

32

16

0

0

100

3

4

5

12. Data points (𝑥, 𝑦) : (2, 4), (4, 8), (8, 12), (10, 0) 2

4

8

12

24

24

10

8

96 136

4

64

184

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(136) − (24)(24) −32 𝑚= = = = −0.2 160 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(184) − 24 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 24 − (−0.2)(24) 28.8 Intercept: 𝑏 = = = = 7.2 𝑛 4 4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −0.2𝑥 + 7.2.

Solutions Section 1.4 Graph:

12 10 8 6 4 2

y

0

2

4

6

𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

1

3

3

1

9

2

4

8

4

16

5

6

30

25

36

8

13

41

30

61

13. a. (1, 3), (2, 4), (5, 6)

8

10

x

12

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(41) − (8)(13) 𝑟= = 2 2 2 2 √3(30) − 8 2√3(61) − 13 2 𝑛(∑ 𝑥 ) − (∑ 𝑥) ⋅ 𝑛(∑ 𝑦 ) − (∑ 𝑦) √ √ 19 ≈ ≈ 0.9959 19.078784

b. (0, −1), (2, 1), (3, 4) 𝒙 0

𝒚

−1

𝒙𝒚

𝒙𝟐

𝒚𝟐

0

0

1

2

1

2

4

1

3

4

12

9

16

5

4

14

13

18

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(14) − (5)(4) 𝑟= = 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅ 𝑛(∑ 𝑦 2) − (∑ 𝑦) 2 √3(13) − 5 2√3(18) − 4 2 √ √ 22 ≈ ≈ 0.9538 23.0651252

c. (4, −3), (5, 5), (0, 0) 𝒙

𝒚

𝒙𝒚

Solutions Section 1.4 𝒙𝟐

𝒚𝟐

16

9

−3

−12

5

5

25

25

25

0

0

0

0

0

9

2

13

41

34

4

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(13) − (9)(2) 𝑟= = 2 2 2 2 √3(41) − 9 2√3(34) − 2 2 𝑛(∑ 𝑥 ) − (∑ 𝑥) ⋅ 𝑛(∑ 𝑦 ) − (∑ 𝑦) √ √ 21 ≈ ≈ 0.3273 64.1560597

The value of 𝑟 in part (a) has the largest absolute value. Therefore, the regression line for the data in part (a) is the best fit. The value of 𝑟 in part (c) has the smallest absolute value. Therefore, the regression line for the data in part (c) is the worst fit. Since 𝑟 is not ±1 for any of these lines, none of them is a perfect fit. 14. a. (1, 3), (−2, 9), (2, 1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

1

3

3

9

-2

9

−18

1 4

81

2

1

2

4

1

1

13

9

91

−13

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(−13) − (1)(13) 𝑟= = 2 2 2 2 √3(9) − 1 2√3(91) − 13 2 𝑛(∑ 𝑥 ) − (∑ 𝑥) ⋅ 𝑛(∑ 𝑦 ) − (∑ 𝑦) √ √ −52 = = −1 (Best, perfect fit) 52

b. (0, 1), (1, 0), (2, 1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

0

1

0

0

1

1

0

0

1

0

2

1

2

4

1

3

2

2

5

2

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(2) − (3)(2) 𝑟= = = 0 (Worst) 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅ 𝑛(∑ 𝑦 2) − (∑ 𝑦) 2 √3(5) − 3 2√3(2) − 2 2 √ √ Solutions Section 1.4

c. (0, 0), (5, −5), (2, −2.1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

0

0

0

0

0

25

25

4

4.41

29

29.41

−5

5

−25

−2.1 −4.2

2

−7.1 −29.2

7

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(−29.2) − (7)(−7.1) 𝑟= = 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅ 𝑛(∑ 𝑦 2) − (∑ 𝑦) 2 √3(29) − 7 2√3(29.41) − (−7.1) 2 √ √ −37.9 ≈ ≈ −0.9997 37.9098932

15. Data points (𝑥, 𝑦) : (0, 3.9), (4, 4.8), (8, 5.7) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

4.8

19.2

16

14.4

64.8

80

0

3.9

8

5.7

4 12

0

45.6

0

64

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(64.8) − (12)(14.4) 21.6 𝑚= ≈ 0.23 = = 96 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(80) − 12 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 14.4 − 0.225(12) 11.7 Intercept: 𝑏 = = = = 3.9 𝑛 3 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ 0.23𝑥 + 3.9. Since 2018 corresponds to 𝑥 = 6, the prediction for 2018 is 𝑦 = 0.23(6) + 3.9 ≈ 5.3 million subscribers.

Solutions Section 1.4 16. Data points (𝑥, 𝑦) : (0, 35), (10, 33), (20, 32) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

33

330

100

100

970

500

0

35

20

32

10 30

0

640

0

400

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(970) − (30)(100) −90 𝑚= = = = −0.15 600 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(500) − 30 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 100 − (−0.15)(30) 104.5 Intercept: 𝑏 = ≈ 34.83 = = 𝑛 3 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ −0.15𝑥 + 34.83. Since 2022 corresponds to 𝑥 = 22, the prediction for 2022 is 𝑦 = −0.15(22) + 34.83. ≈ 31.5 million subscribers.

17. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). Data points (𝑝, 𝑞) : (6, 1.2), (5, 1.4), (4.5, 1.5) 𝒑

𝒒

𝒑𝒒

𝒑𝟐

6

1.2

7.2

4.5

1.5

6.75 20.25

5

15.5

1.4 4.1

7

36 25

20.95 81.25

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑝𝑞) − (∑ 𝑝)(∑ 𝑞) 3(20.95) − (15.5)(4.1) −0.7 𝑚= = = = −0.2 3.5 𝑛(∑ 𝑝 2) − (∑ 𝑝) 2 3(81.25) − 15.5 2 ∑ 𝑞 − 𝑚(∑ 𝑝) 4.1 − (−0.2)(15.5) 7.2 Intercept: 𝑏 = = = = 2.4 𝑛 3 3

Thus, the regression line is 𝑞 = 𝑚𝑝 + 𝑏 = −0.2𝑝 + 2.4. When the selling price is $550, 𝑝 = 5.5, and so 𝑞 ≈ −0.2(5.5) + 2.4 = 1.3 billion smartphones.

Solutions Section 1.4 18. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). Data points (𝑝, 𝑞) : (4, 0.7), (3, 1), (2.5, 1.5) 𝒑

𝒒

𝒑𝒒

𝒑𝟐

4

0.7

2.8

16

2.5

1.5

3.75

6.25

3

9.5

1

3.2

3

9

9.55 31.25

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑝𝑞) − (∑ 𝑝)(∑ 𝑞) 3(9.55) − (9.5)(3.2) −1.75 𝑚= = = = −0.5 3.5 𝑛(∑ 𝑝 2) − (∑ 𝑝) 2 3(31.25) − 9.5 2 ∑ 𝑞 − 𝑚(∑ 𝑝) 3.2 − (−0.5)(9.5) 7.95 Intercept: 𝑏 = ≈ 2.7 = = 𝑛 3 3

Thus, the regression line is 𝑞 = 𝑚𝑝 + 𝑏 ≈ −0.5𝑝 + 2.7. When the selling price is $350, 𝑝 = 3.5, and so 𝑞 ≈ −0.5(3.5) + 2.7 = 0.95 billion, or 950 million smartphones. 19. Following is the table we use to compute the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

20

3

60

400

40

6

240

1,600

80

9

720

6,400

100

15

1,500 10,000

240

33

2,520 18,400

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(2,520) − (240)(33) 2,160 Slope: 𝑚 = = = = 0.135 16,000 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(18,400) − 240 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 33 − (0.135)(240) 0.6 = = = 0.15 𝑛 4 4

The regression model is therefore 𝑦 = 𝑚𝑥 + 𝑏 = 0.135𝑥 + 0.15. 𝑦(50) = 0.135(50) + 0.15 = 6.9 million jobs

Solutions Section 1.4 20. Following is the table we use to compute the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

10

200

2,000

100

40

900

36,000

1,600

50

1,000 50,000

2,500

80

2,000 160,000 6,400

180 4,100 248,000 10,600 (The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(248,000) − (180)(4,100) 254,000 Slope: 𝑚 = = = = 25.4 10,000 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(10,600) − 180 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 4100 − (25.4)(180) −472 = = = −118 𝑛 4 4

The regression model is therefore 𝑦 = 𝑚𝑥 + 𝑏 = 25.4𝑥 − 118. 𝑦(70) = 25.4(70) − 118 = $1,660 billion

21. a. Data points (𝑅, 𝑃 ) : (190, 4), (240, 12), (300, 12), (420, 27) 𝑹

𝑷

𝑹𝑷

240

12

2,880

420

27

11,340 176,400

190 300 1,150

4

12 55

𝑹𝟐

760

36,100

3,600

90,000

57,600

18,580 360,100

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑅𝑃 ) − (∑ 𝑅)(∑ 𝑃 ) 4(18,580) − (1,150)(55) 11,070 𝑚= ≈ 0.094 = = 2 2 2 117,900 𝑛(∑ 𝑅 ) − (∑ 𝑅) 4(360,100) − 1,150 ∑ 𝑃 − 𝑚(∑ 𝑅) 55 − 0.09389(1,150) −52.9771 Intercept: 𝑏 = ≈ ≈ −13.244 = 𝑛 4 4

Thus, the regression line is 𝑃 = 𝑚𝑅 + 𝑏 ≈ 0.094𝑅 − 13.244.

Solutions Section 1.4 Graph:

P

30 25 20 15 10 5 0 R 100 200 300 400 500 (Independent variable is 𝑅 and dependent variable is 𝑃 ) b. The units of measurement of the slope are units of profit per unit of revenue: billions of dollars of profit per billion dollars of revenue, or just dollars of profit per dollar of revenue. Thus, Amazon earned $0.094 in profit per additional $1 in revenue. c. 𝑃 = 0.094𝑅 − 13.244, and we are given 𝑃 = 15. Substituting gives 15 = 0.094𝑅 − 13.244 Solving for 𝑅 gives 28.244 𝑅= ≈ 300 0.094 Thus, the company would need to earn about $300 billion in revenue. d. The graph shows a good fit, so the linear model seems reasonable. 22. a. Data points (𝑅, 𝐼) : (190, 5), (240, 15), (300, 14), (420, 28) 𝑹

𝑰

240

15

3,600

420

28

11,760 176,400

190 300 1,150

5

14 62

𝑹𝑰

𝑹𝟐

950

36,100

4,200

90,000

57,600

20,510 360,100

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑅𝐼) − (∑ 𝑅)(∑ 𝐼) 4(20,510) − (1,150)(62) 10,740 𝑚= ≈ 0.091 = = 117,900 𝑛(∑ 𝑅 2) − (∑ 𝑅) 2 4(360,100) − 1,150 2 ∑ 𝐼 − 𝑚(∑ 𝑅) 62 − 0.09109(1,150) −42.75827 Intercept: 𝑏 = ≈ ≈ −10.69 = 𝑛 4 4

Thus, the regression line is 𝐼 = 𝑚𝑅 + 𝑏 ≈ 0.091𝑅 − 10.69.

Solutions Section 1.4 Graph:

I

30 25 20 15 10 5 0 R 180 230 280 330 380 430 (Independent variable is 𝑅 and dependent variable is 𝐼) b. The units of measurement of the slope are units of operating income per unit of revenue: billions of dollars of operating operating income per billion dollars of revenue, or just dollars of operating income per dollar of revenue. Thus, Amazon had operating income of $0.091 per $1 in revenue. c. 𝐼 = 0.091𝑅 − 10.69, and we are given 𝐼 = 10. Substituting gives 10 = 0.091𝑅 − 10.69 Solving for 𝑅 gives 20.69 𝑅= ≈ 227 0.091 Thus, the company would need to earn approximately $227 billion in revenue. d. The graph shows a good fit, so the linear model seems reasonable. 23. a. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown. Regression equation: 𝐿 = 39.29𝑛 + 528.71

b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Applied Calculus is growing at a rate of 39.29 pages per edition. 24. a. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown. Regression equation: 𝐿 = 25.43𝑛 + 552.24

Solutions Section 1.4

b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Finite Mathematics is growing at a rate of 25.43 pages per edition.

25. a. Since production is a function of cultivated area, we take 𝑥 as cultivated area, and 𝑦 as production: 𝑥 𝑦

25

30

32

40

52

15

25

30

40

60

See the technology note accompanying Example 2 for the use of technology to obtain regression lines. We obtained the following regression line and plot in Excel. (coefficients rounded to two decimal places):𝑦 = 1.62𝑥 − 23.87

b. To interpret the slope 𝑚 = 1.62, recall that units of 𝑚 are units of 𝑦 per unit of 𝑥; that is, millions of tons of production of soybeans per million acres of cultivated land. Thus, production increases by 1.62 million tons of soybeans per million acres of cultivated land. More simply, each acre of cultivated land produces about 1.62 tons of soybeans.

26. a. Since production is a function of cultivated area, we take 𝑥 as cultivated area, and 𝑦 as production: 𝑥 𝑦

30

42

69

59

74

74

20

33

55

57

83

88

See the technology note accompanying Example 2 for the use of technology to obtain regression lines. We obtained the following regression line and plot in Excel. (coefficients rounded to two decimal places):𝑦 = 1.38𝑥 − 24.04

b. To interpret the slope 𝑚 = 1.38, recall that units of 𝑚 are units of 𝑦 per unit of 𝑥; That is, millions of tons of production of soybeans per million acres of cultivated land. Thus, production increases by 1.38 million tons of soybeans per million acres of cultivated land. More simply, each acre of cultivated land produces about 1.38 tons of soybeans. 27. a. 𝑦 = Continental net income as a function of 𝑥 = Price of oil. See the technology notes

Solutions Section 1.4 accompanying Example 2 and 3 for the use of technology to obtain regression lines and correlation coefficients. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown:

Regression equation: 𝑦 = −11.85𝑥 + 797.71 Correlation coefficient: 𝑟 ≈ −0.414b. As |𝑟| ≈ 0.414 is significantly less than 0.8, the values of 𝑥 and 𝑦 are not strongly correlated, so that Continental's net income does not appear correlated to the price of oil.c. The points in the graph are nowhere near the regression line, confirming the conclusion in (b).

28. a. 𝑦 = Continental net income as a function of 𝑥 = Price of oil. See the technology notes accompanying Examples 2 and 3 for the use of technology to obtain regression lines and correlation coefficients. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown:

Regression equation: 𝑦 = −28.90𝑥 + 1208.01 Correlation coefficient: 𝑟 ≈ −0.408b. As |𝑟| ≈ 0.408 is significantly less than 0.8, the values of 𝑥 and 𝑦 are not strongly correlated, so that American's net income does not appear correlated to the price of oil.c. The points in the graph are nowhere near the regression line, confirming the conclusion in (b).

29. a. Using 𝑥 = Number of natural science doctorates and 𝑦 = Number of engineering doctorates gives us the following table of values: 𝑥 5,000 5,400 5,900 6,200 6,300

6,700

𝑦 7,600 8,400 9,600 9,500 10,200 10,800

Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = 1.85𝑥 − 1570

Solutions Section 1.4 Graph:

y

12000 10000 8000 6000 4000 2000 0 x 4500 5500 6500 b. To interpret the slope, recall that units of the slope are units of 𝑦 (engineering doctorates) per unit of 𝑥 (natural science doctorates). Thus, 𝑚 = 1.85 engineering doctorates per natural science doctorate, indicating that there are around 1.85 additional doctorates in engineering per additional doctorate in the natural sciences. c. Using the technology method of Example 3, we can use technology to show the value of 𝑟 2 : 𝑟 2 ≈ 0.9649 𝑟 = √𝑟 2 ≈ √0.9649 ≈ 0.982 Since 𝑟 is close to 1, the correlation between 𝑥 and 𝑦 is a strong one. d. Yes; the data points are close to and randomly scattered about the regression line.

30. a. Using 𝑥 = Number of social science doctorates and 𝑦 = Number of education doctorates gives us the following table of values: 𝑥 7,900 8,500 8,700 9,000 8,900 8,900 𝑦 5,300 4,800 4,800 5,100 4,800 4,700

Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = −0.353𝑥 + 7970 Graph:

5500

y

5000

4500 x 7500 8000 8500 9000 9500 b. To interpret the slope, recall that units of the slope are units of 𝑦 (education doctorates) per unit of 𝑥 (social science doctorates). Thus, 𝑚 ≈ −0.35 education doctorates per social science doctorate, indicating that there are about 0.35 fewer doctorates in education per additional doctorate in the social sciences. c. Using the technology method of Example 3, we can use technology to show the value of 𝑟 2 : 𝑟 2 ≈ 0.3914 𝑟 = √𝑟 2 ≈ √0.3914 ≈ 0.626 Since 𝑟 is not close to 1, the correlation between 𝑥 and 𝑦 is a poor one. d. No; the data points suggest no relationship in particular; they are far from the regression line and haphazardly scattered.

Solutions Section 1.4 31. a. As 𝑡 is time in years since 2010, we use the following set of data for the regression: 𝑡

0

2

4

6

8

10

𝑦 5,000 5,400 5,900 6,200 6,300 6,700 Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = 164𝑡 + 5100 Graph:

7000

y

6000 5000 4000

0

2

4

6

t

8

10

𝑟 ≈ 0.985 b. Units of the slope are units of 𝑦 (natural science doctorates) per unit of 𝑡 (years); thus, doctorates per year. So, the number of natural science doctorates has been increasing at a rate of about 164 per year. c. The slopes of successive pairs of points do not show an increasing nor decreasing trend as we go from left to right, so the number of natural science doctorates is increasing at a more-or-less constant rate. d. Yes: If 𝑟 had been equal to 1, then the points would lie exactly on the regression line, which would indicate that the number of doctorates is growing at a constant rate. 32. a. As 𝑡 is time in years since 2010, we use the following set of data for the regression: 𝑡

0

2

4

6

8

10

𝑦 7,900 8,500 8,700 9,000 8,900 8,900 Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = 92.9𝑡 + 8190 Graph:

10000

y

9000 8000 7000

t

0 2 4 6 8 10 𝑟 ≈ 0.916, indicating a reasonably good fit. b. Units of the slope are units of 𝑦 (social science doctorates) per unit of 𝑡 (years); thus, doctorates per year. Thus, the number of social science doctorates has been increasing at a rate of about 93 per year. c. The data points suggest a concave-down curve rather than a straight line, indicating that the number of

Solutions Section 1.4 doctorates has been growing at a slower and slower rate (the slopes of successive pairs of points increase as we go from left to right). d. No: If 𝑟 had been equal to 1, then the points would lie exactly on the regression line, which would indicate that the number of doctorates is growing at a constant rate. 33. a. More-or-less constant rate; Exercise 29 suggests a roughly linear relationship between the number of natural science doctorates and the number of engineering doctorates, and Exercise 31 suggests that the number of natural science doctorates has been increasing at a more-or less constant rate. Therefore, the number of engineering doctorates is also increasing at a more-or-less constant rate. b. No; 𝑟 = 1 in Exercise 29 would indicate an exactly linear relationship between the number of natural science doctorates and the number of engineering doctorates, and so the conclusion would be the same. c. No; 𝑟 = 1 in Exercise 31 would indicate that the number of natural science doctorates has been increasing at a constant rate, and so the conclusion would be the same. 34. a. Haphazard; The graph in Exercise 30 suggests that the number of education doctorates has no particular relationship to the number of social science doctorates, and Exercise 32 suggests that the number of social science doctorates has been increasing with time. Therefore, the number of education doctorates would be expected to have no particular relationship to time either; that is, to behave haphazardly with respect to time as well. b. Yes; 𝑟 = 1 in Exercise 30 would indicate an exactly linear relationship between the number of social science doctorates and the number of education doctorates, and so the conclusion would be that the number of education doctorates increased at a faster and faster rate. c. No; 𝑟 = 1 in Exercise 32 would indicate that the number of social science doctorates has been increasing at a constant rate, and so the conclusion would be the same, as the number of education doctorates would still be haphazard. 35. a. Using the method of Example 3, we obtain the following regression line and plot (coefficients rounded to two decimal places): 𝑝 = 0.13𝑡 + 0.22; 𝑟 ≈ 0.97 Graph:

b. The first and last points lie above the regression line, while the central points lie below it, suggesting a curve. c. Here is a worksheet showing the computation of the residuals (based on Example 1 in the text):

Notice that the residuals are positive at first, become negative, and then become positive, confirming the impression from the graph.

Solutions Section 1.4 36. a. Using the method of Example 3, we obtain the following regression line and plot (coefficients rounded to two decimal places): 𝑐 = 35.5𝑡 − 21; 𝑟 ≈ 0.92 Graph:

b. The first and last points lie above the regression line, while the central points lie below it, suggesting a curve. c. Here is a worksheet showing the computation of the residuals (based on Example 1 in the text):

Notice that the residuals are positive at first, become negative, and then become positive, confirming the impression from the graph. 37. The regression line is defined to be the line that gives the lowest sum-of-squares error, SSE. If we are given two points, (𝑎, 𝑏) and (𝑐, 𝑑) with 𝑎 ≠ 𝑐, then there is a line that passes through these two points, giving SSE = 0. Since 0 is the smallest value possible, this line must be the regression line. 38. SSE = 0; the straight line passing through the given points has predicted values equal to the observed values. Hence, the residuals are zero, giving SSE = 0.

39. If the points (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ), ..., (𝑥𝑛 , 𝑦𝑛 ) lie on a straight line, then the sum-of-squares error, SSE, for this line is zero. Since 0 is the smallest value possible, this line must be the regression line. 40. No. The regression line may pass through none of the given points. 41. Calculation of the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

0

0

0

0

−𝑎 𝑎 0

𝑎 𝑎

2𝑎

−𝑎 2

𝑎2

0

2𝑎 2

𝑎2

𝑎2

(The bottom row contains the column sums.) 𝑛 = 3 (number of data points)}

Slope: 𝑚 =

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

3(0) − (0)(2𝑎) =0 3(2𝑎 2) − 0 2

Solutions Section 1.4

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2

Correlation coefficient 𝑟 =

=

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

√ √ we have just seen that this numerator is zero. Hence, 𝑟 = 0.

has the same numerator as 𝑚, and

42. Calculation of the regression line: 𝑥 0 0

𝑎

𝑎

𝑦

𝑥𝑦

𝑥2

−𝑎

0

0

0

0

0

0

0

0

𝑎

𝑎2 𝑎2

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(0) − (𝑎)(0) Slope: 𝑚 = = =0 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3𝑎 2 − 0 2

Correlation coefficient 𝑟 =

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

√ √ we have just seen that this numerator is zero. Hence, 𝑟 = 0.

has the same numerator as 𝑚, and

43. No. The regression line through (−1, 1), (0, 0), and (1, 1) passes through none of these points. 44. A mathematical model may only be valid for a limited range of values of the variables concerned, and extrapolation can lead to absurd results. 45. (Answers may vary.) The data in Exercise 35 give 𝑟 ≈ 0.97, yet the plotted points suggest a curve, not a straight line.

46. (Answers may vary.) If 𝑟 is not close to 1, then the points are not close to the regression line; they may be scattered randomly above and below the line in a manner not suggesting a parabola.

Solutions Chapter 1 Review Chapter 1 Review

1. a. 1 b. −2 c. 0 d. 𝑓(2) − 𝑓(−2) = 0 − 1 = −1

2. a. −1 b. −3 c. 0 d. 𝑓(2) − 𝑓(−2) = 0 − (−1) = 1

3. a. 1 b. 0 c. 0 d. 𝑓(1) − 𝑓(−1) = 0 − 1 = −1

4. a. 2 b. −1 c. 0 d. 𝑓(1) − 𝑓(−1) = 0 − 2 = −2

5. 𝑦 = −2𝑥 + 5 𝑦-intercept = 5, slope = −2

6. 2𝑥 − 3𝑦 = 12 Solving for 𝑦 gives

−3𝑦 = −2𝑥 + 12 𝑦 = 23 𝑥 − 4 : 𝑦-intercept = −4, slope = 23

7. 𝑦 =

1 𝑥 2

{𝑥 − 1

if − 1 ≤ 𝑥 ≤ 1

Solutions Chapter 1 Review

if 1 < 𝑥 ≤ 3

8. 𝑓(𝑥) = 4𝑥 − 𝑥 2 with domain [0, 4] Technology formula: 4*x-x^2

9. The graph of the function has a V-shape, indicating an absolute value function.

10. The graph of the function is a straight line, indicating a linear function.

11. The graph of the function is a straight line, indicating a linear function.

12. In the graph, 𝑦 doubles for each 1-unit increase in 𝑥, indicating an exponential model.

14. In the graph, 𝑦 is halved for each 1-unit increase in 𝑥, indicating an exponential model.

Solutions Chapter 1 Review 13. The parabolic shape of the graph indicates a quadratic model.

15. Through (3, 2) with slope −3 Point: (3, 2) Slope: 𝑚 = −3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − (−3)(3) = 2 + 9 = 11 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −3𝑥 + 11. 16. Through (−2, 4) with slope −1 Point: (−2, 4) Slope: 𝑚 = −2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 4 − (−1)(−2) = 2 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −𝑥 + 2. 17. Through (1, −3) and (5, 2)

𝑦2 − 𝑦1 2 − (−3) 5 = = = 1.25 𝑥2 − 𝑥1 5−1 4 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −3 − (1.25)(1) = −4.25 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 1.25𝑥 − 4.25. Point: (1, −3)

Slope: 𝑚 =

18. Through (−1, 2) and (1, 0)

𝑦2 − 𝑦1 0−2 −2 = = = −1 𝑥2 − 𝑥1 1 − (−1) 2 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − (−1)(−1) = 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −𝑥 + 1. Point: (−1, 2)

Slope: 𝑚 =

Intercept:

19. Through (1, 2) parallel to 𝑥 − 2𝑦 = 2 Point: (1, 2) Slope: Same as slope of 𝑥 − 2𝑦 = 2. To find the slope, solve for 𝑦 : −2𝑦 = −𝑥 − 2 1 1 𝑦 = 𝑥 + 1, so that 𝑚 = . 2 2 1 3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − (1) = 2 2 1 3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 + . 2 2 20. Through (−3, 1) parallel to −2𝑥 − 4𝑦 = 5 Point: (−3, 1)

1 5 Slope: Same as slope of −2𝑥 − 4𝑦 = 5. To find the slope, solve for 𝑦, getting 𝑦 = − 𝑥 − . 2 4 1 Thus, 𝑚 = − . 2 1 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 1 + (−3) = − 2 2 1 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥 − . 2 2

Solutions Chapter 1 Review 21. With slope 4 crossing 2𝑥 − 3𝑦 = 6 at its 𝑥-intercept We need the 𝑥-intercept of 2𝑥 − 3𝑦 = 6. This is given by setting 𝑦 = 0 and solving for 𝑥 : 2𝑥 − 0 = 6 𝑥=3 Thus, the point is (3, 0) because 𝑦 = 0 on the 𝑥-axis. Slope: 𝑚 = 4 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − 4(3) = −12 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 4𝑥 − 12 22. With slope 1/2 crossing 3𝑥 + 𝑦 = 6 at its 𝑥-intercept We need the x-intercept of 3𝑥 + 𝑦 = 6. This is given by setting 𝑦 = 0 and solving for 𝑥 : 3𝑥 + 0 = 6 𝑥=2 Thus, the point is (2, 0) because 𝑦 = 0 on the 𝑥-axis. 1 Slope: 𝑚 = 2 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (2) = −1 2 Thus, the equation is 1 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 − 1 2 23. 𝑦 = −𝑥∕2 + 1 : 𝑥

−1

Observed 𝑦 Predicted 𝑦 Residual2 1

1.5

0.25

1

2

0.5

2.25

2

0

0

0

SSE:

2.5

𝑦 = −𝑥∕4 + 1 : 𝑥

−1

Observed 𝑦 Predicted 𝑦 Residual2 1

1.25

0.0625

1

2

0.75

1.5625

2

0

0.5

0.25

SSE:

1.875

The second line, 𝑦 = −𝑥∕4 + 1, is a better fit.

24. 𝑦 = 𝑥 + 1 : 𝑥

−2

Solutions Chapter 1 Review

Observed 𝑦 Predicted 𝑦 Residual2 −1

−1

1

0

1

0

1

1

0

1

2

2

0

2

4

3

1

3

3

4

1

SSE:

3

−1

𝑦 = 𝑥∕2 + 1 :

0

𝑥

Observed 𝑦 Predicted 𝑦 Residual2

−1

0

1

1

0.5

0.25

0

1

1

0

1

2

1.5

0.25

2

4

2

4

3

3

2.5

0.25

SSE:

5.75

−2

−1

The first line, 𝑦 = 𝑥 + 1, is the better fit. 25.

𝑥

𝑦

−1

𝑥𝑦

−1

1

𝑥2

𝑦2

1

1

1

2

2

1

4

2

0

0

4

0

2

3

1

6

5

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(1) − (2)(3) −3 Slope: 𝑚 = ≈ −0.214 = = 14 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(6) − 2 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 3 − (−0.214)(2) ≈ 1.14 = 𝑛 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −0.214𝑥 + 1.14. 𝑟=

√

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

√

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

=

3(1) − (2)(3)

√3(6) − 2 2√3(5) − 3 2

≈ −0.33

Solutions Chapter 1 Review 26.

𝑥

𝑦

−2

−1

−1

1

𝑥𝑦

𝑥2

𝑦2

2

−1

4

1

1

1

0

1

0

0

1

1

2

2

1

4

2

4

8

4

16

3

3

9

9

9

3

10

20

19

32

(The bottom row contains the column sums.)

𝑛 = 6 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 6(20) − (3)(10) 90 Slope: 𝑚 = ≈ 0.857 = = 2 2 2 105 𝑛(∑ 𝑥 ) − (∑ 𝑥) 6(19) − 3 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 10 − (0.857)(3) ≈ 1.24 = 𝑛 6

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = 0.857𝑥 + 1.24. 𝑟=

√

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

√

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

=

6(20) − (3)(10)

√6(19) − 3 2√6(32) − 10 2

≈ 0.92

27. a. Graph:

Since the data definitely suggests a curve, we rule out a linear function, leaving us with a choice of quadratic or exponential. Of the two, an exponential function would fit best, given the leveling off we see on the left; the graph of a quadratic function would not flatten out, but instead form a low point and begin rising again toward the left.b. The ratios (rounded to 1 decimal place) are: 𝑉 (1)∕𝑉 (0) 𝑉 (2)∕𝑉 (1) 300 =3 100

𝑉 (3)∕𝑉 (2)

𝑉 (4)∕𝑉 (3)

𝑉 (5)∕𝑉 (4)

𝑉 (6)∕𝑉 (5)

1,000 3,300 10,500 33,600 107,400 ≈ 3.3 ≈ 3.2 ≈ 3.2 ≈ 3.2 = 3.3 300 1,000 3,300 10,500 33,600

They are close to 3.2. c. The data suggest that website traffic is increasing by a factor of around 3.2 per year, so the prediction for next year (year 6) would be around 3.2×107,400 ≈ 343,700 visits per day.

Solutions Chapter 1 Review 28. a. Graph:

Since the data definitely suggest a curve, we rule out a linear function, leaving us with a choice of quadratic or exponential. Of the two, a quadratic function would fit best, given the parabolic shape of the graph. b. The differences (rounded to 1 decimal place) are: 𝐶(1) − 𝐶(0)

−0.32 ≈ −0.3

𝐶(2) − 𝐶(1) 𝐶(3) − 𝐶(2) 𝐶(4) − 𝐶(3) 𝐶(5) − 𝐶(4) −0.1

0.12 ≈ 0.1

0.28 ≈ 0.3

0.48 ≈ 0.5

The rounded differences increase linearly with slope 0.2. c. Assuming the linear trend of differences continue, the next difference 𝐶(6) − 𝐶(5) will be around 0.7, so that the cost of a paperback will be about $5.88 + 0.70 = $6.58.

0.03𝑥 + 2 if 0 ≤ 𝑥 ≤ 50 {0.05𝑥 + 1 if 𝑥 > 50 Notice that 𝑥 is thousands of visit per day, so 10,000 visits corresponds to 𝑥 = 10, and the servers will crash an average of 𝑐(10) = 0.03(10) + 2 = 2.3 times per day. (We used the first formula because 10 is in the interval [0, 50].) For 50,000 visitors, 𝑐(50) = 0.03(50) + 2 = 3.5 crashes per day. (We again used the first formula because 50 is still in the interval [0, 50].) For 100,000 visitors, 𝑐(100) = 0.05(100) + 1 = 6 crashes per day. (We used the second formula because 100 is in the interval (50, +∞).) b. The coefficient 0.03 is the slope of the first formula, indicating that, for website traffic of up to 50,000 visits per day (0 ≤ 𝑥 ≤ 50), the number of crashes is increasing by 0.03 per additional thousand visits. c. To experience 8 crashes in a day, we desire 𝑐(𝑥) = 8. If we try the first formula, we get 0.03𝑥 + 2 = 8 giving 𝑥 = (8 − 2)∕0.03 = 200 , which is not in the domain of the first formula. So, we try the second formula: 0.05𝑥 + 1 = 8 7 0.05𝑥 = 7, so 𝑥 = = 140, 0.05 which is in the domain of the second formula. Thus, we estimate that there were 140,000 visitors that day. 29. a. 𝑐(𝑥) =

1.55𝑥 if 0 ≤ 𝑥 ≤ 100 {1.75𝑥 − 20 if 100 < 𝑥 ≤ 250 𝑠(60) = 1.55(60) = 93 books per day (We used the first formula, since 60 is in [0, 100].) 𝑠(100) = 1.55(100) = 155 books per day (We used the first formula, since 100 is in [0, 100].) 𝑠(160) = 1.75(160) − 20 = 260 books per day (We used the second formula, since 160 is in (100, 250].) b. The coefficient 1.75 is the slope of the second formula, measured in books sold per thousand visitors. Thus, book sales are increasing at a rate of 1.75 books per thousand new visitors when the number of visitors is between 100,000 and 250,000 per day. c. To sell an average of 300 books per day, we desire 𝑛(𝑥) = 300. If we try the first formula, we get 1.55𝑥 = 300, giving 𝑥 ≈ 194, which is not in the domain of the first formula. So, we try the second formula: 30. a.

𝑠(𝑥) =

1.75𝑥 − 20 = 300

Solutions Chapter 1 Review

320 ≈ 182.9 thousand visitors, 1.75 which is in the domain of the second formula. Thus, about 182,900 visitors per day will result in average sales of 300 books per day. 1.75𝑥 = 320, so 𝑥 =

31.

𝑡

1

𝑛(𝑡) 12.5

2

3

4

5

6

37.5

62.5

72.0

74.5

75.0

(a) Technology formulas: (A): 300/(4+100*5^(-t)) (C): -2.3*t^2+30.0*t-3.3

(B): 13.3*t+8.0 (D): 7*3^(0.5*t)

Here are the values for the four given models (rounded to 1 decimal place): 𝑡

1

2

3

4

5

6

(A) 12.5

37.5

62.5

72.1

74.4

74.9

(B) 21.3

34.6

47.9

61.2

74.5

87.8

(C) 24.4

47.5

66.0

79.9

89.2

93.9

(D) 12.1

21.0

36.4

63.0 109.1 189.0

Model (A) gives an almost perfect fit, whereas the other models are not even close. b. Looking at the table, we see the following behavior as 𝑡 increases: (A) Leveling off (B) Rising (C) Rising (begins to fall after 7 months, however) (D) Rising 32.

𝑡

1

2

𝑛(𝑡) 1,330 520

3

4

5

520 1,340 2,980

(a) Technology formulas: (A): 3000/(1+12*2^(-t)) (C): 300*1.6^t

(B): 2000/(4.2-0.7*t) (D): 100*(4.1*t^2-20.4*t+29.5)

Here are the values for the three given models (rounded to the nearest integer): 𝑡

1

2

(A)

429

750 1,200 1,714 2,182

(B)

571

714

(C)

480

768 1,229 1,966 3,146

(D) 1,320 510

3

952

520

4

5

1,429 2,857

1,350 3,000

Model (D) gives a close fit, whereas the other models are not even close. b. If you extrapolate the models, you find the following behavior: (A) Leveling off (B) Becomes undefined and then negative (C) Rising (D) Rising

Solutions Chapter 1 Review 33. a. Using 𝑣(𝑐) = −0.000005𝑐 + 0.085𝑐 + 1,750, we get 2

𝑣(5,000) = −0.000005(5,000) 2 + 0.085(5,000) + 1,750 = −125 + 425 + 1,750 = 2,050 𝑣(6,000) = −0.000005(6,000) 2 + 0.085(6,000) + 1,750 = −180 + 510 + 1,750 = 2,080

Thus, increasing monthly advertising from $5,000 to $6,000 per month would result in 2,080 − 2,050 = 30 more visits per day. b. The following table shows the result of increasing expenditure by steps of $1,000: Tech formula: -0.000005*x^2+0.085*x+1750 𝑐 5,000 6,000 7,000 8,000 9,000 10,000

𝑣(𝑐) 2,050 2,080 2,100 2,110 2,110 2,100

The successive changes in the numbers of visits are: 2,080 − 2,050 = 30; 2,100 − 2,080 = 20; 2,110 − 2,100 = 10; 2,110 − 2,110 = 0; 2,100 − 2,110 = −10, showing that the numbers of visits would increase at a slower and slower rate and then begin to decrease. c. Here is a portion of the graph of 𝑣 :

For 𝑐 = 8,500 or larger, we see that website traffic is projected to decrease as advertising increases, and then drop toward zero. Thus, the model does not appear to give a reasonable prediction of traffic at expenditures larger than $8,500 per month. 34. a. Using 𝑐(𝑛) = 0.0008𝑛 2 − 72𝑛 + 2,000,000, we get

𝑐(20,000) = 0.0008(20,000) 2 − 72(20,000) + 2,000,000 = 880,000 𝑐(30,000) = 0.0008(30,000) 2 − 72(30,000) + 2,000,000 = 560,000

Thus, increasing the run size from 20,000 to 30,000 per month would result in a savings of 880, 000 − 560, 000 = 320,000 dollars. b. The following table shows the result of increasing run size in steps of 10,000: Tech formula: 0.0008*x^2-72*x+2000000 𝑛 20,000

30,000

40,000

50,000

60,000

70,000

𝑐(𝑛) 880,000 560,000 400,000 400,000 560,000 880,000 Change

-320,000 -160,000

0

160,000 320,000

The table shows that the cost decreases at a slower and slower rate and then begins to increase. Going from 30,000 to 40,000 decreases the cost by $160,000—considerably less than going from 20,000 to 30,000.

Solutions Chapter 1 Review

c. Here is a portion of the graph of 𝑣 :

The graph shows that the cost is a minimum for a print run size of around 45,000.

𝑣2 − 𝑣1 2,100 − 2,050 50 = = = 0.05 𝑐2 − 𝑐1 6,000 − 5,000 1,000 Intercept: 𝑏 = 𝑣1 − 𝑚𝑐1 = 2,050 − (0.05)(5,000) = 1,800 Thus, the equation is 𝑣 = 𝑚𝑐 + 𝑏 = 0.05𝑐 + 1,800. b. A budget of $7,000 per month for banner ads corresponds to 𝑣 = 7,000. 𝑣(7,000) = 0.05(7,000) + 1,800 = 2,150 new visitors per day c. We are given 𝑣 = 2,500 and want 𝑐. 2,500 = 0.05𝑐 + 1,800 0.05𝑐 = 2,500 − 1,800 = 700 700 Thus, 𝑐 = = $14,000 per month. 0.05 35. a. Point: (5,000, 2,050)

Slope: 𝑚 =

𝑐2 − 𝑐1 550,000 − 880,000 330,000 = =− = −16.5 𝑛2 − 𝑛1 40,000 − 20,000 20,000 Intercept: 𝑏 = 𝑐1 − 𝑚𝑛1 = 880,000 − (−16.5)(20,000) = 1,210,000 Thus, the equation is 𝑐 = 𝑚𝑛 + 𝑏 = −16.5𝑛 + 1,210,000. b. 𝑐(25,000) = −16.5(25,000) + 1,210,000 = $797,500 c. We are given 𝑐 = 418,000 and want 𝑛. 418,000 = −16.5𝑛 + 1,210,000 −16.5𝑛 = 418,000 − 1,210,000 = −792,000 −792,000 Thus, 𝑛 = = 48,000. −16.5 36. a. Point: (20,000, 880,000)

Slope: 𝑚 =

𝑑2 − 𝑑1 93.5 − 74.5 19 = = = 0.95 𝑤2 − 𝑤1 90 − 70 20 Intercept: 𝑏 = 𝑑1 − 𝑚𝑤1 = 74.5 − (0.95)(70) = 8 Thus, the equation is 𝑑 = 𝑚𝑤 + 𝑏 = 0.95𝑤 + 8. OHagan dropped 90 m, so 𝑑 = 90, and we want 𝑤. 90 = 0.95𝑤 + 8 0.95𝑤 = 90 − 8 = 82 82 Thus, 𝑤 = ≈ 86 kg. .95 37. Point: (𝑤, 𝑑) = (70, 74.5)

Slope: 𝑚 =

𝑇2 − 𝑇1 75 − 80 −5 = = = 0.25 𝑟2 − 𝑟1 120 − 140 −20 Intercept: 𝑏 = 𝑇1 − 𝑚𝑟1 = 80 − (0.25)(140) = 45 Thus, the equation is 𝑇 = 𝑚𝑟 + 𝑏 = 0.25𝑟 + 45. 𝑇 = 65, and we want 𝑟. 65 = 0.25𝑟 + 45 0.25𝑟 = 65 − 45 = 20 38. Point: (𝑟, 𝑇 ) = (140, 80)

Slope: 𝑚 =

Thus, 𝑟 =

20 = 80 chirps/min. 0.25

Solutions Chapter 1 Review

39. a. Cost function: 𝐶 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed cost = $500 per week, and 𝑚 = marginal cost = $5.50 per album Thus, the linear cost function is 𝐶 = 5.5𝑥 + 500. Revenue function: 𝑅 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed revenue = 0, and 𝑚 = marginal revenue = $9.50 per album Thus, the linear revenue function is 𝑅 = 9.5𝑥. Profit function: 𝑃 = 𝑅 − 𝐶 𝑃 = 9.5𝑥 − (5.5𝑥 + 500) = 4𝑥 − 500 b. For breakeven, 𝑃 = 0 4𝑥 − 500 = 0 4𝑥 = 500 500 𝑥= = 125 albums per weekTo make a profit, the company should sell more than this number. 4 c. 𝑅 = 8.00𝑥 𝑃 = 8𝑥 − (5.5𝑥 + 500) = 2.5𝑥 − 500 For breakeven, 500 2.5𝑥 − 500 = 0, so 𝑥 = = 200 2.5 To make a profit, the company should sell more than this number.

40. a. Cost function: 𝐶 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed cost = $900 per month, and 𝑚 = marginal cost = $4 per novel Thus, the linear cost function is 𝐶 = 4𝑥 + 900. Revenue function: 𝑅 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed revenue = 0, and 𝑚 = marginal revenue = $5.50 per novel Thus, the linear revenue function is 𝑅 = 5.50𝑥. Profit function: 𝑃 = 𝑅 − 𝐶 𝑃 = 5.50𝑥 − (4𝑥 + 900) = 5.50𝑥 − 4𝑥 − 900 = 1.50𝑥 − 900b. For breakeven, 𝑃 = 0 1.50𝑥 − 900 = 0 1.50𝑥 = 900 900 𝑥= = 600 novels per monthc. 𝑅 = 5.00𝑥 1.50 𝑃 = 5.00𝑥 − (4𝑥 + 900) = 5𝑥 − 4𝑥 − 900 = 𝑥 − 900 For breakeven, 𝑥 − 900 = 0, so 𝑥 = 900 novels per month 41. a. Demand: We are given two points: (𝑝, 𝑞) = (7, 500) and (9.5, 300) 𝑞 − 𝑞1 300 − 500 −200 Slope: 𝑚 = 2 = = = −80 𝑝2 − 𝑝1 9.5 − 7 2.5 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 500 − (−80)(7) = 500 + 560 = 1,060 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −80𝑝 + 1,060. b. When 𝑝 = $12, the demand is 𝑞 = −80(12) + 1,060 = 100 albums per week c. From Exercise 39, the cost function is 𝐶 = 5.5𝑞 + 500 = 5.5(−80𝑝 + 1,060) + 500 = −440𝑝 + 5,830 + 500 𝐶 = −440𝑝 + 6,330

We use 𝑞 for the monthly sales rather than 𝑥.

We want everything expressed in terms of 𝑝, so we used the demand equation.

To compute the profit in terms of price, we need the revenue as well:

Solutions Chapter 1 Review 𝑅 = 𝑝𝑞 = 𝑝(−80𝑝 + 1,060) = −80𝑝 2 + 1,060𝑝Profit: 𝑃 = 𝑅 − 𝐶𝑃 = −80𝑝 2 + 1,060𝑝 − (−440𝑝 + 6,330) = −80𝑝 2 + 1,500𝑝 − 6,330 Now we compare profits for the three prices: 𝑃 (7.00) = −80(7) 2 + 1,500(7) − 6,330 = $250 𝑃 (9.50) = −80(9.5) 2 + 1,500(9.5) − 6,330 = $700 𝑃 (12) = −80(12) 2 + 1,500(12) − 6,330 = $150 Thus, charging $9.50 will result in the largest weekly profit of $700. 42. a. Demand: We are given two points: (𝑝, 𝑞) = (10, 350) and (5.5, 620) 𝑞 − 𝑞1 620 − 350 270 Slope: 𝑚 = 2 = = = −60 𝑝2 − 𝑝1 5.5 − 10 −4.5 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 350 − (−60)(10) = 350 + 600 = 950 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −60𝑝 + 950. b. When 𝑝 = $15, the demand is 𝑞 = −60(15) + 950 = −900 + 950 = 50 novels per month c. From Exercise 40, the cost function is 𝐶 = 4𝑞 + 900 = 4(−60𝑝 + 950) + 900

= −240𝑝 + 3,800 + 900 𝐶 = −240𝑝 + 4,700

We use 𝑞 for the monthly sales rather than 𝑥.

We want everything expressed in terms of 𝑝, so we used the demand equation.

To compute the profit in terms of price, we need the revenue as well: 𝑅 = 𝑝𝑞 = 𝑝(−60𝑝 + 950) = −60𝑝 2 + 950𝑝 Profit: 𝑃 = 𝑅 − 𝐶 𝑃 = −60𝑝 2 + 950𝑝 − (−240𝑝 + 4,700) = −60𝑝 2 + 1,190𝑝 − 4,700 Now we compare profits for the three prices: 𝑃 (5.50) = −60(5.5) 2 + 1,190(5.5) − 4,700 = $30 𝑃 (10) = −60(10) 2 + 1,190(10) − 4,700 = $1,200 𝑃 (15) = −60(15) 2 + 1,190(15) − 4,700 = −$350 (loss) Thus, charging $10 will result in the largest monthly profit of $1,200. 43. a. Calculation of the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

8

440

3,520

64

8.5

380

3,230

72.25

10

250

2,500

100

11.5

180

2,070 132.25

38

1,250 11,320 368.5

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(11,320) − (38)(1,250) −2,220 Slope: 𝑚 = = = = −74 30 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(368.5) − 38 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 1,250 − (−74)(38) = = 1,015.5 𝑛 4

Solutions Chapter 1 Review Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −74𝑥 + 1,015.5. Using the variable names 𝑝 and 𝑞 makes this equation 𝑞 = −74𝑝 + 1,015.5. b. 𝑞(10.50) = −74(10.50) + 1,015.5 = 238.5 ≈ 239 albums per week 44. a. Calculation of the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

5.5

620

3,410

30.25

10

350

3,500

100

11.5

350

4,025 132.25

12

300

3,600

39

1,620 14,535 406.5

144

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(14,535) − (39)(1,620) −5,040 Slope: 𝑚 = = = = −48 105 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(406.5) − 39 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 1,620 − (−48)(39) = = 873 𝑛 4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −48𝑥 + 873. Using the variable names 𝑝 and 𝑞 makes this equation 𝑞 = −48𝑝 + 873. b. 𝑞(8) = −48(8) + 873 = 489 novels per month

Solutions Chapter 1 Case Study Chapter 1 Case Study

1. Here is the given data, with 𝑡 = 0 representing 2020: 𝑡

𝑦

0

1

2

3

4

5

6

0

0.3

1.5

2.6

3.4

4.3

5.0

Using technology, we obtain the following linear regression model: 𝑦 = 0.8893𝑡 − 0.225; 𝑟 ≈ 0.9948 Since 𝑚 ≈ 0.889, MeTube viewership is increasing at a rate of about 0.889 billion views per year. 2. Graph:

The graph does not suggest a quadratic model because the plotted points do not suggest a parabola. 3. Graph with regression parabola:

Regression equation: 𝑦 = −0.001190𝑡 2 + 0.8964𝑡 − 0.2310 The parabola appears to fit no better than the regression line, suggesting that the quadratic model is not appropriate. 4. Here is the tabulated data together with the result of =LINEST(A2:A8,B2:C8,,TRUE):

𝑝 is computed using =TDIST(ABS(E1/E2),F4,2), and returns the value 𝑝 ≈ 0.9662. There is a very low degree of confidence, 1 − 𝑝 ≈ 0.0338, or 3.38%, we can have in asserting that the coefficient of 𝑥 2 is not zero (or that a quadratic model is needed) and so a quadratic model is not appropriate.

Solutions Section 2.1 Section 2.1

1. 𝑓(𝑥) = 2𝑥 2 − 𝑥 − 2 a. 𝑎 = 2, 𝑏 = −1, 𝑐 = −2 b. Table of values: 𝑥

𝑓(𝑥)

−3 19

−2 8

−1 1

0

1

−2

−1

0

1

2

3

4

13

2

3

c. 𝑓(𝑎 + ℎ) = 2(𝑎 + ℎ) 2 − (𝑎 + ℎ) − 2 = 2(𝑎 2 + 2𝑎ℎ + ℎ 2) − (𝑎 + ℎ) − 2 = 2𝑎 2 + 4𝑎ℎ + 2ℎ 2 − 𝑎 − ℎ − 2 d. 2x^2-x-2 2. 𝑓(𝑥) = −2𝑥 2 + 𝑥 + 2 a. 𝑎 = −2, 𝑏 = −1, 𝑐 = 2 b. Table of values: 𝑥

−3

𝑓(𝑥) −19

−2 −8

−1 −1

2

1

−4

−13

1

2

3

c. 𝑓(𝑎 + ℎ) = −2(𝑎 + ℎ) 2 + (𝑎 + ℎ) + 2 = −2(𝑎 2 + 2𝑎ℎ + ℎ 2) + (𝑎 + ℎ) + 2 = −2𝑎 2 − 4𝑎ℎ − 2ℎ 2 + 𝑎 + ℎ + 2 d. -2x^2+x+2 3. 𝑓(𝑥) = 10𝑥 2 − 5𝑥 a. 𝑎 = 10, 𝑏 = −5, 𝑐 = 0 b. Table of values: 𝑥

−3

𝑓(𝑥) 105

−2 50

−1 15

0 0

5

30

75

1

2

3

c. 𝑓(𝑎 + ℎ) = 10(𝑎 + ℎ) 2 − 5(𝑎 + ℎ) = 10(𝑎 2 + 2𝑎ℎ + ℎ 2) − 5(𝑎 + ℎ) = 10𝑎 2 + 20𝑎ℎ + 10ℎ 2 − 5𝑎 − 5ℎ d. 10x^2-5x 4. 𝑓(𝑥) = −𝑥 2 − 50 a. 𝑎 = −1, 𝑏 = 0, 𝑐 = −50 b. Table of values: 𝑥

−3

𝑓(𝑥) −59

−2

−54

−1

−51

0

−50

−51

−54

−59

c. 𝑓(𝑎 + ℎ) = −(𝑎 + ℎ) 2 − 50 = −(𝑎 2 + 2𝑎ℎ + ℎ 2) − 50 = −𝑎 2 − 2𝑎ℎ − ℎ 2 − 50 d. -(x^2)-50 (See the margin note next to Quick Example 2 in the text as to the reason for the parentheses.) 5. 𝑓(𝑥) = −𝑥 2 − 𝑥 − 1 a. 𝑎 = −1, 𝑏 = −1, 𝑐 = −1

Solutions Section 2.1 b. Table of values: 𝑥

𝑓(𝑥)

−3 −7

−2 −3

−1 −1

0

1

2

3

−1

−3

−7

−13

0

1

2

3

c. 𝑓(𝑎 + ℎ) = −(𝑎 + ℎ) 2 − (𝑎 + ℎ) − 1 = −(𝑎 2 + 2𝑎ℎ + ℎ 2) − (𝑎 + ℎ) − 1 = −𝑎 2 − 2𝑎ℎ − ℎ 2 − 𝑎 − ℎ − 1 d. -(x^2)-x-1 (See margin note next to Quick Example 2 in the textbook as to the reason for the parentheses.) 6. 𝑓(𝑥) = −3𝑥 2 + 3𝑥 − 1 a. 𝑎 = −3, 𝑏 = 3, 𝑐 = −1 b. Table of values: 𝑥

−3

𝑓(𝑥) −37

−2

−19

−1 −7

−1

−1

−7

−19

c. 𝑓(𝑎 + ℎ) = −3(𝑎 + ℎ) 2 + 3(𝑎 + ℎ) − 1 = −3(𝑎 2 + 2𝑎ℎ + ℎ 2) + 3(𝑎 + ℎ) − 1 = −3𝑎 2 − 6𝑎ℎ − 3ℎ 2 + 3𝑎 + 3ℎ − 1 d. -3x^2+3x-1

7. 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 2. 𝑎 = 1, 𝑏 = 3, 𝑐 = 2; −𝑏∕(2𝑎) = −3∕2, 𝑓(−3∕2) = −1∕4, so: vertex: (−3∕2, −1∕4), 𝑦-intercept = 𝑐 = 2. 𝑥 2 + 3𝑥 + 2 = (𝑥 + 1)(𝑥 + 2), so: 𝑥-intercepts: −2, −1. 𝑎 > 0 so the parabola opens upward.

8. 𝑓(𝑥) = −𝑥 2 − 𝑥. 𝑎 = −1, 𝑏 = −1, 𝑐 = 0; −𝑏∕(2𝑎) = −1∕2, 𝑓(−1∕2) = 1∕4, so: vertex: (−1∕2, 1∕4), 𝑦-intercept = 𝑐 = 0. −𝑥 2 − 𝑥 = −𝑥(𝑥 + 1), so: 𝑥-intercepts: −1, 0. 𝑎 < 0 so the parabola opens downward.

9. 𝑓(𝑥) = −𝑥 2 + 4𝑥 − 4. 𝑎 = −1, 𝑏 = 4, 𝑐 = −4; −𝑏∕(2𝑎) = 2, 𝑓(2) = 0, so: vertex: (2, 0), 𝑦-intercept = 𝑐 = −4. −𝑥 2 + 4𝑥 − 4 = −(𝑥 − 2) 2, so: 𝑥-intercept: 2. 𝑎 < 0 so the parabola opens downward.

Solutions Section 2.1

10. 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1. 𝑎 = 1, 𝑏 = 2, 𝑐 = 1; −𝑏∕(2𝑎) = −1, 𝑓(−1) = 0, so: vertex: (−1, 0), 𝑦-intercept = 𝑐 = 1. 𝑥 2 + 2𝑥 + 1 = (𝑥 + 1) 2, so: 𝑥-intercept: −1. 𝑎 > 0 so the parabola opens upward.

11. 𝑓(𝑥) = −𝑥 2 − 40𝑥 + 500. 𝑎 = −1, 𝑏 = −40, 𝑐 = 500; −𝑏∕(2𝑎) = −20, 𝑓(−20) = 900, so: vertex: (−20, 900), 𝑦-intercept = 𝑐 = 500. −𝑥 2 − 40𝑥 + 500 = −(𝑥 + 50)(𝑥 − 10), so: 𝑥-intercepts: −50, 10. 𝑎 < 0 so the parabola opens downward.

12. 𝑓(𝑥) = 𝑥 2 − 10𝑥 − 600. 𝑎 = 1, 𝑏 = −10, 𝑐 = −600; −𝑏∕(2𝑎) = 5; 𝑓(5) = −625, so: vertex: (5, −625), 𝑦-intercept = 𝑐 = −600. 𝑥 2 − 10𝑥 − 600 = (𝑥 + 20)(𝑥 − 30), so: 𝑥-intercepts: −20, 30. 𝑎 > 0 so the parabola opens upward.

13. 𝑓(𝑥) = 𝑥 2 + 𝑥 − 1. 𝑎 = 1, 𝑏 = 1, 𝑐 = −1; −𝑏∕(2𝑎) = −1∕2; 𝑓(−1∕2) = −5∕4, so: vertex: (−1∕2, −5∕4), 𝑦-intercept = 𝑐 = −1. From the quadratic formula: 𝑥-intercepts: −1∕2 ± √5∕2. 𝑎 > 0 so the parabola opens upward.

Solutions Section 2.1

14. 𝑓(𝑥) = 𝑥 2 + √2𝑥 + 1. 𝑎 = 1, 𝑏 = √2, 𝑐 = 1; −𝑏∕(2𝑎) = −√2∕2; 𝑓(−√2∕2) = 1∕2, so: vertex: (−√2∕2, 1∕2), 𝑦-intercept = 𝑐 = 1. 𝑏 2 − 4𝑎𝑐 = −2 < 0, so no 𝑥-intercept. 𝑎 > 0 so the parabola opens upward.

15. 𝑓(𝑥) = 𝑥 2 + 1. 𝑎 = 1, 𝑏 = 0, 𝑐 = 1; −𝑏∕(2𝑎) = 0, 𝑓(0) = 1, so: vertex: (0, 1), 𝑦-intercept = 𝑐 = 1. 𝑏 2 − 4𝑎𝑐 = −4 < 0, so no 𝑥-intercept. 𝑎 > 0 so the parabola opens upward.

16. 𝑓(𝑥) = −𝑥 2 + 5. 𝑎 = −1, 𝑏 = 0, 𝑐 = 5; −𝑏∕(2𝑎) = 0, 𝑓(0) = 5, so vertex: (0, 5), 𝑦-intercept = 𝑐 = 1. From the quadratic formula: 𝑥-intercepts: −√5, √5. 𝑎 < 0 so the parabola opens downward.

17. 𝑞 = −4𝑝 + 100, 𝑅 = 𝑝𝑞 = 𝑝(−4𝑝 + 100) = −4𝑝 2 + 100𝑝; maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $12.50

Solutions Section 2.1

18. 𝑞 = −3𝑝 + 300, 𝑅 = 𝑝𝑞 = 𝑝(−3𝑝 + 300) = −3𝑝 2 + 300𝑝, maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $50

19. 𝑞 = −2𝑝 + 400, 𝑅 = 𝑝𝑞 = 𝑝(−2𝑝 + 400) = −2𝑝 2 + 400𝑝, maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $100

20. 𝑞 = −5𝑝 + 1200, 𝑅 = 𝑝𝑞 = 𝑝(−5𝑝 + 1200) = −5𝑝 2 + 1200𝑝, maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $120

21. 𝑦 = −0.7955𝑥 2 + 4.4591𝑥 − 1.6000 22. 𝑦 = −0.7955𝑥 2 − 4.4591𝑥 − 1.6000

23. 𝑦 = −1.1667𝑥 − 6.1667𝑥 − 3.0000 2

Solutions Section 2.1

24. 𝑦 = −0.3333𝑥 2 + 1.6667𝑥 + 3.0000 25. a. Positive because the data suggest a curve that is concave up. b. The data suggest a parabola that is concave up (𝑎 positive) and with 𝑦-intercept around 1,840. Only choice (C) has both properties. c. −𝑏∕(2𝑎) = 44∕12 ≈ 4, which is 2014. Extrapolating in the positive direction leads one to predict more and more steeply rising military expenditure, which may or may not occur; extrapolating in the negative direction predicts more and more steeply increasing military expenditure as we go back in time, contradicting history — military expenditures have risen and fallen many times. 26. a. Negative because the data suggest a curve that is concave down. b. The data suggest a parabola that is concave down (𝑎 negative) and with 𝑦-intercept around 72. Only choice (B) has both properties. c. −𝑏∕(2𝑎) = 0.7∕0.16 = 4.375, which is closest to 2014. Extrapolating in either direction leads one to predict eventually negative values for the funding, which doesn't make sense. 27. The given function 𝐼(𝑡) = 11𝑡 2 − 170𝑡 + 1,300 is quadratic with graph a concave-up parabola (as 𝑎 = 11 is positive). Thus, its vertex is the lowest point on the graph and occurs when 𝑡 = −𝑏∕(2𝑎) = 170∕22 ≈ 7.73 ≈ 8, corresponding to 2018. At that time, imports were about

𝐼(10.26) = 11(7.73) 2 − 170(7.73) + 1,300 ≈ 640 thousand barrels/day (to two significant digits).

28. The given function 𝑃 (𝑡) = −0.01𝑡 2 + 0.01𝑡 + 2.5 is quadratic with graph a concave-down parabola (as 𝑎 = −0.01 is negative). Thus, its vertex is the highest point on the graph and occurs when 𝑡 = −𝑏∕(2𝑎) = 0.01∕0.02 = 0.5, midway through 2010. At that time, production was approximately 𝑃 (0.5) = −0.01(0.5) 2 + 0.01(0.5) + 2.5 ≈ 2.5 million barrels/day (to two significant digits).

29. a. The vertex of 𝑦 = 0.4𝑥 2 + 𝑥 + 26.5 occurs when 𝑥 = −𝑏∕(2𝑎) = −1∕0.8 = −1.25 ≈ −1, which would correspond to 2019. b. 𝑟(𝑥) = 0.4𝑥 2 + 𝑥 + 26.5 has domain [0, 5], and its value increases as 𝑥 increases from 0 (all the coefficients are positive). So, its lowest value occurs at 𝑥 = 0, corresponding to 2020. (One can also graph the given function to verify that its lowest value occurs at 𝑥 = 0.) c. The 𝑥-coordinate of the vertex in part (a) is outside the domain [0, 5] of 𝑟(𝑥) and so the model does not apply to that value of 𝑥. d. From the answer to part (b), the lowest revenue is 𝑟(0) = 0.4(0) 2 + 0 + 26.5 = 26.5 billion dollars. 30. a. The vertex of 𝑦 = −0.2𝑥 2 + 3𝑥 + 18 occurs when 𝑥 = −𝑏∕(2𝑎) = 3∕0.4 = 7.5 ≈ 8, which would correspond to 2028. b. 𝑟(𝑥) = −0.2𝑥 2 + 3𝑥 + 18 has domain [0, 5], and its graph is seen to increase as 𝑥 increases from 0 to 5. So, its highest value occurs at 𝑥 = 5, corresponding to 2025. c. The 𝑥-coordinate of the vertex in part (a) is outside the domain [0, 5] of 𝑟(𝑥) and so the model does not apply to that value of 𝑥. d. From the answer to part (b), the highest revenue is 𝑟(5) = −0.2(5) 2 + 3(5) + 18 = 28 billion dollars. 31. 𝑞 = −0.5𝑝 + 140. Revenue is 𝑅 = 𝑝𝑞 = −0.5𝑝 2 + 140𝑝. Maximum revenue occurs when 𝑝 = −𝑏∕(2𝑎) = $140; the corresponding revenue is 𝑅 = $9, 800. 32. 𝑞 = −2𝑝 + 320. Revenue is 𝑅 = 𝑝𝑞 = −2𝑝 2 + 320𝑝. Maximum revenue occurs when

Solutions Section 2.1 𝑝 = −𝑏∕(2𝑎) = $80; the corresponding revenue is 𝑅 = $12, 800.

33. The given data points are (𝑝, 𝑞) = (40, 200, 000) and (60, 160, 000). The line passing through these points is 𝑞 = −2000𝑝 + 280, 000. Revenue is 𝑅 = 𝑝𝑞 = −2000𝑝 2 + 280, 000𝑝. Maximum revenue occurs when 𝑝 = −𝑏∕(2𝑎) = 70 houses; the corresponding revenue is 𝑅 = $9, 800, 000. 34. The given data points are (𝑝, 𝑞) = (50, 190, 000) and (70, 170, 000). The line passing through these points is 𝑞 = −1000𝑝 + 240, 000. Revenue is 𝑅 = 𝑝𝑞 = −1000𝑝 2 + 240, 000𝑝. Maximum revenue occurs when 𝑝 = −𝑏∕(2𝑎) = 120 houses; the corresponding revenue is 𝑅 = $14, 400, 000. 35. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏 (𝑥 is the price 𝑝 and 𝑦 is the demand 𝑞). We are given two points on its graph: (3, 28, 000) and (5, 19, 000). Slope: 𝑚 = 𝑝2 −𝑝1 = 19,000−28,000 = −9,000 = −4, 500 5−3 2 𝑞 −𝑞 2

1

Intercept:

𝑏 = 𝑞1 − 𝑚𝑝1 = 28, 000 − (−4, 500)(3) = 28, 000 + 13, 500 = 41, 500

Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −4, 500𝑝 + 41, 500. b. Revenue: 𝑅= 𝑝𝑞 = 𝑝(−4, 500𝑝 + 41, 500)= −4, 500𝑝 2 + 41, 500𝑝

For maximum revenue,

𝑝 = −𝑏 ≈$4.61. = −41,500 2𝑎 −900

The corresponding daily revenue is

𝑅 = −4500(4.61) 2 + 41, 500(4.61) = $95, 680.55.

c. The maximum annual revenue the company could have earned was 95, 680.55 × 365≈$34, 923, 400, which is about $10 million short of what it needed to break even. Therefore, it would not have been possible to break even.

36. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏 (𝑥 is the price 𝑝 and 𝑦 is the demand 𝑞). We are given two points on its graph: (5, 14) and (3, 18). Slope: 4 𝑚 = 𝑝2 −𝑝1 = 18−14 = −2 = −2 3−5 𝑞 −𝑞 2

1

Intercept:

𝑏 = 𝑞1 − 𝑚𝑝1 = 14 − (−2)(5) = 24

Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −2𝑝 + 24. b. Revenue: 𝑅= 𝑝𝑞 = 𝑝(−2𝑝 + 24)= −2𝑝 2 + 24𝑝

For maximum revenue,

𝑝 = −𝑏 = −24 =Ż6. 2𝑎 −4

The corresponding daily revenue is

𝑅 = −2(6) 2 + 24(6) =Ż72 million.

Solutions Section 2.1

c. The maximum annual revenue the company could have earned was 72, 000, 000 × 670 =Ż48,240,000,000, which is about Ż240 million more than it needed to finance the lab. Therefore, it would have been possible to finance the lab.

37. a. The data points are (𝑥, 𝑞) = (2, 280) and (1.5, 560). Thus, 𝑞 = −560𝑥 + 1, 400 and 𝑅 = 𝑥𝑞 = −560𝑥 2 + 1, 400𝑥. b. 𝑃 = 𝑅 − 𝐶 = −560𝑥 2 + 1400𝑥 − 30. The largest monthly profit occurs when 𝑥 = −𝑏∕(2𝑎) = $1.25 and then 𝑃 = $845 per month.

38. a. The data points are (𝑥, 𝑞) = (8, 400) and (4, 600). Thus, 𝑞 = −50𝑥 + 800 and 𝑅 = −50𝑥 2 + 800𝑥. b. 𝑃 = 𝑅 − 𝐶 = −50𝑥 2 + 800𝑥 − 500. The largest weekly profit occurs when 𝑥 = −𝑏∕(2𝑎) = $8 per T-shirt and then 𝑃 = $2700 per week. 39. As a function of 𝑞, 𝐶 = 0.5𝑞 + 20. Substituting 𝑞 = −400𝑥 + 1, 200, we get 𝐶 = 0.5(−400𝑥 + 1, 200) + 20 = −200𝑥 + 620.

The profit is 𝑃 = 𝑅 − 𝐶 = 𝑥𝑞 − 𝐶 = −400𝑥 2 + 1400𝑥 − 620. The profit is largest when 𝑥 = −𝑏∕(2𝑎) =$1.75 per log-on; the corresponding profit is 𝑃 =$605 per month. 40. As a function of 𝑞, 𝐶 = 5𝑞 + 400. Substituting for 𝑞, we get 𝐶 = 5(−40𝑥 + 600) + 400 = −200𝑥 + 3400.

The profit is 𝑃 = 𝑅 − 𝐶 = 𝑥𝑞 − 𝐶 = −40𝑥 2 + 800𝑥 − 3400. The profit is largest when 𝑥 = −𝑏∕(2𝑎) =$10 per T-shirt; the corresponding profit is 𝑃 =$600 per week.

41. a. The data points are (𝑝, 𝑞) = (10, 300) and (15, 250), so 𝑞 = −10𝑝 + 400. b. 𝑅 = 𝑝𝑞 = −10𝑝 2 + 400𝑝 c. 𝐶 = 3𝑞 + 3000 = 3(−10𝑝 + 400) + 3, 000 = −30𝑝 + 4, 200 d. 𝑃 = 𝑅 − 𝐶 = −10𝑝 2 + 430𝑝 − 4, 200. The maximum profit occurs when 𝑝 = −𝑏∕(2𝑎) =$21.50.

42. a. The data points are (𝑝, 𝑞) = (2500, 15) and (2000, 20), so 𝑞 = −0.01𝑝 + 40. b. 𝑅 = 50𝑝𝑞 = 50𝑝(−0.01𝑝 + 40) = −0.5𝑝2 + 2000𝑝 c. (i) 𝐶 = 120, 000 + 80, 000𝑞, so (ii) 𝐶 = 120, 000 + 80, 000(−0.01𝑝 + 40) = −800𝑝 + 3, 320, 000. d. 𝑃 = 𝑅 − 𝐶 = −0.5𝑝 2 + 2800𝑝 − 3, 320, 000. The maximum profit occurs when 𝑝 = −𝑏∕(2𝑎) =$2800 per hour.

Solutions Section 2.1 43. Here is the Excel tabulation of the data, together with the scatter plot and the quadratic (polynomial order 2) trendline (with the option "Display equation on chart" checked):

From the trendline, the quadratic model is 𝑓(𝑡) = 6𝑡 2 − 46𝑡 + 1,840. To estimate world military expenditure in 2017, substitute the corresponding value 𝑡 = 7, to obtain 𝑓(7) = 6(7) 2 − 46(7) + 1,840 = 1,812 representing $1,812 billion. The actual figure (from Exercise 25) is $1,850 billion, so the predicted value is $38 billion lower than the actual value. 44. Here is the Excel tabulation of the data, together with the scatter plot and the quadratic (polynomial order 2) trendline (with the option "Display equation on chart" checked):

From the trendline, the quadratic model is 𝑓(𝑡) = −0.0731𝑡 2 + 0.616𝑡 + 72.08. To estimate the amount of funding in 2015, substitute the corresponding value 𝑡 = 5, to obtain 𝑓(5) = −0.0731(5) 2 + 0.616(5) + 72.08 ≈ 73.33, or $73.33 billion. The actual amount of funding (from Exercise 26) was $73.33 billion which agrees with the predicted figure to two decimal places.

Solutions Section 2.1 45. a. Here is the Excel tabulation of the data, together with a scatter plot and quadratic (polynomial order 2) trendline. The option "Display equation on chart" has been checked.

We round the coefficients to two significant digits to get 𝑆(𝑡) = −0.70𝑡 2 − 6.0𝑡 + 50. b. To predict the sales in 2015 we substitute 𝑡 = 5 and compute 𝑆(5) = −0.70(5) 2 − 6.0(5) + 50 = 2.5 ≈ 3 million units. \\Likewise, the 2016 prediction is 𝑆(6) = −0.70(6) 2 − 6.0(6) + 50 = −11.2 ≈ −11 million units. Even though we expected 2015 sales to be much lower than the 2014 sales, the 2016 prediction, being negative, shows the danger of extrapolating curve-fitting models. 46. a. Here is the Excel tabulation of the data, together with a scatter plot and quadratic (polynomial order 2) trendline. The option "Display equation on chart" has been checked.

We round the coefficients to two significant digits to get 𝑆(𝑡) = −3.2𝑡 2 + 20𝑡 + 22. b. To predict the sales in 2010 we substitute 𝑡 = 5 and compute 𝑆(5) = −3.2(5) 2 + 20(5) + 22 = 42 million units. \\Likewise, the 2011 prediction is 𝑆(6) = −3.2(6) 2 + 20(6) + 22 = 26.8 ≈ 27 million units. These values are far from the figures of 50.4 and 42.6 million units given in the table in the preceding exercise. This shows the danger of shows the danger of extrapolating curve-fitting models.

Solutions Section 2.1 47. If 𝑎 = 0, then 𝑓(𝑥) = 𝑏𝑥 + 𝑐, a linear function, so its graph is a straight line.

48. Since 𝑐 is the 𝑦-intercept of the graph, 𝑐 = 0 implies that the graph has a 𝑦-intercept of zero, meaning that it passes through the origin.

49. Since the curve is concave up, 𝑎 is positive. Since the 𝑦-intercept is negative, 𝑐 is negative. Hence the correct choice is (C). 50. Since the curve is concave down, 𝑎 is negative. Since the 𝑦-intercept is positive, 𝑐 is positive. Hence the correct choice is (B).

51. Positive; The 𝑥-coordinate of the vertex is negative, so −𝑏∕(2𝑎) must be negative. Since 𝑎 is positive (the parabola is concave up), this means that 𝑏 must also be positive to make −𝑏∕(2𝑎) negative. 52. Positive; The 𝑥-coordinate of the vertex is positive, so −𝑏∕2𝑎 must be positive. Since 𝑎 is negative (the parabola is concave down), this means that 𝑏 must be positive to make −𝑏∕2𝑎 positive.

53. The 𝑥-coordinate of the vertex represents the unit price that leads to the maximum revenue, the 𝑦-coordinate of the vertex gives the maximum possible revenue, the 𝑥-intercepts give the unit prices that result in zero revenue, and the 𝑦-intercept gives the revenue resulting from zero unit price (which is obviously zero). 54. The 𝑥-coordinate of the vertex gives the time at which the stone reaches its highest point, the 𝑦-coordinate of the vertex gives the maximum height, the 𝑥-intercepts give the times that the stone is at zero height, while the 𝑦-intercept gives the height of the stone at time zero. 55. Graph the data to see whether the points suggest a curve rather than a straight line. If the curve suggested by the graph is concave up or concave down, then a quadratic model would be a likely candidate. 56. Graphing the data shows that neither a linear model nor a quadratic model is appropriate. The curve suggested by the graph is concave down on [0, 4] and concave up on [4, 8]. This is unlike a parabola, which is either always concave up or concave down. 57. No; The graph of a quadratic function is a parabola. In the case of a concave-up parabola, the curve would unrealistically predict sales increasing extremely fast and becoming unrealistically large in the future. In the case of a concave-down parabola, the curve would predict "negative" sales from some point on. 58. Answers may vary. Some uses are: using the model to interpolate; that is, predict missing values inside the domain; using the model for short-term prediction; using a quadratic model to determine maximum possible revenue or profit.

59. If 𝑞 = 𝑚𝑝 + 𝑏 (with 𝑚 < 0), then the revenue is given by 𝑅 = 𝑝𝑞 = 𝑚𝑝 2 + 𝑏𝑝. This is the equation of a parabola with 𝑎 = 𝑚 < 0 and so is concave down. Thus, the vertex is the highest point on the parabola, showing that there is a single highest value for 𝑅, namely, the 𝑦-coordinate of the vertex. 60. The given equation is the equation of a concave-up parabola, and thus has the vertex as its lowest point. The 𝑦-coordinate of the vertex therefore gives the lowest possible average cost, which occurs when 𝑥 is assigned the value of the 𝑥-coordinate of the vertex.

2

Solutions Section 2.1

61. Since 𝑅 = 𝑝𝑞, the demand must be given by 𝑞 = 𝑅𝑝 = −50𝑝 𝑝+60𝑝 = −50𝑝 + 60. 2

62. Since 𝑅 = 𝑝𝑞, the demand is given by 𝑞 = 𝑅𝑝 = −50𝑝 +60𝑝+50 . This is not linear. = −50𝑝 + 60 + 50 𝑝 𝑝

Solutions Section 2.2 Section 2.2 1. 4^x 𝑥

𝑓(𝑥)

2. 3^x 𝑥

𝑓(𝑥)

−3

−2

−1

0

1 4

16

64

−3

−2

−1

0

1

2 9

3

27

−2

−1

0

1

2 1 9

3

1 27

−2

−1

0

1 1 4

2

1 16

3

1 64

0

1

2 8

3

16

2

3

1 64

1 16

1 27

1 9

3. 3^(-x) 𝑥

𝑓(𝑥)

−3 27

9

4. 4^(-x) 𝑥

𝑓(𝑥)

−3 64

16

1 4

1 3

3

4

5. 2*2^x or 2*(2^x) 𝑥

𝑔(𝑥)

−3

−2

1 4

1 2

−1 1

6. 2*3^x or 2*(3^x) 𝑥

𝑔(𝑥)

−3 2 27

𝑥

−3

ℎ(𝑥) −24

1

1

1

2

3

1 3

4

3

−2

−1

0

1 6

18

54

−2

−1

0

1

2

3

2 9

7. -3*2^(-x)

1

2

−12

2 3

−6

2

−3

−

3 2

−

3 4

−

3 8

Solutions Section 2.2 8. -2*3^(-x) 𝑥

−3

ℎ(𝑥) −54 9. 2^x-1 𝑥

𝑟(𝑥)

−3

−

7 8

10. 2^(-x)+1 𝑥

𝑟(𝑥)

−3 9

11. 2^(x-1) 𝑥

𝑠(𝑥)

−3 1 16

12. 2^(1-x) 𝑥

𝑠(𝑥) 13.

−3 16

0

1

2

3

−2

−1 −6

−2

−

2 3

−

2 9

−

−2 3 4

−1

−

0

1

2

3

−2

−1

0

1

2

3

−2

−1

0

1

2

3

−2

−1

0

1

2

3

−18

−

5

1 8

8

1 2

3

1 4

4

0

2

1 2

2

1

3 2

1

1

3

5 4

2

1 2

2 27

7

9 8

4

1 4 14.

Solutions Section 2.2 15.

16.

17.

18.

19.

𝑥

𝑓(𝑥) 𝑔(𝑥)

−2

0.5 8

−1

1.5 4

0

4.5 2

1

13.5 1

2

40.5 1 2

For every increase in 𝑥 by one unit, the value of 𝑓 is multiplied by 3, so 𝑓 is exponential. Since 𝑓(0) = 4.5, the exponential model is 𝑓(𝑥) = 4.5(3 𝑥). For every increase in 𝑥 by one unit, the value of 𝑔 is multiplied by 1/2, so 𝑔 is exponential. Since 𝑔(0) = 2, the exponential model is 𝑔(𝑥) = 2(1∕2) 𝑥, or 2(2 −𝑥). 20.

𝑥

−2

−1

0

1

2

𝑔(𝑥)

3

0

−1

0

3

𝑓(𝑥)

1 2

1

2

4

8

For every increase in 𝑥 by one unit, the value of 𝑓 is multiplied by 2, so 𝑓 is exponential. The values of 𝑔 decrease and then increase, so 𝑔 is not exponential. Since 𝑓(0) = 2, the exponential model is 𝑓(𝑥) = 2(2 𝑥).

𝑥

−2

𝑔(𝑥)

0.3

𝑥

−2

21.

−1

𝑓(𝑥) 22.5

7.5 0.9

0

Solutions Section 2.2

1

2.5

7.5

2.7

8.1

2

22.5 16.2

When 𝑥 increases from −1 to 0, the value of 𝑓 is multiplied by 1/3, but when 𝑥 is increased from 0 to 1, the value of 𝑓 is multiplied by 3. So 𝑓 is not exponential. When 𝑥 increases from −1 to 0, the value of 𝑔 is multiplied by 3, but when 𝑥 is increased from 1 to 2, the value of 𝑔 is multiplied by 2. So 𝑔 is not exponential. 22.

𝑓(𝑥) 𝑔(𝑥)

−1

0.3

0.9

3

0

1

2.7

2

8.1

24.3

1.5

0.75 0.375 0.1875

−1

0

For every increase in 𝑥 by one unit, the value of 𝑓 is multiplied by 3, so 𝑓 is exponential. Since 𝑓(0) = 2.7, the exponential model is 𝑓(𝑥) = 2.7(3 𝑥). For every increase in 𝑥 by one unit, the value of 𝑔 is multiplied by 0.5, so 𝑔 is exponential. Since 𝑔(0) = 0.75, the exponential model is 𝑔(𝑥) = 0.75(0.5) 𝑥, or 0.75(2 −𝑥). 𝑥

23.

−2

𝑓(𝑥) 100 𝑔(𝑥) 100

200 20

1

400

600

4

2

800

0.8

0.16

1

2

10

2

The values of 𝑓(𝑥) double for every one-unit increase in 𝑥 except when 𝑥 increases from 1 to 2, when the value of 𝑓 is multiplied by 4/3. Hence 𝑓 is not exponential. On the other hand, 𝑔 is multiplied by 0.2 for every increase by one unit in 𝑥, so 𝑔 is exponential. Since 𝑔(0) = 4, the exponential model is 𝑔(𝑥) = 4(0.2) 𝑥. 𝑥

24.

𝑓(𝑥) 𝑔(𝑥)

−2

−1

0.8

0.2

80

40

0

0.1 20

0.05 0.025

The value of 𝑓(𝑥) is multiplied by 1/4 when 𝑥 increases from −2 to −1 but halved 𝑥 when increases from −1 to 0. Hence 𝑓 is not exponential. The value of 𝑔(𝑥) is multiplied by 1/2 when 𝑥 increases from −2 to −1 but multiplied by 0.2 when 𝑥 increases from 1 to 2. Hence 𝑔 is not exponential. 25. 2.718^(-2*x) 𝑥

−3

−2

−1

𝑓(𝑥) 403.2 54.58 7.388

0 1

1

2

3

0.1354 0.01832 0.002480

Solutions Section 2.2 26. 2.718^(x/5) 𝑥

−3

−2

0

−1

𝑔(𝑥) 0.5488 0.6703 0.8187 27. 1.01*2.02^(-4*x) 𝑥

−3

−2

1

1

2

3

1.221 1.492 1.822

−1

0

1

2

3

−1

0

1

2

3

0

1

2

3

ℎ(𝑥) 4 662 280.0 16.82 1.01 0.06066 0.003643 0.0002188 28. 3.42*3^(-x/5) 𝑥

−3

−2

ℎ(𝑥) 6.612 5.308 4.261 3.42 2.745 2.204 1.769 29. 50*(1+1/3.2)^(2*x) 𝑥

−3

−2

−1

𝑟(𝑥) 9.781 16.85 29.02

50

30. 0.043*(4.5-5/1.2)^(-x) 𝑥

−3

−2

−1

86.13 148.4 255.6 0

1

2

3

𝑟(𝑥) 0.001592 0.004778 0.01433 0.043 0.129 0.387 1.161 31. The following solutions also show some common errors you should avoid. 2^(x-1) not 2^x-1 32. 2^(-4*x) not 2^-4*x 33. 2/(1-2^(-4*x)) not 2/1-2^-4*x and not 2/1-2^(-4*x) 34. 2^(3-x)/(1-2^x) or (2^(3-x))/(1-2^x) not 2^3-x/1-2^x and not 2^3-x/(1-2^x) 35. (3+x)^(3*x)/(x+1) or ((3+x)^(3*x))/(x+1)not (3+x)^(3*x)/x+1 and not (3+x^(3*x))/(x+1) 36. 20.3^(3*x)/(1+20.3^(2*x)) or (20.3^(3*x))/(1+20.3^(2*x))not 3^(3*x)/1+20.3^(2*x) and not (20.3^3*x)/(1+20.3^2*x) 37. 2*e^((1+x)/x) or 2*EXP((1+x)/x)not 2*e^1+x/x and not 2*e^(1+x)/x and not 2*EXP(1+x)/x 38. 2*e^(2/x)/x or (2*e^(2/x))/x or 2*EXP(2/x)/x or (2*EXP(2/x))/xnot 2*e^((2/x)/x) and not 2*e^2/x/x and not 2*EXP((2/x)/x)

39. In these solutions, 𝑓1 is black and 𝑓2 is gray.

40.

y1 = 1.6^x y2 = 1.8^x

y1 = 2.2^x y2 = 2.5^x

41. (Note that the 𝑥-axis shown here crosses at 200, not 0.)

42. (Note that the 𝑥-axis shown here crosses at 95.)

y1 = 300*1.1^x y2 = 300*1.1^(2*x)

y1 = 100*1.01^(2*x) y2 = 100*1.01^(3*x)

43.

44.

y1 = 2.5^(1.02*x) y2 = e^(1.02*x) or exp(1.02*x)

y1 = 2.5^(-1.02*x) y2 = e^(-1.02*x) or exp(-1.02*x)

45. (Note that the 𝑥-axis shown here crosses at 900.)

46. (Note that the 𝑥-axis here crosses at 1,100.)

Solutions Section 2.2

y1 = 1000*1.045^(-3*x) y2 = 1000*1.045^(3*x)

y1 = 1202*1.034^(-3*x) y2 = 1202*1.034^(3*x)

47. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 0.5. Also, 𝑓(0) = 500. So, 𝑓(𝑥) = 500(0.5) 𝑥.

Solutions Section 2.2 48. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 2. Also, 𝑓(0) = 500. So, 𝑓(𝑥) = 500(2) 𝑥. 49. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 3. Also, 𝑓(0) = 10. So, 𝑓(𝑥) = 10(3) 𝑥.

50. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 1/3. Also, 𝑓(0) = 90. So, 𝑓(𝑥) = 90(1∕3) 𝑥. 51. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 225∕500 = 0.45. Also, 𝑓(0) = 500. So, 𝑓(𝑥) = 500(0.45) 𝑥.

52. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 3∕5 = 0.6. Also, 𝑓(0) = 5. So, 𝑓(𝑥) = 5(0.6) 𝑥. 53. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = −110 and 𝐴𝑏 2 = −121. Dividing, 𝑏 = −121∕(−110) = 1.1. Substituting, 𝐴(1.1) = −110, so 𝐴 = −100. Thus, 𝑓(𝑥) = −100(1.1) 𝑥.

54. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = −41 and 𝐴𝑏 2 = −42.025. Dividing, 𝑏 = −42.025∕(−41) = 1.025. Substituting, 𝐴(1.025) = −41, so 𝐴 = −41∕1.025 = −40. Thus, 𝑓(𝑥) = −40(1.025) 𝑥. 55. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives 36 = 𝐴𝑏 2 324 = 𝐴𝑏 4

Dividing the second equation by the first gives 324∕36 = 9 = 𝑏 2, so 𝑏 = 3. Substituting into the first equation now gives 36 = 𝐴(3) 2 = 9𝐴, so 𝐴 = 36∕9 = 4. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = 4(3 𝑥). 56. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives −4 = 𝐴𝑏 2 −16 = 𝐴𝑏 4.

Dividing the second equation by the first gives −16∕(−4) = 4 = 𝑏 2, so 𝑏 = 2. Substituting into the first equation now gives −4 = 𝐴(2) 2 = 4𝐴, so 4𝐴 = −14. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = −1(2 𝑥). 57. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives −25 = 𝐴𝑏 −2 −0.2 = 𝐴𝑏.

Dividing the second equation by the first gives −0.2∕(−25) = 0.008 = 𝑏 3, so 𝑏 = 0.2. Substituting into the second equation now gives −0.2 = 0.2𝐴, so 𝐴 = −1. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = −1(0.2 𝑥). 58. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives 1.2 = 𝐴𝑏 0.108 = 𝐴𝑏 3.

Dividing the second equation by the first gives 0.108∕1.2 = 0.09 = 𝑏 2, so 𝑏 = 0.3. Substituting into the first equation now gives 1.2 = 0.3𝐴, so 𝐴 = 4. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = 4(0.3 𝑥). 59. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = 3 and 𝐴𝑏 3 = 6. Dividing, 𝑏 2 = 6∕3 = 2, so 𝑏 = √2 ≈ 1.4142. Substituting, 𝐴√2 = 3, so 𝐴 = 3∕√2 ≈ 2.1213. Thus, 𝑦 = 2.1213(1.4142 𝑥).

3

60. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = 2 and 𝐴𝑏 4 = 6. Dividing, 𝑏 3 = 6∕2 = 3, so 𝑏 = √3 ≈ 1.4422. Solutions Section 2.2

3

3

Substituting, 𝐴 √3 = 2, so 𝐴 = 2∕ √3 ≈ 1.3867. Thus, 𝑦 = 1.3867(1.4422 𝑥).

61. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 2 = 3 and 𝐴𝑏 6 = 2. Dividing, 𝑏 4 = 2∕3, so 𝑏 = √2∕3 ≈ 0.9036. 4

Substituting, 𝐴( √2∕3) 2 = 3, so 𝐴 = 3∕( √2∕3) 2 ≈ 3.6742. Thus, 𝑦 = 3.6742(0.9036 𝑥). 4

4

62. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 −1 = 2 and 𝐴𝑏 3 = 1. Dividing, 𝑏 4 = 1∕2, so 𝑏 = √1∕2 ≈ 0.8409. Substituting, 𝐴∕ √1∕2 = 2, so 𝐴 = 2 √1∕2 ≈ 1.6818. Thus, 𝑦 = 1.6818(0.8409 𝑥). 4

4

4

63. 𝑦 = 1.0442(1.7564) 𝑥 64. 𝑦 = 1.0442(0.5694) 𝑥

65. 𝑦 = 15.1735(1.4822) 𝑥 66. 𝑦 = 0.4782(1.8257) 𝑥

67. We want a model of the form 𝑓(𝑡) = 𝐴𝑏 𝑡. We are given two points on the graph: (0, 300) and (2, 75). Substituting the coordinates, we get 300 = 𝐴𝑏 0 = 𝐴 75 = 𝐴𝑏 . 2

Substitute (0, 300). Substitute (2, 75).

This gives 𝐴 = 300, 75 = 300𝑏 2, so 𝑏 2 = 75∕300 = 0.25, giving 𝑏 = 0.25 1∕2 = 0.5, and the model is 𝑓(𝑡) = 300(0.5) 𝑡. After 5 hours, 𝑓(5) = 300(0.5) 5 = 9.375 mg. 68. We want a model of the form 𝑓(𝑡) = 𝐴𝑏 𝑡. We are given two points on the graph: (0, 200) and (2, 112.5). Substituting the coordinates, we get 200 = 𝐴𝑏 0 = 𝐴 112.5 = 𝐴𝑏 2.

Substitute (0, 200).

Substitute (2, 112.5).

This gives 𝐴 = 200, 112.5 = 200𝑏 2, so 𝑏 2 = 112.5∕200 = 0.5625, giving 𝑏 = 0.5625 1∕2 = 0.75, and the model is 𝑓(𝑡) = 200(0.75) 𝑡. After 4 hours, 𝑓(4) = 200(0.75) 4 ≈ 63 mg/dL. 69. a. Linear model: 𝑆 = 𝑚𝑡 + 𝑏 Points: (0, 320) and (5, 950) 𝑆 − 𝑆1 950 − 320 Slope: 𝑚 = 2 = = 126 𝑡2 − 𝑡1 5−0 Intercept: 𝑏 = 𝑆1 − 𝑚𝑡1 = 320 − (126)(0) = 320 Model: 𝑆 = 𝑚𝑡 + 𝑏 = 126𝑡 + 320 Exponential model: 𝑆 = 𝐴𝑏 𝑡 Substitute (0, 320): 320 = 𝐴𝑏 0 = 𝐴; substitute (5, 950): 950 = 𝐴𝑏 5 = 320𝑏 5. Thus 950 = 𝑏5 320

𝑏 =(

950 1∕5 ≈ 1.24. 320 )

Solutions Section 2.2

Model: 𝑆 = 𝐴𝑏 𝑡 = 320(1.24) 𝑡 The exponential model is applicable: The successive ratios of the values of 𝑆 are not too far from 𝑏 = 1.24. The successive differences climb steadily, making a linear model not appropriate. b. 2020 corresponds to 𝑡 = 4, and so 𝑆 = 𝐴𝑏 𝑡 = 320(1.24) 4 ≈ 757 GW, in reasonable agreement with the actual figure of 770 GW. 70. a. Linear model: 𝑊 = 𝑚𝑡 + 𝑏 Points: (0, 200) and (10, 710) 𝑊 − 𝑊1 710 − 200 Slope: 𝑚 = 2 = = 51 𝑡2 − 𝑡1 10 − 0 Intercept: 𝑏 = 𝑊1 − 𝑚𝑡1 = 200 − (51)(0) = 200 Model: 𝑊 = 𝑚𝑡 + 𝑏 = 51𝑡 + 200 Exponential model: 𝑊 = 𝐴𝑏 𝑡 Substitute (0, 200): 200 = 𝐴𝑏 0 = 𝐴; substitute (10, 710): 710 = 𝐴𝑏 10 = 200𝑏 10. Thus 710 = 3.55 = 𝑏 10 200 𝑏 = 3.55 1∕10 ≈ 1.14.

Model: 𝑊 = 𝐴𝑏 𝑡 = 200(1.14) 𝑡 The exponential model is applicable: The successive ratios of the values of 𝑊 are not too far from 𝑏 2 ≈ 1.30. (Note that 𝑏 2 is the ratio we expect to see every two years, as in the values given.) The successive differences climb fairly steadily, making a linear model not appropriate. b. 2018 corresponds to 𝑡 = 8, and so 𝑊 = 𝐴𝑏 𝑡 = 200(1.14) 8 ≈ 571 GW, in reasonable agreement with the actual figure of 590 GW. 71. a. The successive ratios are 𝑡

Ratios

10

20

30

40

50

60

210 229 252 282 309 331 ≈ 1.12 ≈ 1.09 ≈ 1.10 ≈ 1.12 ≈ 1.10 ≈ 1.07 187 210 229 252 282 309

As the ratios are all close to 1.1 with no systematic change, we conclude that the growth is approximately exponential. b. The 1970 and 1980 data together with the answer to part (a) give us: (10, 210): 𝐴𝑏 10 = 210 (20, 229): 𝐴𝑏 20 = 229 229 𝑏 10 = ≈ 1.09047619 210 𝑏 ≈ 1.09047619 1∕10 ≈ 1.00870 to 6 significant digits 210 210 2 𝐴𝑏 10 = 210 gives 𝐴 = 10 = ≈ 192.576, 229 𝑏

So the model is 𝑈(𝑡) = 192.576(1.00870) 𝑡, and predicts 𝑈(60) = 192.576(1.00870) 60 ≈ 324 million for 2020, reasonably close to the actual figure of 331 million.

Solutions Section 2.2 72. a. The successive ratios are 𝑡

Ratios

10

20

30

40

50

60

3.70 4.46 5.33 6.14 6.96 7.79 ≈ 1.22 ≈ 1.21 ≈ 1.20 ≈ 1.15 ≈ 1.13 ≈ 1.12 3.03 3.70 4.46 5.33 6.14 6.96

As the ratios show a definite downward trend from 1.22 to 1.12, we conclude that the growth is not exponential, as exponential growth would require the ratios to be approximately constant. b. The 1970 and 1980 data together with the answer to part (a) give us: (10, 3.70): 𝐴𝑏 10 = 3.70 (20, 4.46): 𝐴𝑏 20 = 4.46 4.46 𝑏 10 = ≈ 1.205405405 3.70 𝑏 ≈ 1.205405405 1∕10 ≈ 1.01886 to 6 significant digits 3.70 3.70 2 𝐴𝑏 10 = 3.70 gives 𝐴 = 10 = ≈ 3.06951, 4.46 𝑏

So the model is 𝑊 (𝑡) = 3.06951(1.01886) 𝑡, and predicts 𝑊 (60) = 3.06951(1.01886) 60 ≈ 9.42 billion in 2020, more than 20% higher than the actual figure of 7.79 billion. 73. If 𝑦 represents the size of the culture at time 𝑡, then 𝑦 = 𝐴𝑏 𝑡. We are told that the initial size is 1,000, so 𝐴 = 1, 000. We are told that the size doubles every 3 hours, so 𝑏 3 = 2, or 𝑏 = 2 1∕3. Thus, 𝑦 = 1, 000(2 1∕3) 𝑡 = 1, 000(2 𝑡∕3). There will be 1, 000(2 48∕3) = 65, 536, 000 bacteria after 2 days. 74. If 𝑦 represents the size of the culture at time 𝑡, then 𝑦 = 𝐴𝑏 𝑡. We are told that the initial size is 1,000, so 𝐴 = 1, 000. We are told that 𝑦 = 1, 500 when 𝑡 = 2, so 1, 500 = 1, 000𝑏 2, giving 𝑏 = 1.5 1∕2. Thus, 𝑦 = 1, 000(1.5 1∕2) 𝑡 = 1, 000(1.5 𝑡∕2). There will be 1, 000(1.5 48∕2) ≈ 16, 800, 000 bacteria after 2 days.

75. The desired model is 𝐶(𝑡) = 𝐴𝑏 𝑡. At time 𝑡 = 0 (April 1, 2021) the number of active cases was 615,000, so 𝐴 = 615,000. Since the number was increasing by 6.9% each day, that number is multiplied by 1.069 each day, so 𝑏 = 1.069. Hence, the model is 𝐶(𝑡) = 615,000(1.069) 𝑡. As April 21, 2021 corresponds to 𝑡 = 20, the estimated number of cases was about 𝐶(20) = 615,000(1.069) 20 ≈ 2,335,765. 76. The desired model is 𝑆(𝑡) = 𝐴𝑏 𝑡. At time 𝑡 = 0 (April 1, 2003) the number of cases was 1,804, so 𝐴 = 1,804. Since the number was increasing by 4% each day, the number of cases is multiplied by 1.04 each day, so 𝑏 = 1.04. Hence, the model is 𝑆(𝑡) = 1,804(1.04) 𝑡. Since April 30, 2003 corresponds to 𝑡 = 29, the number of cases was about 𝑆(14) = 1,804(1.04) 29 ≈ 5,626. 77. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 5,000, 𝑟 = 0.05∕100 = 0.0005, and 𝑛 = 12. We get the model 𝐴(𝑡) = 5,000(1 + 0.0005∕12) 12𝑡 In July 2028 (𝑡 = 7), the deposit would be worth 5,000(1 + 0.0005∕12) 12(7) ≈ $5,018. 78. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 4,000, 𝑟 = 0.0061, and 𝑛 = 365. We get the model 𝐴(𝑡) = 4,000(1 + 0.0061∕365) 365𝑡

In July 2029 (𝑡 = 8), the deposit would be worth 4,000(1 + 0.0061∕365) 365(8) ≈ $4,200. Solutions Section 2.2

79. Substitute 𝐴 = 15,000, 𝑃 = 10,000, 𝑟 = 0.025, and 𝑛 = 1 into the compound interest formula: 15,000 = 10,000(1 + 0.025) 𝑡, (1.025) 𝑡 = 1.5, 𝑡 = log(1.5)∕ log(1.025) ≈ 16 years.

80. Substitute 𝐴 = 11,000, 𝑃 = 10,000, 𝑟 = 0.00025, and 𝑛 = 1 into the compound interest formula: 11,000 = 10,000(1 + 0.00025) 𝑡, (1.00025) 𝑡 = 1.1, 𝑡 = log(1.1)∕ log(1.00025) ≈ 381 years. 81. Substitute 𝐴 = 20,000, 𝑃 = 10,400, 𝑟 = 0.010, and 𝑛 = 12 into the compound interest formula: 20,000 = 10,400(1 + 0.010∕12) 12𝑡, (1 + 0.010∕12) 12𝑡 = 200∕104, 12𝑡 = log(200∕104)∕ log(1 + 0.010∕12) ≈ 785 months. 82. Substitute 𝐴 = 20,000, 𝑃 = 10,400, 𝑟 = 0.025, and 𝑛 = 12 into the compound interest formula: 20,000 = 10,400(1 + 0.025∕12) 12𝑡, (1 + 0.025∕12) 12𝑡 = 200∕104, 12𝑡 = log(200∕104)∕ log(1 + 0.025∕12) ≈ 314 months.

83. Substitute 𝐴 = 2𝑃 , 𝑟 = 0.035, and 𝑛 = 2 into the compound interest formula: 2𝑃 = 𝑃 (1 + 0.035∕2) 2𝑡, (1.0175) 2𝑡 = 2, 2𝑡 = log 2∕ log 1.0175, 𝑡 = (log 2∕ log 1.0175)∕2 ≈ 20 years.

84. Substitute 𝐴 = 2𝑃 , 𝑟 = 0.0425, and 𝑛 = 2 into the compound interest formula: 2𝑃 = 𝑃 (1 + 0.0425∕2) 2𝑡, (1.02125) 2𝑡 = 2, 2𝑡 = log 2∕ log 1.02125, 𝑡 = (log 2∕ log 1.02125)∕2 ≈ 16 years. 85. If 99.95% has decayed, then 0.05% remains, so 𝐶(𝑡) = 0.0005𝐴. Therefore, 0.0005𝐴 = 𝐴(0.999879) 𝑡, (0.999879) 𝑡 = 0.0005, so 𝑡 = log(0.0005)∕ log(0.999879) ≈ 63,000 years old. 86. If 30% has decayed, then 70% remains, so 𝐶(𝑡) = 0.70𝐴. Therefore, 0.70𝐴 = 𝐴(0.999879) 𝑡, (0.999879) 𝑡 = 0.70, 𝑡 = log(0.70)∕ log(0.999879) ≈ 3,000 years old.

87. a. Let 𝑦 be the number of frogs in year 𝑡, with 𝑡 = 0 representing 2 years ago; we seek a model of the form 𝑦 = 𝐴𝑏 𝑡. We are given the initial value of 𝐴 = 50,000 and are told that 50,000𝑏 2 = 32,000. This gives 𝑏 = (32,000∕50,000) 1∕2 = 0.64 1∕2 = 0.8. Thus, 𝑦 = 50,000(0.8) 𝑡. b. When 𝑡 = 3, 𝑦 = 50,000(0.8) 3 = 25,600 tags. 1,000 c. We want 𝑡 so that 50,000(0.8) 𝑡 = 1,000 so 0.8 𝑡 = = 0.02. Taking the logarithm of both sides 50,000 gives log(0.8 𝑡) = log 0.02 ⟹ 𝑡 log 0.8 = log 0.02, so log 0.02 𝑡= ≈ 17.53 years, log 0.8 so by 𝑡 = 18 (16 years from now) the number will have dropped to less than 1,000.

88. a. Let 𝑦 be the number of flies in year 𝑡, with 𝑡 = 0 representing 3 years ago; we seek a model of the form 𝑦 = 𝐴𝑏 𝑡. We are given the initial value of 𝐴 = 4,000 and we are told that 4,000𝑏 3 = 1,372. This gives 𝑏 = (1,372∕4,000) 1∕3 = 0.343 1∕3 = 0.7. Thus, 𝑦 = 4,000(0.7) 𝑡. b. When 𝑡 = 4, 𝑦 = 4,000(0.7) 4 = 960.4 ≈ 960 flies. 500 c. We want 𝑡 so that 4,000(0.7) 𝑡 = 500 so 0.7 𝑡 = = 0.125. Taking the logarithm of both sides gives 4,000 log(0.7 𝑡) = log 0.125 ⟹ 𝑡 log 0.7 = log 0.125, so log 0.125 𝑡= ≈ 5.8 years, log 0.7 so by 𝑡 = 6 (three years from now) the number will have dropped to less than 500.

Solutions Section 2.2 89. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked):

Note that the displayed model has the form 𝑃 (𝑡) = 0.3394𝑒 0.1563𝑡. 𝑒 is the number 2.71828183...... discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 0.153𝑒 0.0863𝑡 = 0.3394(𝑒 0.1563) 𝑡 ≈ 0.3394(1.169) 𝑡

𝑒 0.1563 is EXP(0.1563) in Excel

so the model is

𝑃 (𝑡) = 0.339(1.169) 𝑡.

b. In 2005, 𝑡 = 11, so the predicted cost is 𝑃 (15 = 0.339(1.169) 11 ≈ $1.9 million.

90. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked):

Note that the displayed model has the form 𝑃 (𝑡) = 0.1533𝑒 0.0863𝑡. 𝑒 is the number 2.71828183...... discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 0.153𝑒 0.0863𝑡 = 0.153(𝑒 0.0863𝑡) 𝑡 ≈ 0.153(1.090) 𝑡

so the model is

𝑒 0.0863 is EXP(0.0863) in Excel

𝑃 (𝑡) = 0.153(1.090) 𝑡.

Solutions Section 2.2

b. In 2005, 𝑡 = 15, so the predicted cost is 𝑃 (15) = 0.153(1.090) 15 ≈ $0.56 million.

91. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked; we have enlarged the graph in order to see more clearly how the points are scattered around the regression curve):

Note that the displayed model has the form 𝑈(𝑡) = 189.17𝑒 0.0096𝑡. 𝑒 is the number 2.71828183 . . . discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 189.17𝑒 0.0096𝑡 = 189.17(𝑒 0.0096) 𝑡 ≈ 189.2(1.010) 𝑡

so the model is

𝑈(𝑡) = 189.2(1.010) 𝑡.

𝑒 0.0096 is EXP(0.0096) in Excel

Solutions Section 2.2 b. Yes; the points are very close to, and scattered randomly around, the regression curve. 92. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked; we have enlarged the graph in order to see more clearly how the points are scattered around the regression curve):

Note that the displayed model has the form 𝑊 (𝑡) = 3.1722𝑒 0.0158𝑡 𝑒 is the number 2.71828183 . . . discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 3.1722𝑒 0.0158𝑡 = 3.1722(𝑒 0.0158) 𝑡 ≈ 3.172(1.016) 𝑡

𝑒 0.0158 is EXP(0.0158) in Excel

so the model is

𝑊 (𝑡) = 3.172(1.016) 𝑡.

b. No; the points suggest a graph that is less curved than the regression curve. 93. (B) An exponential function eventually becomes larger than any polynomial. 94. (B) An exponential decay function eventually becomes smaller than the reciprocal of any polynomial (see Exercise 99).

Solutions Section 2.2 95. Exponential functions of the form 𝑓(𝑥) = 𝐴𝑏 𝑥 (𝑏 > 1) increase rapidly for large values of 𝑥. In reallife situations, such as population growth, this model is reliable only for relatively short periods of growth. Eventually, population growth tapers off because of pressures such as limited resources and overcrowding. 96. If an investment earns 5% compounded continuously, it is as if interest is being added every moment. Thus, the interest for 1 month will have been calculated on a growing amount, rather than on the original fixed (smaller) amount used for interest compounded monthly. 97. Linear functions are better for cost models where there is a fixed cost and a variable cost and for simple interest, where interest is paid only on the original amount invested. Exponential models are better for compound interest and population growth. In both of these latter examples, the rate of growth depends on the current number of items, rather than on a fixed initial quantity. 98. Quadratic models are better for revenue and profit functions where demand depends linearly on the price. Exponential models are better for compound interest and population growth. 99. Take the ratios 𝑦2 ∕𝑦1 and 𝑦3 ∕𝑦2 . If they are the same, the points fit on an exponential curve.

100. For the points to fit on an exponential curve we must have 𝑦2 ∕𝑦1 = 𝑦3 ∕𝑦2 , hence 𝑦3 = 𝑦22∕𝑦1 .

Solutions Section 2.3 Section 2.3

1. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 5,000 and 𝑟 = 0.10, giving 𝑓(𝑡) = 5,000𝑒 0.10𝑡.

2. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 2,000 and 𝑟 = 0.053, giving 𝑓(𝑡) = 2,000𝑒 0.053𝑡.

3. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 1,000 and 𝑟 = −0.063, giving 𝑓(𝑡) = 1,000𝑒 −0.063𝑡.

4. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 10,000 and 𝑟 = −0.60, giving 𝑓(𝑡) = 10,000𝑒 −0.60𝑡. 5. 𝑓(𝑥) = 4𝑒 2𝑥 = 4(𝑒 2) 𝑥 ≈ 4(7.389) 𝑥

6. 𝑓(𝑥) = 2.1𝑒 −0.1𝑥 = 2.1(𝑒 −0.1) 𝑥 ≈ 2.1(0.9048) 𝑥

7. 𝑓(𝑡) = 2.1(1.001) 𝑡 = 2.1𝑒 (ln 1.001)𝑡 ≈ 2.1𝑒 0.0009995𝑡

8. 𝑓(𝑡) = 2.1(0.991) 𝑡 = 23.4𝑒 (ln 0.991)𝑡 ≈ 23.4𝑒 −0.009041𝑡 9. 𝑓(𝑡) = 10(0.987) 𝑡 = 10𝑒 (ln 0.987)𝑡 ≈ 10𝑒 −0.01309𝑡 10. 𝑓(𝑡) = 2.3(2.2) 𝑡 = 2.3𝑒 (ln 2.2)𝑡 ≈ 2.3𝑒 0.7885𝑡

11. 𝑄 = 1,000 when 𝑡 = 0; half-life = 1. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 1,000. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 1, so 𝑘 = ln 2, and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 1,000𝑒 −𝑡 ln 2.

12. 𝑄 = 2,000 when 𝑡 = 0; half-life = 5. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 2,000. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 5, so 2𝑘 = ln 2, giving 𝑘 = (ln 2)∕5 and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 2,000𝑒 −𝑡(ln 2)∕5.

13. 𝑄 = 500 when 𝑡 = 0; half-life = 50. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 500. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 50, so 𝑘 = ln 2∕50, and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 500𝑒 −𝑡 ln 2∕50. 14. 𝑄 = 50 when 𝑡 = 0; half-life = 500. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 50. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 500, so 𝑘 = ln 2∕500, and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 50𝑒 −𝑡 ln 2∕500. 15. 𝑄 = 1,000 when 𝑡 = 0; doubling time = 2. We want a model of the form 𝑄 = 𝑄0 𝑒 𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 1,000. For 𝑘, we use the formula 𝑡𝑑 𝑘 = ln 2 with 𝑡𝑑 = doubling time = 2, so 2𝑘 = ln 2, giving 𝑘 = (ln 2)∕2 and the model is 𝑄 = 𝑄0 𝑒 𝑘𝑡 = 1,000𝑒 𝑡(ln 2)∕2.

Solutions Section 2.3 16. 𝑄 = 2,000 when 𝑡 = 0; doubling time = 5. We want a model of the form 𝑄 = 𝑄0 𝑒 𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 2,000. For 𝑘, we use the formula 𝑡𝑑 𝑘 = ln 2 with 𝑡𝑑 = doubling time = 5, so 5𝑘 = ln 2, giving 𝑘 = (ln 2)∕5 and the model is 𝑄 = 𝑄0 𝑒 𝑘𝑡 = 2,000𝑒 𝑡(ln 2)∕5.

17. 𝑄 = 1,000𝑒 0.5𝑡. Since the exponent is positive, the model represents exponential growth. We use the formula 𝑡𝑑 𝑘 = ln 2, giving 𝑡𝑑 (0.5) = ln 2, giving doubling time = 𝑡𝑑 = (ln 2)∕0.5 = 2 ln 2.

18. 𝑄 = 1,000𝑒 −0.025𝑡. Since the exponent is negative, the model represents exponential decay. We use the formula 𝑡ℎ 𝑘 = ln 2, giving 𝑡ℎ (0.025) = ln 2, giving half-life = 𝑡ℎ = (ln 2)∕0.025 = 40 ln 2. 19. 𝑄 = 100𝑒 −𝑡. Since the exponent is negative, the model represents exponential decay. We use the formula 𝑡ℎ 𝑘 = ln 2, giving 𝑡ℎ (1) = ln 2, giving half-life = 𝑡ℎ = ln 2.

20. 𝑄 = 5 000𝑒 𝑡∕3. Since the exponent is positive, the model represents exponential growth. We use the formula 𝑡𝑑 𝑘 = ln 2, giving 𝑡𝑑 (1∕3) = ln 2, giving doubling time = 𝑡𝑑 = 3 ln 2. 21. 𝑄 = 𝑄0 𝑒 −4𝑡. Since the exponent is negative, the model represents exponential decay. We use the formula 𝑡ℎ 𝑘 = ln 2, giving 𝑡ℎ (4) = ln 2, giving half-life = 𝑡ℎ = (ln 2)∕4. 22. 𝑄 = 𝑄0 𝑒 𝑡. Since the exponent is positive, the model represents exponential growth. We use the formula 𝑡𝑑 𝑘 = ln 2, giving 𝑡𝑑 (1) = ln 2, giving doubling time = 𝑡𝑑 = ln 2.

23. a. From the continuous compounding formula, 1,000𝑒 0.04×10 = $1,491.82, of which $491.82 is interest. b. 𝑃 = 1,000𝑒 0.04𝑡 = 1,000(𝑒 0.04) 𝑡 ≈ 1,000(1.040811) 𝑡, which corresponds to an annually compounded interest rate of 4.08% 24. a. From the continuous compounding formula, 2,000𝑒 0.31×10 = $44,395.90, of which $42,395.90 is interest. b. 𝑃 = 2,000𝑒 0.31𝑡 = 2,000(𝑒 0.31) 𝑡 ≈ 2,000(1.3634) 𝑡, which corresponds to an annually compounded interest rate of 36.34%

25. Substitute 𝐴 = 700, 𝑃 = 500, and 𝑟 = 0.10 in the continuous compounding formula: 700 = 500𝑒 0.10𝑡. Solve for 𝑡: 𝑒 0.10𝑡 = 700∕500 = 7∕5, 0.10𝑡 = ln(7∕5), 𝑡 = 10 ln(7∕5) ≈ 3.36 years. 26. Substitute 𝐴 = 700, 𝑃 = 500, and 𝑟 = 0.15 in the continuous compounding formula: 700 = 500𝑒 0.15𝑡. Solve for 𝑡: 𝑒 0.15𝑡 = 700∕500 = 7∕5, 0.15𝑡 = ln(7∕5), 𝑡 = ln(7∕5)∕0.15 ≈ 2.24 years. 27. Substitute 𝐴 = 3𝑃 and 𝑟 = 0.10 in the continuous compounding formula: 3𝑃 = 𝑃 𝑒 0.10𝑡. Solve for 𝑡: 𝑒 0.10𝑡 = (3𝑃 )∕𝑃 = 3, 0.10𝑡 = ln 3, 𝑡 = 10 ln 3 ≈ 11 years. 28. Substitute 𝐴 = 1,000,000, 𝑃 = 1,000 and 𝑟 = 0.10 in the continuous compounding formula: 1,000,000 = 1,000𝑒 0.10𝑡. Solve for 𝑡: 𝑒 0.10𝑡 = 1,000,000∕1,000 = 1,000, 0.10𝑡 = ln 1,000, 𝑡 = 10 ln 1,000 ≈ 69 years. 29. The bank can afford to pay its customers 𝑟 𝑛𝑡 𝑃 (1 + ) = 𝑃 (1.036) 𝑡 𝑛

= 𝑃 #𝑒 ln 1.036$

Solutions Section 2.3

𝑡

≈ 𝑃 (𝑒 0.035367) 𝑡, corresponding to about 3.54% compounded continuously. 30. The bank can afford to pay its customers 𝑟 𝑛𝑡 𝑃 (1 + ) = 𝑃 (1.0456) 𝑡 𝑛

= 𝑃 #𝑒 ln 1.0456$ ≈ 𝑃 (𝑒 0.04459) 𝑡,

𝑡

corresponding to about 4.46% compounded continuously.

31. The desired model is 𝐶(𝑡) = 𝐴𝑒 𝑘𝑡. At time 𝑡 = 0 (April 1, 2014) the number of cases was 100, so 𝐴 = 100. Because the number was increasing exponentially with a growth constant of 72% per month, 𝑘 = 0.72 per month. Hence, the model is 𝐶(𝑡) = 100𝑒 0.72𝑡. As August 1, 2014 corresponds to 𝑡 = 4, the number of cases was about 𝐶(4) = 100𝑒 0.72×4 ≈ 1,781. 32. The desired model is 𝐷(𝑡) = 𝐴𝑒 𝑘𝑡. At time 𝑡 = 0 (April 1, 2014) the number of deaths was 90, so 𝐴 = 90. Because the number was increasing exponentially with a growth constant of 60% per month, 𝑘 = 0.60. Hence, the model is 𝐷(𝑡) = 90𝑒 0.60𝑡. As October 1, 2014 corresponds to 𝑡 = 6, the number of deaths was about 𝐷(6) = 90𝑒 0.60×6 ≈ 3,294.

33. At time 𝑡 = 0 (April 1, 2021) the number of active cases was 614,649, so 𝐴 = 614,649. Since the number was increasing by 6.9% each day, that number is multiplied by 1.069 each day, so an exponential model for the number of active cases is 𝐶(𝑡) = 614,649(1.069) 𝑡. To convert this to the desired form 𝐴𝑒 𝑘𝑡 we follow the hint and write 𝐶(𝑡) = 614,649(1.069) 𝑡 = 614,649#𝑒 ln 1.069$

𝑡

≈ 614,649(𝑒 0.06672363) 𝑡 = 614,649𝑒 0.06672363𝑡

Thus, 𝑘 ≈ 0.06672 per day to 4 significant digits, and the model is 𝐶(𝑡) = 614,649𝑒 0.06672𝑡. As April 21, 2021 corresponds to 𝑡 = 20, the estimated number of cases was about 𝐶(20) = 614,649𝑒 0.06672(20) ≈ 2,334,000. The figure with the fewest significant digits we were given was the rate of increase of 6.9% per day, so we should not trust the answer to more than two significant digits. 34. At time 𝑡 = 0 (April 1, 2003) the number of cases was 1,804, so 𝐴 = 1,804. Since the number was increasing by 4% each day, the number of cases is multiplied by 1.04 each day, so an exponenial model for the number of cases is 𝑆(𝑡) = 1,804(1.04) 𝑡. To convert this to the desired form 𝐴𝑒 𝑘𝑡 we follow the hint and write 𝑆(𝑡) = 1,804(1.04) 𝑡 = 1,804#𝑒 ln 1.04$

𝑡

≈ 1,804(𝑒 0.0392207) 𝑡 = 1,804𝑒 0.0392207𝑡

Thus, 𝑘 ≈ 0.03922 per day to 4 significant digits, and the model is 𝑆(𝑡) = 1,804𝑒 0.03922𝑡. As April 30, 2003 corresponds to 𝑡 = 29, the number of cases was about 𝑆(29) = 1,804𝑒 0.03922(29) ≈ 5,626.

35. We want a model of the form 𝐴 = 𝑃 𝑒 𝑟𝑡. For the rate of interest 𝑟, we use the formula 𝑡𝑑 𝑟 = ln 2 with

Solutions Section 2.3 𝑡𝑑 = doubling time = 3, so 3𝑟 = ln 2, giving 𝑟 = (ln 2)∕3 ≈ 0.231, or 23.1%.

36. We want a model of the form 𝐴 = 𝑃 𝑒 −𝑟𝑡. For the rate of interest 𝑟, we use the formula 𝑡ℎ 𝑟 = ln 2 with 𝑡ℎ = half-life = 2, so 2𝑟 = ln 2, giving 𝑟 = (ln 2)∕2 ≈ 0.347, or 34.7%. The investment is depreciating continuously at a rate of 34.7% per year. 37. The growth constant is 𝑘 = 0.025 per year, so the doubling time is 𝑡𝐷 = (ln 2)∕0.025 ≈ 27.73 years. 38. The growth constant is 𝑘 = 0.035 per year, so the doubling time is 𝑡𝐷 = (ln 2)∕0.035 ≈ 19.80 years. 39. For interest compounded twice per year, 0.00025 2𝑡 𝐴(𝑡) = 𝑃 (1 + = 𝑃 (1.000125) 2𝑡 = 𝑃 𝑒 (ln 1.000125)⋅2𝑡 2 ) ≈ 𝑃 𝑒 0.000249984𝑡, which would be 0.0249984% per year. 40. For interest compounded quarterly, 0.01 4𝑡 𝐴(𝑡) = 𝑃 (1 + = 𝑃 (1.0025) 4𝑡 = 𝑃 𝑒 (ln 1.0025)⋅4𝑡 4 ) ≈ 𝑃 𝑒 0.0099875𝑡, which would be 0.99875% per year.

41. a. The years 1950, 2000, 2050, and 2100 correspond to 𝑡 = 200, 250, 300, and 350 respectively. Technology formula: 280*e^(0.00146x) year 1950 2000 2050 2100 𝑡 200

𝐶(𝑡) parts per million 375

250

300

350

403

434

467

b. The level surpasses 450 when

450 = 280𝑒 0.00146𝑡 450 𝑒 0.00146𝑡 = 280 450 0.00146𝑡 = ln ( = ln 450 − ln 280 280 ) 1 𝑡= (ln 450 − ln 280) ≈ 324.97. 0.00146

Thus, the level passes 450 shortly before the end of 2074 (𝑡 = 324.97; we should not round, as the 450 level is predicted to be passed shortly before 2075 (𝑡 = 325)). 42. a. The years 1950, 2000, 2050, and 2100 correspond to 𝑡 = 200, 250, 300, and 350 respectively. Technology formula: 720*e^(0.00355x) year 1950 2000 2050 2100 𝑡

200

250

300

350

𝐶(𝑡) parts per billion 1,460 1,750 2,090 2,490 b. The level surpasses 2,000 when

2,000 = 720𝑒 0.00355𝑡 2,000 𝑒 0.00355𝑡 = 720 2,000 0.00355𝑡 = ln ( = ln 2,000 − ln 720 720 ) 1 𝑡= (ln 2,000 − ln 720) ≈ 287.79. 0.00355

Solutions Section 2.3

Thus, the level passes 2,000 around three-quarters way through 2037 (𝑡 = 287.79; we should not round, as the 2,000 level is predicted to be passed before 2038 (𝑡 = 288)). 43. The decay constant is 𝑘 = 0.000433. Thus, ln 2 ln 2 𝑡ℎ = ≈ 1,600 years (to the nearest 100 years). = 𝑘 0.000433 44. The decay constant is 𝑘 = 0.0861. Thus, ln 2 ln 2 𝑡ℎ = ≈ 8.05 days (to two decimal places). = 𝑘 0.0861 45. a. 𝑡ℎ 𝑘 = ln 2 5𝑘 = ln 2 𝑘 = (ln 2)∕5 ≈ 0.139 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡 ≈ 𝑄0 𝑒 −0.139𝑡 b. One third has decayed when 2/3 is left: 2 𝑄 = 𝑄0 𝑒 −0.139𝑡 3 0 2 = 𝑒 −0.139𝑡 3

ln(2∕3) = −0.139𝑡 𝑡 = ln(2∕3)∕(−0.139) ≈ 3 years.

46. a. 𝑡ℎ 𝑘 = ln 2 28𝑘 = ln 2 𝑘 = (ln 2)∕28 ≈ 0.0248 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡 ≈ 𝑄0 𝑒 −0.0248𝑡 b. Three fifths has decayed when 2/5 is left: 2 𝑄 = 𝑄0 𝑒 −0.0248𝑡 5 0 2 = 𝑒 −0.0248𝑡 5

ln(2∕5) = −0.0248𝑡 𝑡 = ln(2∕5)∕(−0.0248) ≈ 37 years.

47. Let 𝑄(𝑡) be the amount left after 𝑡 million years. We first find a model of the form 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡. To find 𝑘 use 𝑡ℎ 𝑘 = ln 2: 710𝑘 = ln 2 𝑘 = (ln 2)∕710 ≈ 0.0009763 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡 ≈ 𝑄0 𝑒 −0.0009763𝑡. We now answer the question: 𝑄0 = 10g, 𝑄(𝑡) = 1g. Substituting in the model gives 1 = 10𝑒 −0.0009763𝑡 𝑒 −0.0009763𝑡 = 0.1 −0.0009763𝑡 = ln(0.1) 𝑡 = − ln(0.1)∕0.0009763 ≈ 2,360 million years (rounded to three significant digits).

Solutions Section 2.3 48. Let 𝑄(𝑡) be the amount left after 𝑡 years. We first find a model of the form 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡. To find 𝑘 use 𝑡ℎ 𝑘 = ln 2: 24,400𝑘 = ln 2 𝑘 = (ln 2)∕24,400 ≈ 0.000028408 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡(𝑄0 𝑒 −0.000028408𝑡. We now answer the question: 𝑄0 = 10g, 𝑄(𝑡) = 1g. Substituting in the model gives 1 = 10𝑒 −0.000028408𝑡 𝑒 −0.000028408𝑡 = 0.1 −0.000028408𝑡 = ln(0.1) 𝑡 = − ln(0.1)∕0.000028408 ≈ 81,100 years (rounded to three significant digits). 49. a. By the discussion at the end of the section, Tripling time × Growth constant = ln 3 6𝑘 = ln 3, so ln 3 𝑘= ≈ 0.183. 6 ln 2 b. The doubling time is ≈ 3.8 months. 𝑘 50. a. By the discussion at the end of the section, Quadrupling time × Growth constant = ln 4 8𝑘 = ln 4, so ln 4 𝑘= ≈ 0.1733. 8 ln 3 b. The tripling time is ≈ 6.3 months. 𝑘

51. Let 𝑄(𝑡) be the amount left after 𝑡 hours. We first find a model of the form 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡. To find 𝑘 use 𝑡ℎ 𝑘 = ln 2: 2𝑘 = ln 2 𝑘 = (ln 2)∕2 ≈ 0.3466 𝑄(𝑡) = 𝑄0 𝑒 −𝑘𝑡 ≈ 𝑄0 𝑒 −0.3466𝑡. We now answer the question: 𝑄0 = 300mg, 𝑄(𝑡) = 100mg. Substituting in the model gives 100 = 300𝑒 −0.3466𝑡 𝑒 −0.3466𝑡 = 1∕3 −0.3466𝑡 = ln(1∕3) 𝑡 = − ln(1∕3)∕0.3466 ≈ 3.2 hours.

52. Let 𝑄 = 𝑄0 𝑒 −𝑘𝑡 represent the blood alcohol level at time 𝑡 hours. We are told that 𝑄0 = 200 and that one fourth of the alcohol is removed every hour; so, after 1 hour, three fourths remains, so 0.75𝑄0 = 𝑄0 𝑒 −𝑘. Thus, 𝑒 −𝑘 = 0.75, and so 𝑘 = − ln(0.75) ≈ 0.28768. We wish to find 𝑡 when 𝑄 = 80: 80 = 200𝑒 −0.28768𝑡 80 ln 0.4 𝑒 −0.28768𝑡 = ≈ 3.2 hours. = 0.4, so 𝑡 = 200 −0.28768

53. We first find a model of the form 𝐴(𝑡) = 𝑃 𝑏 𝑡. We are given the data points (0, 1) and (2, 0.7), so 𝑃 = 1 and 𝑏 2 = 0.7, so 𝑏 = (0.7) 1∕2; the model is 𝐴(𝑡) = (0.7) 𝑡∕2 = (√0.7) 𝑡. This model can also be written as 𝑒 𝑟𝑡, where 𝑟 = ln √0.7 ≈ −0.1783, so 𝑘 ≈ 0.1783: ln 2 ln 2 The half life is then ≈ ≈ 3.89 days. 𝑘 0.1783

Solutions Section 2.3 54. We first find a model of the form 𝐴(𝑡) = 𝑃 𝑏 𝑡. We are given the data points (0, 1) (half of the original 2 gm was stolen) and (3, 0.1), so 𝑃 = 1 and 𝑏 3 = 0.1, so 𝑏 = (0.1) 1∕3, and the model is 𝐴(𝑡) = (0.1) 𝑡∕3 = ((0.1) 1∕3) 𝑡, so 𝑏 = (0.1) 1∕3. This model can also be written as 𝑒 𝑟𝑡, where 𝑟 = ln(0.1) 1∕3 ≈ −0.76753, so 𝑘 ≈ 0.76753: ln 2 ln 2 The half life is then ≈ ≈ 0.903 days. 𝑘 0.76753 55. Answers may vary. One example is in the case of continuous compounding; the use of 𝐴𝑏 𝑡 would require us to write the continuous compounding formula 𝐴𝑒 𝑟𝑡 as 𝐴(𝑒 𝑟) 𝑡.

56. Answers may vary. One example is in the case of compound interest, where the use of 𝐴𝑒 𝑘𝑡 would 𝑛

require us to write the compound interest formula as 𝐴𝑒 (ln(1+𝑟∕𝑛) )𝑡.

57. This reasoning is suspect. The bank need not use its computer resources to update all the accounts every minute but can instead use the continuous compounding formula to calculate the balance in any account at any time as needed. 58. In the first minute your balance has grown to around $10,000.00228, which our spreadsheet rounds to $10,000, and so that repeats evey minute, leaving you wih no interest. 59. The article was published about a year before the "housing bubble" burst in 2006, whereupon, contrary to the prediction of the graph, house prices started to fall and continued to drop for several years. This shows the danger of using any mathematical model to extrapolate. However, the blogger was cautious in the choice of words, claiming only to be estimating what the future U.S. median house price "might be."

60. Suppose for the sake of argument that the model had been constructed around 𝑡 = 20 (1993). Then the resulting exponential model would have predicted house prices reasonably accurately for around 12 more years. It is just coincidence that the particular model was created a year before prices started to fall. In general, exponential models are very good at extrapolating current trends; it is only when things go awry that they become unreliable.

Solutions Section 2.4 Section 2.4

1. 𝑁 = 7, 𝐴 = 6, 𝑏 = 2; 7/(1+6*2^-x)

2. 𝑁 = 4, 𝐴 = 0.333, 𝑏 = 4; 4/(1+0.333*4^-x)

3. 𝑁 = 10, 𝐴 = 4, 𝑏 = 0.3; 10/(1+4*0.3^-x)

4. 𝑁 = 100, 𝐴 = 5, 𝑏 = 0.5; 100/(1+5*0.5^-x)

5. 𝑁 = 4, 𝐴 = 7, 𝑏 = 1.5; 4/(1+7*1.5^-x)

Solutions Section 2.4 6. 𝑁 = 8.5, 𝐴 = 3.25, 𝑏 = 1.05; 8.5/(1+3.25*1.05^-x)

7. 𝑁 = 200, the limiting value. 10 = 𝑓(0) = 𝑁∕(1 + 𝐴) = 200∕(1 + 𝐴), so 𝐴 = 19. For small values of 𝑥, 𝑓(𝑥) ≈ 10𝑏 𝑥; to double with every increase of 1 in 𝑥 we must therefore have 𝑏 = 2. This gives 200 𝑓(𝑥) = . 1 + 19(2 −𝑥) 8. 𝑁 = 10, the limiting value. 1 = 𝑓(0) = 𝑁∕(1 + 𝐴) = 10∕(1 + 𝐴), so 𝐴 = 9. For small values of 𝑥, 𝑓(𝑥) ≈ 𝑏 𝑥; to grow by 50% with every increase of 1 in 𝑥 we must therefore have 𝑏 = 1.5. This gives 10 𝑓(𝑥) = . 1 + 9(1.5) −𝑥

9. 𝑁 = 6, the limiting value. 3 = 𝑓(0) = 𝑁∕(1 + 𝐴) = 6∕(1 + 𝐴), so 𝐴 = 1. 4 = 𝑓(1) = 6∕(1 + 𝑏 −1), 6 1 + 𝑏 −1 = 6∕4 = 3∕2, 𝑏 −1 = 1∕2, 𝑏 = 2. So, 𝑓(𝑥) = . 1 + 2 −𝑥

10. 𝑁 = 4, the limiting value. 1 = 𝑓(0) = 𝑁∕(1 + 𝐴) = 4∕(1 + 𝐴), so 𝐴 = 3. 2 = 𝑓(1) = 4∕(1 + 3𝑏 −1), 4 1 + 3𝑏 −1 = 4∕2 = 2, 𝑏 −1 = 1∕3, 𝑏 = 3. So, 𝑓(𝑥) = . 1 + 3(3 −𝑥)

11. B: The limiting value is 9, eliminating (A). The initial value is 2 = 9∕(1 + 3.5), not 9∕(1 + 0.5), so the answer is (B), not (C). 12. A: The limiting value is 8, eliminating (C). The initial value is 1 = 8∕(1 + 7) as in (A).

13. B: The graph decreases, so 𝑏 < 1, eliminating (C). The 𝑦-intercept is 2 = 8∕(1 + 3) as in (B).

14. C: The graph decreases, so 𝑏 < 1, eliminating (A). The 𝑦-intercept is 2.5 = 10∕(1 + 3) as in (C).

15. C: The initial value is 2, as in (A) or (C). If 𝑏 were 5, an increase in 1 in 𝑥 would multiply the value of 𝑓(𝑥) by approximately 5 when 𝑥 is small. However, increasing 𝑥 from 0 to 10 does not quite double the value. Hence 𝑏 must be smaller, as in (C). 16. C: The initial value is 2, as in (A) or (C). If 𝑏 were 15, an increase in 1 in 𝑥 would multiply the value of 𝑓(𝑥) by approximately 15 when 𝑥 is small. However, increasing 𝑥 from 0 to 20 does not quite double the value. Hence 𝑏 must be smaller, as in (C).

17. 𝑦 =

7.2 1 + 2.4(1.04) −𝑥

19. 𝑦 =

97 1 + 2.2(0.942) −𝑥

21. 𝑓(𝑥) = log4 𝑥. We plot some points and draw the graph: 𝑥 1/16

𝑓(𝑥)

−2

1/4 −1

1/2

−1∕2

18. 𝑦 =

10 1 + 2.5(1.04) −𝑥

20. 𝑦 =

75 1 + 1.5(0.980) −𝑥

Solutions Section 2.4

1

2

4

0

1/2

1

22. 𝑓(𝑥) = log5 𝑥. We plot some points and draw the graph: 𝑥 1/25

𝑓(𝑥)

−2

1/5 −1

1

5

0

1

23. 𝑓(𝑥) = log1∕4 𝑥. We plot some points and draw the graph:

24. 𝑓(𝑥) = log1∕5 𝑥. We plot some points and draw the graph:

Solutions Section 2.4

𝑥 1/16

𝑓(𝑥)

2

1/4

1/2

1

1

1/2

0

2

−1∕2

4

−1

25. 𝑓(𝑥) = log4 𝑥 + 1. We plot some points and draw the graph: 𝑥 1/16

𝑓(𝑥)

−1

1/4 0

1/2

1∕2

1

2

4

1

3/2

2

27. 𝑓(𝑥) = log4 (𝑥 − 1). We plot some points and draw the graph: 𝑥 17/16

𝑓(𝑥)

−2

5/4 −1

3/2

−1∕2

2

3

5

0

1/2

1

𝑥 1/25

𝑓(𝑥)

2

1/5

1

1

0

5

−1

26. 𝑓(𝑥) = log5 𝑥 − 1. We plot some points and draw the graph: 𝑥 1/25

𝑓(𝑥)

−3

1/5 −2

1

5 0

−1

28. 𝑓(𝑥) = log5 (𝑥 + 1). We plot some points and draw the graph: 𝑥 −24∕25 −4∕5

𝑓(𝑥)

−2

−1

0

4

0

1

29. a. We can eliminate (C) because 𝑏 = 0.8 is less than 1, giving a decreasing function of 𝑡. We can eliminate (D) because 𝑏 = 13.313, indicating an initial rate of growth of 1,331.3% per month—far faster than suggested by the data. (Alternatively, we can calculate ln 𝐴∕ ln 𝑏 ≈ 2 and notice that the point of inflection is further to the right than that.) The value 𝑁 = 44.6 in (B) predicts a leveling-off value of 44.6%, which is clearly too small. Thus, we are left with choice (A). b. Mid-January is in the initial period of the model, when the rate of growth is governed by exponential

Solutions Section 2.4 growth with base 𝑏 (𝐴 = 142 is much larger than 1). Since 𝑏 = 3.066, the number was being multiplied by 3.066 = 1 + 2.066 each month, so growing by around 207% per month at that time.

30. a. We can eliminate (B) because 𝑏 = 0.931 < 1, giving a decreasing function of 𝑡. We can eliminate (A) because 𝑏 = 3.83, indicating that the curve initially behaves like an exponential function with base 3.8, almost quadrupling every month. This is not what is suggested in the data. (Alternatively, we can calculate ln 𝐴∕ ln 𝑏 ≈ 3 and notice that the point of inflection is significantly further to the right than that.) The value 𝑁 = 60.32 in (C) predicts a horizontal asymptote of 60.32, which is clearly too small. Thus, we are left with choice (D). b. Mid-February is in the initial period of the model, when the rate of growth is governed by exponential growth with base 𝑏 (𝐴 = 104 is much larger than 1). Since 𝑏 = 2.88, the number was being multiplied by 2.88 = 1 + 1.88 each month, so growing by around 188% per month at that time.

31. a. We can eliminate (B) and (D) because 𝑏 = 0.8 is less than 1, giving a decreasing function of 𝑡. The value 𝑁 = 2.0 in (C) predicts a leveling-off value of around 2%, which is clearly too small. Thus, we are left with choice (A). ln 𝐴 ln 8.6 b. The logistic curve is steepest when 𝑡 = . This occurs when 𝑡 = ≈ 3.7 years since 2000; during ln 𝑏 ln 1.8 2003. 32. a. We can eliminate (D) because 𝑏 = 0.8 is less than 1, giving a decreasing function of 𝑡. The leveling-off value in the graph is around 𝑁 = 1, 300, so we can eliminate (A) and (C) Thus, we are left with choice (B). ln 𝐴 ln 4.2 b. The logistic curve is steepest when 𝑡 = . This occurs when 𝑡 = ≈ 2.7 years since 2000; during ln 𝑏 ln 1.7 2002. 33. a. The limiting value of a logistic function with 𝑏 < 1 is 0, so the long-term projected emission would be zero. 𝑁 49.2 b. For large 𝑡, 𝐶(𝑡) ≈ 𝑏 𝑡 = (0.957) 𝑡 ≈ 223.6(0.957) 𝑡. 𝐴 0.22 c. Set 𝐶(𝑡) = 20 and solve for 𝑡: 49.2 20 = 1 + 0.22(0.957) −𝑡 49.2 1 + 0.22(0.957) −𝑡 = = 2.46 20 log((2.46 − 1)∕0.22) 𝑡=− ≈ 43, log(0.957) representing 2063. 34. a. For 𝑏 < 1 the parameter 𝑁 is the value of the logistic function for large negative values of 𝑡, so, if the model were to be extrapolated, it would indicate that sulfur emissions in the distant past were approximately 90 metric tons above the 47-ton baseline. 𝑁 89.79 b. For large 𝑡, 𝑆(𝑡) ≈ 𝑏 𝑡 = (0.91) 𝑡 ≈ 498.83(0.91) 𝑡. 𝐴 0.18 c. Set 𝑆(𝑡) = 25 and solve for 𝑡: 89.79 25 = 1 + 0.18(0.91) −𝑡 89.79 1 + 0.18(0.91) −𝑡 = = 3.5916 25 log((3.5916 − 1)∕0.18) 𝑡=− ≈ 28, log(0.91) representing 2048.

Solutions Section 2.4 35. 𝑁 = 10, 000, the susceptible population. 1, 000 = 10, 000∕(1 + 𝐴), so 𝐴 = 9. In the initial stages, the rate 10, 000 of increase was 25% per day, so 𝑏 = 1.25. Thus, 𝑁(𝑡) = . 𝑁(7) ≈ 3, 463 cases. 1 + 9(1.25) −𝑡 36. 𝑁 = 10, 000, the susceptible population. 1 = 10, 000∕(1 + 𝐴), so 𝐴 = 9, 999. In the initial stages, the rate 10, 000 of increase was 40% per day, so 𝑏 = 1.4. Thus, 𝑁(𝑡) = ; 𝑁(21) ≈ 1, 049 cases. 1 + 9, 999(1.4) −𝑡

37. 𝑁 = 3, 000, the total available market. 100 = 3, 000∕(1 + 𝐴), so 𝐴 = 29. Sales are initially doubling every 3, 000 5 days, so 𝑏 5 = 2, or 𝑏 = 2 1∕5. Thus, 𝑁(𝑡) = . Set 𝑁(𝑡) = 700 and solve for 𝑡: 1 + 29(2 1∕5) −𝑡 700 = 3, 000∕[1 + 29(2 1∕5) −𝑡], 1 + 29(2 1∕5) −𝑡 = 30∕7, (2 1∕5) −𝑡 = 23∕(29 × 7), 𝑡 = − log[23∕(29 × 7)]∕ log(2 1∕5) ≈ 16 days. 38. 𝑁 = 100, the saturation level. 2 = 100∕(1 + 𝐴), so 𝐴 = 49. Sales are initially doubling every 2 years, so 100 𝑏 2 = 2, or 𝑏 = √2. Thus, 𝑁(𝑡) = . Set 𝑁(𝑡) = 50 and solve for 𝑡: 1 + 49(√2) −𝑡 50 = 100∕[1 + 49(√2) −𝑡], 1 + 49(√2) −𝑡 = 2, (√2) −𝑡 = 1∕49, 𝑡 = − log(1∕49)∕ log(√2) ≈ 11 years, or sometime in 2004. 39. a. You can use the Evaluator and Grapher app (or the version online at the Website) as follows: 1. In the "Points and curve-fit" window (the curve with plotted points icon in the app) enter the data and choose the logistic model either by pressing "Examples: a few times or typing it in directly: $1/(1+$2*$3^(-x)) 2. Press "Fit curve." 3. To see the graph and points in a convenient window, in "Graph settings" window (the "f(x)" icon in the app) put XMin = 0 and xMax = 100, and then press "Plot graphs" (or the graph icon in the app)

The resulting regression equation, as a function of 𝑥, shown in the "Enter function of x" window (f(x) button in the app) is, with coefficients rounded to four significant digits: 𝐶(𝑡) =

15.71 1 + 82.88(1.079) −𝑡

b. To predict the 2018 value, use 𝑡 = 48 (or enter 48 as 𝑥 in the "Evaluator" window (table icon in the app) and press "Evaluate"): 𝐶(45) =

15.71 ≈ 5.0 million dollars, slightly lower than the observed $5.2 million value. 1 + 82.88(1.079) −48

c. The long-term prediction of the price is 𝑁 ≈ $15.7 million. d. The answer to part (c) of this exercise is significantly higher than the prediction based on the data in the next exercise. The given data does not extend beyond the apparent point of inflection of the logistic curve, which is a significant distance from the point where the curve levels off, so we should not expect a reliable prediction of the long-term value. 40. a. You can use the Evaluator and Grapher app (or the version online at the Website) as follows: 1. In the "Points and curve-fit" window (the curve with plotted points icon in the app) enter the data and

Solutions Section 2.4 choose the logistic model either by pressing "Examples: a few times or typing it in directly: $1/(1+$2*$3^(-x)) 2. Press "Fit curve." 3. To see the graph and points in a convenient window, in "Graph settings" window (the "f(x)" icon in the app) put XMin = 0 and xMax = 100, and then press "Plot graphs" (or the graph icon in the app)

The resulting regression equation, as a function of 𝑥, shown in the "Enter function of x" window (f(x) button in the app) is, with coefficients rounded to four significant digits: 𝐶(𝑡) =

11.34 1 + 54.25(1.094) −𝑡

b. To predict the 2018 value, use 𝑡 = 43 (or enter 43 as 𝑥 in the "Evaluator" window (table icon in the app) and press "Evaluate"): 𝐶(43) =

11.34 ≈ 5.3 million dollars, slightly higher than the observed $5.2 million value. 1 + 54.25(1.094) −43

c. The long-term prediction of the price is 𝑁 ≈ $11.3 million. d. The answer to part (c) of this exercise is significantly lower than the $15.7 million prediction based on the data in the preceding exercise. The given data does not extend beyond the apparent point of inflection of the logistic curve, which is a significant distance from the point where the curve levels off, so we should not expect a reliable prediction of the long-term value.

41. a. Following Example 2, if using Excel, start with initial rough estimates of 𝑁, 𝐴, and 𝑏. (Their exact value is not important.) 𝑁 = leveling-off value ≈ 1, 100. 𝑏 = 1.1 (slightly greater than 1, since 𝑁 is increasing with 𝑡). 𝐴: Use 𝑦-intercept = 𝑁∕(1 + 𝐴): 1, 100 767 = 1+𝐴 1 + 𝐴 = 1, 100∕767 ≈ 1.4. So, 𝐴 ≈ 0.4. Solver then gives the following solution: 𝑁 ≈ 1, 090, 𝐴 ≈ 0.410, 𝑏 ≈ 1.09. Thus, the regression model is 1, 090 𝐵(𝑡) = . 1 + 0.410(1.09) −𝑡 The model predicts that the number will level off around 𝑁 = 1, 090 teams. ln 𝐴 ln 0.410 b. The logistic curve is steepest when 𝑡 = . This occurs when 𝑡 = ≈ −10.3. This means that, ln 𝑏 ln 1.09 according to the model, the number of teams was rising fastest about 10.3 years prior to 1990; that is, sometime during 1979. c. Since the coefficient of 𝑏 is 1.09, the number of men's basketball teams was growing by about 9% per year in the past, well before 1979. 42. a. Following Example 2, if using Excel, start with initial rough estimates of 𝑁, 𝐴, and 𝑏. (Their exact value is not important.) 𝑁 = leveling-off value ≈ 1, 050. 𝑏 = 1.1 (slightly greater than 1, since 𝑁 is increasing with 𝑡). 𝐴: Use 𝑦-intercept = 𝑁∕(1 + 𝐴): 1, 050 782 = 1+𝐴 1 + 𝐴 = 1, 050∕782 ≈ 1.3. So, 𝐴 ≈ 0.3. Solver then gives the following solution: 𝑁 ≈ 1, 080, 𝐴 ≈ 0.401, 𝑏 ≈ 1.12. Thus, the regression model is 1, 080 𝐵(𝑡) = . 1 + 0.401(1.12) −𝑡

Solutions Section 2.4 The model predicts that the number will level off around 𝑁 = 1, 080 teams. ln 𝐴 ln 0.401 b. The logistic curve is steepest when 𝑡 = . This occurs when 𝑡 = ≈ −8.06. This means that, ln 𝑏 ln 1.12 according to the model, the number of teams was rising fastest about 8.06 years prior to 1990; that is, sometime during 1981. c. Since the coefficient of 𝑏 is 1.12, the number of women's basketball teams was growing by about 12% per year in the past, well before 1981. 43. For a limiting value of 4.5 million, we take 𝑁 = 4, 500. Here is the set of values (𝑥, 𝑁∕𝑦 − 1) with 𝑁 = 4, 500: 𝑥

0

10

20

30

40

55

60

65

70

75

50

𝑁∕𝑦 − 1 1.02703 0.90678 0.65441 0.40187 0.27119 0.17801 𝑥

𝑁∕𝑦 − 1 0.15090 0.12782 0.11111 0.10294 0.09756

The exponential regression equation is 𝑦 ≈ 1.1466(0.96551) 𝑥 so that 𝐴 = 1.1466. 𝑏 −1 = 0.96551, giving 𝑏 = 1∕0.96551 ≈ 1.0357. Thus the logistic model is 4, 500 𝑦= , 1 + 1.1466(1.0357) −𝑡 To predict when 𝑦 = 4, 000, we set 4, 500 4, 000 = 1 + 1.1466(1.0357) −𝑡 and solve for 𝑡: 4, 500 1 + 1.1466(1.0357) −𝑡 = = 1.125 4, 000 so (1.0357) −𝑡 = (1.125 − 1)∕1.1466 ≈ 0.1090, giving ln(0.1090) 𝑡=− ≈ 63, or 2013. ln(1.0357)

44. For a limiting value of 110,000 million, we take 𝑁 = 110. Here is the set of values (𝑥, 𝑁∕𝑦 − 1) with 𝑁 = 110: 𝑥

0

10

20

30

40

55

60

65

70

75

50

𝑁∕𝑦 − 1 3.07407 2.33333 0.74603 0.12245 0.05769 0.03774 𝑥

𝑁∕𝑦 − 1 0.01852 0.01852 0.02804 0.02804 0.01852

The exponential regression equation is 𝑦 ≈ 2.3596(0.92877) 𝑥 so that 𝐴 = 2.3596. 𝑏 −1 = 0.92877, giving 𝑏 ≈ 1.07670. Thus the logistic model is 110 𝑦= . 1 + 2.3596(1.0767) −𝑡 To predict when 𝑦 = 80, we set 110 80 = 1 + 2.3596(1.0767) −𝑡 and solve for 𝑡: 110 1 + 2.3596(1.0767) −𝑡 = = 1.12 80 so (1.0767) −𝑡 = (1.125 − 1)∕2.3596 ≈ 0.0530, giving ln(0.0530) 𝑡=− ≈ 40, or 1990. ln(1.0767)

Solutions Section 2.4

45. a. Substitute: 2 7.9 = log 𝑀0 − 10.7. 3 Solve for log 𝑀0 : 3 log 𝑀0 = (7.9 + 10.7) = 27.9, 2 𝑀0 = 10 27.9 ≈ 7.943 × 10 27 ergs. b. Compute the energy released as in (a): 2 9.0 = log 𝑀0 − 10.7 3 3 log 𝑀0 = (9.0 + 10.7) = 29.55 2 𝑀 = 10 29.55 ≈ 3.548 × 10 29 ergs. Comparing, the energy released in 1906 was 7.943 × 10 27 ≈ 2.24% of the seismic moment in 2011. 3.548 × 10 29 c. Start with the given equation: 2 𝑅 = log 𝑀0 − 10.7 3 log 𝑀0 = 1.5(𝑅 + 10.7). In exponent form, this is 𝑀0 = 10 1.5(𝑅+10.7). d. Applying part (c) to each of the two earthquakes gives 𝑀1 = 10 1.5(𝑅1 +10.7) 𝑀2 = 10 1.5(𝑅2 +10.7). Dividing gives 𝑀2 10 1.5(𝑅2 +10.7) = = 10 1.5(𝑅2 −𝑅1 ). 𝑀1 10 1.5(𝑅1 +10.7) e. If 𝑅2 − 𝑅1 = 2, then 𝑀2 ∕𝑀1 = 10 1.5(2) = 10 3 = 1,000.

46. a. By substitution we find: Whisper: 21 dB, TV: 75 dB, Loud music: 120 dB, Jet aircraft: 140 dB b. Loud music and jet aircraft: These are the ones greater than 90 dB. 𝐼 c. 𝐷 = 10 log 𝐼0 𝐼 log = 0.1𝐷 𝐼0 𝐼 = 10 0.1𝐷 𝐼0 𝐼 = 𝐼0 10 0.1𝐷 d. From part (c) 𝐼1 = 𝐼0 10 0.1𝐷1 and 𝐼2 = 𝐼0 10 0.1𝐷2 . This gives 𝐼2 ∕𝐼1 = 10 0.1𝐷2 ∕10 0.1𝐷1 = 10 0.1(𝐷2 −𝐷1 ). e. If 𝐷2 − 𝐷1 = 1, then 𝐼2 ∕𝐼1 = 10 0.1 ≈ 1.259.

47. a. By substitution: 75 dB, 69 dB, 61 dB b. 𝐷 = 10 log(320 × 10 7) − 10 log 𝑟 2 ≈ 95 − 20 log 𝑟 c. Solve 0 = 95 − 20 log 𝑟: log 𝑟 = 95∕20, 𝑟 = 10 95∕20 ≈ 57, 000 feet (rounding up so that we're beyond the point where the decibel level is 0 dB). 48. a. By substitution: Blood: 7.4; Milk: 6.3; Soap solution: 11; Black coffee: 6.9 b. Substitute, then solve for 𝐻 +: 5.0 = − log(𝐻 +), 𝐻 + = 10 −5 moles. c. Solve for 𝐻 + in terms of the pH: 𝐻 + = 10 −𝑝𝐻 . So, if the pH increases by 1, 𝐻 + decreases to 10 −1 = 1∕10

Solutions Section 2.4 of the original amount. 49. a. To obtain the logarithmic regression equation for the data in Excel, do a scatter plot and add a Logarithmic trendline with the option "Display equation on chart" checked. On the TI-83/84 Plus, use [STAT], select CALC, and choose the option LnReg. The resulting regression equation with coefficients rounded to 3 digits is 𝑃 (𝑡) = 32.1 ln 𝑡 − 36.0. b. The over-65 population reaches 80 million when 32.1 ln 𝑡 − 36.0 = 80 116 ln 𝑡 = 32.1 𝑡 = 𝑒 116∕32.1 ≈ 37.1,

which is during 2037. c. The logarithm increases without bound (choice (A)). 50. a. To obtain the logarithmic regression equation for the data in Excel, do a scatter plot and add a Logarithmic trendline with the option "Display equation on chart" checked. On he TI-83/84 Plus, use [STAT], select CALC, and choose the option LnReg. The resulting regression equation with coefficients rounded to 3 digits is 𝑃 (𝑡) = 8.06 ln 𝑡 − 15.7. b. The over-85 population reaches 16 million when 8.06 ln 𝑡 − 15.7 = 16 31.7 ln 𝑡 = 8.06 𝑡 = 𝑒 31.7∕8.06 ≈ 51.1,

which is during 2051. c. The logarithm term eventually becomes much larger than the constant term, so the larger coefficient on the logarithm will eventually give a higher predicted percentage (choice (C)). 51. Following is the Excel tabulation of the data, together with the scatter plot and the logarithmic trendline (with the option "Display equation on chart" checked):

From the regression output, the model is 𝑆(𝑡) = 42.2 ln 𝑡 + 84.3 (coefficients rounded to three significant digits). Extrapolating in the positive direction results in a prediction of ever-increasing R&D expenditures by industry. This is reasonable to a point, as expenditures cannot reasonably be expected to increase without bound. Extrapolating in the negative direction eventually leads to negative values, which do not model reality. 52. Following is the Excel tabulation of the data, together with the scatter plot and the logarithmic trendline (with the option "Display equation on chart" checked):

Solutions Section 2.4

From the regression output, the model is 𝑆(𝑡) = 5.76 ln 𝑡 + 3.46 (coefficients rounded to three significant digits). Extrapolating in the positive direction results in a prediction of ever-increasing R&D expenditures by the government. This is reasonable to a point, as expenditures cannot reasonably be expected to increase without bound. Extrapolating in the negative direction eventually leads to negative values, which do not model reality. 53. a. To obtain the logarithmic regression equation for the data in a spreadsheet like Excel, do a scatter plot and add a Logarithmic trendline with the option "Display equation on chart" checked. On the TI-83/84 Plus, use [STAT], select CALC, and choose the option LnReg. The resulting regression equation with coefficients rounded to 3 digits is 𝑈(𝑡) = −8.25 ln 𝑡 + 50.9. 2030 corresponds to 𝑡 = 30, and 𝑈(30) = −8.25 ln 30 + 50.9 ≈ 22.8% b. For the logistic curve, 184 𝑉 (30) = 26.9 + ≈ 26.9%. 1 + 2.09(0.746) −30 The graph of the U.S. share appears to level off between 25% and 30%, which agrees with the logistic curve projection for 2030. The logarithmic model projects a significantly lower value, and so appears less reasonable. c. As the logarithm curve increases without bound as 𝑡 gets large, the negative sign causes it to decrease without bound, so the values of 𝑈(𝑡) will eventually become larger and larger negative. On the other hand, 𝑉 (𝑡) approaches 26.9 + 0 = 26.9% (using the propery for the long-term behavior of a logistic function with 𝑏 < 1), confirming the conclusion in part (b). 54. a. To obtain the logarithmic regression equation for the data in a spreadsheet like Excel, do a scatter plot and add a Logarithmic trendline with the option "Display equation on chart" checked. On the TI-83/84 Plus, use [STAT], select CALC, and choose the option LnReg. The resulting regression equation with coefficients rounded to 3 digits is 𝐶(𝑡) = 16.3 ln 𝑡 − 20.7. 2030 corresponds to 𝑡 = 30, and 𝐶(30) = 16.3 ln 30 − 20.7 ≈ 34.7% b. For the logistic curve, 28.6 𝐷(30) = ≈ 28.5%. 1 + 10.4(1.30) −30 The graph of the China share appears to level off between 25% and 30%, which agrees with logistic curve projection. The logarithmic model projects a considerably higher value, and so appears less reasonable. c. The logarithm curve increases without bound as 𝑡 gets large, so the values of 𝐶(𝑡) will eventually become larger and larger without bound. On the other hand, 𝐷(𝑡) approaches 28.6% (using the propery for the longterm behavior of a logistic function with 𝑏 > 1), confirming the conclusion in part (b). 55. Just as diseases are communicated via the spread of a pathogen (such as a virus), new technology is communicated via the spread of information (such as advertising and publicity). Further, just as the spread of a disease is ultimately limited by the number of susceptible individuals, so the spread of a new technology is

Solutions Section 2.4 ultimately limited by the size of the potential market. 56. Exponential functions grow large without bound and fail to take into account the limited size of the susceptible population. Logistic functions correctly predict a slowing in the spread of an epidemic as the number of cases grows toward the total susceptible population. 57. It can be used to predict where the sales of a new commodity might level off. 58. Answers may vary. Life expectancy figures (Example 5 should (optimistically) continue to increase without a definite bound as predicted by a logistic model. Similarly, the total mount of funding on R&D (Exercises 51 and 52) could conceivably continue to increase without bound as does a logarithmic model. 59. The curve is still a logistic curve, but decreases when 𝑏 > 1 and increases when 𝑏 < 1.

60. If 𝐴 = 0, then the function is constant: 𝑃 (𝑡) = 𝑁. If 𝐴 < 0, then the function will have a vertical asymptote at 𝑡 = ln 𝐴∕ ln 𝑏; to the left of the asymptote it approaches 0; to the right it decreases to 𝑁. ln 𝐴 in the formula for 𝑦 gives ln 𝑏 𝑁 𝑁 𝑦= . = ln 𝐴∕ ln 𝑏 − 1 + 𝐴𝑏 1 + 𝐴𝑏 − ln𝑏 𝐴 1 1 But 𝑏 − ln𝑏 𝐴 = = , so 𝑏 ln𝑏 𝐴 𝐴 𝑁 𝑁 𝑦= = = 𝑁∕2. 1 + 𝐴(1∕𝐴) 1 + 1 61. Substituting 𝑡 =

62. For 𝑡 = ln 𝐴∕ ln 𝑏 to be positive, the numerator and denominator must have the same sign. Case 1: Both positive: In this case, both 𝐴 and 𝑏 must be > 1. Case 2: Both negative: In this case, both 𝐴 and 𝑏 must be < 1. For 𝑡 = ln 𝐴∕ ln 𝑏 to be negative, the numerator and denominator must have opposite sign. That is, either 𝐴 > 1 and 𝑏 < 1, or 𝐴 < 1 and 𝑏 > 1. For 𝑡 = ln 𝐴∕ ln 𝑏 to be zero, ln 𝐴 must be zero; that is, 𝐴 = 1. 63. Graph:

The lowest graph on the right is 𝑦 = ln 𝑥. The middle graph is 𝑦 = 2 ln 𝑥, and the uppermost graph is 𝑦 = 2 ln 𝑥 + 0.5. Multiplying by 𝐴 stretches the graph in the 𝑦-direction by a factor of 𝐴. Adding 𝐶 moves the graph 𝐶 units vertically up.

Solutions Section 2.4 64. Graph:

The top graph on the right is 𝑦 = ln 𝑥, the one ending below that is − ln 𝑥. The second lowest graph is 𝑦 = −2 ln 𝑥, and the lowest graph is 𝑦 = −2 ln 𝑥 − 0.5. Multiplying by a negative 𝐴 flips the graph vertically and stretches the graph in the 𝑦-direction by a factor of |𝐴|. Adding a negative 𝐶 moves the graph |𝐶| units vertically down. 65. The logarithm of a negative number, were it defined, would be the power to which a base must be raised to give that negative number. But raising a base to a power never results in a negative number, so there can be no such number as the logarithm of a negative number. 66. They can be used to solve equations analytically for an exponent, as in the calculation of the time an investment should be held to yield a given return.

67. Any logarithmic curve 𝑦 = log𝑏 𝑡 + 𝐶 will eventually surpass 100% and hence will not be suitable as a long-term predictor of market share. 68. Any logarithmic curve 𝑦 = log𝑏 𝑡 + 𝐶 will become negative as 𝑡 gets close to zero, and hence not be suitable for indefinite backward extrapolation. 69. Time is increasing logarithmically with population: Solving 𝑃 = 𝐴𝑏 𝑡 for 𝑡 gives 𝑡 = log𝑏 (𝑃 ∕𝐴) = log𝑏 𝑃 − log𝑏 𝐴, which is of the form 𝑡 = log𝑏 𝑃 + 𝐶.

70. Time is increasing exponentially with population: Solving 𝑃 = log𝑏 𝑡 + 𝐶 for 𝑡 gives 𝑡 = 𝑏 (𝑃 −𝐶) = 𝑏 −𝐶 𝑏 𝑃 , which is of the form 𝑡 = 𝐴𝑏 𝑃 .

71. For the two function, write 𝑄1 = 𝐴1 ln 𝑡 + 𝐵1 𝑄2 = 𝐴2 ln 𝑡 + 𝐵2 . Adding gives 𝑄1 + 𝑄2 = 𝐴1 ln 𝑡 + 𝐵1 + 𝐴2 ln 𝑡 + 𝐵2 = (𝐴1 + 𝐴2 ) ln 𝑡 + (𝐵1 + 𝐵2 ), which is of the form Constant× ln 𝑡+Constant, and is thus a logarithmic function. 72. The units in the first model are millions of dollars and in the second billions of dollars, so we multiply the second by 1,000 to convert it to millions of dollars as well. Using more significant digits in the two regression models, their sum is 42.181 ln 𝑡 + 84.304 + 5,757.6 ln 𝑡 + 3,457.4 ≈ 5,799.8 ln 𝑡 + 3,541.7. The regression model of the sum of the data (again, multiplying the second set of data by 1,000) is 5,799.8 ln 𝑡 + 3,541.7, which agrees with the sum to five significant digits. This suggests that the sum of logarithmic regression

Solutions Section 2.4 models is the regression model of the sum.

Solutions Chapter 2 Review Chapter 2 Review

1. 𝑎 = 1, 𝑏 = 2, 𝑐 = −3, so −𝑏∕(2𝑎) = −1; 𝑓(−1) = −4, so the vertex is (−1, −4). 𝑦-intercept: 𝑐 = −3. 𝑥 2 + 2𝑥 − 3 = (𝑥 + 3)(𝑥 − 1), so the 𝑥-intercepts are −3 and 1. 𝑎 > 0, so the parabola opens upward.

2. 𝑎 = −1, 𝑏 = −1, 𝑐 = −1, so −𝑏∕(2𝑎) = −1∕2; 𝑓(−1∕2) = −3∕4, so the vertex is (−1∕2, −3∕4). 𝑦-intercept: 𝑐 = −1. 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no 𝑥-intercepts. 𝑎 < 0, so the parabola opens downward.

3. For every increase in 𝑥 by 1 unit, the value of 𝑓 is multiplied by 1∕2, so 𝑓 is exponential. Since 𝑓(0) = 5, the exponential model is 𝑓(𝑥) = 5(1∕2) 𝑥, or 5(2 −𝑥). 𝑔(2) = 0, whereas exponential functions are never zero, so 𝑔 is not exponential. 4. The values of 𝑓 decrease by 2 for every increase in 𝑥 by 1 unit, so 𝑓 is linear (not exponential). For every increase in 𝑥 by 1 unit, the value of 𝑔 is multiplied by 2, so 𝑔 is exponential. Since 𝑔(0) = 3, the exponential model is 𝑔(𝑥) = 3(2 𝑥). 5. We use the table of values shown on the left to obtain the graph shown on the right:

Solutions Chapter 2 Review

𝑥

-3

-2

-1

0

1

2

3

𝑔(𝑥) 27/2

1/18

1/6

1/2

3/2

9/2

27/2

9/2

3/2

1/2

1/6

1/18

1/54

𝑓(𝑥) 1/54

6. We use the table of values shown on the left to obtain the graph shown on the right:

𝑥

-3

-2

-1

0

1

2

3

𝑔(𝑥) 128

1/8

1/2

2

8

32

128

32

8

2

1/2

1/8

1/32

𝑓(𝑥) 1/32

7. The technology formulas for the two functions are TI-83/84 Plus:\quad e^x; e^(0.8*x) Excel:\quad =EXP(x); =EXP(0.8*x)

8. The technology formulas for the two functions are TI-83/84 Plus and Excel:\quad 2*1.01^x ; 2*0.99^x Technology gives us the following graphs:

Technology gives us the following graphs:

9. Use 𝐴 = 𝑃 !1 +

𝑟 𝑚𝑡 0.03 60 " ; 𝑃 = 3, 000, 𝑟 = 0.034; 𝑚 = 12; 𝑡 = 5. So, 𝐴 = 3, 000!1 + " ≈ $3, 484.85. 𝑚 12

𝑟 " ; 𝑃 = 10, 000, 𝑟 = 0.025; 𝑚 = 4; 𝑡 = 10. So, 𝑚 0.025 40 𝐴 = 10, 000!1 + " ≈ $12, 830.27. 4 10. Use 𝐴 = 𝑃 !1 +

𝑚𝑡

Solutions Chapter 2 Review

𝑟 𝑚𝑡 " ; 𝐴 = 5, 000, 𝑟 = 0.03; 𝑚 = 12; 𝑡 = 10. Substituting gives 𝑚 0.03 120 5, 000 = 𝑃 !1 + " ⇒𝑃 = 5, 000∕(1 + 0.03∕12) 120 ≈ $3, 705.48. 12

11. Use 𝐴 = 𝑃 !1 +

𝑟 𝑚𝑡 " ; 𝐴 = 10, 000, 𝑟 = 0.025; 𝑚 = 4; 𝑡 = 10. Substituting gives 𝑚 0.025 40 10, 000 = 𝑃 !1 + " ⇒𝑃 = 10, 000∕(1 + 0.025∕4) 40 ≈ $7, 794.07. 4

12. Use 𝐴 = 𝑃 !1 +

13. Use 𝐴 = 𝑃 𝑒 𝑟𝑡; 𝑃 = 3, 000, 𝑟 = 0.03; 𝑡 = 5. So, 𝐴 = 3, 000𝑒 0.15 ≈ $3, 485.50.

14. Use 𝐴 = 𝑃 𝑒 𝑟𝑡; 𝑃 = 10, 000, 𝑟 = 0.025; 𝑡 = 10. So, 𝐴 = 10, 000𝑒 0.25 ≈ $12, 840.25.

15. Increasing 𝑥 by one half unit triples the value of 𝑓. Therefore, increasing 𝑥 by 1 unit multiplies the value of 𝑓 by 9, giving 𝑏 = 9. 𝐴 = 𝑓(0) = 4.5. Therefore, 𝑓(𝑥) = 𝐴𝑏 𝑥 = 4.5(9 𝑥).

16. Increasing 𝑥 by 1 unit decreases the value of 𝑓 by 75%. That is, it reduces the value of 𝑓 to 25% of its original value. Therefore, 𝑏 = 0.25. 𝐴 = 𝑓(0) = 5. Therefore, 𝑓(𝑥) = 𝐴𝑏 𝑥 = 5(0.25 𝑥). 17. 2 = 𝐴𝑏 1 and 18 = 𝐴𝑏 3. Dividing, 𝑏 2 = 9, so 𝑏 = 3. Then 2 = 3𝐴, so 𝐴 = 2∕3: 𝑓(𝑥) =

2 𝑥 3 . 3

18. 10 = 𝐴𝑏 1 and 5 = 𝐴𝑏 3. Dividing, 𝑏 2 = 1∕2, so 𝑏 = 1∕√2. Then 10 = 𝐴∕√2, so 𝐴 = 10√2: 1 𝑥 𝑓(𝑥) = 10√2! " . √2 19. We use the following table of values: 𝑥 1/27

𝑓(𝑥) 𝑔(𝑥)

1/9

1/3

1

3

9

27

-3

-2

-1

0

1

2

3

3

2

1

0

-1

-2

-3

Graphing these gives the following curves:

Solutions Chapter 2 Review 20. We use the following table of values: 𝑥 1/1,000 1/100 1/10

𝑓(𝑥) 𝑔(𝑥)

1

10

100 1,000

-3

-2

-1

0

1

2

3

3

2

1

0

-1

-2

-3

Graphing these gives the following curves:

21. 𝑄0 = 5 (given) 𝑡ℎ 𝑘 = ln 2 100𝑘 = ln 2 𝑘 = (ln 2)∕100 ≈ 0.00693 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 5𝑒 −0.00693𝑡

22. 𝑄0 = 10, 000 (given) 𝑡ℎ 𝑘 = ln 2 5𝑘 = ln 2 𝑘 = (ln 2)∕5 ≈ 0.139 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 10, 000𝑒 −0.139𝑡

23. 𝑄0 = 2.5 (given) 𝑡𝑑 𝑘 = ln 2 2𝑘 = ln 2 𝑘 = (ln 2)∕2 ≈ 0.347 𝑄 = 𝑄0 𝑒 𝑘𝑡 = 2.5𝑒 0.347𝑡

24. 𝑄0 = 10, 000 (given) 𝑡𝑑 𝑘 = ln 2 15𝑘 = ln 2 𝑘 = (ln 2)∕15 ≈ 0.0462 𝑄 = 𝑄0 𝑒 𝑘𝑡 = 10, 000𝑒 0.0462𝑡

25. 3, 000 = 2, 000(1 + 0.04∕12) 12𝑡 𝑡 = [log(3∕2)∕ log(1 + 0.04∕12)]∕12 ≈ 10.2 years

26. 3, 000 = 2, 000(1 + 0.0675∕365) 365𝑡 𝑡 = [log(3∕2)∕ log(1 + 0.0675∕365)]∕365 ≈ 6 years 27. 3, 000 = 2, 000𝑒 0.0375𝑡 𝑡 = ln(3∕2)∕0.0375 ≈ 10.8 years

28. 1, 200 = 1, 000(1 + 1∕4) 4𝑡 = 1, 000(1.25 4𝑡) 𝑡 = [log(1, 200∕1, 000)∕ log(1.25)]∕4 ≈ 0.2 years

29. 𝑁 = 900 is given, 100 = 900∕(1 + 𝐴) gives 𝐴 = 8, initially increasing 50% per unit increase in 𝑥 gives 𝑏 = 1.5: 900 𝑓(𝑥) = . 1 + 8(1.5) −𝑥 30. 𝑁 = 25 is given, 5 = 25∕(1 + 𝐴) gives 𝐴 = 4, 𝑏 = 1.1 is given in the initial exponential form: 25 𝑓(𝑥) = . 1 + 4(1.1) −𝑥

Solutions Chapter 2 Review 31. 𝑁 = 20 is given, 20∕(1 + 𝐴) = 5, so 𝐴 = 3, decreasing at a rate of 20% per unit of 𝑥 near 0 means 𝑏 = 1 − 0.2 = 0.8: 20 𝑓(𝑥) = . 1 + 3(0.8) −𝑥 32. 𝑁 ≈ 10 and 𝑏 = 0.8 are given, and for small 𝑥, 𝑁∕(1 + 𝐴) = 10∕(1 + 𝐴) = 2, so 𝐴 = 4, giving 10 𝑓(𝑥) = . 1 + 4(0.8) −𝑥

33. a. The largest volume will occur at the vertex: −𝑏∕(2𝑎) = 0.085∕(2 × 0.000005) ≈ $8, 500 per month; substituting 𝑐 = 8, 500 gives ℎ = an average of approximately 2,100 hits per day. b. Solve ℎ = 0 using the quadratic formula: 𝑐 ≈ $29, 049 per month (the other solution given by the quadratic formula is negative). c. The fact that −0.000005, the coefficient of 𝑐 2, is negative. 34. a. Since 𝑡 is time in years since the start of 2000 and the data are from the start of 2001 to the end of 2003, the domain is 1 ≤ 𝑡 ≤ 4, or [1, 4]. b. The growth rate is a minimum at the vertex: 𝑏 −6 𝑡=− =− = 1.5, 2𝑎 4 midway through 2001. c. A zero rate of growth would correspond to an intercept of the 𝑡-axis. However, the discriminant 𝑏 2 − 4𝑎𝑐 = 36 − 4(2)(12) = −60 is negative, so there are no 𝑡-intercepts, and hence no zero rate of growth. d. The fact that the coefficient of 𝑡 2 is positive. e. No. What was decreasing was the number of new broadband users. Put another way, the number of broadband users was growing at a declining rate in the first half of 2001. f. The beginning of 2013 is 𝑡 = 13; 𝑛(13) = 2(13) 2 − 6(13) + 12 = 272 million new users per year at the start of 2013 𝑛(14) = 2(14) 2 − 6(14) + 12 = 320 million new users per year at the start of 2014. These numbers are unreasonably large (recall that they represent new users per year). In fact, their sum exceeds the entire U.S. population.

35. a. 𝑅 = 𝑝𝑞 = −60𝑝 2 + 950𝑝. The maximum revenue occurs at the vertex: 𝑝 = −𝑏∕(2𝑎) = 950∕(2 × 60) = $7.92 per novel. At that price the monthly revenue is 𝑅 = $3, 760.42. b. 𝐶 = 900 + 4𝑞 = 900 + 4(−60𝑝 + 950) = −240𝑝 + 4, 700, so 𝑃 = 𝑅 − 𝐶 = −60𝑝 2 + 950𝑝 − (−240𝑝 + 4, 700) = −60𝑝 2 + 1, 190𝑝 − 4, 700. The maximum monthly profit occurs at the vertex: 𝑝 = −𝑏∕(2𝑎) = 1, 190∕(2 × 60) = $9.92 per novel. At that price, the monthly profit is 𝑃 = $1, 200.42. 36. a. Demand Function: The given points are (𝑝, 𝑞) = (0, 2, 000) and (0.10, 1, 000). 𝑞 − 𝑞1 1, 000 − 2, 000 Slope: 𝑚 = 2 = = −10, 000 𝑝2 − 𝑝1 0.10 − 0 Intercept: 𝑏 = 2, 000 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −10, 000𝑝 + 2, 000. b. Revenue = 𝑝𝑞 = 𝑝(−10, 000𝑝 + 2, 000) = −10, 000𝑝 2 + 2, 000𝑝. The maximum revenue occurs at the vertex: 𝑝 = −𝑏∕(2𝑎) = 2, 000∕(20, 000) = $0.1, or 10¢ per newspaper. At that price the monthly revenue is 𝑅 = −10, 000(0.1) 2 + 2, 000(0.1) = $100. c. 𝐶 = 200 + 0.04𝑞 = 200 + 0.04(−10, 000𝑝 + 2, 000) = −400𝑝 + 280, so 𝑃 = 𝑅 − 𝐶 = −10, 000𝑝 2 + 2, 000𝑝 − (−400𝑝 + 280) = −10, 000𝑝 2 + 2, 400𝑝 − 280.

Solutions Chapter 2 Review The maximum monthly profit occurs at the vertex: 𝑝 = −𝑏∕(2𝑎) = 2, 400∕20, 000 = $0.12 per newspaper. At that price, the monthly profit is 𝑃 = −10, 000(0.12) 2 + 2, 400(0.12) − 280 = −136, a loss of $136. 37. a. 1997 corresponds to 𝑡 = 0, and so the harvest was 𝑛 = 9.1(0.81 0) = 9.1 million pounds. The value of the base 𝑏 = 0.81 of the exponential model tells us that the value each year is 0.81 times, or 81% of the value the previous year. In other words, the value decreases by 100 − 81 = 19 each year. b. 2013 corresponds to 𝑡 = 16, and 𝑛(16) = 9.1(0.81 16) ≈ 0.31, or about 310,000 pounds. 38. Let 𝑆 be the stock price and 𝑡 be the time in hours since the IPO. Model the stock price by 𝑆 = 𝐴𝑏 𝑡. Then 𝐴 = 10, 000 and 𝑏 3 = 2 (since the stock is doubling in price every 3 hours), so 𝑏 = 2 1∕3. After 8 hours, 𝑆 = 10, 000(2 1∕3) 8 = $63, 496.04. 39. The model for the lobster harvest from Exercise 37 is 𝑛(𝑡) = 9.1(0.81 𝑡) million pounds. We need to find the value of 𝑡 when it first dips below 200,000, which is 0.2 million pounds: 0.2 = 9.1(0.81 𝑡) 0.2∕9.1 = 0.81 𝑡 log(0.2∕9.1) = 𝑡 log(0.81) 𝑡 = log(0.2∕9.1)∕ log(0.81) ≈ 18.1. Thus, the harvest is still above 200,000 lobsters at 𝑡 = 18 (2015), but will be below 200,000 at 𝑡 = 19 (2016). 40. Solve 50, 000 = 10, 000(2 𝑡∕3): 𝑡 = 3 log 5∕ log 2 ≈ 7.0 hours.

41. In 2010 (𝑡 = 13), the harvest was 9.1(0.81 13) ≈ 0.5880. From that point on, the model is 0.5880(1.24) 𝑡 where 𝑡 is now time in years since 2010. In 2013, the size of the harvest is 0.5880(1.24) 3 ≈ 1.12 million pounds.

42. After 10 hours the stock is worth 10, 000(2 10∕3) = $100, 793.68. From this level it follows a new exponential curve with 𝐴 = 100, 793.68 and 𝑏 4 = 2∕3 (it loses 1/3 of its value every 4 hours), hence 𝑏 = (2∕3) 1∕4. Now solve 10, 000 = 100, 793.68(2∕3) 𝑡∕4 for 𝑡 = 4 log(10, 000∕100, 793.68)∕ log(2∕3) ≈ 22.8. Adding the first 10 hours, the stock will be worth $10,000 again 32.8 hours after the IPO. 43. We use the data shown in the following table: 𝑡

0

𝑛 11.6

2

4

6

8

10

12

8.3

3.0

1.4

1.6

1.2

0.8

The regression model is Excel: 𝑛(𝑡) = 9.5841𝑒 −0.224𝑡 = 9.5841(𝑒 −0.224) 𝑡 ≈ 9.5841(0.7993) 𝑡 TI-83/84 Plus and Website: 9.5841(0.79967) 𝑡 Rounding to 2 digits gives 𝑛(𝑡) = 9.6(0.80 𝑡) million pounds of lobster.

Solutions Chapter 2 Review 44. We use the data shown in the following table: 𝑡

0

𝑃 1.00

1

2

3

4

5

6

7

8

1.30

1.75

2.05

2.60

3.00

3.95

5.00

6.70

The regression model is Excel: 𝑃 (𝑡) = 1.0321𝑒 0.2276𝑡 = 1.0321(𝑒 0.2276) 𝑡 ≈ 1.03(1.26) 𝑡 TI-83/84 Plus and Web Site: 𝑦 = 1.03206(1.25564 𝑡) Rounding to 2 digits gives 𝑃 (𝑡) = 1.03(1.26) 𝑡. To predict numerically when the price passes $10 we can use a table of values in the TI-83/84 Plus, Excel, or the Website with y = 1.03*(1.26^x) 𝑥

1.03*(1.26^x)

0

1.03

1

1.2978

2

1.6352

3

2.0604

4

2.5961

5

3.2711

6

4.1215

7

5.1932

8

6.5434

9

8.2446

10

10.3883

Thus, the stock price first passes $10 at the end of the 10th hour.

45. C: (A) is true because 𝐿1 = 𝐿0 𝑒 −1∕𝑡 < 𝐿0 . (B) is true because 𝐿3𝑡 = 𝐿0 𝑒 −3 > 0. (D) is true because increasing 𝑡 decreases 𝑥∕𝑡, which increases 𝑒 −𝑥∕𝑡 (makes it closer to 1), hence increases 𝐿𝑥 for every 𝑥. (E) is true because 𝑒 −𝑥∕𝑡 is never 0. On the other hand, (C) is false because 𝐿𝑡 = 𝐿0 𝑒 −1 ≈ 0.37𝐿0 < 𝐿0 ∕2. 46. a. The given function has the form Constant + logistic function. The logistic function part levels off at 𝑁 = 4, 470. Therefore, adding the constant gives a leveling of at 4, 470 + 6, 050 = 10, 520. b. We solve 4, 470 10, 000 = 6, 050 + 1 + 14(1.73 −𝑡) 4, 470 3, 950 = 1 + 14(1.73 −𝑡) 1 + 14(1.73 −𝑡) = 4, 470∕3, 950 14(1.73 −𝑡) = 4, 470∕3, 950 − 1 1.73 −𝑡 = (4, 470∕3, 950 − 1)∕14 −𝑡 log(1.73) = log((4, 470∕3, 950 − 1)∕14) 𝑡 = − log((4, 470∕3, 950 − 1)∕14)∕ log(1.73) ≈ 8.5, about 8.5 weeks. c. Sales are rising most rapidly when 𝑡 = ln 𝐴∕ ln 𝑏 = ln 14∕ ln 1.73 ≈ 5 to the nearest week.

Solutions Chapter 2 Case Study Chapter 2 Case Study 1. Using, for example, a spreadsheet logarithmic regression, we obtain 𝑁(𝑡) ≈ 0.1895 ln 𝑡 + 19.665 2029: 𝑁(10) ≈ 0.1895 ln 10 + 19.665 ≈ 20.1013 2030: 𝑁(11) ≈ 0.1895 ln 11 + 19.665 ≈ 20.1194 2031: 𝑁(12) ≈ 0.1895 ln 12 + 19.665 ≈ 20.1359 2032: 𝑁(13) ≈ 0.1895 ln 13 + 19.665 ≈ 20.1511 2029–2030: 𝑁(11) − 𝑁(10) ≈ 0.0181: 18,100 students 2030–2031: 𝑁(12) − 𝑁(11) ≈ 0.0165: 16,500 students 2032–2031: 𝑁(13) − 𝑁(12) ≈ 0.0152: 15,200 students 2. The value of a logistic function 𝑁(𝑡) = shifted logistic function 𝑁(𝑡) = model is

𝑁(𝑡) =

𝑀 for large values of 𝑡 is approximately 𝑀. Thus, for a 1 + 𝐴𝑏 −𝑡

𝑀 + 𝐶, its value for large 𝑡 is approximately 𝑀 + 𝐶. Here, the 1 + 𝐴𝑏 −𝑡

0.407 + 19.7, 1 + 37.0(2.058 −𝑡)

so the value for large 𝑡 is approximately

𝑀 + 𝐶 = 0.407 + 19.7 ≈ 20.1 million students

3. The values of a logarithmic function 𝑁(𝑡) = 𝐴 ln 𝑥 + 𝐶 grow without bound for larger and larger values of 𝑡, so the logarithmic model 𝑁(𝑡) ≈ 0.1895 ln 𝑡 + 19.665 grows larger and larger without bound. 4. Using the current parameters as the guess, we obtain 𝑁(𝑡) =

0.964 + 19.65 1 + 13.4(1.394 −𝑡)

The new estimates for 2027–2029 are (to three significant digit): 0.964 2027: 𝑁(8) = + 19.65 ≈ 20.1 1 + 13.4(1.394 −8) 0.964 2028: 𝑁(9) = + 19.65 ≈ 20.2 1 + 13.4(1.394 −9) 0.964 2029: 𝑁(10) = + 19.65 ≈ 20.3 1 + 13.4(1.394 −10) Long term projection: 𝑀 + 𝐶 ≈ 0.964 + 19.65 ≈ 20.6 million students. 5. Using a spreadsheet with the smaller set of data, we obtain 𝑁(𝑡) = 0.1596 ln 𝑡 + 19.689

The new estimates for 2027–2029 to three significant digits: 2027: 𝑁(8) = 0.1596 ln 8 + 19.689 ≈ 20.0 2028: 𝑁(9) = 0.1596 ln 9 + 19.689 ≈ 20.0 2029: 𝑁(10) = 0.1596 ln 10 + 19.689 ≈ 20.1 Long term projection: Enrollment increases without bound. 6. Answers may vary: If, for example, the college-age population in the U.S. were to eventually level off without significant increases, then a (shifted) logistic model would be more realistic, as a logarithmic

Solutions Chapter 2 Case Study model predicts ever-increasing college populations. On the other hand, if the college-age population in the U.S. continues to grow, but at a smaller and smaller rate, then a logarithmic model would be more realistic for fairly long-term projections (even though the logarithmic model eventually increases without bound the time it needs to make measureable increases grows exponentially).

Solutions Section 3.1 Section 3.1

1. = 2,000, = 0.06, = 1 = = 2,000(0.06)(1) = $120 = + = 2,000 + 120 = $2,120

2. = 1,000, = 0.04, = 10 = = 1,000(0.04)(10) = $400 = + = 1,000 + 400 = $1,400

3. = 4,000, = 0.005, = 8 = = 4,000(0.005)(8) = $160 = + = 4,000 + 160 = $4,160

4. = 2,000, = 0.001, = 12 = = 2,000(0.001)(12) = $24 = + = 2,000 + 24 = $2,024

5. = 20,200, = 0.05, = 1 2 = = 20,200(0.05)(1 2) = $505 = + = 20,200 + 505 = $20,705

6. = 10,100, = 0.11, = 1 4 = = 10,100(0.11)(1 4) = $277.75 = + = 10,100 + 277.75 = $10,377.75

7. = 10,000, = 0.03, = 10 12 = = 10,000(0.03)(10 12) = $250 = + = 10,000 + 250 = $10,250

8. = 6,000, = 0.09, = 5 12 = = 6,000(0.09)(5 12) = $225 = + = 6,000 + 225 = $6,225

9. = 12,000, = 0.0005, = 10 = = 12,000(0.0005)(10) = $60 = + = 12,000 + 60 = $12,060

10. = 8,000, = 0.0003, = 5 = = 8,000(0.0003)(5) = $12 = + = 8,000 + 12 = $8,012

11. = (1 + ) = 10,000 (1 + 0.02 × 5) = $9,090.91

12. = (1 + ) = 20,000 (1 + 0.05 × 2) = $18,181.82

13. = (1 + ) = 1,000 (1 + 0.07 × 0.5) = $966.18

14. = (1 + ) = 5,000 (1 + 0.01 × 3) = $4,854.37

15. = (1 + ) = 15,000 (1 + 0.0003 × 15) = $14,932.80

16. = (1 + ) = 30,000 (1 + 0.06 × 20 12) = $27,272.73

17. = (1 + ) = 5,000(1 + 0.08 × 0.5) = $5,200

18. = (1 + ) = 10,000(1 + 0.01 × 15) = $11,500

19. = (1 + ) = 1,000 (1 + 0.0002 × 12) = $997.61

20. = (1 + ) = 30,360 (1 + 0.095 × 4) = $22,000

Solutions Section 3.1 21. We are given the interest and asked to compute . = 250, = 1,000, = 5 = 250 = 1,000 × × 5 = 5,000 250 = = 0.05, or 5% 5,000

22. We are given the interest and asked to compute . = 2,800, = 10,000, = 4 = 2,800 = 10,000 × × 4 = 40,000 2,800 = = 0.07, or 7% 40,000

23. Interest every six months: = = (10,000)(0.08750)(0.5) = $437.50 Total interest over the 10-year life of the bond: = = (10,000)(0.08750)(10) = $8,750

24. Interest every six months: = = (5,000)(0.03250)(0.5) = $81.25 Total interest over the 7-year life of the bond: = = (5,000)(0.03250)(7) = $1,137.50

25. =

26. =

4,083 = = $3,000 1 + 1 + (.019)(10) 2,875 = = $2,500 1 + 1 + (.0375)(4)

27. Goldman Sachs: = = (5,000)(0.0375)(4) = $750 Wells Fargo: = = (5,000)(0.032)(5) = $800 Wells Fargo would pay the most total interest, $800. 28. Ford: = = (2,000)(0.03250)(7) = $455 Verizon: = = (2,000)(0.04125)(6) = $495 Verizon would pay the most total interest, $495.

29. We are given present and future values, and want to compute . = 4,640, = 4,000, = 0.08 = (1 + ) 4,640 = 4,000(1 + 0.08 ) 4,640 1 + 0.08 = = 1.160 4,000 0.08 = 0.160 0.160 = = 2 years 0.08

30. We are given and , and want to compute . = 1,000, = 640, = 0.08 = 640 = 1,000 × 0.08 = 80 640 = = 8 years 80

31. = 50 = 1,000 × 4 = 4,000 ; = 50 4,000 = 0.0125, which is 1.25% weekly interest. Annual rate = 52 × 0.0125 = 0.65, or 65%

32. = 60 = 1,500 × 3 = 4,500 ; = 60 4,500 = 0.0133, which is 1.33% weekly interest. Annual rate = 52 × 60 4,500 0.6933, or 69.33%

Solutions Section 3.1 33. You will pay 5,000 × 0.09 × 2 = $900 in interest on the loan. Adding the $100 fee, you pay the bank a total of $1,000 over the two years. To find the effective rate, we solve 1,000 = 5,000 × 2 = 10,000 ; = 1,000 10,000 = 0.1, so the rate is 10%.

34. You will pay 7,000 × 0.08 × 3 = $1,680 in interest on the loan. Adding the $100 fee, you pay the bank a total of $1,780 over the two years. To find the effective rate, we solve 1,780 = 7,000 × 3 = 21,000 ; = 1,780 21,000 0.08476, so the rate is 8.476%. 35. 5 = 69 12; = 12 × 5 69 36. 10 = 99 6; = 6 × 10 99

0.86957 or 86.957% 0.60606 or 60.606%

37. Use = (1 + ) with = 13,800, = 61,400, = 12 (months) 61,400 = 13,800(1 + 12 ) = 13,800 + 165,600 = (61,400 13,800) 165,600 0.2874, or 28.74%

38. Use = (1 + ) with = 13,800, = 32,600, = 6 (months; selling in April 2021) 32,600 = 13,800(1 + 6 ) = 13,800 + 82,800 = (32,600 13,800) 82,800 0.2271, or 22.71% 39. Selling in any month subsequent to April 1 would have given you a profit. We could calculate the monthly returns for each of those six possible months. However, notice from Quick Example 6 that the interest per month for each of these six alternatives would be the slope of the line joining the April 1 point and the corresponding point on the graph. Of these six lines from the April 1 point, the line to the May 1 point is the steepest, so gives the largest monthly interest, and hence the largest monthly interest rate. Monthly interest (April 1 to May 1) = Slope = Monthly interest rate =

52,100

Monthly interest 19,500 = 32,600

1

32,600

0.5982,

= 19,500.

or 59.82% 40. Selling on July 1, August 1, or September 1 would have resulted in a loss. We could calculate the monthly (negative) returns for each of those three possible months. However, notice from Quick Example 6 that the loss per month for each of these six alternatives would be the slope of the line joining the May 1 point and the corresponding point on the graph. Of the three lines from the May 1 point, the line to the July 1 point is the steepest negative, so gives the largest monthly loss, and hence the largest monthly percentage loss. Monthly loss (May 1 to July 1) = Slope = Monthly (negative) interest rate =

41,500

2

52,100

Monthly loss 5,300 = 52,100

=

5,300.

0.1017,

or a 10.17% loss. 41. Yes. Simple interest increase is linear. The slope of the line through the points for January 1 and February 1 and that of the line through the points for February 1 and March 1 are equal, (13,200 in each case) and therefore the three points are on the same straight line. 42. No. Simple interest increase is linear. The graph is visibly not linear in that time period; the slopes of the

Solutions Section 3.1 lines through the successive pairs of marked points are quite different.

43. 1950 population: = 500,000; 2020 population: = 3,300,000 = = 3,300,000 500,000 = 2,800,000 = 2,800,000 = 500,000 (70) = 35,000,000 2,800,000 = = 0.08 or 8% 35,000,000 44. 1950 population: = 500,000; 1990 population: = 2,500,000 = = 2,500,000 500,000 = 2,000,000 = 2,000,000 = 500,000 (40) = 20,000,000 2,000,000 = 0.10 or 10% 20,000,000

45. We are given: = 1950 population = 500,000; = 0.08 (from Exercise 43); = 80 (years to 2030) = (1 + ) = 500,000(1 + 0.08 × 80) = 3,700,000

46. We are given: = 1950 population = 500,000; = 0.10 (from Exercise 44); = 80 (years to 2030) = (1 + ) = 500,000(1 + 0.10 × 80) = 4,500,000

47. = 1950 population = 500,000; = 0.08 (from Exercise 43). After years, the population will be = (1 + ) = 500,000(1 + 0.08 ) = 500,000 + 40,000 or 500 + 40 thousand ( = time in years since 1950). Graph:

48. = 1950 population = 500,000; = 0.10 (from Exercise 44). After years, the population will be = (1 + ) = 500,000(1 + 0.10 ) = 500,000 + 50,000 or 500 + 50 thousand ( = time in years since 1950).

Solutions Section 3.1 Graph:

49. Actual discount = 0.0025 2 = 0.00125. Selling price = 5,000 = $4,993.75, = $5,000, = 0.5 = (1 + ) 5,000 = 4,993.75(1 + 0.5 ) 1 + 0.5 = 5,000 4,993.75 0.5 = 5,000 4,993.75 1 = 2(5,000 4,993.75 1) 0.002503, or 0.2503%

50. Actual discount = 0.0006 4 = 0.00015. Selling price = 15,000 = $14,997.75, = $15,000, = 0.25 = (1 + ) 15,000 = 14,997.75(1 + 0.25 ) 1 + 0.25 = 15,000 14,997.75 0.25 = 15,000 14,997.75 1 = 4(15,000 14,997.75 1) 0.00060009, or 0.060009%

(0.00125)(5,000) = $4,993.75.

(0.00015)(15,000) = $14,997.75.

51. To say that the discount rate is 3.705% means that its selling price ( ) is 3.705% lower than its maturity value ( ). To simplify the calculation, let us use a T-bill with a maturity value of $10,000: = 10,000 10,000(0.03705 2) = 9814.75 10,000 = 9,814.75(1 + 0.5 ) = 9,814.75 + 4,907.375 = (10,000 9,814.75) 4,907.375 = 0.03775, or 3.775%

52. To say that the discount rate is 3.470% means that its selling price ( ) is 3.470% lower than its maturity value ( ). To simplify the calculation, let us use a T-bill with a maturity value of $10,000: = 10,000 10,000(0.03470 4) = 9,913.25 10,000 = 9,913.25(1 + 0.25 ) = 9,913.25 + 2,478.3125 = (10,000 9,913.25) 2,478.3125 = 0.03500, or 3.500%

Solutions Section 3.1 53. Graph (A) is the only possible choice, because the equation = (1 + ) = + gives the future value as a linear function of time. 54. = (1 + ) = + ( ) , which is a linear equation with slope . Thus, the slope is Slope = Interest rate × Present value. 55. = (1 + ) = + . Since = 1,000 + 0.5 , we have = 1,000 = 1,000 = 0.5 so 0.5 = = 0.0005, or 0.05% 1,000 56. = (1 + ) = + . Since = 400 + 5 , we have = 400 = 400 = 5 so 5 = = 0.0125, or 1.25% 400

57. Wrong. In simple interest growth, the change each year is a fixed percentage of the starting value, and not the preceding year's value. (Also see the next exercise.) 58. The economy last year was larger than the year before; 1% of last year's economy is larger than 1% of the year before's.

59. Simple interest is always calculated on a constant amount, . If interest is paid into your account, then the amount on which interest is calculated does not remain constant. 60. To say that the discount rate is after one year, or = (1

)

One has, for a single year

= (1 + ) = (1 ) (1 + ) 1 = (1 )(1 + ) 1 1+ = 1 giving 1 = 1 1

means that its selling price ( ) is (1

) times its maturity value ( )

Solutions Section 3.2 Section 3.2

1. We use = (1 + ) . = $10,000, = 0.002, = 1, = 15

2. We use = (1 + ) . = $10,000, = 0.0005, = 1, = 6

Technology: 10000*(1+0.002)^15

Technology: 10000*(1+0.0005)^6

3. We use = (1 + ) . = $10,000, = 0.002, = 1, = 10 × 12 = 120

4. We use = (1 + ) . = $10,000, = 0.0045, = 1, = 20 × 12 = 240

Technology: 10000*(1+0.002)^120

Technology: 10000*(1+0.0045)^240

5. We use = (1 + ) . = $10,000, = 0.03, = 1, = 10

6. We use = (1 + ) . = $10,000, = 0.04, = 1, = 8

Technology: 10000*(1+0.03)^10

Technology: 10000*(1+0.04)^8

7. We use = (1 + ) . = $10,000, = 0.025, = 4, = 5

8. We use = (1 + ) . = $10,000, = 0.015, = 52, = 5

= (1 + ) = 10,000(1 + 0.002) 15 = 10,000(1.002) 15 $10,304.24

= (1 + ) = 10,000(1 + 0.002) 120 = 10,000(1.002) 120 $12,709.44

= (1 + ) = 10,000(1 + 0.03) 10 = 10,000(1.03) 10 $13,439.16

= (1 + ) = 10,000(1 + 0.025 4) 4×5 = 10,000(1.00625) 20 $11,327.08

Technology: 10000*(1+0.025/4)^(4*5)

= (1 + ) = 10,000(1 + 0.0005) 6 = 10,000(1.0005) 6 $10,030.04

= (1 + ) = 10,000(1 + 0.0045) 240 = 10,000(1.0045) 240 $29,375.54

= (1 + ) = 10,000(1 + 0.04) 8 = 10,000(1.04) 8 $13,685.69

= (1 + ) 0.015 52×5 = 10,000 1 + 52

$10,778.72

Technology: 10000*(1+0.015/52)^(52*5)

Solutions Section 3.2

9. We use = (1 + ) . = $10,000, = 0.065, = 365, = 10

10. We use = (1 + ) . = $10,000, = 0.112, = 12, = 12

Technology: 10000* (1+0.065/365)^(365*10)

Technology: 10000* (1+0.112/12)^(12*12)

11. We use = (1 + ) = $1,000, = 10, = 0.05

12. We use = (1 + ) = $1,000, = 5, = 0.06

Technology: 1000*(1+0.05)^(-10)

Technology: 1000*(1+0.06)^(-5)

13. We use = (1 + ) = $1,000, = 5, = 0.042, = 52

14. We use = (1 + ) = $1,000, = 10, = 0.053, = 4

Technology: 1000*(1+0.042/52)^(-52*5)

Technology: 1000*(1+0.053/4)^(-4*10)

15. We use = (1 + ) = $1,000, = 4, = 0.05

16. We use = (1 + ) = $1,000, = 5, = 0.04

Technology: 1000*(1-0.05)^(-4)

Technology: 1000*(1-0.04)^(-5)

= (1 + ) 0.065 365×10 = 10,000 1 + 365 0.065 3,650 $19,154.30 = 10,000 1 + 365

= (1 + ) = 1,000(1 + 0.05) 10 = 1,000(1.05) 10 $613.91

= (1 + ) 0.042 52×5 = 1,000 1 + 52 0.042 260 $810.65 = 1,000 1 + 52

= (1 + ) = 1,000(1 0.05) 4 = 1,000(0.95 4) $1,227.74

= (1 + ) 0.112 12×12 = 10,000 1 + 12 0.112 144 $38,105.24 = 10,000 1 + 12

= (1 + ) = 1,000(1 + 0.06) 5 = 1,000(1.06) 5 $747.26

= (1 + ) 0.053 4×10 = 1,000 1 + 4 0.053 40 $590.66 = 1,000 1 + 4

= (1 + ) = 1,000(1 0.04) 5 = 1,000(0.96 5) $1,226.43

Solutions Section 3.2

17. nom = 0.05, = 4

18. nom = 0.05, = 12

e = (1 + nom ) 1 0.05 4 1 = 1 + 4 0.0509, or 5.09%

e = (1 + nom ) 1 0.05 12 1 = 1 + 12 0.0512, or 5.12%

Technology: (1+0.05/4)^4-1

Technology: (1+0.05/12)^12-1

19. nom = 0.10, = 12

20. nom = 0.10, = 365

Technology: (1+0.10/12)^12-1

Technology: (1+0.10/365)^365-1

21. nom = 0.10, = 365 × 24 = 8,760

22. nom = 0.10, = 365 × 24 × 60 = 525,600

e = (1 + nom ) 1 0.10 12 1 = 1 + 12 0.1047, or 10.47%

e = (1 + nom ) 1 0.10 365 1 = 1 + 365 0.1052, or 10.52%

e = (1 + nom ) 1 0.10 8,760 1 = 1 + 8,760 0.1052, or 10.52%

e = (1 + nom ) 1 0.10 525,600 = 1 + 525,600 0.1052, or 10.52%

1

Technology: (1+0.10/8760)^8760-1

Technology: (1+0.10/525600)^525600-1

23. = $1,000, = 0.06, = 4, = 4

24. = $10,000, = 0.02, = 4, = 5

The deposit will have grown by $1,268.99 $1,000 = $268.99. Technology: 1000*(1+0.06/4)^(16)

The deposit will have grown by $11,048.96 $10,000 = $1,048.96 Technology: 10000*(1+0.02/4)^(20)

25. = $3,000, =

26. = $5,000, =

= (1 + ) = 1,000(1 + 0.06 4) 4×4 = 1,000(1.015) 16 $1,268.99

0.376, = 1, = 3

= (1 + ) = 3,000(1 0.376) 3

$728.91

Technology: 3000*(1-0.376)^3

= (1 + ) = 1,000(1 + 0.02 4) 4×5 = 1,000(1.005) 20 $11,048.96

0.42, = 1, = 4

= (1 + ) = 5,000(1 0.42) 4

$565.82

Technology: 5000*(1-0.42)^4

Solutions Section 3.2

27. = $5,000, = 0.055, = 1, = 10

28. = $10,000, = 0.0625, = 1, = 15

Technology: 5000*(1+0.055)^-10

Technology: 10000*(1+0.0625)^-15

29. Gold: = $5,000, = 0.10, = 1, = 10

30. Munis: = $5,000, = 0.06, = 1, = 10

CDs: = $5,000, = 0.05, = 2, = 10

CDs: = $5,000, = 0.03, = 6, = 10

Combined value after 10 years = $12,968.71 + $8,193.08 = $21,161.79 Technology: 5000*(1+0.10)^10+5000* (1+0.05/2)^(2*10)

Combined value after 10 years = $8,954.24 + $6,774.24 = $15,698.49 Technology: 5000*(1+0.06)^10+5000* (1+0.03/6)^(6*10)

31. = $200,000, =

32. = $40,000, =

= (1 + ) = 5,000(1 + 0.055) 10 = 5,0001.055 10 $2,927.15

= (1 + ) = 5,000(1 + 0.10) 10 = 5,000(1.10) 10 $12,968.71

= (1 + ) = 5,000(1 + 0.05 2) 2×10 = 5,000(1.025) 20 $8,193.08

0.02, = 1, = 10

= (1 + ) = 200,000(1 0.02) 10 = 200,000(0.98) 10 $163,414.56

= (1 + ) = 10,000(1 + 0.0625) 15 = 10,0001.0625 15 $4,027.78

= (1 + ) = 5,000(1 + 0.06) 10 = 5,000(1.06) 10 $8,954.24

= (1 + ) = 5,000(1 + 0.03 6) 6×10 = 5,000(1.005) 60 $6,774.24

0.051, = 1, = 12

= (1 + ) = 40,000(1 0.051) 12 = 40,000(0.949) 12 $21,342.95

Technology: 200000*(1-0.02)^10

Technology: 40000*(1-0.051)^12

33. = $1,000,000, = 0.06, = 1, = 30

34. = $1,000,000, = 0.07, = 1, = 40

Technology: 1000000*(1+0.06)^-30

Technology: 1000000*(1+0.07)^-40

= (1 + ) = 1,000,000(1 + 0.06) 30 = 1,000,000(1.06) 30 $174,110

= (1 + ) = 1,000,000(1 + 0.07) 40 = 1,000,000(1.07) 40 $66,780

35. = $297.91, = = 6 × 3 = 18,

0.05, = 1,

Solutions Section 3.2

= (1 + ) = 297.91(1 0.05) 18 = 297.91(0.95) 18 $750.00

36. = 100, = 0.06, = 1, = 2 × 5 = 10 = (1 + ) = 100(1 + 0.06) 10 = 100(1.06) 10 179 kits per month

Technology: 100*(1+0.06)^10

Technology: 297.91*(1-0.05)^-18 37. = (1 + ) = 7,445.37(1 + 0.02544) 19 38. = (1 + ) = 8,368.30(1 + 0.01836) 4

$12,000

$9,000

39. = (1 + ) = 10,000(1 + 0.03045) 10 = 10,000(1.03045) 10 40. = (1 + ) = 5,000(1 + 0.03258) 7 = 5,000(1.03258) 7

$7,408.51

$3,994.88

41. Total interest on $1 during the 6-year life of the bonds is = = (1 + 0.01551) 6 1 $0.09674 Thus, the monthly simple interest rate would be 0.09674 (6 × 12)

0.0013, or 0.13%.

42. Total interest on $1 during the 5-year life of the bonds is = = (1 + 0.02486) 5 1 $0.13064 Thus, the biannual simple interest rate would be 0.13064 (5 × 2)

0.0131, or 1.31%.

43. = $100,000, = 0.04, = 1, = 15

44. = $80,000, = 0.05, = 1, = 10

Technology: 100000*(1+0.04)^-15

Technology: 80000*(1+0.05)^-10

45. = $30,000, = 0.02, = 1, = 5

46. = $30,000, = 0.01, = 1, = 5 × 2 = 10

Technology: 30000*(1+0.02)^-5

Technology: 30000*(1+0.01)^-10

= (1 + ) = 100,000(1 + 0.04) 15 = 100,000(1.04) 15 $55,526.45 per year

= (1 + ) = 30,000(1 + 0.02) 5 = 30,000(1.02) 5 $27,171.92

= (1 + ) = 80,000(1 + 0.05 1) 10 = 80,000(1.05) 10 $49,113.06 per year

= (1 + ) = 30,000(1 + 0.01) 10 = 30,000(1.01) 10 $27,158.61

Solutions Section 3.2

47. = $200,000, = 0.06, = 1, = 10 = (1 + ) = 200,000(1 + 0.06) 10 = 200,000(1.06) 10 $111,678.96

Technology: 200000*(1+0.06)^-10

48. = $200,000, = 0.005, = 1, = 10 × 12 = 120 = (1 + ) = 200,000(1 + 0.005) 120 = 200,000(1.005) 120 $109,926.55

Technology: 200000*(1+0.005)^-120

49. Step 1: Calculate the future value of the investment: = $1,000, = 0.05, = 1, = 2 = (1 + ) = 1,000(1 + 0.05) 2 = 1,000(1.05) 2 $1,102.50

Technology: 1000*(1+0.05)^2 Step 2: Discount this value using inflation: = $1,102.50, = 0.03, = 1, = 2 = (1 + ) = 1,102.50(1 + 0.03) 2 = 1,102.50(1.03) 2 $1,039.21

Technology: 1102.50*(1+0.03)^-2

50. Step 1: Calculate the future value of the investment: = $10,000, = 0.08, = 12, = 2 = (1 + ) = 10,000(1 + 0.08 12) 12×2

$11,728.88

Technology: 10000*(1+0.08/12)^24 Step 2: Discount this value using inflation: = $11,728.88, = 0.01, = 1, = 24 = (1 + ) = 11,728.88(1 + 0.01) 24 = 11,728.88(1.01) 24 $9,237.27

Technology: 11728.88*(1+0.01)^-24

Solutions Section 3.2 51. Compare the effective yields of the two investments: First Investment: nom = 0.12, = 1 e = nom = 0.12, or 12%

52. Compare the effective yields of the three investments: First Investment: nom = 0.15, = 1 e = nom = 0.15, or 15%

Second Investment: nom = 0.119, = 12

Second Investment: nom = 0.145, = 4

The better investment is the second.

Technology: (1+0.145/4)^4-1

e = (1 + nom ) 1 0.119 12 1 = 1 + 12 0.1257, or 12.57%

e = (1 + nom ) 1 0.145 4 1 = 1 + 4 0.1531, or 15.31%

Third Investment: nom = 0.14, = 12

e = (1 + nom ) 1 0.14 12 1 = 1 + 12 0.1493, or 14.93%

Technology: (1+0.14/12)^12-1 The best investment is the second: 14.5% compounded quarterly. 53. = $24, = 0.063, = 1, = 2021 1626 = 395

54. = $24, = 0.062, = 1, = 2021 1626 = 395

This is more than the 2021 estimated market value of $508,176 million. Thus, the Lenape could have bought the island back in 2021. Technology: 24*(1+0.063)^395

This is less than the 2021 estimated market value of $508,176 million. Thus, the Lenape could not have bought the island back in 2021. Technology: 24*(1+0.062)^395

55. = 100 reals, = 0.0968, = 1, = 5

56. = 1,000 pesos, = 0.5140, = 1, = 5

Technology: 100*(1+0.0968)^5

Technology: 1000*(1+0.5140)^5

= (1 + ) = 24(1 + 0.063) 395 = 24(1.063) 395 $725,856 million

= (1 + ) = 100(1 + 0.0968) 5 = 100(1.0968) 5 159 reals

= (1 + ) = 24(1 + 0.062) 395 = 24(1.062) 395 $500,490 million

= (1 + ) = 1,000(1 + 0.5140) 5 = 1,000(1.5140) 5 7,955 pesos

Solutions Section 3.2

57. = 1,000 bolivianos, = 0.0020, = 1, = 10 = (1 + ) = 1,000(1 + 0.0020) 10 = 1,000(1.0020) 10 980 bolivianos

58. = 20,000 pesos, = 0.0559, = 1, = 10 = (1 + ) = 20,000(1 + 0.0559) 10 = 20,000(1.0559) 10 11,609 pesos

Technology: 20000*(1+0.0559)^-10

Technology: 1000*(1+0.0020)^-10 59. Step 1: Calculate the future value of the investment: = 1,000 pesos, = 0.05, = 2, = 10

60. Step 1: Calculate the future value of the investment: = 1,000 pesos, = 0.05, = 2, = 10

Technology: 1000*(1+0.05/2)^20

Technology: 1000*(1+0.05/2)^20

Step 2: Discount this amount using the inflation rate: = 1,638.62, = 0.0480, = 1, = 10

Step 2: Discount this amount using the inflation rate: = 1,638.62, = 0.0759, = 1, = 10

Technology: 1638.62*(1+0.0480)^-10

Technology: 1638.62*(1+0.0759)^-10

= (1 + ) 0.05 2×10 = 1,000 1 + 2 = 1,000(1.025) 20 1,638.62 pesos

= (1 + ) = 1,638.62(1 + 0.0480) 10 = 1,638.62(1.0480) 10 1,025 pesos

= (1 + ) 0.05 2×10 = 1,000 1 + 2 = 1,000(1.025) 20 1,638.62 pesos

= (1 + ) = 1,638.62(1 + 0.0759) 10 = 1,638.62(1.0759) 10 788 pesos

Solutions Section 3.2 61. We compare the future values of 1 unit of the currency for a 1-year period: Mexico: = 1 peso, = 0.06, = 1, = 1

62. We compare the future values of 1 unit of the currency for a 1-year period: Brazil: = 1 real, = 0.10, = 1, = 1

Now discount this using inflation: = 1.06, = 0.0559, = 1, = 1

Now discount this using inflation: = 1.10, = 0.0968, = 1, = 1

= (1 + ) = 1(1 + 0.06) 1 = 1.06 pesos

= (1 + ) = 1.06(1 + 0.0559) 1

= (1 + ) = 1(1 + 0.10) 1 = 1.10 reals

= (1 + ) = 1.10(1 + 0.0968) 1 1.0029 reals

1.0039 pesos

Nicaragua: = 1 córdoba, = 0.045, = 2, = 1

Uruguay: = 1 peso, = 0.075, = 2, = 1

Now discount this using inflation: = 1.0455, = 0.0412, = 1, = 1

Now discount this using inflation: = 1.0764, = 0.0759, = 1, = 1

The investment in Nicaragua is better.

The investment in Brazil is better.

= (1 + ) 0.045 2 = 1 1 + 2 = 1.0455 córdobas

= (1 + ) = 1.0455(1 + 0.0412) 1 = 1.0041 córdobas

= (1 + ) 0.075 2×1 = 1 1 + 2 1.0764 pesos

= (1 + ) 1.0764(1 + 0.0759) 1 1.0005 pesos

63. = 58,900, = 61,400, = 12, = 4 months = 1 3 year, so = 12 3 = 4 61,400 = 58,900(1 + 12) 4, so solving for gives = 12[(61,400 58,900) 1 4 1] 12.54% 64. = 32,600, = 61,400, = 12, = 6 months = 1 2 year, so = 12 2 = 6 61,400 = 32,600(1 + 12) 6, so solving for gives = 12[(61,400 32,600) 1 6 1] 133.54%

0.1254 or 1.3354 or

65. Calculating the annual returns as in the preceding exercises (or by using one of the TVM utilities), we get the following figures: Aug

Sep

Oct

159.04% 27.16% 167.38% The largest annual return would have been 167.38% if you had sold on October 1, 2021.

Solutions Section 3.2 66. Calculating the annual returns as in the preceding exercises (or by using one of the TVM utilities), we get the following figures: Jan

Feb

Mar

144.83% 306.40% 319.69% The largest annual return would have been 319.69% if you had sold on March 1, 2021. 67. No. Compound interest increase is exponential, and hence a curve that is concave upwards. However, the three points coresponing to January 1, February 1 and March 1 2021 are on a straight line (see Exercise Exercise ?? in Section __._) and therefore not a curve. 68. No. Compound interest increase is exponential, and hence a curve that is concave upwards.However, the three points coresponing to April 1, May 1, and June 1 2021 define a curve that is concave down and therefore not exponentially increasing. 69. My investment: = $5,000, = 0.054, = 2 0.054 2 = (1 + ) = 5,000 1 + = 5,000(1.027) 2 2 Friend's investment: = $6,000, = 0.048, = 2 0.048 2 = (1 + ) = 6,000 1 + = 6,000(1.024) 2 2 Solution via logarithms: 5,000(1.027) 2 = 6,000(1.024) 2 (1.027) 2 = 1.2(1.024) 2 2 log 1.027 = log[1.2(1.024) 2 ] = log 1.2 + 2 log 1.024 2 (log 1.027 log 1.024) = log 1.2 log 1.2 = 31 2(log 1.027 log 1.024) Solution via graphing: Technology Formulas: Y1 = 5000*1.027^(2*x) Graph and zoomed-in view:

Y2 = 6000*1.024^(2*x)

The graphs cross around 31 years. The value of the investment then is 5,000(1.027) 2×31 $26,100 (rounded to 3 significant digits) 70. = $3,000, = 0.05, = 365

= (1 + ) = 3,000 1 +

Logarithms:

0.05 365 365

0.05 365 3,000 1 + = 10,000 365 0.05 365 10 = 1 + 365 3 0.05 10 365 log 1 + = log 365 3 log(10 3) = 24 365 log(1 + 0.05 365)

Solutions Section 3.2

Technology: Y1 = 3000*(1+0.05/365)^(365*x) Graph and zoomed-in view:

Y2 = 10000

24 years

71. = 40,000, = 1.0 (a 100% increase per period), = 1, in half-years = (1 + ) = 40,000(1 + 1.0) = 40,000(2) Logarithms: 40,000(2) = 1,000,000 2 = 25 log 2 = log 25 log 25 = 4.65 half-years log 2 Technology: Y1 = 40000*2^x Graph and zoomed-in view:

Y2 = 1000000

4.65. Since measures 6-month periods, this corresponds to 4.65 2

72. = $4,354, = 0.05, = 1, in half-years = (1 + ) = 4,354(1 0.05) = 4,354(0.95) Logarithms: 4,354(0.95) = 50

2.3 years.

0.95 = 50 4,354 log 0.95 = log(50 4,354) log(50 4,354) 87.1 = log 0.95 Technology: Y1 = 4354*0.95^x Graph and zoomed-in view:

Solutions Section 3.2

Y2 = 50

87.1. Since measures 6-month periods, this corresponds to 87.1 2

44 years.

73. a. The amount you pay for the bond is its present value. = $100,000, = 0.15, = 1, = 30 = (1 + ) = 100,000(1 + 0.15) 30 $1,510.31 b. Because the future value remains fixed at $100,000, the value of the bond at any time is its present value at that time, given the prevailing interest rate. Since it will pay $100,000 in 13 years' time, we have: = $100,000, = 0.0475, = 1, = 13 = (1 + ) = 100,000(1 + 0.0.475) 13 $54,701.29 c. By part (a) the bond cost you $1,510.31 and, by part (b), was worth $54,701.29 after 17 years. = $1,510.31, = $54,701.29, = 1, = 17 = (1 + ) 54,701.29 = 1,510.31(1 + ) 17 54,701.29 = (1 + ) 17 1,510.31 54,701.29 1 17 1+ = � � 1,510.31 54,701.29 1 17 = � 1 0.2351, or 23.51% � 1,510.31 Technology: (54701.29/1510.31)^(1/17)-1 74. a. The amount you pay for the bond is its present value. = $100,000, = 0.05, = 1, = 30 = (1 + ) = 100,000(1 + 0.05) 30 $23,137.74 b. The value of the bond at any time is its present value at that time, given the prevailing interest rate. Since it will pay $100,000 in 15 years' time, we have: = $100,000, = 0.12, = 1, = 15 = (1 + ) = 100,000(1 + 0.12) 15 $18,269.63 c. By part (a), the bond cost you $23,137.74 and, by part (b), was worth $18,269.63 after 15 years. = $23,137.74, = $18,269.63, = 1, = 15 = (1 + )

Solutions Section 3.2 18,269.63 = 23,137.74(1 + ) 15 18,269.63 = (1 + ) 15 23,137.74 18,269.63 1 15 1+ = � � 23,137.74 18,269.63 1 15 = � 1 0.0156, or 1.56 � 23,137.74 You will have lost 1.56% per year. Technology: (18269.63/23137.74)^(1/15)-1

75. The function = (1 + ) is not a linear function of , but an exponential function. Thus, its graph is not a straight line. 76. Wrong. The second investment earns more, because the second investment pays interest on a larger amount after 6 months.

77. Wrong. Its growth can be modeled by 0.01(1 + 0.10) = 0.01(1.10) . This is an exponential function of , not a linear one. 78. After one interest period: The calculations are the same for a single interest period. 79. A compound interest investment behaves as though it were being compounded once a year at the effective rate. Thus, if two equal investments have the same effective rates, they grow at the same rate and the graphs of future value will be the same. 80. The effective rate equals the nominal rate when the interest is compounded once a year because, by definition, the effective rate is the annually compounded rate that would result in the same future value. 81. The effective rate exceeds the nominal rate when the interest is compounded more than once a year because then interest is being paid on interest accumulated during each year, resulting in a larger effective rate. Conversely, if the interest is compounded less often than once a year, the effective rate is less than the nominal rate. 82. No, it will not return to the same value it had originally. The simplest way to see this is to note the following lack of symmetry: The interest earned in year 5 is slightly less than the depreciation in year 6, since the depreciation is calculated on a larger value. A direct (two-step) calculation of the future value after 10 years gives (1.1 5)(0.9 5) 0.95099 , slightly less than . 83. Compare their future values in constant dollars. That is, compute their future values, and then discount each for inflation. The investment with the larger future value is the better investment. 84. First compute the value in 2011 dollars, using the 2011 inflation rate, then use the 2010 inflation rate to convert the answer to 2010 dollars, and so on, down to 2000 dollars. 85. = 100, = 0.10, = 1, 10, 100, … Y1 = 100*(1.10)^x Y2 = 100*(1+ 0.10/10)^(10*x) Y3 = 100*(1+ 0.10/100)^(100*x) …

Solutions Section 3.2

The graphs are approaching a particular curve (shown darker) as gets larger, approximately the curve given by the largest value of . 86. Since the graph gets arbitrarily close to the -axis as gets larger, the future value gets arbitrarily close to zero as time goes on, so it will eventually dip below any value that is set, like $1.

Solutions Section 3.3 Section 3.3

1. = $100, = interest paid each period = 0.05 12, = total number of periods = 12 × 10 = 120 (1 + 0.05 12) 120 1 (1 + ) 1 = $15,528.23 = 100 0.05 12 Technology: 100*((1+0.05/12)^120-1)/(0.05/12) 2. = $150, = 0.03 12, = 12 × 20 = 240 (1 + 0.03 12) 240 1 (1 + ) 1 = $49,245.30 = 150 0.03 12 Technology: 150*((1+0.03/12)^240-1)/(0.03/12)

3. = $1,000, = 0.07 4, = 4 × 20 = 80 (1 + 0.07 4) 80 1 (1 + ) 1 = $171,793.82 = 1,000 0.07 4 Technology: 1000*((1+0.07/4)^80-1)/(0.07/4) 4. = $2,000, = 0.07 4, = 4 × 10 = 40 (1 + 0.07 4) 40 1 (1 + ) 1 = $114,468.27 = 2,000 0.07 4 Technology: 2000*((1+0.07/4)^40-1)/(0.07/4) 5. = $5,000, = $100, = 0.05 12, = 12 × 10 = 120

We need to calculate the sum of = (1 + ) and =

(1 + )

1

(1 + )

1

.

(1 + 0.05 12) 120 1 $23,763.28 0.05 12 Technology: 5000*(1+0.05/12)^120+100*((1+0.05/12)^120-1)/(0.05/12) = 5,000(1 + 0.05 12) 120 + 100

6. = $10,000, = $150, = 0.03 12, = 12 × 20 = 240

We need to calculate the sum of = (1 + ) and =

.

(1 + 0.03 12) 240 1 $67,452.85 0.03 12 Technology: 10000*(1+0.03/12)^240+150*((1+0.03/12)^240-1)/(0.03/12) = 10,000(1 + 0.03 12) 240 + 150

7. = $10,000, = 0.05 12, = 12 × 5 = 60 0.05 12 = = 10,000 (1 + ) 1 (1 + 0.05 12) 60 1 Technology: 10000*0.05/12/((1+0.05/12)^60-1)

$147.05

9. = $75,000, = 0.06 4, = 4 × 20 = 80 0.06 4 = = 75,000 (1 + ) 1 (1 + 0.06 4) 80 1 Technology: 75000*0.06/4/((1+0.06/4)^80-1)

$491.12

8. = $20,000, = 0.03 12, = 12 × 10 = 120 0.03 12 = $143.12 = 20,000 (1 + ) 1 (1 + 0.03 12) 120 1 Technology: 20000*0.03/12/((1+0.03/12)^120-1)

Solutions Section 3.3 10. = $100,000, = 0.07 4, = 4 × 20 = 80 0.07 4 $582.09 = = 100,000 (1 + ) 1 (1 + 0.07 4) 80 1 Technology: 100000*0.07/4/((1+0.07/4)^80-1)

11. We first account for the future value of the $10,000 already in the account: = $10,000, = 0.05 12, = 12 × 5 = 60 = (1 + ) = 10,000(1 + 0.05 12) 60 We subtract this from $20,000 to get the future value of the payments, so: = (1 + ) 1 where = $20,000 10,000(1 + 0.05 12) 60 =

[20,000

10,000(1 + 0.05 12) 60](0.05 12)

$105.38 (1 + 0.05 12) 60 1 Technology: (20000-10000*(1+0.05/12)^60)*0.05/12/((1+0.05/12)^60-1) 12. We first account for the future value of the $10,000 already in the account: = $10,000, = 0.03 12, = 12 × 10 = 120 = (1 + ) = 10,000(1 + 0.03 12) 120 We subtract this from $30,000 to get the future value of the payments, so: = (1 + ) 1 where = $30,000 10,000(1 + 0.03 12) 120 =

[30,000

10,000(1 + 0.03 12) 120](0.03 12)

$118.12 (1 + 0.03 12) 120 1 Technology: (30000-10000*(1+0.03/12)^120)*0.03/12/((1+0.03/12)^120-1) 13. = $500, = 0.03 12, = 12 × 20 = 240 1 (1 + 0.03 12) 240 1 (1 + ) = $90,155.46 = 500 0.03 12 Technology: 500*(1-(1+0.03/12)^(-240))/(0.03/12)

14. = $1,000, = 0.05 12, = 12 × 15 = 180 1 (1 + 0.05 12) 180 1 (1 + ) = $126,455.24 = 1000 0.05 12 Technology: 1000*(1-(1+0.05/12)^(-180))/(0.05/12) 15. = $1,500, = 0.06 4, = 4 × 20 = 80 1 (1 + 0.06 4) 80 1 (1 + ) = $69,610.99 = 1,500 0.06 4 Technology: 1500*(1-(1+0.06/4)^(80))/(0.06/4) 16. = $2,000, = 0.04 4 = 0.01, = 4 × 20 = 80 1 (1 + ) 1 (1 + 0.01) 80 = = 2,000 0.01 Technology: 2000*(1-(1+0.01)^(-80))/0.01

$109,776.41

Solutions Section 3.3 17. = $10,000, = $500, = 0.03 12, = 12 × 20 = 240 We need to fund both the future value and the payments, so the present value is the sum (1 + ) 1 (1 + 0.03 12) 240 = 10,000(1 + 0.03 12) 240 + 500 0.03 12

= (1 + ) +

1

$95,647.68

Technology: 10000*(1+0.03/12)^(-240)+500*(1-(1+0.03/12)^(-240))/(0.03/12)

18. = $10,000, = $1,000, = 0.05 12, = 12 × 15 = 180 We need to fund both the future value and the payments, so the present value is the sum (1 + ) 1 = 10,000(1 + 0.05 12) 180 + 1,000

= (1 + ) +

1

(1 + 0.05 12) 180 0.05 12

$135,917.31

Technology: 20000*(1+0.05/12)^(-180)+1000*(1-(1+0.05/12)^(-180))/(0.05/12) 19. = $100,000, = 0.03 12, = 12 × 20 = 240 0.03 12 = $554.60 = 100,000 1 (1 + ) 1 (1 + 0.03 12) 240 Technology: 100000*(0.03/12)/(1-(1+0.03/12)^(-240))

20. = $150,000, = 0.05 12, = 12 × 15 = 180 0.05 12 = $1,186.19 = 150,000 1 (1 + ) 1 (1 + 0.05 12) 180 Technology: 150000*(0.05/12)/(1-(1+0.05/12)^(-180)) 21. = $75,000, = 0.04 4 = 0.01, = 4 × 20 = 80 0.01 = = 75,000 1 (1 + ) 1 (1 + 0.01) 80 Technology: 75000*0.01/(1-(1+0.01)^(-80))

22. = $200,000, = 0.06 4 = 0.015, = 4 × 15 = 60 0.015 = = 200,000 1 (1 + ) 1 (1 + 0.015) 60 Technology: 200000*0.015/(1-(1+0.015)^(-60))

$1,366.41

$5,078.69

23. = $100,000, = $10,000, = 0.03 12, = 12 × 20 = 240 Part of the present value has to fund the future value of $10,000: (1 + ) = 10,000(1 + 0.03 12) 240 is the amount required; we subtract this from the present value and use = 100,000 10,000(1 + 0.03 12) 240 in the payment formula. (0.03 12)[100,000 10,000(1 + 0.03 12) 240] = $524.14 = 1 (1 + ) 1 (1 + 0.03 12) 240 Technology: (0.03/12)*(100000-10000*(1+0.03/12)^(-240))/(1-(1+0.03/12)^(-240)) 24. = $150,000, = $20,000, = 0.05 12, = 12 × 15 = 180

Solutions Section 3.3 Part of the present value has to fund the future value of $20,000: (1 + ) = 20,000(1 + 0.05 12) 180 is the amount required; we subtract this from the present value and use = 150,000 20,000(1 + 0.05 12) 180 in the payment formula. (0.05 12)[150,000 20,000(1 + 0.05 12) 180] = $1,111.37 = 1 (1 + ) 1 (1 + 0.05 12) 180 Technology: (0.05/12)*(150000-20000*(1+0.05/12)^(-180))/(1-(1+0.05/12)^(-180)) 25. = $10,000, = 0.09 12 = 0.0075, = 4 × 12 = 48 0.0075 = $248.85 = 10,000 1 (1 + ) 1 (1 + 0.0075) 48 Technology: 10000*0.0075/(1-(1+0.0075)^(-48))

26. = $20,000, = 0.08 12, = 12 × 5 = 60 0.08 12 = $405.53 = 20,000 1 (1 + ) 1 (1 + 0.08 12) 60 Technology: 20000*(0.08/12)/(1-(1+0.08/12)^(-60))

27. = $100,000, = 0.05 4, = 4 × 20 = 80 0.05 4 = $1,984.65 = 100,000 1 (1 + ) 1 (1 + 0.05 4) 80 Technology: 100000*(0.05/4)/(1-(1+0.05/4)^(-80))

28. = $1,000,000, = 0.04 4, = 4 × 10 = 40 0.04 4 = $30,455.60 = 1,000,000 1 (1 + ) 1 (1 + 0.04 4) 40 Technology: 1000000*(0.04/4)/(1-(1+0.04/4)^(-40)) 29. = 100,000, = 0.043 12, = 12 × 30 = 360 0.043 12 = = 100,000 = $494.87 1 (1 + ) 1 (1 + 0.043 12) 360 Technology: 100000*(0.043/12)/(1-(1+0.043/12)^(-360))

30. = 250,000, = 0.062 12, = 12 × 15 = 180 0.062 12 = = 250,000 = $2,136.75 1 (1 + ) 1 (1 + 0.062 12) 180 Technology: 250000*(0.062/12)/(1-(1+0.062/12)^(-180))

31. = 1,000,000, = 0.054 12, = 12 × 30 = 360 0.054 12 = = 1,000,000 = $5,615.31 1 (1 + ) 1 (1 + 0.054 12) 360 Technology: 1000000*(0.054/12)/(1-(1+0.054/12)^(-360))

32. = 2,000,000, = 0.045 12, = 12 × 15 = 180 0.045 12 = = 2,000,000 = $15,299.87 1 (1 + ) 1 (1 + 0.045 12) 180 Technology: 2000000*(0.045/12)/(1-(1+0.045/12)^(-180))

Solutions Section 3.3 33. We calculated = $494.87 in Exercise 29. = 0.043 12, = 12 × 30 = 360, = 12 × 10 = 120, so = 240 1 (1 + ) ( ) = 1 (1 + 0.043 12) 240 = 494.87 = $79,573.29 0.043 12 Technology: 494.87*(1-(1+0.043/12)^(-240))/(0.043/12) 34. We calculated = $2,136.75 in Exercise 30. = 0.062 12, = 12 × 15 = 180, = 12 × 10 = 120, so = 60 1 (1 + ) ( ) = 1 (1 + 0.062 12) 60 = 2,136.75 = $109,994.70 0.062 12 Technology: 2136.75*(1-(1+0.062/12)^(-60))/(0.062/12)

35. We calculated = $5,615.31 in Exercise 31. = 0.054 12, = 12 × 30 = 360, = 12 × 5 = 60, so = 300 1 (1 + ) ( ) = 1 (1 + 0.054 12) 300 = 5,615.31 = $923,373.42 0.054 12 Technology: 5615.31*(1-(1+0.054/12)^(-300))/(0.054/12) 36. We calculated = $15,299.87 in Exercise 32. = 0.045 12, = 12 × 15 = 180, = 12 × 8 = 96, so = 84 1 (1 + ) ( ) = 1 (1 + 0.045 12) 84 = 15,299.87 = $1,100,697.30 0.045 12 Technology: 15299.87*(1-(1+0.045/12)^(-84))/(0.045/12) 37. First, we calculate the monthly payments: = 50,000, = 0.085 12, = 12 × 200 = 2,400 0.085 12 = = 50,000 = $354.17 1 (1 + ) 1 (1 + 0.085 12) 2,400 Technology: 50000*(0.085/12)/(1-(1+0.085/12)^(-2400)) Now calculate the outstanding balance based on the above payments: = 12 × 20 = 240, so = 2,160 1 (1 + ) ( ) = 1 (1 + 0.085 12) 2,160 = 354.17 = $50,000.46 0.085 12 Technology: 354.17*(1-(1+0.085/12)^(-2160))/(0.085/12) This is more than the original value of the loan. In a 200-year mortgage, the fraction of the initial payments going toward reducing the principal is so small that the rounding upward of the payment makes it appear that, for many years at the start of the mortgage, more is owed than the original amount borrowed. 38. First, we calculate the monthly payments: = 100,000, = 0.096 12, = 12 × 200 = 2,400

Solutions Section 3.3 0.096 12 = = 100,000 = $800 1 (1 + ) 1 (1 + 0.096 12) 2,400 Technology: 100000*(0.096/12)/(1-(1+0.096/12)^(-2400)) Now calculate the outstanding balance based on the above payments: = 12 × 20 = 240, so = 2,160 1 (1 + ) ( ) = 1 (1 + 0.096 12) 2,160 = 800 = $100,000.00 0.096 12 Technology: 800*(1-(1+0.096/12)^(-2160))/(0.096/12) The outstanding principal has not changed from the original amount. In a 200-year mortgage, the fraction of the initial payments going toward reducing the principal is so small that rounding the payment amount to the nearest cent erases that contribution in this case. 39. The periodic payments are based on a 4.875% annual payment. For payments twice a year, this is = 1,000(0.04875 2) = $24.375 Since the bond yield is 4.880%, = 0.0488 2 = 0.0244; = 2 × 10 = 20. The present value comes from the future value of $1,000 and the payments, which we treat like an annuity: (1 + ) 1 (1 + 0.0244) 20 = 1,000(1 + 0.0244) 20 + 24.375 0.0244 1 1.0244 20 $999.61 = 1,000(1.0244) 20 + 24.375 0.0244

= (1 + ) +

1

Technology: 1000*1.0244^(-20)+24.375*(1-1.0244^(-20))/0.0244 Online Time Value of Money Utility:

40. The periodic payments are based on a 5.375% annual payment. For payments twice a year, this is = 1,000(0.05375 2) = $26.875 Since the bond yield is 5.460%, = 0.0546 2 = 0.0273; = 2 × 30 = 60. The present value comes from the future value of $1,000 and the payments, which we treat like an annuity: (1 + ) 1 (1 + 0.0273) 60 = 1,000(1 + 0.0273) 60 + 26.875 0.0273 60 1 1.0273 $987.53 = 1,000(1.0273) 60 + 26.875 0.0273

= (1 + ) +

1

Technology: 1000*1.0273^(-60)+26.875*(1-1.0273^(-60))/0.0273

Solutions Section 3.3 Online Time Value of Money Utility:

41. The periodic payments are based on a 3.625% annual payment. For payments twice a year, this is = 1,000(0.03625 2) = $18.125 Since the bond yield is 3.705%, = 0.03705 2 = 0.018525; = 2 × 2 = 4. The present value comes from the future value of $1,000 and the payments, which we treat like an annuity: (1 + ) 1 = 1,000(1 + 0.018525) 4 + 18.125

= (1 + ) +

1

(1 + 0.018525) 4 0.018525 1 1.018525 4 $998.47 = 1,000(1.018525) 4 + 18.125 0.018525

Technology: 1000*1.018525^(-4)+18.125*(1-1.018525^(-4))/0.018525 Online Time Value of Money Utility:

42. The periodic payments are based on a 4.375% annual payment. For payments twice a year, this is = 1,000(0.04375 2) = $21.875 Since the bond yield is 4.475%, = 0.04475 2 = 0.022375; = 2 × 5 = 10. The present value comes from the future value of $1,000 and the payments, which we treat like an annuity: (1 + ) 1 = 1,000(1 + 0.022375) 10 + 21.875

= (1 + ) +

1

(1 + 0.022375) 10 0.022375 10 1 1.022375 $995.56 = 1,000(1.022375) 10 + 21.875 0.022375

Solutions Section 3.3 Technology: 1000*1.022375^(-10)+21.875*(1-1.022375^(-10))/0.022375 Online Time Value of Money Utility:

43. The periodic payments are based on a 5.5% annual payment. For payments twice a year, this is = 1,000(0.055 2) = $27.5 Since the bond yield is 6.643%, = 0.06643 2 = 0.033215; = 2 × 10 = 20. The present value comes from the future value of $1,000 and the payments, which we treat like an annuity: (1 + ) 1 (1 + 0.033215) 20 = 1,000(1 + 0.033215) 20 + 27.5 0.033215 20 1 1.033215 $917.45 = 1,000(1.033215) 20 + 27.5 0.033215

= (1 + ) +

1

Technology: 1000*1.033215^(-20)+27.5*(1-1.033215^(-20))/0.033215 Online Time Value of Money Utility:

44. The periodic payments are based on a 6.25% annual payment. For payments twice a year, this is = 1,000(0.0625 2) = $31.25 Since the bond yield is 33.409%, = 0.33409 2 = 0.167045; = 2 × 10 = 20. The present value comes from the future value of $1,000 and the payments, which we treat like an annuity: (1 + ) 1 (1 + 0.167045) 20 = 1,000(1 + 0.167045) 20 + 31.25 0.167045 20 1 1.167045 $224.08 = 1,000(1.167045) 20 + 31.25 0.167045

= (1 + ) +

1

Solutions Section 3.3 Technology: 1000*1.167045^(-20)+31.25*(1-1.167045^(-20))/0.167045 Online Time Value of Money Utility:

45. = $400, = 0.3031 12, = 12 × 20 = 240 (1 + 0.3031 12) 240 1 (1 + ) 1 = $6,288,642.68 = 400 0.3031 12 Technology: 400*((1+0.3031/12)^240-1)/(0.3031/12) 46. = $380, = 0.1798 12, = 12 × 25 = 300 (1 + 0.1798 12) 300 1 (1 + ) 1 = $2,171,731.44 = 380 0.1798 12 Technology: 380*((1+0.1798/12)^300-1)/(0.1798/12) 47. Current value: = $500, = 0.0363 12, = 12 × 15 = 180 (1 + 0.0363 12) 180 1 (1 + ) 1 = $119,393.41 = 500 0.0363 12 Technology: 500*((1+0.0363/12)^180-1)/(0.0363/12) Value in 20 years: = $500, = 0.0310 12, = 12 × 20 = 240 (1 + ) 1 = (1 + ) + (1 + 0.0310 12) 240 1 = 119,393.41(1 + 0.0310 12) 240 + 500 0.0310 12 $387,723.01 Technology: 119393.41*(1+0.0310/12)^240 + 500*((1+0.0310/12)^240-1)/(0.0310/12) 48. Current value: = $500, = 0.0310 12, = 12 × 15 = 180 (1 + 0.0310 12) 180 1 (1 + ) 1 = $114,398.69 = 500 0.0310 12 Technology: 500*((1+0.0310/12)^180-1)/(0.0310/12) Value in 20 years: = $500, = 0.0363 12, = 12 × 20 = 240 (1 + ) 1 = (1 + ) + (1 + 0.0363 12) 240 1 = 114,398.69(1 + 0.0363 12) 240 + 500 0.0363 12 $412,135.79 Technology: 114398.69*(1 + 0.0363/12)^240 + 500*((1+0.0363/12)^240-1)/(0.0363/12) 49. = $1,500,000, = 0.0363 12, = 12 × 30 = 360

Solutions Section 3.3 0.0363 12 = $2,307.49 = 1,500,000 (1 + ) 1 (1 + 0.0363 12) 360 1 Technology: 1500000*0.0363/12/((1+0.0363/12)^360-1)

50. = $1,000,000, = 0.0310 12, = 12 × 25 = 300 0.0310 12 = $2,210.96 = 1,000,000 (1 + ) 1 (1 + 0.0310 12) 300 1 Technology: 1000000*0.0310/12/((1+0.0310/12)^300-1)

51. Take to be the amount you deposit each month into the stock fund. Then the deposit into the bond fund is four times that amount. The future value of the account is the sum of the future values of two investments: $ per month at 30.31% for 12 × 25 = 300 months, and $4 per month at 3.10% for 300 months: (1 + 0.3031 12) 300 1 (1 + 0.0310 12) 300 1 = + 4 0.3031 12 0.0310 12 70,362.85 + 1,809.17 = 72,172.02 2,000,000 We need this amount to equal two million: 72,172.02 = 2,000,000, so = 27.712. 72,172.02 Thus, $27.72 should be invested in the stock fund each month (rounding up to ensure you save enough). The amount in the bond fund is four times that amount: 4 × 27.72 = $110.88 per month.

52. Take to be the amount deposited each month into the stock fund. Then the deposit into the bond fund is three times that amount. The future value of the account is the sum of the future values of two investments: $ per month at 17.98% for 12 × 25 = 300 months, and $3 per month at 3.63% for 300 months: (1 + 0.1798 12) 300 1 (1 + 0.0363 12) 300 1 = + 3 0.1798 12 0.0363 12 5,715.08 + 1,462.54 = 7,177.62 We need this amount to equal one million: 7,177.62 = 1,000,000, so 1,000,000 = 139.322 139.33 (rounding up to ensure sufficient savings). This is the amount 7,177.62 that should be in the stock fund. The total investment per month is four times that amount: 4 × 139.33 = $557.32 per month. Note that rounding up to three decimal places instead of two results in a slightly smaller value, $557.29, for 4 , but in that case your deposits in the stock fund would be slighly more than 25% of the total. 53. Funding the $5,000 withdrawals: = 5,000, = 0.3031 12, = 12 × 10 = 120 1 (1 + 0.3031 12) 120 1 (1 + ) = $188,033.18 = 5,000 0.3031 12 Technology: 5000*(1-(1 + 0.3031/12)^(-120))/(0.3031/12) To fund the lump sum of $30,000 after 10 years, we need the present value of $30,000 under compound interest: = 30,000(1 + 0.3031 12) 120 = $1,503.58. (We rounded up to ensure that we won't fall a little short.)

So, the total in the trust should be $188,033.18 + $1,503.58 = $189,536.76. However, you can check on a TVM calculator that $189,536.75 would also be sufficient to fund the 54. Funding the $2,000 withdrawals: = 2,000, = 0.1798 12, = 12 × 20 = 240

Solutions Section 3.3 1 (1 + 0.1798 12) 240 (1 + ) = $129,720.83 (We rounded up to = 2,000 0.1798 12 ensure that we won't fall a little short.) Technology: 2000*(1-(1 + 0.1798/12)^(-240))/(0.1798/12) To fund the lump sum of $100,000 after 20 years, we need the present value of $100,000 under compound interest: 1

= 100,000(1 + 0.1798 12) 240 = $2,817.49. (We are rounding up to ensure that won't fall a little short.)

So, the total in the trust should be $129,720.83 + $2,817.49 = $132,538.32. Notice that we rounded up twice to get the total. However, using the TVM solver will show that payments of one penny less, $132,538.31, are also sufficient to fund the trust. 55. This is a two-stage process: (1) An accumulation stage to build the retirement fund (2) An amortization stage depleting the fund during retirement Stage 1: Building the retirement fund. = $1,200, = 0.04 4 = 0.01, = 4 × 40 = 160. (1 + ) 1 160 1.01 1 = 1,200 0.01 $469,659.16

=

Technology: 1200*(1.01^160-1)/0.01 Stage 2: Depleting the fund. = $469,659.16, = 0.04 4 = 0.01, = 4 × 25 = 100. 1 (1 + ) 469,659.16 × 0.01 = 1 1.01 100 $7,451.49

=

Technology: 469659.16*0.01/(1-1.01^(-100)) 56. This is a two-stage process: (1) An accumulation stage to build the retirement fund (2) An amortization stage depleting the fund during retirement Stage 1: Building the retirement fund. = $300, = 0.05 12, = 12 × 45 = 540. (1 + ) 1 (1 + 0.05 12) 540 = 300 0.05 12 $181,630.10

=

1

Technology: 300*((1+0.05/12)^540-1)/(0.05/12) Stage 2: Depleting the fund. = $607,931.19, = 0.05 12, = 12 × 20 = 240. =

1

(1 + )

=

607,931.19 × 0.05 12

Solutions Section 3.3

1 (1 + 0.05 12) 240 $4,012.08

Technology: 607931.19*(0.05/12)/(1-(1+0.05/12)^(-240)) 57. This is a two-stage process: (1) An accumulation stage to build the retirement fund (2) An amortization stage depleting the fund during retirement As in the text, we work backward, starting with Stage 2, where we have = $5,000, = 0.03 12 = 0.0025, and = 20 × 12 = 240, and we need to calculate the starting value . (1 + ) 1 (1 + 0.0025) 240 = 5,000 0.0025 1 (1.0025) 240 = 5,000 0.0025 $901,554.57

=

1

Technology: 5000*(1-1.0025^(-240))/0.0025 Thus, in Stage 1, you need to accumulate $901,554.57 in the annuity: = $901,554.57, = 0.03 12 = 0.0025, = 40 × 12 = 480. (1 + ) 1 901,554.57 × 0.0025 = 1.0025 480 1 $973.54 per month

=

Technology: 901554.57*0.0025/(1.0025^480-1) 58. This is a two-stage process: (1) An accumulation stage to build the retirement fund (2) An amortization stage depleting the fund during retirement (1 + ) 1 (1 + 0.0125) 100 = 12,000 0.0125 1 (1.0125) 100 = 12,000 0.0125 $682,816.07

=

1

Technology: 12000*(1-1.0125^(-100))/0.0125 Thus, in Stage 1, Meg needs to accumulate $682,816.07 in the annuity: = $682,816.07, = 0.05 4 = 0.0125, = 4 × 45 = 180. (1 + ) 1 682,816.07 × 0.0125 = 1.0125 180 1 $1,021.40 per quarter

=

Solutions Section 3.3 Technology: 682816.07*0.0125/(1.0125^180-1)

59. = $1,000,000, = 0.048 12 = 0.004, = 12 × (87 30) = 684 0.004 = $278.92 = 1,000,000 (1 + ) 1 1.004 684 1 Technology: 1000000*0.004/(1.004^684-1) 60. = $1,000,000, = 0.048 12 = 0.004, = 12 × (76 30) = 552 0.004 = $496.43 = 1,000,000 (1 + ) 1 1.004 552 1 Technology: 1000000*0.004/(1.004^552-1) 61. Take to be the current age of an insured person. = $500,000, = 0.048 12 = 0.004; 0.004 = = 500,000 (1 + ) 1 1.004 1 0.004 Men: = = 500,000 12(75 ) (1 + ) 1 1.004 1 0.004 Women: = = 500,000 (1 + ) 1 1.004 12(80 ) 1 Technology: 500000*0.004/(1.004^(12*(75-x))-1) 500000*0.004/(1.004^(12*(80-x))-1) Result: Age

Men

Women

30

$261.99

$200.59

50

$864.98

$623.33

70 $7,389.87 $3,254.53 62. Take to be the current age of an insured person. = $800,000, = 0.048 12 = 0.004; 0.004 = = 800,000 (1 + ) 1 1.004 1 0.004 Men: = = 800,000 12(73 ) (1 + ) 1 1.004 1 0.004 Women: = = 800,000 (1 + ) 1 1.004 12(79 ) 1 Technology: 800000*0.004/(1.004^(12*(73-x))-1) 800000*0.004/(1.004^(12*(79-x))-1) Result: Age

Men

Women

20

$274.30

$201.46

40

$829.22

$584.26

60 $3,703.46 $2,155.21

Solutions Section 3.3 63. While in Mexico: = $750,000, = 0.048 12 = 0.004, = 12 × (73 22) = 612 0.004 = $285.47 = 750,000 (1 + ) 1 1.004 612 1 When he moved to Canada eight years later, = 12 × 8 = 96, so the value of the policy was (1 + ) 1 1.004 96 1 = $33,330.15. = 285.47 0.004 To calculate the required payments in Canada, we use an annuity with a present value of $33,330.15 and want the payments necessary to increase this amount to $750,000 in (80 30) = 50 years ( = 12 × 50 = 600), so we use the formula in the "Before we go on" discussion in Example 1: (1 + ) 1 = (1 + ) + and solve for to obtain = ( (1 + ) ) (1 + ) 1 0.004 = �750,000 33,330.15(1.004) 600� 1.004 600 1 $154.19. So, the premiums decreased by 285.47 154.19 = $131.28. 64. While in the U.K.: = $800,000, = 0.048 12 = 0.004, = 12 × (83 25) = 696 0.004 = $212.00 = 800,000 (1 + ) 1 1.004 696 1 When she moved to India ten years later, = 12 × 10 = 120, so the value of the policy was (1 + ) 1 1.004 120 1 = $32,569.98. = 212.00 0.004 To calculate the required payments in India, we use an annuity with a present value of $32,569.98 and want the payments necessary to increase this amount to $800,000 in (72 35) = 37 years ( = 12 × 37 = 444), so we use the formula in the "Before we go on" discussion in Example 1: (1 + ) 1 = (1 + ) + and solve for to obtain = ( (1 + ) ) (1 + ) 1 0.004 = �800,000 32,569.98(1.004) 444� 1.004 444 1 $498.08. So, the premiums increased by 498.08 212.00 = $286.08. 65. We know the payments and need to calculate the present value: = 0.0303 12, = 12 × 30 = 360, = 600 = 1 (1 + ) 0.0303 12 600 = 1 (1 + 0.0303 12) 360 1 (1 + 0.0303 12) 360 so = 600 $141,768.99. 0.0303 12 This is the amount you can afford to finance. Including the down payment gives a total of $141,768.99 + 20,000 = $161,768.99. 66. We know the payments and need to calculate the present value:

Solutions Section 3.3 = 0.0230 12, = 12 × 15 = 180, = 900 = 1 (1 + ) 0.0230 12 900 = 1 (1 + 0.0230 12) 180 1 (1 + 0.0230 12) 180 so = 900 $136,899.74. 0.0230 12 This is the amount you can afford to finance. Including the down payment gives a total of $136,899.74 + 50,000 = $186,899.74

67. Your hunch: Wait a month for the price to go down to $140,000: = $140,000, = 0.0310 12, = 12 × 30 = 360 0.0310 12 = $597.82 = 140,000 1 (1 + ) 1 (1 + 0.0310 12) 360 Technology: 140000*(0.0310/12)/(1-(1+0.0310/12)^(-360)) Broker's suggestion: Buy now for $150,000: = $150,000, = 0.0303 12, = 12 × 30 = 360 0.0303 12 = $634.84 = 150,000 1 (1 + ) 1 (1 + 0.0303 12) 360 Technology: 150000*(0.0303/12)/(1-(1+0.0303/12)^(-360)) Conclusion: Wait a month and pay $634.84 597.82 = $37.02 less per month. 68. Your friends' hunch: Wait a month for the price to go down to $297,000: = $297,000, = 0.0250 12, = 12 × 15 = 180 0.0250 12 = $1,980.36 = 297,000 1 (1 + ) 1 (1 + 0.0250 12) 180 Technology: 297000*(0.0250/12)/(1-(1+0.0250/12)^(-180)) Broker's suggestion: Buy now for $300,000: = $300,000, = 0.0230 12, = 12 × 15 = 180 0.0230 12 = $1,972.25 = 300,000 1 (1 + ) 1 (1 + 0.0230 12) 180 Technology: 300000*(0.0230/12)/(1-(1+0.0230/12)^(-180)) Conclusion: The agent was right; buying now would have saved your friends $1,980.36 1,972.25 = $8.11 per month.

69. We first calculate the payments: = $150,000, = 0.0303 12, = 12 × 30 = 360 0.0303 12 = $634.84 = 150,000 1 (1 + ) 1 (1 + 0.0303 12) 360 Technology: 150000*(0.0303/12)/(1-(1+0.0303/12)^(-360)) Now calculate the outstanding principal: = 12 × 15 = 180, = 360 180 = 180, = 634.84 1 (1 + ) ( ) Outstanding principal = 1 (1 + 0.0303 12) 180 $91,736.52 = 634.84 0.0303 12 Technology: 634.84*(1-(1+0.0303/12)^-180)/(0.0303/12) 70. We first calculate the payments: = $300,000, = 0.0230 12, = 12 × 15 = 180 0.0230 12 = $1,972.25 = 300,000 1 (1 + ) 1 (1 + 0.0230 12) 180 Technology: 300000*(0.0230/12)/(1-(1+0.0230/12)^(-180))

Solutions Section 3.3 Now calculate the outstanding principal: = 12 × 8 = 96, = 180 96 = 84, = 1,972.25 1 (1 + ) ( ) Outstanding principal = 1 (1 + 0.0230 12) 84 $152,885.46 = 1,972.25 0.0230 12 Technology: 1972.25*(1-(1+0.0230/12)^-84)/(0.0230/12) 71. We first calculate the payments on the original mortgage: = $250,000, = 0.0303 12, = 12 × 30 = 360 0.0303 12 = $1,058.06 = 250,000 1 (1 + ) 1 (1 + 0.0303 12) 360 Technology: 250000*(0.0303/12)/(1-(1+0.0303/12)^(-360)) Now calculate the outstanding principal: = 12 × 10 = 120, = 360 120 = 240, = 1,058.06 1 (1 + ) ( ) Outstanding principal = 240 1 (1 + 0.0303 12) $190,264.14 = 1,058.06 0.0303 12 Technology: 1058.06*(1-(1+0.0303/12)^-240)/(0.0303/12) Now calculate the mortgage payments at the new rate on the the outstanding principal plus prepayment fee: = $190,264.14 × 1.04 $197,874.71, = 0.0200 12, = 12 × 20 = 240 0.0200 12 = $1,001.02 = 197,874.71 1 (1 + ) 1 (1 + 0.0200 12) 240 Technology: 197874.71*(0.0200/24)/(1-(1+0.0200/24)^(-240)) Saving = $1,058.06 1,001.02 = $57.04 72. We first calculate the payments on the original mortgage: = $500,000, = 0.0230 12, = 12 × 15 = 180 0.0230 12 = $3,287.08 = 500,000 1 (1 + ) 1 (1 + 0.0230 12) 180 Technology: 500000*(0.0230/12)/(1-(1+0.0230/12)^(-180)) Now calculate the outstanding principal: = 12 × 5 = 60, = 180 60 = 120, = 3,287.08 1 (1 + ) ( ) Outstanding principal = 1 (1 + 0.0230 12) 120 $352,074.52 = 3,287.08 0.0230 12 Technology: 3287.08*(1-(1+0.0230/12)^-120)/(0.0230/12) Now calculate the mortgage payments at the new rate on the the outstanding principal plus prepayment fee: = $352,074.52 × 1.03 $362,636.76, = 0.0150 12, = 12 × 10 = 120 0.0150 12 = $3,256.17 = 362,636.76 1 (1 + ) 1 (1 + 0.0150 12) 120 Technology: 362636.76*(0.0150/12)/(1-(1+0.0150/12)^(-120)) Saving = $3,287.08 3,256.17 = $30.91 73. We first calculate the payments: = $40,000, = 0.0920 12, = 12 × 5 = 60 0.0920 12 = = 40,000 1 (1 + ) 1 (1 + 0.0920 12) 60

$834.22

Solutions Section 3.3 Technology: 40000*(0.0920/12)/(1-(1+0.0920/12)^(-60)) Now calculate the amounts you could have financed at the other rates: At 9.00%: = 0.0900 12, = 60, = 834.22 = 1 (1 + ) 0.0900 12 834.22 = 1 (1 + 0.0900 12) 60 1 (1 + 0.0900 12) 60 so = 834.22 $40,187.19. 0.0900 12 At 9.50%: = 0.0950 12, = 60, = 834.22 = 1 (1 + ) 0.0950 12 834.22 = 1 (1 + 0.0950 12) 60 1 (1 + 0.0950 12) 60 so = 834.22 $39,721.24. 0.0950 12 74. We first calculate the payments: = $20,000, = 0.0945 12, = 12 × 3 = 36 0.0945 12 = $640.19 = 20,000 1 (1 + ) 1 (1 + 0.0945 12) 36 Technology: 20000*(0.0945/12)/(1-(1+0.0945/12)^(-36)) Now calculate the amounts you could have financed at the other rates: At 9.20%: = 0.0920 12, = 36, = 640.19 = 1 (1 + ) 0.0920 12 640.19 = 1 (1 + 0.0920 12) 36 1 (1 + 0.0920 12) 36 so = 640.19 $20,073.12. 0.0920 12 At 9.75%: = 0.0975 12, = 36, = 640.19 = 1 (1 + ) 0.0975 12 640.19 = 1 (1 + 0.0975 12) 36 1 (1 + 0.0975 12) 36 so = 640.19 $19,912.63. 0.0975 12 75. Treat the card as a loan: = 5,000, = 0.1399 12, = 12 × 10 = 120 = 1 (1 + ) 0.1399 12 $77.60. = 5,000 1 (1 + 0.1399 12) 120 Technology: 5000*(0.1399/12)/(1-(1+0.1399/12)^(-120)) 76. Treat the card as a loan: = 8,000, = 0.1399 12, = 12 = 1 (1 + ) 0.1399 12 = 8,000 1 (1 + 0.1399 12) 12

$718.26.

Solutions Section 3.3 Technology: 8000*(0.1399/12)/(1-(1+0.1399/12)^(-12))

77. Solid Savings & Loan: = $10,000, = 0.09 12 = 0.0075, = 12 × 4 = 48. 10,000 × 0.0075 = $248.85 = 1 (1 + ) 1 1.0075 48 Technology: 10000*0.0075/(1-1.0075^(-48)) Fifth Federal Bank & Trust: = $10,000, = 0.07 12, = 12 × 3 = 36. 10,000 × 0.07 12 = $308.77 = 1 (1 + ) 1 (1 + 0.07 12) 36 Technology: 10000*(0.07/12)/(1-(1+0.07/12)^(-36)) Answer: You should take the loan from Solid Savings & Loan: It will have payments of $248.85 per month. The payments on the other loan would be more than $300 per month.

78. 10% Loan: = $20,000, = 0.10 12, = 12 × 5 = 60 20,000 × 0.10 12 = $424.94 = 1 (1 + ) 1 (1 + 0.10 12) 60 Technology: 20000*(0.10/12)/(1-(1+0.10/12)^(-60)) 9% Loan: = $20,000, = 0.09 12 = 0.0075, = 12 × 4 = 48 20,000 × 0.0075 = $497.70 = 1 (1 + ) 1 1.0075 48 Technology: 20000*0.0075/(1-1.0075^(-48)) The first loan will have lower monthly payments. Total Interest Payments: 10% Loan: You pay a total of 60 × $424.94 = $25,496.40. Of this, $25,496.40 20,000 = $5,496.40 is interest. 9% Loan: You pay a total of 48 × $497.70 = $23,889.60. Of this, $23,889.60 20,000 = $3,889.60 is interest. Thus, the first loan will have a larger total interest payment. 79. We first calculate the original payments: = $96,000, = 0.0975 12, = 12 × 30 = 360 0.0975 12 = $824.79 = 96,000 1 (1 + ) 1 (1 + 0.0975 12) 360 Technology: 96000*(0.0975/12)/(1-(1+0.0975/12)^(-360)) Now calculate the outstanding principal: = 12 × 4 = 48, = 360 48 = 312, = 824.79 1 (1 + ) ( ) Outstanding principal = 1 (1 + 0.0975 12) 312 $93,383.71 = 824.79 0.0975 12 Technology: 824.79*(1-(1+0.0975/12)^(-312))/(0.0975/12) Total Interest Paid During the First Loat = Sum of payments − Reduction in principal = 48 × 824.79 (96,000 93,383.71) = $36,973.63 Now calculate the mortgage payments at the new rate on the outstanding principal: = $93,383.71, = 0.06875 12, = 12 × 30 = 360 0.06875 12 = $613.46 = 93,383.71 1 (1 + ) 1 (1 + 0.06875 12) 360 Technology: 93383.71*(0.06875/12)/(1-(1+0.06875/12)^(-360)) Total Interest Paid During the New Loat = Sum of payments − Reduction in principal = 360 × 613.46 93,383.71 = $127,461.89 Thus, the total interest paid over the duration of the two loans is $36,973.63 + 127,461.89 = $164,435.52. Had the mortgage not been refinanced, the total interest would have been

Solutions Section 3.3 Total Interest Paid = Sum of payments − Reduction in principal = 360 × 824.79 96,000 = $20,0924.40.Thus, the saving on interest is $200,924.40 164, 435.52 = $36, 488.88 80. We first calculate the original payments: = $120,000, = 0.10 12, = 12 × 30 = 360 0.10 12 = $1,053.09 = 120,000 1 (1 + ) 1 (1 + 0.10 12) 360 Technology: 120000*(0.10/12)/(1-(1+0.10/12)^(-360)) Now calculate the outstanding principal: = 12 × 3 = 36, = 360 36 = 324, = 1,053.09 1 (1 + ) ( ) Outstanding principal = 1 (1 + 0.10 12) 324 $117,782.44 = 1,053.09 0.10 12 Technology: 1053.09*(1-(1+0.10/12)^(-324))/(0.10/12) Total Interest Paid = Sum of payments − Reduction in principal = 36 × 1,053.09 (120,000 117,782.44) = $35,693.68 Now calculate the mortgage payments at the new rate on the outstanding principal: = $117,782.44, = 0.065 12, = 12 × 15 = 180 0.065 12 = $1,026.01 = 117,782.44 1 (1 + ) 1 (1 + 0.065 12) 180 Technology: 117,782.44*(0.065/12)/(1-(1+0.065/12)^(-180)) Total Interest Paid During the New Loat = Sum of payments − Reduction in principal = 180 × 1,026.01 117,782.44 = $66,899.36 Thus, the total interest paid over the duration of the loan is $35,693.68 + 66,899.36 = $102,593.04. Had the mortgage not been refinanced, the total interest would have been Total Interest Paid = Sum of payments − Reduction in principal = 360 × 1,053.09 120,000 = $259,112.40.Thus, the saving on interest is $259,112.40 102,593.04 = $156,519.36

Solutions Section 3.3 81. We can construct an amortization table using the technique outlined in the Technology Guides. For example, using Excel, we might set it up as follows:

Adding the payments on principal (Column C) and interest payments (Column D) for each year will give the following table: Year

Interest

Payment on Principal

1

$3,934.98

$1,798.98

2

$3,785.69

$1,948.27

3

$3,623.97

$2,109.99

4

$3,448.84

$2,285.12

5

$3,259.19

$2,474.77

6

$3,053.77

$2,680.19

7

$2,831.32

$2,902.64

8

$2,590.39

$3,143.57

9

$2,329.48

$3,404.48

10

$2,046.91

$3,687.05

11

$1,740.88

$3,993.08

12

$1,409.47

$4,324.49

13

$1,050.54

$4,683.42

14

$661.81

$5,072.15

15

$240.84

$5,491.80

Solutions Section 3.3 82. We can create an amortization table as in Exercise 81. We will get the following: Year

Interest

Payment on Principal

1

$9,238.08

$556.32

2

$9,181.34

$613.06

3

$9,118.83

$675.57

4

$9,049.93

$744.47

5

$8,974.02

$820.38

6

$8,890.36

$904.04

7

$8,798.17

$996.23

8

$8,696.56

$1,097.84

9

$8,584.62

$1,209.78

10

$8,461.26

$1,333.14

11

$8,325.30

$1,469.10

12

$8,175.50

$1,618.90

13

$8,010.39

$1,784.01

14

$7,828.46

$1,965.94

15

$7,627.96

$2,166.44

Adding up the payments on principal gives a total of $17,955.22 paid on principal over the first 15 years, so $95,000 17,955.22 = $77, 044.78 is still owed on the mortgage. 83. The payments on the loan, ignoring the fee, are 0.09 12 = 5,000 = $228.42. 1 (1 + 0.09 12) 24 Add to each payment 100/24 = $4.17 to get a new payment of $232.59. Now use technology to find the interest rate being charged on a 2-year $5,000 loan with this payment. For example, we can use the Online Time Value of Money Utility:

We see that the interest rate is 10.81%. 84. The payments on the loan, ignoring the fee, are 0.08 12 = 7,000 = $219.35. 1 (1 + 0.08 12) 36 Add to each payment 100/36 = $2.78 to get a new payment of $222.13. Now use technology to find the

Solutions Section 3.3 interest rate being charged on a 3-year $7,000 loan with this payment. For example, we can use the Online Time Value of Money Utility:

We see that the interest rate is 8.86%. 85. TI-83/84 Plus:

This gives 153.5 12 13 years to retirement. Online Time Value of Money Utility:

86. TI-83/84 Plus:

This gives 63.72 4 16 years to retirement. Online Time Value of Money Utility:

Solutions Section 3.3 87. TI-83/84 Plus:

This gives 55.798 12 4.5 years to repay the debt. Online Time Value of Money Utility:

88. Since we are not told what "eventually" means, let us try a 50-year schedule. We get TI-83/84 Plus:

In other words, if we pay $25.01 it will take us 50 years to pay off the debt. Larger values of give amounts fractionally larger than $25.00. To be assured of eventually paying off the debt, you should therefore pay $25.01. Online Time Value of Money Utility:

Another way of seeing this: The monthly interest on $2,000 at 15% is $2000 × 0.15 12 = $25.00. So, if you pay only $25.00, all you are paying is interest and you are not reducing the $2,000 principal. 89. Graph the future value of both accounts using the formula for the future value of a sinking fund. Your account: = 0.045 12 = 0.00375, = $100. (1 + ) 1 1.00375 1 = = 100 0.00375 To graph this, use the technology formula 100*(1.00375^x-1)/0.00375. Lucinda's account: = 0.065 12, = $75.

Solutions Section 3.3 (1 + 0.065 12) 1 (1 + ) 1 = = 75 (0.065 12) To graph this, use the technology formula 75*((1+0.065/12)^x-1)/(0.065/12) Graph:

The graphs appear to cross around = 300, so we zoom in there:

The graphs cross around = 287, so the number of years is approximately = 287.5 12

24 years.

90. If you purchased the car with an 8% per year loan, the length of the loan would be given as follows: TI-83/84 Plus:

To the nearest year, this is 61 5 so the answer is 5 years or less.

5 years. Leasing the car for less time would result in lower payments,

91. He is wrong because his estimate ignores the interest that will be earned by your annuity—both while it is increasing and while it is decreasing. Your payments will be considerably smaller (depending on the interest earned). 92. She is incorrect. For example, compare Option 1, making a single payment of $10,000 and letting it earn interest for the next 10 years, with Option 2, making yearly payments of $1,000 for 10 years. All $10,000 earns interest for all 10 years in Option 1, while most of your money earns interest for a shorter period of time (and hence earns less interest) in Option 2. 93. Wrong; the split investment earns more. For instance, after ten years it earns $31,056.46 + $32,775.87 = $63,832.33, which is more than the $63,803.03 earned by the single investment. (Mathematically, the future value does not depend linearly on the interest rate.) 94. Wrong; the combined investments earn net interest. For instance, after 10 years the value is: $62,112.91 + $37,833.85 = $99,946.76, which is more than the accumulated payments: $96,000.

Solutions Section 3.3 95. He is not correct. For instance, the payments on a $100,000 10-year mortgage at 12% are $1,434.71, while for a 20-year mortgage at the same rate, they are $1,101.09, which is a lot more than half the 10year mortgage payment. 96. He is correct. For instance, the payments on a $100,000 10-year mortgage at 12% are $1,434.71, while for a $200,000 10-year mortgage at the same rate, they are twice that amount: $2,869.42.

97. = maturity value = $1,000, = 1,000(0.035 2) = $17.50, = selling price = $994.69, = 2 × 5 = 10. Using technology, we compute the interest rate to be approximately 3.617%. (The online utility rounds this to two decimal places.) TI-83/84 Plus:

Online Time Value of Money Utility ("Years" mode):

98. = $1,000, = 1,000(0.03375 2) = $16.875, = $991.20, = 2 × 10 = 20. Using technology, we compute the interest rate to be approximately 3.480%. (The online utility rounds this to two decimal places.) TI-83/84 Plus:

Online Time Value of Money Utility ("Years" mode):

99. = (1 + ) =

(1 + )

1

(1 + ) =

1

(1 + )

100. There are several possible explanations, including the following: Consider two funds with identical interest rates and compounding: The accumulating fund (beginning with no money) and the payment fund, beginning with the present value necessary to fund the payments. Move each payment from the payment fund to the accumulating fund. Since the funds pay the same interest, the net effect is a single account paying compound interest, with no payments or withdrawals. Therefore, the present value of the payment fund must give the future value of the accumulating fund using the future value formula for compound interest.

Solutions Chapter 3 Review Chapter 3 Review

1. = 6,000(1 + 0.0475 × 5) = $7,425.00

2. = 10,000(1 + 0.0525 × 2.5) = $11,312.50

3. = 6,000(1 + 0.0475 12) 60 = $7,604.88

4. = 10,000(1 + 0.0525 2) 5 = $11,383.24

5. = 100

(1 + 0.0475 12) 60 0.0475 12

1

= $6,757.41

6. = 2,000

(1 + 0.0525 2) 5 0.0525 2

1

= $10,538.96

7. = 6,000 (1 + 0.0475 × 5) = $4,848.48

8. = 10,000 (1 + 0.0525 × 2.5) = $8,839.78

9. = 6,000(1 + 0.0475 12) 60 = $4,733.80

10. = 10,000(1 + 0.0525 2) 5 = $8,784.85

11.

= 100

1

(1 + 0.0475 12) 0.0475 12

13.

= 12,000

15.

= 6,000

17.

0.0475 12

(1 + 0.0475 12) 60

1

= 10,000

60

1

0.0475 12

= $5,331.37

1

(1 + 0.0475 12) 60 0.0475 12

= $177.58

= $112.54

(1 + 0.0475 12) 60

= $187.57

12. = 2,000

1

14. = 20,000

16. = 10,000

(1 + 0.0525 2) 5 = $9,258.32 0.0525 2 0.0525 2

(1 + 0.0525 2) 5

18. = 15,000

19. 10,000 = 6,000(1 + 0.0475 ) = 6,000 + 285 . = (10,000

20. 15,000 = 10,000(1 + 0.0525 ) = 10,000 + 525 . = (15,000

0.0525 2

1

1

= $3,795.44

(1 + 0.0525 2) 5

1

0.0525 2

(1 + 0.0525 2) 5

6,000) 285 = 14.0 years.

10,000) 525 = 9.5 years

21. 10,000 = 6,000(1 + 0.0475 12) 12 . To solve algebraically requires logarithms: log(10,000 6,000) = 10.8 years 12 log(1 + 0.0475 12) We could also find this using, for example, the TI-83/84 Plus TVM Solver. 22. 15,000 = 10,000(1 + 0.0525 2) 2 . To solve algebraically requires logarithms: log(15,000 10,000) = 7.8 years 2 log(1 + 0.0525 2) We could also find this using, for example, the TI-83/84 Plus TVM Solver.

= $2,160.22

= $3,240.33

Solutions Chapter 3 Review (1 + 0.0475 12) 12 1 23. 10,000 = 100 . To solve algebraically requires logarithms: 0.0475 12 log[0.0475 × 10,000 (100 × 12) + 1] = 7.0 years 12 log(1 + 0.0475 12) We could also find this using, for example, the TI-83/84 Plus TVM Solver. (1 + 0.0525 2) 2 1 . To solve algebraically requires logarithms: 0.0525 2 log[0.0525 × 15,000 (2,000 × 2) + 1] = 3.5 years 12 log(1 + 0.0525 2) We could also find this using, for example, the TI-83/84 Plus TVM Solver. 24. 15,000 = 2,000

25. Each interest payment is 10,000 × 0.06 2 = $300. For an annuity earning 7% and paying $300 every 6 months for 5 years, the present value is 1 (1 + 0.07 2) 10 = 300 = $2,494.98. 0.07 2 The present value of the $10,000 maturity value is = 10,000(1 + 0.07 2) 10 = $7,089.19. The total price is $2,494.98 + 7,089.19 = $9,584.17. 26. Each interest payment is 10,000 × 0.06 2 = $300. For an annuity earning 5% and paying $300 every 6 months for 5 years, the present value is 1 (1 + 0.05 2) 10 = 300 = $2,625.62. 0.05 2 The present value of the $10,000 maturity value is = 10,000(1 + 0.05 2) 10 = $7,811.98. The total price is $2,625.62 + 7,811.98 = $10,437.60. 27. 5.346% (using, for example, the TI-83/84 Plus TVM Solver) 28. 4.662% (using, for example, the TI-83/84 Plus TVM Solver) 29. = 3.28, = 45.74, = 9 12 years 45.74 = 3.28(1 + 9 12) 1 + 9 12 = 45.74 3.28 = (45.74 3.28 1) (9 12) 17.2602 = 1,726.02%

30. = 33.95, = 12.36, = 22 12 years 12.36 = 33.95(1 + 22 12) 1 + 22 12 = 12.36 33.95 = (12.36 33.95 1) (22 12) 0.3469 =

34.69%

31. The only dates on which she would have gotten an increase rather than a decrease were November 2019, February 2020, and August 2020. Calculating the annual returns as in the preceding exercises, we get the following figures: Jan. 2017–Nov. 2019: 60.45% Jan. 2017–Feb. 2020: 85.28% Jan. 2017–Aug. 2020: 75.37% The largest annual return was 85.28% if she sold in February 2020. 32. The only dates on which he would have gotten a loss rather than an increase were December 2015, January 2017, March 2018, and October 2018. Calculating the annual returns as in the preceding exercises, we get the following

Solutions Chapter 3 Review figures: Aug. 2014–Dec. 2015: 40.79% Aug. 2014–Jan. 2017: 10.02% Aug. 2014–Mar. 2018: 15.81% Aug. 2014–Oct. 2018: 7.17% The largest annual loss was 40.79% if he sold in December 2015. 33. No. Simple interest increase is linear. We can compare slopes between successive points to see whether the slope remained roughly constant: From December 2012 to August 2014 the slope was (16.31 3.28) (20 12) = 7.818, while from August 2014 to March 2015 the slope was (33.95 16.31) (7 12) = 30.24. These slopes are quite different. 34. First calculate the annual interest rate from February 2020 through August 2020: = (45.74 44.86 1) (6 12) 0.03923 Now use the simple interest formula to get the price in December 2021: 44.86(1 + 0.03923 × 22 12) 48.09

35. Use the compound interest formula: = (1 + ) , where = $150,000, = 0.20, = 1, = 1, 2, 3, 4, 5. 2010: = 150,000(1.20) = $180,000 2011: = 150,000(1.20) 2 = $216,000 2012: = 150,000(1.20) 3 = $259,200 2013: = 150,000(1.20) 4 = $311,040 2014: = 150,000(1.20) 5 = $373,248 Revenues first surpass $300,000 in 2013. 36. = $20,000, = 0.0375, = 1, = 7 (quarters) = (1 + ) = 20,000(1 0.0375) 7 $15,305.06

37. After the first day of trading, the value of each share will be $6. Thereafter, the shares appreciate in value by 8% per month for 6 months, and O'Hagan desires a future value of at least $500,000. = 500,000, = 0.08, = 1, = 6 = (1 + ) = 500,000 × 1.08 6 $315,084.81 Therefore, since each share will be worth $6 after the first day, the number of shares they must sell is at least 315,084.81 52,514.14. 6 Since they can offer only a whole number of shares, we must round this up to get the minimum desired future value. Thus, they should offer at least 52,515 shares. 38. After the first day of trading, the value of each share will be $3(1 in value by 10% per week for 5 weeks, and so the future value is = (1 + ) = 1.20(1 0.10) 5 $0.71 per share, so the approximate market value is 600,000 × 0.71 = $426,000.

39. = 250,000, = 0.095 12, = 12 × 10 = 120 250,000 × 0.095 12 = = 1 (1 + ) 1 (1 + 0.095 12) 120 40. = 250,000, = 0.065 12, = 12 × 8 = 96 250,000 × 0.065 12 = = 1 (1 + ) 1 (1 + 0.065 12) 96

$3,234.94 $3,346.56

0.6) = $1.20. Thereafter, the shares depreciate

Solutions Chapter 3 Review 41. = 0.095 12, = 3,000, = 12 × 10 = 120 1 (1 + 0.095 12) 120 1 (1 + ) = $231,844 (to the nearest dollar) = 3,000 (0.095 12) 42. = 0.065 12, = 3,000, = 12 × 8 = 96 1 (1 + 0.065 12) 96 1 (1 + ) = = 3,000 (0.065 12)

$224,111 (to the nearest dollar)

43. = 250,000, = 0, = 3,000, = 12 × 10 = 120, and we are seeking the interest rate. Online Time Value of Money Utility:

The interest rate would be 7.75%.

44. = 250,000, = 0, = 3,000, = 12 × 8 = 96, and we are seeking the interest rate. Online Time Value of Money Utility:

The interest rate would be 3.59%.

45. = 50,000, = 1,000 + 800 (company contribution) = 1,800, = 0.073 12, = 1, = 12 × 10 = 120. Considering the contribution of the present $50,000 as well as the payments, we get (1 + 0.073 12) 120 1 (1 + ) 1 = (1 + ) + $420,275 = 50,000(1 + 0.073 12) 120 + 1,800 0.073 12 (to the nearest dollar). Technology: 50000*(1+0.073/12)^120+1800*((1+0.073/12)^120-1)/(0.073/12) 46. = 60,000, = 950 + 800(company contribution) = 1,750, = 0.073 12, = 1, = 12 × 8 = 96. Considering the contribution of the present $60,000 as well as the payments, we get (1 + 0.073 12) 96 1 (1 + ) 1 = (1 + ) + $334,670 = 60,000(1 + 0.073 12) 96 + 1,750 0.073 12 (to the nearest dollar). Technology: 60000*(1+0.073/12)^96+1750*((1+0.073/12)^96-1)/(0.073/12)

Solutions Chapter 3 Review 47. For the company's contribution, take = 800, = 0.073 12, = 12 × 10 = 120. (1 + 0.073 12) 120 1 (1 + ) 1 = $140,778 = 800 0.073 12 (to the nearest dollar). Technology: 800*((1+0.073/12)^120-1)/(0.073/12)

48. For the company's contribution, take = 800, = 0.073 12, = 12 × (8 + 5) = 156. (1 + 0.073 12) 156 1 (1 + ) 1 = $207,217 = 800 0.073 12 (to the nearest dollar). Technology: 800*((1+0.073/12)^156-1)/(0.073/12)

49. We first take out the effect of the initial $50,000: = 50,000, = 0.073 12, = 1, = 12 × 10 = 120. = (1 + ) = 50,000(1 + 0.073 12) 120 Thus, the payments have to result in a future value of = 500,000 50,000(1 + 0.073 12) 120 396,475.163 We can now use the payment formula to determine the necessary payments: 396,475.163 × 0.073 12 = $2,253.06 (1 + ) 1 (1 + 0.073 12) 120 1 Since $800 of this is contributed by the company, Callahan's payments should be $2,253.06 800 = $1,453.06.

50. We first take out the effect of the initial $60,000: = 50,000, = 0.073 12, = 1, = 12 × 8 = 96. = (1 + ) = 60,000(1 + 0.073 12) 96 Thus, the payments have to result in a future value of = 600,000 60,000(1 + 0.073 12) 96 $492,598.3656 We can now use the payment formula to determine the necessary payments: 492,598.3656 × 0.073 12 = $3,793.08 (1 + ) 1 (1 + 0.073 12) 96 1 Since $800 of this is contributed by the company, Egan's payments should be $3,793.08 800 = $2,993.08.

51. First compute the amount Callahan needs at the start of retirement: = 5,000, = 0.087 12, = 12 × 30 = 360 1 (1 + 0.087 12) 360 1 (1 + ) = $638,461.93 = 5,000 0.087 12 Technology: 5000*(1-(1+0.087/12)^(-360))/(0.087/12) In order to accumulate this amount, using the information from Exercise 45, we first discount the effect of the current $50,000 in the account: = 50,000, = 638,461.93, = 0.073 12, = 12 × 10 = 120. The initial $50,000 will grow to 50,000(1 + 0.073 12) 120 so the payments need to result in a future value of only 638,461.93 50,000(1 + 0.073 12) 120 534,937.093. Now use the payment formula: 534,937.093 × 0.073 12 = $3,039.90 = (1 + ) 1 (1 + 0.073 12) 120 1 Since $800 of this is contributed by the company, Callahan's payments should be $3,039.90 800 = $2,239.90. 52. First compute the amount Egan needs at the start of retirement: = 6,000, = 0.078 12, = 12 × 25 = 300

Solutions Chapter 3 Review 1 (1 + 0.078 12) 300 (1 + ) = $790,915.68 = 6,000 0.078 12 Technology: 6000*(1-(1+0.078/12)^(-300))/(0.078/12) In order to accumulate this amount, we first discount the effect of the current $60,000 in the account: = 60,000, = 0.073 12, = 12 × 8 = 96. = (1 + ) = 60,000(1 + 0.073 12) 96 Thus, the payments have to result in a future value of = 790,915.68 60,000(1 + 0.073 12) 96 $683,514.04. Now use the payment formula: 683,514.04 × 0.073 12 = $5,263.17 (1 + ) 1 (1 + 0.073 12) 96 1 Since $800 of this is contributed by the company, Egan's payments should be $5,263.17 800 = $4,463.17. 1

53. The bond will pay interest every 6 months amounting to 50,000(0.072 2) = $1,800. For someone purchasing this bond after one year, there will be 9 years to maturity. Think of the bond as an investment that will pay the owner $1,800 every 6 months for 9 years, at which time it will pay $50,000. This is exactly the behavior of an annuity paired with an investment with future value $50,000 = 50,000, = 1,800, = 0.063 2 = 0.0315, = 1, = 2 × 9 = 18 The present value has contributions both from the investment and the annuity: 1 (1 + ) 1 1.0315 180 = (1 + ) + $53,055.66 = 50,000(1.0315) 18 + 1,800 0.0315 Technology: 50000*1.0315^(-18)+1800*(1-1.0315^(-18))/0.0315 54. This is similar to Exercise 53, but with = 50,000, = 1,800, = 0.06 2 = 0.03, = 2 × 8.5 = 17 1 (1 + ) 1 1.03 170 = (1 + ) + $53,949.84 = 50,000(1.03) 17 + 1,800 0.03 Technology: 50000*1.03^(-17)+1800*(1-1.03^(-17))/0.03

55. Here, = 50,000, = 54,000, = 1,800, = 2 × 8.5 = 17, and we are seeking the interest rate. Online Time Value of Money Utility:

The interest rate would have to be 5.99%.

56. Here, = 50,000, = 52,000, = 1,800, = 2 × 8.5 = 17, and we are seeking the interest rate.

Solutions Chapter 3 Review Online Time Value of Money Utility:

The interest rate would have to be 6.58%.

Solutions Chapter 3 Case Study Chapter 3 Case Study 1. The start of the sixth year is the critical time; if the Wongs can afford payments then, they will continue to be able to afford them throughout the loan as one can verify by glancing at the worksheet. To solve with Excel, use one of the three TVM utilities discussed in the book or, in Excel, use =RATE(12*25,-2271.09,343700.55)*12 This will return approximately 0.06267892, or 6.27% per year. Since this is 5% above the Fed discount rate, the latter would have to be 1.27%. 2. This is similar to Exercise 1, and we use =RATE(12*25,-2271.09,380000)*12 which returns approximately 0.0522, or 5.22%, meaning that the Fed rate would have to be 0.22%. 3. This is similar to Exercise 1, and we use =RATE(12*25,-2271.09,400640.49)*12 which returns approximately 0.0469, or 4.69%, meaning that the Fed rate would have to be negative, which is impossible; the Wongs could never make the payments regardless of the Fed rate in this case. 4. Adjusting the price in cell K3 gives affordable payments in the most expensive scenario for a $195,000 mortgage. Adding the $20,000 down payment, the Wongs could afford a $215,000 home. 5. Adjusting the price in cell K3 gives affordable payments in the most expensive scenario for a $170,000 mortgage (to the nearest $5,000). Adding the $20,000 down payment, the Wongs could afford a $190,000 home. 6. Of the three types of mortgage, the hybrid is the most affordable. and the first steep payment (in worstcase scenario 3) is the $4,402.22. 28% of their income first passes that level in year 23 (see the spreadsheet), meaning that they would have to wait until year 23 5 = 18, or 17 more years, before being able to afford such a home.

Solutions Section 4.1 Section 4.1

1. 2𝑥 − 𝑦 = 1 Solving for 𝑦 gives 𝑦 = 2𝑥 − 1, so the general solution is (𝑥, 𝑦) = (𝑥, 2𝑥 − 1); 𝑥 arbitrary. This is the solution parameterized by 𝑥. We get particular solutions by choosing values for 𝑥, such as 𝑥 = −1 ⇒ (𝑥, 𝑦) = (𝑥, 2𝑥 − 1) = (𝑥, 2(−1) − 1) = (−1, −3) 𝑥 = 0 ⇒ (𝑥, 𝑦) = (𝑥, 2𝑥 − 1) = (0, 2(0) − 1) = (0, −1) 𝑥 = 1 ⇒ (𝑥, 𝑦) = (𝑥, 2𝑥 − 1) = (1, 2(1) − 1) = (1, 1) 1 For the solution parameterized by 𝑦, solve the original equation for 𝑥 to get 𝑥 = (𝑦 + 1), so the general 2 solution parameterized by 𝑦 is (𝑥, 𝑦) = ! 12 (𝑦 + 1), 𝑦"; 𝑦 arbitrary

2. 𝑥 + 3𝑦 = 3

1 Solving for 𝑦 gives 𝑦 = − 𝑥 + 1, so the general solution is 3 1 (𝑥, 𝑦) = (𝑥, − 𝑥 + 1); 𝑥 arbitrary. 3 This is the solution parameterized by 𝑥. We get particular solutions by choosing values for 𝑥, such as 1 1 𝑥 = −3 ⇒ (𝑥, 𝑦) = (𝑥, − 𝑥 + 1) = (−3, − (−3) + 1) = (−3, 2) 3 3 1 1 𝑥 = 0 ⇒ (𝑥, 𝑦) = (𝑥, − 𝑥 + 1) = (0, − (0) + 1) = (0, 1) 3 3 1 1 𝑥 = 3 ⇒ (𝑥, 𝑦) = (𝑥, − 𝑥 + 1) = (3, − (3) + 1) = (3, 0) 3 3 For the solution parameterized by 𝑦, solve the original equation for 𝑥 to get 𝑥 = −3𝑦 + 3, so the general solution parameterized by 𝑦 is (𝑥, 𝑦) = (−3𝑦 + 3, 𝑦); 𝑦 arbitrary. 3. 3𝑥 + 4𝑦 = 2

3 1 Solving for 𝑦 gives 𝑦 = − 𝑥 + , so the general solution is 4 2 3 1 (𝑥, 𝑦) = (𝑥, − 𝑥 + ); 𝑥 arbitrary. 4 2 This is the solution parameterized by 𝑥. We get particular solutions by choosing values for 𝑥, such as 3 1 3 1 𝑥 = −2 ⇒ (𝑥, 𝑦) = (𝑥, − 𝑥 + ) = (−2, − (−2) + ) = (−3, 2) 4 2 4 2 3 1 3 1 1 𝑥 = 0 ⇒ (𝑥, 𝑦) = (𝑥, − 𝑥 + ) = (0, − (0) + ) = (0, ) 4 2 4 2 2 3 1 3 1 𝑥 = 2 ⇒ (𝑥, 𝑦) = (𝑥, − 𝑥 + ) = (2, − (2) + ) = (2, −1) 4 2 4 2 1 For the solution parameterized by 𝑦, solve the original equation for 𝑥 to get 𝑥 = (−4𝑦 + 2), so the 3 general solution parameterized by 𝑦 is 1 (𝑥, 𝑦) = ( (−4𝑦 + 2), 𝑦); 𝑦 arbitrary. 3 4. 4𝑥 − 3𝑦 = 6

1 Solving for 𝑦 gives 𝑦 = (4𝑥 − 6), so the general solution is 3 1 (𝑥, 𝑦) = (𝑥, (4𝑥 − 6)); 𝑥 arbitrary. 3

Solutions Section 4.1 This is the solution parameterized by 𝑥. We get particular solutions by choosing values for 𝑥, such as 1 1 𝑥 = −3 ⇒ (𝑥, 𝑦) = (𝑥, (4𝑥 − 6)) = (−3, (4(−3) − 6)) = (−3, −6) 3 3 1 1 𝑥 = 0 ⇒ (𝑥, 𝑦) = (𝑥, (4𝑥 − 6)) = (0, (4(0) − 6)) = (0, −2) 3 3 1 1 𝑥 = 3 ⇒ (𝑥, 𝑦) = (𝑥, (4𝑥 − 6)) = (3, (4(3) − 6)) = (3, 2) 3 3 1 For the solution parameterized by 𝑦, solve the original equation for 𝑥 to get 𝑥 = (3𝑦 + 6), so the general 4 solution parameterized by 𝑦 is 1 (𝑥, 𝑦) = ( (3𝑦 + 6), 𝑦); 𝑦 arbitrary. 4 5. 4𝑥 = −5 We cannot solve for 𝑦, so the general solution cannot be parameterized by 𝑥.

5 For the solution parameterized by 𝑦, solve the original equation for 𝑥 to get 𝑥 = − , so the general 4 solution parameterized by 𝑦 is 5 (𝑥, 𝑦) = (− , 𝑦); 𝑦 arbitrary. 4 We get particular solutions by choosing values for 𝑦 : (−5∕4, −1), (−5∕4, 0), (−5∕4, 1). 6. −3𝑦 = 2

2 For the solution parameterized by 𝑥, solve the original equation for 𝑦 to get 𝑦 = − , so the general 3 solution parameterized by 𝑥 is 2 (𝑥, 𝑦) = (𝑥, − ); 𝑥 arbitrary. 3 We get particular solutions by choosing values for 𝑥 : (−1, −2∕3), (0, −2∕3), (1, −2∕3); We cannot solve for 𝑥, so the general solution cannot be parameterized by 𝑦. 7. 𝑥 − 𝑦 = 0 𝑥+𝑦=4 Adding gives 2𝑥 = 4 ⇒ 𝑥 = 2. Substituting 𝑥 = 2 in the first equation: 2 − 𝑦 = 0 ⇒ 𝑦 = 2. Solution: (2, 2) Graph: 𝑦 = 𝑥; 𝑦 = 4 − 𝑥

8. 𝑥 − 𝑦 = 0 𝑥 + 𝑦 = −6 Adding gives 2𝑥 = −6 ⇒ 𝑥 = −3. Substituting 𝑥 = −3 in the first equation: −3 − 𝑦 = 0 ⇒ 𝑦 = −3. Solution: (−3, −3) Graph: 𝑦 = 𝑥; 𝑦 = −6 − 𝑥

9. 𝑥 + 𝑦 = 4 𝑥−𝑦=2 Adding gives 2𝑥 = 6 ⇒ 𝑥 = 3. Substituting 𝑥 = 3 in the first equation: 3 + 𝑦 = 4 ⇒ 𝑦 = 4 − 3 = 1. Solution: (3, 1) Graph: 𝑦 = 4 − 𝑥; 𝑦 = 𝑥 − 2

10. 2𝑥 + 𝑦 = 2 −2𝑥 + 𝑦 = 2 Adding gives 2𝑦 = 4 ⇒ 𝑦 = 2. Substituting 𝑦 = 2 in the first equation: 2𝑥 + 2 = 2 ⇒ 𝑥 = 0. Solution: (0, 2) Graph: 𝑦 = 2 − 2𝑥; 𝑦 = 2𝑥 + 2

Solutions Section 4.1

11. 3𝑥 − 2𝑦 = 6 2𝑥 − 3𝑦 = −6 Multiply the first equation by 2 and the second by −3 : 6𝑥 − 4𝑦 = 12 −6𝑥 + 9𝑦 = 18. Adding gives 5𝑦 = 30 ⇒ 𝑦 = 6. Substituting 𝑦 = 6 in the first equation: 3𝑥 − 12 = 6 ⇒ 3𝑥 = 18 ⇒ 𝑥 = 6. Solution: (6, 6) 3 2 Graph: 𝑦 = 𝑥 − 3; 𝑦 = 𝑥 + 2 2 3

12. 2𝑥 + 3𝑦 = 5 3𝑥 + 2𝑦 = 5 Multiply the first equation by 2 and the second by −3 : 4𝑥 + 6𝑦 = 10 −9𝑥 − 6𝑦 = −15. Adding gives −5𝑥 = −5 ⇒ 𝑥 = 1. Substituting 𝑥 = 1 in the first equation gives 2 + 3𝑦 = 5 ⇒ 𝑦 = 1. Solution: (1, 1) 2 5 3 5 Graph: 𝑦 = − 𝑥 + ; 𝑦 = − 𝑥 + 3 3 2 2

13. 0.5𝑥 + 0.1𝑦 = 0.7 0.2𝑥 − 0.2𝑦 = 0.6 Multiply both equations by 10: 5𝑥 + 𝑦 = 7 2𝑥 − 2𝑦 = 6. Divide the second equation by 2: 5𝑥 + 𝑦 = 7 𝑥 − 𝑦 = 3. Adding gives 10 5 6𝑥 = 10 ⇒ 𝑥 = = . 6 3 5 Substituting 𝑥 = in 𝑥 − 𝑦 = 3 gives 3 5 5 4 −𝑦=3⇒𝑦= −3=− . 3 3 3 5 4 Solution: ( , − ) 3 3 Graph: 𝑦 = −5𝑥 + 7; 𝑦 = 𝑥 − 3

14. −0.3𝑥 + 0.5𝑦 = 0.1 0.1𝑥 − 0.1𝑦 = 0.4 Multiply both equations by 10 : 𝑥 − 𝑦 = 4. −3𝑥 + 5𝑦 = 1 Multiply the second equation by 3: 3𝑥 − 3𝑦 = 12. −3𝑥 + 5𝑦 = 1 Adding gives 13 2𝑦 = 13 ⇒ 𝑦 = = 6.5. 2 Substituting 𝑦 = 6.5 in 𝑥 − 𝑦 = 4 gives 𝑥 − 6.5 = 4 ⇒ 𝑥 = 10.5. Solution: (10.5, 6.5) Graph: 𝑦 = 0.6𝑥 + 0.2; 𝑦 = 𝑥 − 4

Solutions Section 4.1

𝑥 𝑦 𝑥 − =1 + 𝑦 = −2 3 2 4 Multiply the first equation by 6 and the second by 4: 2𝑥 − 3𝑦 = 6 𝑥 + 4𝑦 = −8. Multiply the second equation by −2 : 2𝑥 − 3𝑦 = 6 −2𝑥 − 8𝑦 = 16. Adding gives 22 −11𝑦 = 22 ⇒ 𝑦 = = −2. −11 Substituting 𝑦 = −2 into 𝑥 + 4𝑦 = −8 gives 𝑥 + 4(−2) = −8 ⇒ 𝑥 − 8 = −8 ⇒ 𝑥 = 0. Solution: (0, −2) Graph: 𝑦 = (2∕3)𝑥 − 2; 𝑦 = −𝑥∕4 − 2 15.

2𝑥 𝑦 1 𝑥 3 + =− −𝑦=− 3 2 6 4 4 Multiply the first equation by 6 and the second by 4: 𝑥 − 4𝑦 = −3. −4𝑥 + 3𝑦 = −1 Multiply the second equation by 4: −4𝑥 + 3𝑦 = −1 ⇒ 4𝑥 − 16𝑦 = −12. Adding gives −13𝑦 = −13 ⇒ 𝑦 = 1. Substituting into 𝑥 − 4𝑦 = −3 gives 𝑥 − 4 = −3 ⇒ 𝑥 = 1. Solution: (1, 1) Graph: 𝑦 = (4∕3)𝑥 − 1∕3; 𝑦 = (1∕4)𝑥 + 3∕4 16. −

3𝑦 1 =− 2 2 Multiply the second equation by 2: 2𝑥 + 3𝑦 = 1 −2𝑥 − 3𝑦 = −1. Adding gives 0 = 0, indicating that the system is redundant: The graphs are the same, so there are infinitely many solutions. We obtain the solutions by solving either equation for 𝑦 (or 𝑥). 2𝑥 + 3𝑦 = 1 ⇒ 3𝑦 = −2𝑥 + 1 ⇒ 𝑦 = (−2𝑥 + 1)∕3 Solution: (𝑥, (−2𝑥 + 1)∕3); 𝑥 arbitrary Graph: 𝑦 = (−2𝑥 + 1)∕3

18. 2𝑥 − 3𝑦 = 1 6𝑥 − 9𝑦 = 3 Divide the second equation by −3 : 2𝑥 − 3𝑦 = 1 −2𝑥 + 3𝑦 = −1. Adding gives 0 = 0, indicating that the system is redundant: There are infinitely many solutions, which we can obtain by solving either equation for 𝑦 (or 𝑥): 2𝑥 − 3𝑦 = 1 ⇒ 3𝑦 = 2𝑥 − 1 ⇒ 𝑦 = (2𝑥 − 1)∕3. Solution: (𝑥, (2𝑥 − 1)∕3); 𝑥 arbitrary Graph: 𝑦 = (2𝑥 − 1)∕3

3𝑦 1 =− 2 2 Multiply the second equation by 2: 2𝑥 + 3𝑦 = 2 −2𝑥 − 3𝑦 = −1. Adding gives 0 = 1, indicating that the system is inconsistent. There is no solution; the lines are parallel. Graph: 𝑦 = (−2𝑥 + 2)∕3; 𝑦 = (−2𝑥 + 1)∕3

20. 2𝑥 − 3𝑦 = 2 6𝑥 − 9𝑦 = 3 Divide the second equation by −3 : 2𝑥 − 3𝑦 = 2 −2𝑥 + 3𝑦 = −1. Adding gives 0 = 1, indicating that the system is inconsistent. There is no solution; the lines are parallel. Graph: 𝑦 = (2𝑥 − 2)∕3; 𝑦 = (2𝑥 − 1)∕3

17. 2𝑥 + 3𝑦 = 1

19. 2𝑥 + 3𝑦 = 2

−𝑥 −

−𝑥 −

Solutions Section 4.1

Parallel lines; no solution Parallel lines; no solution

21. 2𝑥 + 8𝑦 = 10 𝑥+𝑦=5 To graph these, solve for 𝑦 : 𝑦 = (−2𝑥 + 10)∕8 𝑦 = −𝑥 + 5. Graph, with vertical and horizontal scales of 0.1:

22. 2𝑥 − 𝑦 = 3 𝑥 + 3𝑦 = 5 To graph these, solve for 𝑦 : 𝑦 = 2𝑥 − 3 𝑦 = (−𝑥 + 5)∕3. Graph, with vertical and horizontal scales of 0.1:

The grid point closest to the intersection of the lines gives the approximate solution (in this case the exact solution). Solution: (5, 0)

The grid point closest to the intersection of the lines gives the approximate solution (in this case the exact solution). Solution: (2, 1)

23. 3.1𝑥 − 4.5𝑦 = 6 4.5𝑥 + 1.1𝑦 = 0 To graph these, solve for 𝑦 : 𝑦 = (3.1𝑥 − 6)∕4.5 𝑦 = −4.5𝑥∕1.1. Graph, with vertical and horizontal scales of 0.1:

24. 0.2𝑥 + 4.5𝑦 = 1 1.5𝑥 + 1.1𝑦 = 2 To graph these, solve for 𝑦 : 𝑦 = (−0.2𝑥 + 1)∕4.5 𝑦 = (−1.5𝑥 + 2)∕1.1. Graph, with vertical and horizontal scales of 0.1:

Solutions Section 4.1

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (0.3, −1.1) 25. 10.2𝑥 + 14𝑦 = 213 4.5𝑥 + 1.1𝑦 = 448 To graph these, solve for 𝑦 : 𝑦 = (−10.2𝑥 + 213)∕14 𝑦 = (−4.5𝑥 + 448)∕1.1. Graph, with vertical and horizontal scales of 0.1:

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (116.6, −69.7)

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (1.2, 0.2) 26. 100𝑥 + 4.5𝑦 = 540 1.05𝑥 + 1.1𝑦 = 0 To graph these, solve for 𝑦 : 𝑦 = −100𝑥 + 540)∕4.5 𝑦 = −1.05𝑥∕1.1. Graph, with vertical and horizontal scales of 0.1:

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (5.6, −5.4)

27. Line through (0, 1) and (4.2, 2) : 𝑦 − 𝑦1 2−1 1 Point: (0, 1) Slope: 𝑚 = 2 Intercept: 𝑏 = 1 = = 𝑥2 − 𝑥1 4.2 − 0 4.2 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 + 1. 4.2 Line through (2.1, 3) and (5.2, 0) : 𝑦 − 𝑦1 0−3 3 Point: (5.2, 0) Slope: 𝑚 = 2 Intercept: = =− 𝑥2 − 𝑥1 5.2 − 2.1 3.1 3 15.6 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (− )5.2 = 3.1 3.1 3 15.6 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥+ . 3.1 3.1 Solutions Section 4.1

(Given)

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (3.3, 1.8) 28. Line through (2.1, 3) and (4, 2) : 𝑦 − 𝑦1 2−3 1 Point: (2.1, 3) Slope: 𝑚 = 2 = =− 𝑥2 − 𝑥1 4 − 2.1 1.9 1 7.8 𝑏 = 𝑦1 − 𝑚𝑥1 = 3 − (− )2.1 = 1.9 1.9 1 7.8 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥+ . 1.9 1.9 Line through (3.2, 2) and (5.1, 3) : 𝑦 − 𝑦1 3−2 1 Point: (3.2, 2) Slope: 𝑚 = 2 = = 𝑥2 − 𝑥1 5.1 − 3.2 1.9 1 0.6 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − ( )3.2 = 1.9 1.9 1 0.6 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥+ . 1.9 1.9

Intercept:

Intercept:

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (3.6, 2.2) 29. Line through (0, 0) and (5.5, 3) : 𝑦 − 𝑦1 3−0 3 Point: (0, 0) Slope: 𝑚 = 2 = = 𝑥2 − 𝑥1 5.5 − 0 5.5 3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥. 5.5

Intercept: 0 (Given)

Line through (5, 0) and (0, 6) : 𝑦 − 𝑦1 6−0 6 Point: (0, 6) Slope: 𝑚 = 2 = =− 𝑥2 − 𝑥1 0 − 5 5 6 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥 + 6. 5

Solutions Section 4.1 Intercept: 6 (Given)

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (3.4, 1.9) 30. Line through (4.3, 0) and (0, 5) : 𝑦 − 𝑦1 5−0 5 Point: (0, 5) Slope: 𝑚 = 2 Intercept: 5 (Given) = =− 𝑥2 − 𝑥1 0 − 4.3 4.3 5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥 + 5. 4.3 Line through (2.1, 2.2) and (5.2, 1) : 𝑦 − 𝑦1 1 − 2.2 1.2 Point: (5.2, 1) Slope: 𝑚 = 2 = =− 𝑥2 − 𝑥1 5.2 − 2.1 3.1 1.2 6.24 9.34 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 1 − (− )5.2 = 1 + = 3.1 3.1 3.1 1.2 9.34 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥+ . 3.1 3.1

The grid point closest to the intersection of the lines gives the approximate solution. Approximate solution: (2.6, 2.0)

31. Let 𝑥 = the number of soccer fans, and let 𝑦 = the number of football fans. Reword the given statement using the phrases "the number of soccer fans," which is 𝑥, and "the number of football fans," which is 𝑦 : The number of soccer fans is twice the number of football fans. Translate the phrases into symbols: 𝑥 = 2𝑦. In standard form, this becomes 𝑥 − 2𝑦 = 0. 32. Let 𝑥 = the number of hockey players, and let 𝑦 = the number of lacrosse players. Reword the given statement using the phrases "the number of hockey players," which is 𝑥, and "the number of lacrosse players," which is 𝑦 : The number of hockey players is 90% of the number of lacrosse players. (No rewording necessary in this case.) Translate the phrases into symbols:

90 𝑦 = 0.90𝑦. 100 In standard form, this becomes 𝑥 − 0.90𝑦 = 0. 𝑥=

Solutions Section 4.1

33. Let 𝑥 = the number of new clients, and let 𝑦 = the number of old clients. Reword the given statement using the phrases "the number of new clients," which is 𝑥, and "the number of old clients," which is 𝑦 : The number of new clients is 110% the number of old clients. (No rewording necessary in this case.) Translate the phrases into symbols: 110 𝑥= 𝑦 = 1.10𝑦. 100 In standard form, this becomes 𝑥 − 1.10𝑦 = 0.

34. Let 𝑥 = the number of bondholders, and let 𝑦 = the number of stockholders. Reword the given statement using the phrases "the number of bondholders," which is 𝑥, and "the number of stockholders," which is 𝑦 : The number of bondholders is half the number of stockholders. Translate the phrases into symbols: 1 𝑥 = 𝑦. 2 We can rewrite this equation in standard form in various ways, such as 𝑥 − 12 𝑦 = 0 or 2𝑥 − 𝑦 = 0. 35. Let 𝑥 = the number of gas giants, and let 𝑦 = the number of rocky planets. We are given two pieces of information: 1. There are three times as many gas giants as rocky planets. 2. Their total is 12. Thus, we expect to obtain two equations. Reword the given information using the phrases "the number of gas giants," which is 𝑥, and "the number of rocky planets," which is 𝑦 : 1. The number of gas giants is three times the number of rocky planets. 2. The number of gas giants and the number of rocky planets add up to 12. Translate the phrases into symbols: 1. 𝑥 = 3𝑦, or 𝑥 − 3𝑦 = 0 2. 𝑥 + 𝑦 = 12

36. Let 𝑥 = the number of planets that support life, and let 𝑦 = the number of lifeless planets. We are given two pieces of information: 1. Their total is 25. 2. There are four times as many planets that support life as rocky planets. Thus, we expect to obtain two equations. Reword the given information using the phrases "the number of planets that support life," which is 𝑥, and "the number of lifeless planets," which is 𝑦 : 1. The number of planets that support life and the number of lifeless planets add up to 25 2. The number of planets that support life is four times the number of lifeless planets. Translate the phrases into symbols: 1. 𝑥 + 𝑦 = 25 2. 𝑥 = 4𝑦, or 𝑥 − 4𝑦 = 0 37. Let 𝑥 = the number of ordinary shares, and let 𝑦 = the number of preferred shares. We are given two pieces of information: 1. 20% of the total are preferred shares. 2. There are 15 more ordinary shares than preferred shares. Reword the given information using the phrases "the number of ordinary shares," which is 𝑥, and "the

Solutions Section 4.1 number of preferred shares," which is 𝑦 : 1. The number of preferred shares is 20% of the total number (of ordinary and preferred shares). 2. The number of ordinary shares is 15 more than the number of preferred shares. Translate the phrases into symbols: 20 1 1. 𝑦 = (𝑥 + 𝑦) = (𝑥 + 𝑦) 100 5 Simplify by multiplying both sides by 5: 5𝑦 = 𝑥 + 𝑦 ⇒ 𝑥 = 4𝑦, or 𝑥 − 4𝑦 = 0 2. 𝑥 = 15 + 𝑦 or 𝑥 − 𝑦 = 15

38. Let 𝑥 = the number of paid-up customers, and let 𝑦 = the number customers who owe money. We are given two pieces of information: 1. 75% of the total customer base is paid up. 2. There are 14,000 more paid-up customers than customers who owe money. Reword the given information using the phrases "the number of paid-up customers," which is 𝑥, and "the number of customers who still owe money," which is 𝑦 : 1. The number of paid-up customers is 75% of the total number (of paid-up customers and customers who owe money). 2. The number of paid-up customers is 14,000 more than the number of customers who owe money. Translate the phrases into symbols: 75 3 1. 𝑥 = (𝑥 + 𝑦) = (𝑥 + 𝑦) 100 4 Simplify by multiplying both sides by 4: 4𝑥 = 3(𝑥 + 𝑦) ⇒ 𝑥 = 3𝑦, or 𝑥 − 3𝑦 = 0 2. 𝑥 = 14,000 + 𝑦, or 𝑥 − 𝑦 = 14,000 39. Unknowns: 𝑥 = the number of quarts of Creamy Vanilla; 𝑦 = the number of quarts of Continental Mocha Arrange the given information in a table with unknowns across the top: Vanilla (𝑥) Mocha (𝑦) Available

Eggs

2

1

500

Cream

3

3

900

We can now set up an equation for each of the items listed on the left: Eggs: Cream:

2𝑥 + 𝑦 = 500 3𝑥 + 3𝑦 = 900.

Multiply the first equation by −1 and divide the second by 3: 𝑥 + 𝑦 = 300. −2𝑥 − 𝑦 = −500 Adding gives −𝑥 = −200 ⇒ 𝑥 = 200 quarts of vanilla. Substituting 𝑥 = 200 in the first equation gives 2(200) + 𝑦 = 500 ⇒ 400 + 𝑦 = 500 ⇒ 𝑦 = 500 − 400 = 100 quarts of mocha. Solution: Make 200 quarts of vanilla and 100 quarts of mocha.

40. Unknowns: 𝑥 = the number of sections of Finite Math; 𝑦 = the number of sections of Applied Calculus Since there are a total of 110 sections, 𝑥 + 𝑦 = 110. Information about enrollment gives 60𝑥 + 50𝑦 = 6, 000 . Multiply the first equation by 5 and divide the second equation by −10 : 5𝑥 + 5𝑦 = 550 −6𝑥 − 5𝑦 = −600.

Solutions Section 4.1

Adding gives −𝑥 = −50 ⇒ 𝑥 = 50 sections of Finite Math. Substituting 𝑥 = 50 in the first equation gives 50 + 𝑦 = 110 ⇒ 𝑦 = 60 sections of Applied Calculus. Solution: Offer 50 sections of Finite Math, 60 sections of Applied Calculus.

41. Unknowns: 𝑥 = the number of servings of Multigrain Cereal; 𝑦 = the number of servings of 2nd Foods Mango Arrange the given information in a table with unknowns across the top: Cereal (𝑥) Mango (𝑦) Desired

Calories

60

80

220

Carbs (g)

11

19

49

We can now set up an equation for each of the items listed on the left: 60𝑥 + 80𝑦 = 220 11𝑥 + 19𝑦 = 49.

Calories: Carbs:

Divide the first equation by 20: 3𝑥 + 4𝑦 = 11 11𝑥 + 19𝑦 = 49. Multiply the first equation by −11 and the second by 3: 33𝑥 + 57𝑦 = 147. −33𝑥 − 44𝑦 = −121 Adding gives 13𝑦 = 26 ⇒ 𝑦 = 2 servings of 2nd Foods Mango. Substituting 𝑦 = 2 in the equation 3𝑥 + 4𝑦 = 11 gives 3𝑥 + 4(2) = 11 ⇒ 3𝑥 = 3 ⇒ 𝑥 = 1 serving of Multigrain Cereal. Solution: Use 1 serving of Multigrain Cereal and 2 servings of 2nd Foods Mango

42. Unknowns: 𝑥 = the number of ounces of cereal; 𝑦 = the number of ounces of fruit Arrange the given information in a table with unknowns across the top: Cereal (𝑥) Fruit (𝑦) Desired

Protein

0.5

0.2

1

Iron

1

2

5

We can now set up an equation for each of the items listed on the left: Cereal: Fruit:

0.5𝑥 + 0.2𝑦 = 1 𝑥 + 2𝑦 = 5.

Multiply the first equation by 10: 5𝑥 + 2𝑦 = 10 𝑥 + 2𝑦 = 5. Subtracting gives 4𝑥 = 5 ⇒ 𝑥 = 1.25 oz of cereal. Substituting this value of x into the second equation gives 1.25 + 2𝑦 = 5 ⇒ 2𝑦 = 3.75 ⇒ 𝑦 = 1.875 oz of fruit. Solution: Mix 1.25 oz of cereal and 1.875 oz of fruit. 43. a. One-third of the U.S. RDA for protein is 20 g. Unknowns: 𝑥 = the number of servings of Vegetarian Baked Beans; 𝑦 = the number of slices of whole wheat bread Arrange the given information in a table with unknowns across the top:

Beans (𝑥) Bread (𝑦) Desired

Solutions Section 4.1

Protein (g)

7

3

20

Carbs (g)

30

12

84

We can now set up an equation for each of the items listed on the left: Protein: Carbs:

7𝑥 + 3𝑦 = 20 30𝑥 + 12𝑦 = 84.

Protein: Carbs:

7𝑥 + 3𝑦 = 30 30𝑥 + 12𝑦 = 84.

Multiply the first equation by −4: 30𝑥 + 12𝑦 = 84. −28𝑥 − 12𝑦 = −80 Adding gives 2𝑥 = 4 ⇒ 𝑥 = 2 servings of beans. Substituting 𝑥 = 2 in the equation 7𝑥 + 3𝑦 = 20 gives 7(2) + 3𝑦 = 20 ⇒ 14 + 3𝑦 = 20 ⇒ 3𝑦 = 6 ⇒ 𝑦 = 2 slices of bread. Solution: Prepare 2 servings of Vegetarian Baked Beans and 2 slices of whole wheat bread. b. Increasing the protein level to half the RDA gives:

Multiply the first equation by −4: 30𝑥 + 12𝑦 = 84. −28𝑥 − 12𝑦 = −120 Adding gives 2𝑥 = −36 ⇒ 𝑥 = −18 servings of beans. Since it is impossible to prepare a negative number of servings of beans, we conclude that it is not possible to prepare such a meal.

44. a. 𝑥 = the number of servings of Vegetarian Baked Beans; 𝑦 = the number of slices of sourdough bread Arrange the given information in a table with unknowns across the top: Pork & Beans (𝑥) Bread (𝑦) Desired

Protein

7

2

12

Carbs.

30

8

50

We can now set up an equation for each of the items listed on the left: Protein: Carbs:

7𝑥 + 2𝑦 = 12 30𝑥 + 8𝑦 = 50.

Multiply the first equation by −4 : 30𝑥 + 8𝑦 = 50. −28𝑥 − 8𝑦 = −48 Adding gives 2𝑥 = 2 ⇒ 𝑥 = 1 serving of beans. Substituting 𝑥 = 1 in the equation 7𝑥 + 2𝑦 = 12 gives 7(1) + 2𝑦 = 12 ⇒ 2𝑦 = 5 ⇒ 𝑦 = 52 slices of bread.

Solution: Prepare 1 serving of beans and 5/2 slices of bread. b. The new equations are Protein: Carbs:

7𝑥 + 2𝑦 = 12 30𝑥 + 8𝑦 = 48.

Multiply the first equation by −4 :

Solutions Section 4.1 30𝑥 + 8𝑦 = 48.

−28𝑥 − 8𝑦 = −48 Adding gives 𝑥 = 0. Substituting in the first equation gives 𝑦 = 6. Solution: Use no beans and 6 slices of bread.

45. Unknowns: 𝑥 = number of servings of Designer Whey; 𝑦 = number of servings of Muscle Milk Arrange the given information in a table with unknowns across the top: Designer Whey Muscle Milk Desired Protein

20

16

152

Carbs

6

10

56

Cost

$0.80

$0.90

We can now set up an equation for each nutrient listed on the left: 20𝑥 + 16𝑦 = 152 6𝑥 + 10𝑦 = 56.

Protein: Carbs:

Multiply the first equation by 5, the second by −8, and add: 100𝑥 + 80𝑦 = 760 −48𝑥 − 80𝑦 = −448 52𝑥 = 312 ⇒ 𝑥 = 6 servings of Designer Whey. Substitute into the second equation to obtain 36 + 10𝑦 = 56 ⇒ 10𝑦 = 20 ⇒ 𝑦 = 2 servings of Muscle Milk. Cost = 6(0.80) + 2(0.90) = $6.60. Solution: Mix 6 servings of Designer Whey and 2 servings of Muscle Milk for a cost of $6.60.

46. Unknowns: 𝑥 = number of servings of Muscle Milk; 𝑦 = number of servings of 100% Whey Gold Standard Arrange the given information in a table with unknowns across the top: Muscle Milk 100% Whey Gold Standard Desired Protein

16

24

320

Sodium

80

210

2,500

Cost

$0.90

$1.00

We can now set up an equation for each nutrient listed on the left: Protein: Sodium:

16𝑥 + 24𝑦 = 320 80𝑥 + 210𝑦 = 2, 500.

Divide the first equation by 8 and the second equation by 10 to obtain 2𝑥 + 3𝑦 = 40 8𝑥 + 21𝑦 = 250. Take 7 times the first − second: 6𝑥 = 30 ⇒ 𝑥 = 5 servings of Muscle Milk. Substitute into the first equation to obtain 80 + 24𝑦 = 320 ⇒ 𝑦 = 10 servings of 100% Whey Gold Standard. Cost = 5(0.90) + 10(1.00) = $14.50. Solution: Mix 5 servings of Muscle Milk and 10 servings of 100% Whey Gold Standard for a cost of $14.50. 47. We first determine how many servings of each kind of supplement it contains: Unknowns: 𝑥 = number of servings of Designer Whey; 𝑦 = number of servings of 100% Whey Gold

Solutions Section 4.1 Standard. Arrange the given information in a table with unknowns across the top: Designer Whey 100% Whey Gold Standard Total Protein

20

24

488

Cost

0.80

1.00

20

Carbs

6

3

We can now set up an equation for protein content and one for cost: 20𝑥 + 24𝑦 = 488 0.80𝑥 + 1.00𝑦 = 20.

Protein: Cost:

Divide the first equation by 4 and multiply the second by 5: 5𝑥 + 6𝑦 = 122 4𝑥 + 5𝑦 = 100. 5 times the first minus 6 times the second gives 𝑥 = 10 servings of Designer Whey. Substituting in the first equation gives 24𝑦 = 288 ⇒ 𝑦 = 12 servings of 100% Whey Gold Standard. From the table, Total amount of carbohydrates = 10(6) + 12(3) = 96 g. 48. We first determine how many servings of each kind of supplement it contains: Unknowns: 𝑥 = number of servings of Designer Whey; 𝑦 = number of servings of Muscle Milk. Arrange the given information in a table with unknowns across the top: Designer Whey Muscle Milk Total Carbs

6

10

190

Cost

0.80

0.90

21

Protein

20

16

We can now set up an equation for carbohydrate content and for cost: 6𝑥 + 10𝑦 = 190 0.80𝑥 + 0.90𝑦 = 21.

Carbs: Cost:

Divide the first equation by 2 and multiply the second by 10: 3𝑥 + 5𝑦 = 95 8𝑥 + 9𝑦 = 210. 9 times the first minus 5 times the second gives −13𝑥 = −195 ⇒ 𝑥 = 15 servings of Designer Whey. Substituting in the first gives 𝑦 = 10 servings of Muscle Milk. From the table, Total supply of protein = 15(20) + 10(16) = 460 g.

49. Unknowns: 𝑥 = number of shares of TWTR; 𝑦 = number of shares of MSFT Arrange the given information in a table with unknowns across the top: TWTR (𝑥) MSFT (𝑦) Total

Sep 1

66

302

43,400

Sep 30

60

282

40,200

We can now set up equations for the start and end of September: Sep 1:

66𝑥 + 302𝑦 = 43,400

60𝑥 + 282𝑦 = 40,200.

Solutions Section 4.1

Sep 30:

Divide the first equation by 2 and the second by 6: 33𝑥 + 151𝑦 = 21,700 10𝑥 + 47𝑦 = 6,700. Multiply the first by −10 and the second by 33: 330𝑥 + 1,551𝑦 = 221,100. −330𝑥 − 1,510𝑦 = −217,000 Adding gives 41𝑦 = 4,100 ⇒ 𝑦 = 100 shares of MSFT. Substituting 𝑦 = 100 into 10𝑥 + 47𝑦 = 6,700 gives 10𝑥 + 47(100) = 6,700 ⇒ 10𝑥 = 2,000 ⇒ 𝑥 = 200 shares of TWTR. Solution: You bought 200 shares of TWTR and 100 shares of MSFT.

50. Unknowns: 𝑥 = number of shares of HES; 𝑦 = number of shares of XOM Arrange the given information in a table with unknowns across the top: HES (𝑥) XOM (𝑦) Total

Jul 1

88

64

18,400

Sep 30

78

60

16,800

We can now set up equations for the start of November and end of January: Jul 1: Sep 30:

88𝑥 + 64𝑦 = 18,400 78𝑥 + 60𝑦 = 16,800.

Divide the first equation by 8 and the second by 6 : 11𝑥 + 8𝑦 = 2,300 13𝑥 + 10𝑦 = 2,800. Multiply the first by 5 and the second by −4: 55𝑥 + 40𝑦 = 11,500 −52𝑥 − 40𝑦 = −11,200. Adding gives 3𝑥 = 300 ⇒ 𝑥 = 100 shares of HES. Substituting 𝑥 = 100 into 78𝑥 + 60𝑦 = 16,800 gives 78(100) + 60𝑦 = 16,800 ⇒ 60𝑦 = 9,000 ⇒ 𝑦 = 150 shares of XOM. Solution: You bought 100 shares of HES and 150 shares of XOM.

51. Unknowns: 𝑥 = number of shares of TD; 𝑦 = number of shares of CNA The total investment was $25,800 ⇒ 66𝑥 + 42𝑦 = 25,800. You earned $969 in dividends: TD dividend = 4% of 66𝑥 invested = 0.04(66𝑥) = 2.64𝑥 CNA dividend = 3.5% of 42𝑦 invested = 0.035(42𝑦) = 1.47𝑦. Thus, 2.64𝑥 + 1.47𝑦 = 969. We therefore have the following system: 66𝑥 + 42𝑦 = 25,800 2.64𝑥 + 1.47𝑦 = 969. Divide the first equation by 6 and multiply the second equation by 100/3: 11𝑥 + 7𝑦 = 4,300 88𝑥 + 49𝑦 = 32,300. Multiply the first equation by 8 and subtract: 7𝑦 = 2,100 ⇒ 𝑦 = 300 shares of CNA. Substituting 𝑦 = 300 in 11𝑥 + 7𝑦 = 4,300 gives 11𝑥 + 7(300) = 4,300 ⇒ 11𝑥 = 2,200 ⇒ 𝑥 = 200 shares of TD. Solution: You purchased 200 shares of TD and 300 shares of CNA.

52. Unknowns: 𝑥 = number of shares of PAA; 𝑦 = number of shares of CVX The total investment was $30,000 ⇒ 10𝑥 + 100𝑦 = 30,000. You earned $1,700 in dividends:

Solutions Section 4.1 PAA dividend = 7% of 10𝑥 invested = 0.07(10𝑥) = 0.7𝑥 CVX dividend = 5% of 100𝑦 invested = 0.05(100𝑦) = 5𝑦. Thus, 0.7𝑥 + 5𝑦 = 1,700. We therefore have the following system: 10𝑥 + 100𝑦 = 30,000 0.7𝑥 + 5𝑦 = 1,700. Divide the first equation by 2 and multiply the second by 10: 5𝑥 + 50𝑦 = 15,000 7𝑥 + 50𝑦 = 17,000. Subtracting the first from the second gives 2𝑥 = 2,000 ⇒ 𝑥 = 1,000 shares of PAA. Substituting 𝑥 = 1,000 in 10𝑥 + 100𝑦 = 30,000 gives 10,000 + 100𝑦 = 30,000 ⇒ 100𝑦 = 20,000 ⇒ 𝑦 = 200 shares of CVX. Solution: You purchased 1,000 shares of PAA and 200 shares of CVX.

53. Unknowns: 𝑥 = number of members voting in favor; 𝑦 = number of members voting against We are given two pieces of information: (1) Total number of votes is 435 𝑥 + 𝑦 = 435 (2) 49 more members voted in favor than against. Rephrasing this gives: The number of members voting in favor exceeded the number of members voting against by 49. 𝑥 − 𝑦 = 49 Thus we have two equations: 𝑥 + 𝑦 = 435 𝑥 − 𝑦 = 49. Adding gives 2𝑥 = 484 ⇒ 𝑥 = 242. Substituting 𝑥 = 242 in the first equation gives 242 + 𝑦 = 435 ⇒ 𝑦 = 435 − 242 = 193. Solution: 242 voted in favor and 193 voted against. 54. Unknowns: 𝑥 = number of senators voting in favor; 𝑦 = number of senators voting against We are given two pieces of information: (1) Total number of votes is 100. 𝑥 + 𝑦 = 100 (2) For a supermajority, twice as many senators must vote in favor of the bill as vote against it. Rephrasing: The number voting in favor is twice the number voting against. 𝑥 = 2𝑦 Thus we have two equations: 𝑥 + 𝑦 = 100 𝑥 − 2𝑦 = 0. Subtracting gives 3𝑦 = 100 ⇒ 𝑦 = 33⅓. Substituting this value in the first equation gives 𝑥 = 66⅔. For a supermajority, at least 66⅔ senators must vote in favor. Since there are no fractional votes, the solution is: At least 67 votes in favor. 55. Unknowns: 𝑥 = number of soccer games won; 𝑦 = number of football games won We are given two pieces of information: (1) The total number of games was 12 ⇒ 𝑥 + 𝑦 = 12 (2) Total number of points earned was 38 ⇒ 2𝑥 + 4𝑦 = 38 (Two points per soccer game and 4 per football game) Thus we have two equations: 𝑥 + 𝑦 = 12 2𝑥 + 4𝑦 = 38. Dividing the second by −2 : 𝑥 + 𝑦 = 12 ⇒ −𝑥 − 2𝑦 = −19. Adding gives

−𝑦 = −7 ⇒ 𝑦 = 7 football games. Substituting 𝑦 = 7 in the first equation gives 𝑥 + 7 = 12 ⇒ 𝑥 = 5 soccer games. Solution: Lombardi House won 5 soccer games and 7 football games. Solutions Section 4.1

56. Unknowns: 𝑥 = number of exposés on the football team; 𝑦 = number of exposés on the film society treasurer We are given two pieces of information: (1) The total number of exposés was 10 ⇒ 𝑥 + 𝑦 = 10. (2) Lawsuit damages amounted to $37 million ⇒ 4𝑥 + 3𝑦 = 37 ($4 million per football team lawsuit, $3 million per film society lawsuit). Thus we have two equations: 𝑥 + 𝑦 = 10 4𝑥 + 3𝑦 = 37. Multiplying the first by −4 : −4𝑥 − 4𝑦 = −40 ⇒ 4𝑥 + 3𝑦 = 37. Adding gives −𝑦 = −3 ⇒ 𝑦 = 3 exposés on the film society treasurer. Substituting in the first equation gives 𝑥 + 3 = 10 ⇒ 𝑥 = 7 exposés on the football team. Solution: There were 7 exposés on the football team, 3 exposés on the film society treasurer. 57. Unknowns: 𝑥 = number of brand 𝑋 pens; 𝑦 = number of brand 𝑌 pens We are given two pieces of information: (1) The total number of pens is 12 ⇒ 𝑥 + 𝑦 = 12. (2) The total amount spent was $42 ⇒ 4𝑥 + 2.8𝑦 = 42 ($4 per band 𝑋 pen and $2.80 per brand 𝑌 pen). Thus we have two equations: 𝑥 + 𝑦 = 12 4𝑥 + 2.8𝑦 = 42. Multiply the second by 10: 𝑥 + 𝑦 = 12 ⇒ 40𝑥 + 28𝑦 = 420. Multiply the first by 10 and divide the second by −4 : 10𝑥 + 10𝑦 = 120 ⇒ −10𝑥 − 7𝑦 = −105. Adding, 3𝑦 = 15 ⇒ 𝑦 = 5 brand 𝑌 pens. Substituting this value in the first equation 𝑥 + 𝑦 = 12 gives 𝑥 + 5 = 12 ⇒ 𝑥 = 7 brand 𝑋 pens. Solution: Elena purchased 7 brand 𝑋 pens and 5 brand 𝑌 pens.

58. Unknowns: 𝑥 = number of reams of yellow paper; 𝑦 = number of reams of white paper We are given two pieces of information: (1) A total of 100 reams are purchased ⇒ 𝑥 + 𝑦 = 100 (2) The total budget is $560 ⇒ 5𝑥 + 6.5𝑦 = 560. Multiply the first equation by 10 and the second by −2 : 10𝑥 + 10𝑦 = 1, 000 ⇒ −10𝑥 − 13𝑦 = −1, 120. Adding gives −3𝑦 = −120 ⇒ 𝑦 = 40 reams of white paper. Substituting in the first equation gives 𝑥 + 40 = 100 ⇒ 𝑥 = 60 reams of yellow paper. Solution: Earl should order 60 reams of yellow paper and 40 reams of white paper. 59. If we want demand to equal supply, we can simply set the two functions equal: −60𝑝 + 150 = 80𝑝 − 60 140𝑝 = 210

210 = $1.50 per pet chia. 140 (We do not need to solve for 𝑞.) 𝑝=

Solutions Section 4.1

60. Set demand equal to supply: −10,000𝑝 + 2,000 = 4,000𝑝 + 600 14,000𝑝 = 1,400 𝑝 = $0.10, or 10¢.

61. Demand: 𝐷 = 85 − 5𝑃 Supply: 𝑆 = 25 + 5𝑃 For the equilibrium, we can equate the supply and demand: Demand = Supply ⇒ 85 − 5𝑃 = 25 + 5𝑃 ⇒ −10𝑃 = −60 ⇒ 𝑃 = $6 per widget. Substituting 𝑃 = $6 in the demand curve gives 𝐷 = 85 − 5(6) = 85 − 30 = 55 widgets. 62. Demand: 𝑄𝐷 = 100 − 10𝑃 Supply: 𝑄𝑆 = 20 + 5𝑃 Since the government sets 𝑃 = 7, we compute 𝑄𝐷 = 100 − 10(7) = 30 𝑄𝑆 = 20 + 5(7) = 55, giving a surplus of 55 − 30 = 25.

63. Demand: The given points are (𝑝, 𝑞) = (3, 16) and (4.5, 10). 𝑞 − 𝑞1 10 − 16 −6 𝑚= 2 = = = −4 𝑝2 − 𝑝1 4.5 − 3 1.5 𝑏 = 𝑞1 − 𝑚𝑝1 = 16 − (−4)(3) = 16 + 12 = 28 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −4𝑝 + 28. Supply: The given points are (𝑝, 𝑞) = (3, 10) and (4.5, 16). 𝑞 − 𝑞1 16 − 10 6 𝑚= 2 = = =4 𝑝2 − 𝑝1 4.5 − 3 1.5 𝑏 = 𝑞1 − 𝑚𝑝1 = 10 − (4)(3) = 10 − 12 = −2 Thus, the supply equation is 𝑞 = 𝑚𝑝 + 𝑏 = 4𝑝 − 2. For the equilibrium price, we can equate the supply and demand:

Supply = Demand ⇒ −4𝑝 + 28 = 4𝑝 − 2 ⇒ −8𝑝 = −30 ⇒ 𝑝 =

30 = 3.75 KMW/kg. 8

64. Demand: The given points are (𝑝, 𝑞) = (4.5, 12) and (2.5, 20). 𝑞 − 𝑞1 20 − 12 8 𝑚= 2 = = = −4 𝑝2 − 𝑝1 2.5 − 4.5 −2 𝑏 = 𝑞1 − 𝑚𝑝1 = 12 − (−4)(4.5) = 30 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −4𝑝 + 30. Supply: The given points are (𝑝, 𝑞) = (4.5, 15) and (2.5, 11). 𝑞 − 𝑞1 11 − 15 −4 𝑚= 2 = = =2 𝑝2 − 𝑝1 2.5 − 4.5 −2 𝑏 = 𝑞1 − 𝑚𝑝1 = 15 − (2)(4.5) = 6 Thus, the supply equation is 𝑞 = 𝑚𝑝 + 𝑏 = 2𝑝 + 6. For the equilibrium price, we can equate the supply and demand: Supply = Demand ⇒ −4𝑝 + 30 = 2𝑝 + 6 ⇒ 6𝑝 = 24 ⇒ 𝑝 = 4 KMW/kg. 65. Unknowns: 𝑥 = number of pairs of dirty socks; 𝑦 = number of T-shirts

Solutions Section 4.1 We are given two pieces of information: (1) A total of 44 items were washed: ⇒ 𝑥 + 𝑦 = 44. (2) There were three times as many pairs of dirty socks as T-shirts. Rephrasing this gives: The number of pairs of dirty socks was three times the number of T-shirts, or 𝑥 = 3𝑦. Thus we have two equations: 𝑥 + 𝑦 = 44 𝑥 − 3𝑦 = 0. Subtracting (or multiplying the second by −1 and adding) gives 𝑦 = 44 ⇒ 𝑦 = 11 T-shirts. Substituting 𝑦 = 11 in the first equation gives 𝑥 + 11 = 44 ⇒ 𝑥 = 44 − 11 = 33 pairs of dirty socks. Solution: Joe's roommate threw out 33 pairs of dirty socks and 11 T-shirts.

66. Unknowns: 𝑥 = number of pieces of raw salmon. 𝑦 = number of pieces of raw tuna Total number of pieces consumed: ⇒ 𝑥 + 𝑦 = 3 Protein intake: Salmon: 50% of 1 oz = 0.5 oz per piece Tuna: 40% of 1.25 = 0.5 oz per piece 0.5𝑥 + 0.5𝑦 = 1.5 Multiplying by 2 gives 𝑥 + 𝑦 = 3, the same equation as the first one. Thus, the system is redundant and has infinitely many solutions. The customer could have consumed any combination of tidbits totaling three.

67. Unknowns: 𝑥 = size of raise for each full-time employee; 𝑦 = size of raise for each part-time employee We are given two pieces of information: (1) Total budget = $6,000. There are 4 full-time employees each getting a raise of 𝑥 and 2 part-time employees each getting a raise of 𝑦. Thus, 4𝑥 + 2𝑦 = 6,000. (2) The raise received by each full-time employee is twice the raise that each of the part-time employees receives. 𝑥 = 2𝑦 Thus we have two equations: 4𝑥 + 2𝑦 = 6,000 𝑥 − 2𝑦 = 0. Adding gives 5𝑥 = 6,000 ⇒ 𝑥 = $1,200 per full-time employee. (We are not asked for the value of 𝑦.) 68. Unknowns: 𝑥 = number of paperback copies sold; 𝑦 = number of hardback copies sold after the paperback appeared We are given two pieces of information: (1) A total of 441,000 − 36,000 = 405,000 copies were sold: ⇒ 𝑥 + 𝑦 = 405,000 (2) The number of paperback copies is 9 times the number of hardbacks sold since the paperback appeared. ⇒ 𝑥 = 9𝑦 ⇒ 𝑥 − 9𝑦 = 0 Subtracting this from the first equation gives 10𝑦 = 405,000 ⇒ 𝑦 = 40,500 hardback copies. Substituting in the first equation gives 𝑥 + 40,500 = 405,000 ⇒ 𝑥 = 364,500 paperback copies.

69. The three lines in a plane must intersect in a single point for there to be a unique solution. This can happen in two ways: (1) The three lines intersect in a single point, or (2) two of the lines are the same, and the third line intersects it in a single point. 70. There are three ways this can happen: (1) The lines are all parallel; (2) two of the lines are parallel; or (3) the three lines intersect in three different points (forming a triangle).

Solutions Section 4.1 71. The equilibrium price occurs at the point where the demand and supply lines cross. Even if two lines have negative slope, they will still intersect if the slopes differ. Therefore, there can be an equilibrium price. 72. The given system of equations cannot be satisfied by any actual number of rocks and pebbles. 73. You cannot round both of them up, since there will not be sufficient eggs and cream. Rounding both answers down will ensure that you will not run out of ingredients. It may be possible to round one answer down and the other up and still have sufficient eggs and cream, and this should be tried. 74. Wrong. Since your factory can only produce whole numbers of gallons, 𝑦 must be a whole number between 0 and 200, giving 201 possible solutions.

75. Since multiplying both sides of an equation by a nonzero number has no effect on its solutions, the graph (which represents the set of all solutions) is unchanged: (B). 76. The graphs are the same: (A). 77. The associated system has no solutions, and so the lines do not intersect. Thus, they must be parallel: (B). 78. Since we obtain a unique solution, the graphs are not parallel: (D). 79. Answers will vary. 80. Answers will vary. 81. Choosing two lines at random gives two random slopes (which are numbers). Since two randomly selected numbers are unlikely to be the same, it follows that two randomly chosen straight lines are very unlikely to be parallel (or the same line). Thus, the two lines are very likely to intersect in a point, giving a unique solution. 82. It is very unlikely. Three randomly chosen lines are unlikely to meet in a single point.

Solutions Section 4.2 Section 4.2 1.

1

1

1

4

1 2 2 1

1

0

3

0 1 1 2

→

1

1

4

0 2 2 (1 2) 2

→

1 0 3

1

1

0

→

0 1 1

1

Translating back to equations gives the solution: = 3, = 1. 2.

2

1 2

2 1 2 2 + 1

2 0 0 (1 2) 1

2 1 2

0 2 4 (1 2) 2

→

1 0 0

2

1

0

→

1

0 1 2 0 1 2 Translating back to equations gives the solution: = 0, = 2. 3.

3

3

→

6

2

3 6 3 2 2 1

2

18 (1 3) 1

0

1

→

0

3 2

6

0 5 30 (1 5) 2 6

4.

2

2

3 5

2 5 2 2 3 1

3

0

2

(1 2) 1

3

1

2

3

→

2

→

1 0 6

5

0 5 5 (1 5) 2

→

1

0

1

→

2

5.

2

1 3 2 1 2 2 2

1 3 2 1 2 0

0

→

0

2

→

3

2 3 1 2 + 1

2

3

5

0

1

6

6

1 2 2

1 + 3 2

1

→

2 3 1 (1 2) 1

1 0 1

1

3

1

→

→

0

0 1 1 2 0 1 1 0 1 1 Translating back to equations gives the solution: = 1, = 1. →

1

1 + 2

2 1 2

0 1 6 2 0 1 6 0 1 6 Translating back to equations gives the solution: = 6, = 6. →

4

0 0 0

→

→

→

Translating back to equations gives: + 32 = 12 . Solve for : = 12 32

General Solution: = 12 32 ; is arbitrary or, in coordinate form, � 12 (1 3 ), �; arbitrary 6.

2 3 1

0

0

6 9 3 (1 3) 2

1 3 2 1 2 0

→

2 2

3 1

3 1 2 1

Translating back to equations gives: 32 = 12 . Solve for : = 12 + 32

→

2 3 1 (1 2) 1 0

0

0

→

General Solution: = 12 + 32 = 12 (1 + 3 ); is arbitrary or, in coordinate form, � 12 (1 + 3 ), �; arbitrary

7.

2

3

2

2

3

2

Solutions Section 4.2 →

2 3 2

2 3

2

1 3 2 1 2 2 2

→

2 3 1 2 + 1

2 3 2

2

3 2

0 0 1

Since Row 2 translates to the false statement 0 = 1, there is no solution. 8.

6 9 3 (1 3) 2

3 1 2 1

2

→

→

0

0

1

Since Row 2 translates to the false statement 0 = 1, there is no solution.

9.

1

1

3

1

3 2 3 1

1

1 0 1 4 0

1

1

4

0

0 1 3 4 0 0

1

2 3 1 → 0

0

1

1

4

1

3

4 1 + 2

3 3 2

4

10. 3 2 1 0 0

1

5 1 1 5 5 3 1

1

5

4

1

1

→

5 1 + 2

4 3 2

5

1 3

1

2 1 2 3 1

25 5 5

1

0

1

1

3 25 1 (1 5) 1

0

1

→ 0

11.

3

1

1

1

2

1

0

1

0.5

0.1

1

0

0

1

1

5

4

0 30 24 (1 6) 3

→ 0 5 4 (1 5) 2 → 0 1 4 5 0 0 0 0 0 0 1

2 3 1 → 0

1 1 3 1

0

1

4 2

0

1

4 1 + 2

1 2 3 2

4

0

1

→ 0 4 0

0

Since Row 3 translates to the false statement 0 = 3, there is no solution. 12. 3 2 4

2

5 1 1 5 5 3

→

0 8 5

1 3

2

1

2 2 2 3 1

25 5

1

3 25 1

1

→ 0

2

8

1

13. 0.1 0.1 1

1

0.3

10 1

10 2 →

11 3 3 3

5 1 3

1

1 3

17

3 5 2 1

11 5 3 3 1

1

3

5

4 1 + 2

0 55 24 8 3 55 2

0 0 83 Since Row 3 translates to the false statement 0 = 83, there is no solution. 1.7

→

1 0 1 5

Translating back to equations gives the solution: = 15 , = 45 . 0

(1 4) 1

1

→ 0 4 3 (1 4) 2 →

Translating back to equations gives the solution: = 14 , = 34 . 1

0

5

1

17

→

→ 0 6 2 (1 2) 2 → 0 12

4

(1 4) 3

5 0 0

17 3 1 + 2

1

3

1

3

3 + 2

1

15

0

1 0 10 3

3 1 (1 3) 2 → 0 1 0 0 0 0

0

→

50 (1 15) 1

Solutions Section 4.2

0

Translating back to equations gives the solution: = 10 , = 13 . 3

14. 3 0

0.1 10 1

0.5

0.3 1

4

1

1 17 1 17

1

17 3 →

0

2

13

1

2

1 0

1

0 1 3

0

13 3 2

2

1

→

1 2 1 5 2

5

5

3

1

4 3 2 + 1 →

1

1

17 3 3 + 1

1

6 0 63 (1 6) 1 0

13

(1 2) 2

1

2

2 0

0

1 3 0

3 5 0

2 13

8 52 (1 4) 3

0

1

1

1

2

0

0

0

1 4 3 + 2 1

0 2 3 →

1

1

0 2 1

2

→

1

0

0

0 0

0

0

1 0

4

3 4

0 4 1

0

16.

0 1

1 0 0 1

2

0

1 0 4 2 → 0 1 0 4 0 0 1 4 0 1 4

1

2

0

6

1 1 0

1 0 0

6

3 2 5 3 1

0

1

0 0 1

1

3

1

6

→ 0 0

1 1

→ 0 1 0 1

0

4 1 2 2

1 0

2 1 3 2

1 0

17. 1 3 1 3 2 3 1 0

1 2 1

6

0

1

2

0 1 0

3

0

1

1 0 0

4

1

0 0 1 1 2

1

1

1

0

3 4 3 + 4 2

4 1 + 2 3

0 2 3

1

1

6

1

1

0

0

1

0 0

1

1

→

0 0 1 1

1 1

0

1 1 + 2 2

2

1 0 1 + 2 2

1

→ 0 1

Translating back to equations gives the solution: = 6, = 1, = 1. 1

4

1 0 0 4

0

4

1

0

0 0 1 4 0 Translating back to equations gives the solution: = 4, = 4, = 4. 1

2

1

0 (1 3) 2 →

3

3

→

1 0 21 2 → 0 1 13 2

Translating back to equations gives the solution: = 21 , = 13 . 2 2

15.

1

3 2

3 2 → 2 3

1

1 1 1

0

→ 0 1 0

0

1

6 1

1 2 0

4

2 2

2

1

2

1

3

0

2 1 → 0 2 4 3 1

1 + 2 3 2 3

0 1 4

→ 0 1 0

Translating back to equations gives the solution: = 1, = 3, = 12 .

0

0

4

(1 2) 2 →

3

2

1

2 1 (1 2) 3

→

→

1

18. 1 3 1 2

6 0

2

1 0

0

0 3

0

1 2 1 3

1 3

(1 2) 1

12 12 1

0

2

3 0

→ 0 3

0 0 8 24 (1 8) 3 0 1

3 2 →

2

1 2 1 2 3

0

2

2 1

0

Solutions Section 4.2

0 0

1 0 0 1

→ 0 1 0 2

0 0 1 3 3

2

1

1

6

1 1

2

1

0

1

0

1

6

2 1

2 2 1 → 0

1 2 2 3 1

0

1 + 3

3 0

0

12 2 + 2 3 → 0 3

3

1 2 1 2 0 2 3

1

1 0 1 + 2 2

2

1 0

1

1

0

0 2 2 →

1

19. 1 2 1 2

1

0

1 0 3 + 2

→

0

2 1

1 2

20. 1 2

0

3

1

1 0

0

1 2 1 2 2 → 1

2 0 2 4 (1 2) 1 0 1 2 4

2

12

3 1 + 2

2 4 3 3 2

(1 3) 2 →

0 0 1 3

2

1 1

1 0 2

1

0 2 1

1 1 0 3 + 2 1

2

0

1 0 1

2

1

→

0

1 0 1 0

1 1 0 2 → 0 1 1 0 0 0 0 0 0 0 0

2

1 3

0

0

1 2 2 2 1

0

1 1 2 2 3 3 1

1 0 1 2 → 0 1 2 4

2

1

3

0

3

2 1

→ 0

1

0

1

1 0 3

0 (1 3) 2 →

0

0

3 0 (1 3) 3

1 + 2

2 4

2 4 3 2

0 0 0 0 0 0 0 0 Translating back to equations gives: = 2, 2 = 4. Thus, the general solution is = 2, = 2 4, arbitrary, or ( 2, 2 4, ); arbitrary.

21.

1

1

2

2

2 2

1 2

3 5 3 5 3 5 2 5 5 3

1 1 0 0

2

0 0

1 1 + 2 2 2

1

5

3

3 3 2

1 1 2 1

→ 2 2 2 3 3 3

1 1

0

→ 0 0 1 0 0

3

2

2 2

(1 2) 2 →

0

1

1

1 2

1 1 3

1 2 1

1 1 3 3

Since Row 3 translates to the false statement 0 = 1, there is no solution.

22.

1

1

2 7

1

1 0

1 7

→

(1 3) 1

3 6

0 Translating back to equations gives: = 0, = 0. Thus, the general solution is = , = , arbitrary, or ( , , ); arbitrary. 1

1

0

0 0 1 3

1 2 0 2 1

1

0

0

Translating back to equations gives the solution: = 1, = 2, = 3. 1 2

3

0

2 0

8 7 14 7 3

1

→ 1 1 7 2

0

0

8 98 (1 2) 3

→

1 1 1

1

1 2

2 1

3 3 1

1 2

1 7 0

0

→

2 1 →

4 49 3 1

→

1

1

1

1 2

0 2 6

1

(1 2) 2 → 0

2

0 1 3 51

1

0

1 2 1 + 2

Solutions Section 4.2

1

3

1

1

1

1

0

4 1

→ 0 1 3

3 51 3 2

0

0

0

Since Row 3 translates to the false statement 0 = 50, there is no solution.

23.

0.5 0.5 0.5 1.5 2 1 4.2

2.1 2.1

0.2

0

1 1 1 3 0

5 3

0

0

1 2 3

1 0 0 1 1 0

1 0 1

0

1

0

1

1 0 0 1

→ 0 1 0

0

1

0

1 3 1 2

1

1

3 3 6 (1 3) 2 →

1

1

1 2

1

3

10 2 → 42 21 21 0 (1 21) 2 →

0

0.2

1

2 3 3 2

→

1 0 0 0

1 0

2

0

1

1

1

1

1

50

1 1 3

1 1 0 2 + 2 1 → 0 1 0 3 + 1

2 2 3 →

1

0 0 1 1 0 0 1 1 Translating back to equations gives the solution: = 1, = 1, = 1.

24.

0.25 0.5 0.2 0.5

1 2 0

3

0 1

0.2

1.5

0

0

0

1

1

0.5

3 1 + 2 2 3 3 + 2

3 0 2 6 1 + 2 3 0 3 1 3 2 + 3 0 0

1

4 1

0.2 0.6 5 2 →

1 3 2

0

0

1 0 0 2

2 3

1

2

1

1

1

3

3 0 2 6

→ 0 3 1 3 0 0

5

0 0 1

0

0

0

0

1 3 2 1 → 2

1

3 1

(1 5) 3

→

3 0 0 6 (1 3) 1

→ 0 3 0 3 (1 3) 2 →

0 1 0 1

0 0 1 0 Translating back to equations gives the solution: = 2, = 1, = 0. 25.

2

3

1 0

1 1 4

1 1 5 2 2 3 1

0

1

→

2 1 0

1

1

4

1 2

1 + 2

→

(1 2) 1

→

3 1 1 0 1 + 2

→

2 0

0

2

0 1 1 2

0 1 1 2 Translating back to equations gives: = 1, = 2. Thus, the general solution is = 1, = 2, arbitrary, or (1, 2, ); arbitrary. 26.

3

1

1 1 0 1

1

4 3 2 1

3 0 0 3 (1 3) 1 0 1 1 3

→

→

3 1 1 0

1 0 0 1 0 1 1 3

4

4

0

12 (1 4) 2

→

0

1

1

3

Solutions Section 4.2 Translating back to equations gives: = 1, + = 3. Thus, the general solution is = 1, = + 3, arbitrary, or (1, + 3, ); arbitrary. 27.

0.75 0.75 1 4 4 1

0

0

1

3 3 4

4

1

16

0

16 16 (1 16) 2

3 3 0 12 (1 3) 1

→

3 1

1

4

0 3 2 1

3 3 4 16 1 + 4 2 0

→

3 4 16

0

1 1 0

1

1

4

→

→

0 0 1 1 0 0 1 1 Translating back to equations gives: = 4, = 1. Thus, the general solution is = + 4, arbitrary, = 1, or ( + 4, , 1); arbitrary. 28.

2

1

1

→

4

1 0.5 0.5 1.5 2 2

2

1

1

4

1 3 2 + 1

1

2

→

2 1 1 4 0

→

0

Since Row 2 translates to the false statement 0 = 7, there is no solution.

0 7

29. 3 1 1 12 (1 3) 1 → 1 1 3 1 3 4

Translating back to equations gives: + 13 13 = 4.

Thus, the general solution is = 4 3 + 3, arbitrary, arbitrary, or (4 3 + 3, , ); , arbitrary.

30. The augmented matrix 1 1 2 21 is already reduced. Translating to equations gives + 3 = 21. General solution: (21 + 3 , , ); , arbitrary 1

31.

1

2

2

0.75 0.75 1

1 1

2

0 0 1 0 0 2 0 1

0

0 0

0

1 0

0

0 0 1

2

2

1

2

0.25 4 3

0

2

4

(1 2) 3

2

2

→

1

2

1 2

20

17 0

1

1 21

→

1 1

→

0 0 0 0 0 1

2 3

0 0 0

2

2

1

2

3

2

4

1

1 0 2 21 2

1 1 0

1 0 0 17 0 0 1

1

2 0

(1 2) 2

1 1 + 2 2 2

2

20

3 2

1

→

1

0 0 0

1 0

32.

1

1

1

7

0

0.75 0.5 0.25 14 4 3 1

1

1

4

1

→

1 3 1

3

1 4

0 1

0

1 2

1 7 2 1

1

1

0

2 1

56 3 3 1 4

4 1

1 1

2 1

3 3 1

1 0

0

3 1 4

2

0

20

4 → 2 → 3 → 4 \quad → \quad

1

1

1 0 2 21 4 + 1

0 1 0 20 0 1 0 20 Translating back to equations gives the solution: = 17, = 20, = 2. 1

2

1

3

→

1

1

→

→ 1 0 0 17 0 1 0 0 0 1 0 0 0

20

2 0

→

1

1

1 2

0 2 6

0 5

2

4

62

0 1 3

1

0

0

0

0

1

0

0

0

0

0

1

0

2

6

4 1

(1 2) 2 (1 2) 4

1

0 11

3

0

1

1

0

3

→

62 3 5 2

4

0

1

3

0

0

0

11

0 1 0 10

2 + 3 3

1

1

0

0 0 1

→

4 1 1 + 4 3

0 1 3

1 0 0

0 1 0 10 2

1

3

5

0

→

1 2 1 + 2

1

0

→

57 (1 19) 3

19

1

1

Solutions Section 4.2

3

1

4 3

3

→

3

0 0 0 0 0 0 0 0 Translating back to equations gives the solution: = 11, = 10, = 3. 1

33.

0 1 1

1 0 0 1

1 5 0 1

1 2 1 1

3 7 2 2 3 1 1 5 1 1 4 1

3 2

0 0

2

0 1

0

0 0

2 0

0

0

0

1 1

1

0

1

1 0

0

→

2 1 + 3 3 2 + 3

0 3 (1 2) 1 0

0

0 0 2 0 1 (1 2) 3 0 0

1

1

0

5 0 1 1 2

1

0

1

0

2 1 1

2 2 1 3 2 2

2

0

0

0 1 0

2 0

0

0 1

0

0 0 2

→

0 0

0

2 3 1 + 2 4 1

0 1 0 0

0

1

1 0

0

0 0 1 0

1 2

1

0

0 0 0 1

2 4

0

1 0 0 0 3 2 →

→

→

0

Translating back to equations gives the solution: = 32 , = 0, = 12 , = 0. 1

1

0

4

1

34.

2

2 3 2 1 2 2 1

4

0

3

0 2

4 4

6 9

10

1 0

1

4 6

4 2 1

0 4 3

6

3 2 + 3

4

5

2

0 0

0 0 0

3

5

15 22

0

1 5

4 5 3

0 0

0 0

3 0

5 1

1

0

→

1 5 4

0 4 0 11 2 2 + 11 4 3 + 5 4

0 4

0

4

0

2

0

4

1

3 6 3 6

1

4

9

8

5

1

5

2

0 4 0 11 2 0 0

4 →

4

0

→

1 + 3

1

0 0

0

3 0

0 0

3

2

0

4

4 1 + 2 3 + 2

2 4 + 2

(1 3) 4

(1 2) 1

0 4 0 0 2 (1 2) 2 0 0

0 0

3 0 0 1

1 0

→

→

→

2

0

0 0

(1 2) 1

1

0 2 0 0 1 (1 2) 2 0 0

0

3 0

0

(1 3) 3

1

0 1

0

Solutions Section 4.2

1 0 0 0 1 2 0 1 0 0 1 2 0 0 1 0 1 3

→

0 0 0 1

0

Translating back to equations gives the solution: = 12 , = 12 , = 13 , = 0. 1

35.

1 5 0 1

0

1 2 1 1

1 5 1 1 3 1

1 1

2 7 2 2 4 1

1

1 0 3 1 0 1 + 3 0 1 2

1 2 3

1

0 0 0

1

0 0 0

0

0 4 3

1

0

→

1

0

→

5 0 1 1 2

1

2 1 1

0

0

0 1 0

2 2 1 4 2

1

→

1 0 3 0 0 0 1 2 0 1 0 0 0 1 0 0 0 0 0 0

Translating back to equations gives: + 3 = 0, + 2 = 1, = 0. General solution: = 3 , = 1 2 , arbitrary, = 0 or ( 3 , 1 2 , , 0); arbitrary 1

1

0

4

1

36.

2

2 3 2 1 2 2 1

4

0

3 10

0

4

3

6

3

1

3

7

1

4 4

4 3 1

1 + 3

0 4 3 6 3 2 + 3 0 0

0

3

0

3

1 0 0

1

5

4 3

1

5

5 4

1 2

0

0

0 1 0 11 4 1 2

→

1

1

0

4

1

→

0

4

3 6 3

4

0

0

5

0

4

0

0

6

1

3

5

5

1

2

4

1

4 1 + 2 3 + 2

(1 4) 1

0 4 0 11 2 (1 4) 2 0

0

0

3

0

0

0

0

→

(1 3) 3

→

0 0 1 5 3 1 3 0 0 0

Translating back to equations gives: + 54 = 12 , + 11 = 12 , 53 = 13 . 4

General solution: � 12 54 , 12 11 , 13 + 53 , �, arbitrary 4

37.

1

1

2

2 1 4 1 3

1

7

2

2 2 1

8 2 3 3 2 1

5 0 17 8 7 (1 5) 1 0 5 0 0

6 0

6 0

→ 0

5

0

5

6

6

6

6

1

1 3 2

→

1 0 17 5 8 5 7 5

1 (1 5) 2 → 0 1 0

1 2 1 4 1 5 1 + 2 2

0 0

6 5 0

6 5 0

1 5 0

Translating back to equations gives: + 17 85 = 75 , + 65 + 65 = 15 . 5

General solution: = 7 5 17 5 + 8 5, = 1 5 6 5 6 5, , arbitrary, or (7 5 17 5 + 8 5, 1 5 6 5 6 5, , ); , arbitrary,

Solutions Section 4.2

or � 15 (7 17 + 8 ), 15 (1 6 6 ), , �; , arbitrary

38.

1

3 2 1 1

1

2

3

0

1

2 2 1

2

2 0 1 1 3 (1 2) 1 0 6 0 0

3 0

→ 0

3 3 2 1

1

1

1 3 2 1 1 2 1 + 2 0

3

6

3

1 0 1 2 1 2 3 2

3 1 (1 6) 2 → 0 1 0 0

6

1 2

0 0

1 2 1 6

0

0

3

1

1 3 2

3

→

0

Translating back to equations gives: 12 + 12 = 32 , + 12 + 12 = 16 .

General solution: = 3 2 + 2 2, = 1 6 2 2, , arbitrary or � 12 (3 + ), 16 (1 3 3 ), , �; , arbitrary 1 0

39. 0 0

0

1

1

1

1

15 1 2

1

1 1 1 2

0

0

0 0

1 0

1

1

1 1 1 0

1

0

0 0 0

1 1

2

0 0 0

0

0 0 1

1 1

0 1 1

1 0 0 0 0 1 0 1 0 0 0 2

5

7 1 + 2 4

1 0 0 2 0 1 0

12

10 2 2 4 12 3 4 5

1 0

2

0

2

17 1 2 3

0 1 1 1 1 2 2 + 3

→ 0 0 0 0 0 0

1

1

0

1

12

1 1 1

0

0

1

→

5

1 0 0 0 2 9 1 + 2 5 0 1 0 0

2

12 2 2 5

0 0 0 0

1

5

→ 0 0 1 0

13 3 2 5 →

2

0 0 0 1 1 1 4 + 5

0 0 1 0 0 3 0 0 0 1 0 4

0 0 0 0 1 5 Translating back to equations gives the solution: = 1, = 2, = 3, = 4, = 5. 1 1 1 1 1 1 1 + 2 0

40. 0 0 0

1

1

1

1 2

0

0

0

1 1

0 0

1 1 1 1 0

1

1 1

1 0 0

2

0 1 1 2 4

0 0 0

1

1 1

0 1 0

2

0 1 2 2 4

0 0 1 1 1 1 3 + 4 0 0 0

0

1 1

1 0 0 1

→ 0 0 0 0 0 0

2 1 1 0 0

0 1

2 3 1 2 3 1 2 2 3

1 1 1 1 0

1 1 1 1

1 0 0 0 2 1 1 + 2 5 0 1 0 0 2 1 2 + 2 5

→ 0 0 1 0 0 0 0 1 0 0 0 0

2 1 1

2 1

1

3 2 5 → 4 5

→

1 0 0 0 0 1

Solutions Section 4.2

0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0

0 0 0 0 1 1 Translating back to equations gives the solution: = 1, = 1, = 0, = 0, = 1.

41.

1

1

2

1

0

1

4

1

1

1

1 3 2 4 2 1

1

1

1

1

0

0

1 0

0

0

0

0 0

0

0 4

1

0

0 0 1 0

0

0 2

0

0 8

0

0

0 1 1 1

2

0 0 0 0

0

0

0 0

2

0

0

0 1

0

0

0

0

0

1 1

1 2

2 2 3 1

2 0

1 0

0

(1 4) 4 (1 8) 5

1

1

0

3

3

0 0

0

0

1

0

0 1 1 1

(1 2) 3 → 0 0 0 0

1 0

0

0 0

0

0 0

1

0

→ 0

1 7 6 5 4 1

0 1 1 1 1 2 0 0

1

0 0 0

1

0

1 1 1

0

0 0

11 6 5 3 2

42. 1 0 0

0 1 1 1

1 1 1

1

1 2

1

1 2

1

1

2

1

1

2

1 1 1 1 2 2 2 2 1

1

1

1

1

1

1

1

1

3

2 4 2

5

0 0

1

4 3

→

5 3

0 0

15

1

9 4 1

0 2 2 2 2 6 (1 2) 4

3

0

18 3 1 → 0 6 5 1

15 1 2 3

2 3 2

3

0 0 0 0 0 0 0 0 0 0 0 0 Translating back to equations gives: = 2, + = 2, = 0. General solution: = 2, = 2 + , arbitrary, arbitrary, = 0 or ( 2, 2 + , , .0); , arbitrary 1

2

2 2 3

0 0 2 0 1

3

2 0

1 + 2

1

1

0 1 1 1 0 2

3 → 0 0

0

1 1

1 1

1

1

1 1 1

1 1

0 1 1 1 1 3 5 + 2

1

1

1

1 1 1

15 3 3

0 3 3 3 3 9 (1 3) 5

1 0 0 0 0 12 0 1 1 1 1

3

0 0 0 0 0

0

3 2 → 0 0 0 0 0

0 1 1 1 1 3 4 + 2

1

0 0 0 0 0

0 0

Translating back to equations gives: = 12, + + + = 3. General solution: = 12, = 3 , , , arbitrary, or (12, 3 , , , ); , , arbitrary

→

→

43. + 2 + = 30 2 + 2 = 30 + 3 + 3 4 = 2 2 9 + = 4 Using technology:

Solutions Section 4.2

Matrix #1 x y 1 2 2 0 1 3 2 -9

z -1 -1 3 0

w 1 2 -4 1

30 30 2 4

Matrix #2 x y 1 2 0 -4 0 1 0 -13

z -1 1 4 2

w 1 0 -5 -1

30 -30 -28 -56

Matrix #3 x y 2 0 0 -4 0 0 0 0

z -1 1 17 -5

w 2 0 -20 -4

30 -30 -142 166

Matrix #4 x y 34 0 0 -68 0 0 0 0

z 0 0 17 0

w 14 368 20 -368 -20 -142 -168 2112

Matrix #5 x y 17 0 0 -17 0 0 0 0

z 0 0 17 0

w 7 5 -20 -7

184 -92 -142 88

Matrix #6 x y 17 0 0 -119 0 0 0 0

z 0 0 119 0

w 0 0 0 -7

272 -204 -2754 88

Matrix #7 x y 1 0 0 1 0 0 0 0

z 0 0 1 0

w 0 0 0 1

16 12/7 -162/7 -88/7

Solution: (16, 12 7, 162 7, 88 7)

44. 4 2 + + = 20 3 + 3 4 = 2 2 + 4 = 4 + 3 + 3 = 2 Using technology:

Solutions Section 4.2

Matrix #1 x y 4 -2 0 3 2 4 1 3

z 1 3 0 3

w 1 -4 -1 0

20 2 4 2

Matrix #2 x y 4 -2 0 3 0 10 0 14

z 1 3 -1 11

w 1 -4 -3 -1

20 2 -12 -12

Matrix #3 x y 12 0 0 3 0 0 0 0

z 9 3 -33 -9

w -5 -4 31 53

64 2 -56 -64

Matrix #4 x y 396 0 0 33 0 0 0 0

z 0 0 -33 0

w 114 -13 31 1470

1608 -34 -56 -1608

Matrix #5 x y 66 0 0 33 0 0 0 0

z 0 0 -33 0

w 19 -13 31 245

268 -34 -56 -268

Matrix #6 x y 16170 0 0 8085 0 0 0 0

z 0 0 -8085 0

w 0 0 0 245

70752 -11814 -5412 -268

Matrix #7 x y 1 0 0 1 0 0 0 0

z 0 0 1 0

w 0 0 0 1

1072/245 -358/245 164/245 -268/245

Solution: � 1,072 , 358 , 164 , 268 � 245 245 245 245

45. + 2 + 3 + 4 + 5 = 6 2 + 3 + 4 + 5 + = 5 3 + 4 + 5 + + 2 = 4 4 + 5 + + 2 + 3 = 3 5 + + 2 + 3 + 4 = 2 Using technology:

Solutions Section 4.2

Matrix #1 x y 1 2 2 3 3 4 4 5 5 1

z 3 4 5 1 2

w 4 5 1 2 3

t 5 1 2 3 4

6 5 4 3 2

Matrix #2 x y 1 2 0 -1 0 -2 0 -3 0 -9

z 3 -2 -4 -11 -13

w 4 -3 -11 -14 -17

t 5 -9 -13 -17 -21

6 -7 -14 -21 -28

Matrix #3 x y 1 0 0 -1 0 0 0 0 0 0

z -1 -2 0 -5 5

w -2 -3 -5 -5 10

t -13 -9 5 10 60

-8 -7 0 0 35

Matrix #4 x y 1 0 0 -1 0 0 0 0 0 0

z -1 -2 0 -1 1

w -2 -3 -1 -1 2

t -13 -9 1 2 12

-8 -7 0 0 7

Matrix #5 x y 1 0 0 -1 0 0 0 0 0 0

z -1 -2 0 -1 1

w 0 0 -1 0 0

t -15 -12 1 1 14

-8 -7 0 0 7

Matrix #6 x y 1 0 0 -1 0 0 0 0 0 0

z 0 0 0 -1 0

w 0 0 -1 0 0

t -16 -14 1 1 15

-8 -7 0 0 7

Solutions Section 4.2 Matrix #7 x y 15 0 0 -15 0 0 0 0 0 0

z 0 0 0 -15 0

w 0 0 -15 0 0

t 0 0 0 0 15

Matrix #8 x y z 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0

w 0 0 1 0 0

t 0 0 0 0 1

-8/15 7/15 7/15 7/15 7/15

-8 -7 -7 -7 7

Solution: ( 8 15, 7 15, 7 15, 7 15, 7 15)

46. 2 + 3 4 = 0 2 + 3 4 + = 0 3 4 + 2 = 0 4 + 2 + 3 = 0 2 + 3 4 = 1 Using technology: Matrix #1 x y 1 -2 -2 3 3 -4 -4 0 0 1

z 3 -4 0 1 -2

w -4 0 1 -2 3

t 0 1 -2 3 -4

0 0 0 0 1

Matrix #2 x y 1 -2 0 -1 0 2 0 -8 0 1

z 3 2 -9 13 -2

w -4 -8 13 -18 3

t 0 1 -2 3 -4

0 0 0 0 1

Matrix #3 x y 1 0 0 -1 0 0 0 0 0 0

z -1 2 -5 -3 0

w 12 -8 -3 46 -5

t -2 1 0 -5 -3

0 0 0 0 1

Matrix #4 x y 5 0 0 -5 0 0 0 0 0 0

z 0 0 -5 0 0

w 63 -46 -3 239 -5

t -10 5 0 -25 -3

0 0 0 0 1

Solutions Section 4.2 Matrix #5 x y 1195 0 0 -1195 0 0 0 0 0 0

z 0 0 -1195 0 0

w 0 0 0 239 0

t -815 45 -75 -25 -842

0 0 0 0 239

Matrix #6 x y 239 0 0 -239 0 0 0 0 0 0

z 0 0 -239 0 0

w 0 0 0 239 0

t -163 9 -15 -25 -842

0 0 0 0 239

Matrix #7 x y z w t 201238 0 0 0 0 -38957 0 -2012380 0 0 2151 0 0 -2012380 0 -3585 0 0 0 201238 0 -5975 0 0 0 0 -842 239 Matrix #8 x y 1 0 0 1 0 0 0 0 0 0

z 0 0 1 0 0

w 0 0 0 1 0

t 0 0 0 0 1

-163/842 -9/842 15/842 -25/842 -239/842

9 15 25 Solution: � 163 , 842 , 842 , 842 , 239 � 842 842

47. 1.6 + 2.4 3.2 = 4.4 5.1 6.3 + 0.6 = 3.2 4.2 + 3.5 + 4.9 = 10.1 We use the Excel Matrix Pivot Tool (on the Website):

→

→ Solution (rounded to 1 decimal place): (1.0, 1.4, 0.2)

48. 2.1 + 0.7 1.4 = 2.3 3.5 4.2 4.9 = 3.3 1.1 + 2.2 3.3 = 10.2 We use the Excel Matrix Pivot Tool (on the Website):

→

Solutions Section 4.2

→

→

→ Solution (rounded to 1 decimal place): (0.9, 2.2, 1.9)

49. 0.2 + 0.3 + 0.4 = 4.5 2.2 + 1.1 4.7 + 2 = 8.3 9.2 1.3 = 0 3.4 + 0.5 3.4 = 0.1 We use the Excel Matrix Pivot Tool (on the Website):

→

→

→

→

Solution (rounded to 1 decimal place): ( 5.5, 0.9, 7.4, 6.6) 50. 1.2 0.3 + 0.4 2 = 4.5 1.9 0.5 3.4 = 0.2 12.1 1.3 = 0 3 + 2 1.1 = 9 We use the Excel Matrix Pivot Tool (on the Website):

→

→

→

→

Solution (rounded to 1 decimal place): (5.0, 0.2, 5.8, 1.9) 51. A pivot is an entry in a matrix that is selected to "clear a column"; that is, use the row operations of a certain type to obtain zeros everywhere above and below it. "Pivoting" is the procedure of clearing a column

Solutions Section 4.2 using a designated pivot. 52. Locate the leading entry of each row, and check to see if it is a 1 and its column is cleared. If there is a leading entry that is not a 1 or whose column is not cleared, the matrix is not reduced. Finally, check to see that the rows are arranged so that rows of zeros are at the bottom, and the leading entry in each row is to the right of the leading entry in the row above. 53. 2 1 + 5 4 , or 6 1 + 15 4 (which is less desirable, since it will produce a row in which every entry is divisible by 3) 54. 2 4 3 2 , or 4 4 6 2 (which is less desirable)

55. It will include a row of zeros. (Subtracting the two rows produces a row of zeros.) 56. It will include a row of zeros. 57. The claim is wrong. If there are more equations than unknowns, there can be a unique solution as well as row(s) of zeros in the reduced matrix, as in Example 6. 58. He is correct: There is at most one pivot per row, and since there are only 5 rows, there cannot be more than 5 pivots. So, at least one of the unknowns must be arbitrary. 59. Since there are 5 columns, there are 4 unknowns. (The last column is for the right-hand sides.) Since there are 5 rows of which 3 are zero, that leaves 2 rows with pivots. Thus, there are 2 unknowns that are not parameters. The remaining 2 unknowns are arbitrary (parameters). 60. Three: 6 unknowns minus 3 pivots 61. The number of pivots must equal the number of variables, since no variable will be used as a parameter. 62. The number of pivots must be less than the number of variables, since at least one variable will be used as a parameter. 63. A simple example is = 1; = 1; + = 2.

64. A simple example is: = 1; = 0; = 1. 65. It has to be the zero solution (each unknown is equal to zero): Putting each unknown equal to zero causes each equation to be satisfied because the right-hand sides are zero. Thus, the zero solution is in fact a solution. Because the solution is unique, this solution is the only solution. 66. No; By the preceding exercise, if the solution were unique, then that solution would have to be (0, 0, 0), which it is not.

67. No: As was pointed out in Exercise 65, every homogeneous system has at least one solution (namely, the zero solution) and hence cannot be inconsistent. 68. No: If a system has the zero solution, then substituting zero for each unknown satisfies every equation in the system. But substituting zero for each unknown causes the left-hand sides to be zero, and hence the righthand sides must also be zero, meaning that the system was homogeneous to begin with.

Solutions Section 4.3 Section 4.3

1. Unknowns: = the number of batches of vanilla, = the number of batches of mocha, = the number of batches of strawberryArrange the given information in a table with unknowns across the top: Vanilla Mocha Strawberry Avail. ( ) ( ) ( ) Eggs

2

1

1

350

Milk

1

1

2

350

Cream

2

2

1

400

We can now set up an equation for each of the items listed on the left: 2 + + = 350 + + 2 = 350 2x + 2y + z = 400.

Eggs: Milk: Cream: 2 1 2

1 1 350

2

1 2 350 2 2 1 → 0 2 1 400 3 1

1 0 0 1

0

1 3

350

0

1 3

1 1 1

1 350 1 2

3 350 0

1 0

2 + 3 3 → 0 1

2 0 2

→ 0 1

50 3 2

0 0

0

3

350

→ 0 1 0

50

(1 2) 1

0 0 3 300 (1 3) 3

100

→

1 0 0 100

50

0 0 1 100 3 0 0 1 100 0 0 1 100 = 100, = 50, = 100 Solution: Make 100 batches of vanilla, 50 batches of mocha, and 100 batches of strawberry.

2. Unknowns: = the number of plain burgers, = the number of double cheeseburgers, = the number of regular cheeseburgers Plain Double Cheese Regular Cheese Avail. ( ) ( ) ( ) Beef patties

1

2

1

19

Buns

1

1

1

13

Cheese slices

0

2

1

15

Beef patties: + 2 + = 19 Buns + + = 13 Cheese slices: 2 + = 15 1 1 0

1

2 1 19

1

1 1 13 2 1 → 0 2 1 15

0

0

4

0

2

1 2

1 0 0 4

1 19 1 + 2 2 0 6

1 15 3 + 2 2

1

0

→ 0 1

0 1 0 6 2 → 0 1 0 6 = 4, = 6, = 3

0

0

1

0

1

7

6 3

1 3

0 0 1 3 0 0 1 3 Solution: Make 4 plain burgers, 6 double cheeseburgers, and 3 regular cheeseburgers.

→

Solutions Section 4.3 3. Unknowns: = the number of sections of Finite Math, = the number of sections of Applied Calculus, = the number of sections of Computer Methods We are given three pieces of information: + + = 6. 40 + 40 + 10 = 210. 40,000 + 60,000 + 20,000 = 260,000 40 + 60 + 20 = 260.

(1) There are a total of 6 sections: (2) The total number of students is 210: (3) The total revenue is $260,000: or, working in thousands of dollars,

Thus, we have a system of 3 equations in 3 unknowns: + + =6 40 + 40 + 10 = 210 40 + 60 + 20 = 260. 1 1 1 6 1 1 1 6 40 40 10 210 (1 10) 2 →

4

1

1

40 60 20 260 (1 20) 3 1 1 0 0 0 1

1

1

1

1 0 0 3 0 0 1 1

6

1 1

1 + 2 3 2

→ 0 0

2

0

1 1

1

6

4 1 21 2 4 1 → 0 0 3 3 (1 3) 2 → 3 1 13 3 2 1

1

= 3, = 2, = 1

0

5

1 1 0

2

1 3

0 1 1

1 0

0

3

1

→ 0 0 1 1 2 → 0 1

0

2

0 1 0 2 Solution: Offer 3 sections of Finite Math, 2 sections of Applied Calculus, and 1 section of Computer Methods.

4. Unknowns: = the number of sections of Ancient History, = the number of sections of Medieval History, = the number of sections of Modern History Sections: Students: Professors: 1 2 1

1

+ + = 45 100 + 50 + 200 = 5,000 2 + + 4 = 100 + + 2 = 60

1 1

45

0

10

1

1 4 100 2 2 1 → 0 1 2

0

60 3 1

0

1

1 0

1 0 0 10 → 0 1 0 20 2 0 1 0 20

1 45 1 + 2 2 10

1 15

1

0

→ 0 1 0

0

= 10, = 20, = 15

3

2

1

55 1 3 3

10 2 2 3 →

15

0 0 1 15 0 0 1 15 Solution: Offer 10 sections of Ancient History, 20 of Medieval, and 15 of Modern.

5. Unknowns: = revenue (in millions) earned though subscription streaming, = revenue (in millions) earned through non-subscription streaming, = revenue (in millions) earned through non-streaming sales. Total revenues were $540 million: + + = 540 Subscription streaming brought in $160 million more than non-subscription streaming. Reword this as follows: The revenue earned from subscription streaming was $160 million more than the revenue earned from non-subscription streaming: = + 160, or = 160. Non-subscription streaming brought in nine times as much as non-streaming sales. Reword this as follows: The revenue earned from non-subscription streaming was nine times the revenue earned from

non-streaming sales: = 9 , or 9 = 0. We thus solve the system + + = 540 = 160 9 = 0. 1

1

1

1

→ 2

0

0

1

160 2 1 → 0

1

700

9

0 2

1

0

1

0

1

0

540

Solutions Section 4.3

1

1 0 0 340 0 1 0 180

0

1 + 3

0

1

2 2

1

0

380 2 3 → 0 2 20

0

0

1

540

2 1 + 2

0

680

(1 2) 1

1 380 9 0

1

= 340, = 180, = 20

0

2 3 + 2

2

0

1

0

→ 0 2 0

0

360 (1 2) 2 → 0 1 20

0

0

1

700

1

380

0

340

19 380 (1 19) 3

0

1

180 2 → 20 3

0 0 1 20 Solution: Revenues were $340 million from subscription streaming, $180 million from non-subscription streaming, and $20 million from non-streaming sales.

6. Unknowns: = revenue (in millions) earned though subscription streaming, = revenue (in millions) earned through non-subscription streaming, = revenue (in millions) earned through non-streaming sales. Total revenues were $660 million: + + = 660. Subscription streaming brought in $240 million more than non-subscription streaming. Reword this as follows: The revenue earned from subscription streaming was $240 million more than the revenue earned from non-subscription streaming: = + 240, or = 240. Subscription streaming brought in twice as much as the other two sales avenues combined. Reword this as follows: The revenue earned from subscription streaming was twice the sum of the sales from the other two avenues: = 2( + ), or 2 2 = 0. We thus solve the system + + = 660 = 240 2 2 = 0. 1 1 1 660 1 1 1 660 1 1 1 660 2 1 + 2 1 1

→ 2

1

0

240 2 1 → 0 2 1 420

0

1

900

2 2

0 2 0

0

1

1

1 0 0 440 0 1 0 200

0

3 1

1 + 3

0 3 3 660 (1 3) 3 2

0

420 2 3 → 0 2 20

0

0

0

0

1

= 440, = 200, = 20

880

→ 0

(1 2) 1

0

2

1

1

0

400 (1 2) 2 → 0 1 20

0

0

1 420

1 220 2 3 2 0

0

1

440

200 2 → 20 3

0 0 1 20 Solution: Revenues were $440 million from subscription streaming, $200 million from non-subscription streaming, and $20 million from non-streaming sales.

Solutions Section 4.3 7. Unknowns: = the number of Airbus A330-300s, = the number of Boeing 767-300ERs, = the number of Boeing Dreamliner 787-9s

Passengers: 330 + 270 + 240 = 4,980 Cost: 260 + 220 + 290 = 4,760 The number of Boeings is twice the number of Airbuses: + = 2 , or 2 = 0. Solving: 2 0 2 1 1 0 1 1 330 270 240 4980 (1 30) 2 → 11 260 220 290 4760 (1 10) 3 2 1 1 0 29

0

26 22

2 1 1

27 332

→ 0

0

29 27 332

→

0

29 0 116 (1 29) 2 → 0 1 0 4

0 35

58 0

29

0 0 0

42 476 (1 7) 3

2 332 (1 2) 1

39 312 (1 39) 3

0

29

29 0

5

0

0 174 (1 29) 1

9

0

8

29 476 3 13 1

0

27 332 6

166 2 2 11 1 →

29 1 + 2

68 29 3 5 2

→

1 166 1 + 3

29 27 332 2 27 3 → 0

1 0 0 6

1

8

= 6, = 4, = 8

0 0 1 8 0 0 1 8 Solution: Order 6 Airbus A330-300s, 4 Boeing 767-300ERs, and 8 Dreamliners. 8. Unknowns: = the number of Airbus A330-300s, = the number of Boeing 767-300ERs, = the number of Boeing Dreamliner 787-9s

Passengers: 330 + 270 + 240 = 2,940 Cost: 260 + 220 + 290 = 2,540 The number of Airbuses is equal to the number of Boeings: = + or = 0. 1 0 1 1 1 0 1 1

Solving: 330 270 240 2940 (1 30) 2 → 11 260 220 290 2540 (1 10) 3

1 1 1 20

0

20 19 98 2 19 3 →

20 0

48

0 0

98

20 1 + 2

0 0

19

0

55 254 5 3 12 2

1 98 1 + 3 1

1 0 0 5 0 1 0 3

2

20

→

20 0 0

= 5, = 3, = 2

0 0

0

26 22

0

1 98

20 19 98 0

8

98 2 11 1 →

29 254 3 26 1

47 94 (1 47) 3

→

0 100 (1 20) 1

20 0 0

9

1

60 (1 20) 2 → 2

0 0 1 2 Solution: Order 5 Airbus A330-300s, 3 Boeing 767-300ERs, and 2 Dreamliners.

9. Unknowns: = the number of tons from CCC, = the number of tons from SSS, = the number of tons from BBF

Solutions Section 4.3

Total order of cheese is 100 tons: + + = 100. Total cost = $5,990: 80 + 50 + 65 = 5,990 Same amount from CCC and BBF: = , or = 0 Solving: 1 1

1

100

1

1

100

1

1

66

2

1

1

100

1

100

1

66

80 50 65 5990 (1 5) 2 → 16 10 13 1198 2 16 1 → 1

0

1

1

2

0

2

0

1

0

1

0 6 3 402 (1 3) 2 → 0 0 1 2 100 0 2 1 134 0

0

3

0 2

0

0

0

0

1

1 0 0 22 0 1 0 56

0

0

1

1

2 1 0

1 134 1

112 (1 2) 2 → 0 1

0

44

22

(1 2) 1

0

0

1

0

0

0

= 22, = 56, = 22

3 1

2 1 + 2

2 100 2 3 2

→ 0 2

66 (1 3) 3

0

1 0

→

1 + 3

134 2 3 → 22

22

56 2 →

1 22 3

0 0 1 22 Solution: The store ordered 22 tons from Cheesy Cream, 56 tons from Super Smooth & Sons, and 22 tons from Bagel's Best Friend.

10. Unknowns: = the number of tons from CCC, = the number of tons from SSS, = the number of tons from BBF Total order of cheese is 36 tons: + + = 36. Total cost = $2,310: 80 + 50 + 65 = 2,310 Two more tons of cream cheese came from BBF. than from SSS: = + 2 or + = 2 Solving: 1 1 1 36 1 1 1 36 80 50 65 2310 (1 5) 2 → 16 10 13 462 2 16 1 → 0

1

1

2

1

1

1

36

2

0

1

34

2

0

0

20

0

1

0 6 3 114 (1 3) 2 → 0 0 1

1

2

0 2 1 38 0

0

3

42

(1 3) 3

(1 2) 1

2

0

1

1

1

2 1

0

1

1

2

36

1 38 1

34

2

2 1 + 2 2 3 2

1 3

→ 0 2 1 38 2 + 3 → 1

0

0

0

0

1

10

14

0 2 0 24 (1 2) 2 → 0 1 0 12 2 → 0

0

1

14

0

0

1

14

→

1 0 0 10

Solutions Section 4.3 = 10, = 12, = 14

0 1 0 12

0 0 1 14 Solution: The store ordered 10 tons from Cheesy Cream, 12 tons from Super Smooth & Sons, and 14 tons from Bagel's Best Friend.

11. Unknowns: = the number of evil sorcerers slain, = the number of trolls slain, = the number of orcs slain Total number slain was 560: + + = 560. Total number of sword thrusts was 620: 2 + 2 + = 620. The number of trolls slain is five times the number of evil sorcerers slain: = 5 , or 5 + = 0. Solving: 1 1 1 560 1 1 1 560 1 + 2 2

2 1 620 2 2 1 → 0 0

5 1 0

0

3 + 5 1

0 6

300

(1 6) 3

1 1

0

60

0 0 1 500 0

1 0

0

10

1

0 6 1

5

1

→ 0

0

0

0

500

2800 3 + 5 2

→

50

→

60

1 500

1

0

1 0 0

10

0 0 1 500 2 → 0 0 1 500

1 3

= 10, = 50, = 500

0 1 0 50 0 1 0 50 Solution: Halmar has slain 10 evil sorcerers, 50 trolls, and 500 orcs.

12. Unknowns: = the number of ounces of rose oil, = the number of ounces of fermented prune oil, = the number of ounces of alcohol Total amount in a bottle is 22 ounces: + + = 22. Each bottle contained 4 more ounces of alcohol than oil of fermented prunes: = + 4, or + = 4. The amount of alcohol was equal to the combined volume of the other two ingredients: = + , or + = 0. Solving: 1 1 1 22 1 1 1 22 0

1

0

1

1

1 0 1

1

1 0

0

0 1 0

0

1

1 1

1

4

0 3 1 22 4

1 11

0

0

4

→ 0 1

1 + 2

0

1

0

→ 0 1 0

1 0 0

7 2 → 0 1 0

1 11 3

0

1

0

4

2 22 (1 2) 3

4

2

1

1

26 4

11

→

1 + 2 3 2 + 3

→

7 = 4, = 7, = 11

0 0 1 11

Solution: Each bottle contained 4 oz. rose petal extract, 7 oz. oil of fermented prunes, and 11 oz. alcohol. 13. Unknowns: = amount of money donated to the MPBF, = amount of money donated to the SCN, = amount of money donated to the NY Jets Given information: (1) The society donated twice as much to the NY Jets as to the MPBF. Rephrase this as follows: The

Solutions Section 4.3 amount of money donated to the NY Jets was equal to twice the amount of money donated to the MPBF: = 2 , or 2 = 0. (2) The society donated equal amounts to the first two funds: = , or = 0. (3) Money donated back to the society: + 2 + 2 = 4,200 Solving: 2 0 1 0 2 0 1 0 1 1

1 2

0

2

4200 2 3 1

7

8400 (1 7) 3

2

0

1

2

0

0

0 2 0

0

2 2 1 → 0

0

1

0

0

(1 2) 1

1200

2

0

0

→ 0 2 0

0

1

0

2 4

1 1

1

0

1

5

0

0

8400 3 + 2 2

→

1 + 3

2 3 →

0

1200

600

0 2 0 1200 (1 2) 2 → 0 1 0 600 2 → 0

0

1 0 0 0 1 0

1

1200

0

600 600

0

1 1200

0 0 1 1200 Solution: It donated $600 to each of the MPBF and the SCN, and $1,200 to the Jets.

14. Unknowns: = number of good reviews, = number of bad reviews, = number of mediocre reviews, = number of reviews left blank Given information: There are 280 more bad reviews than good ones: = 280. There are half as many blank reviews as bad ones: = 1 2 , or 2 = 0. The good reviews and blank reviews together total 170: + = 170. Removing 280 of the bad reviews leaves him with a total of 400 reviews of all types: + ( 280) + + = 400, or + + + = 680. 1 0

1 0

0

1 0 2

280 0

0 0

1

170 3 + 1

1 0 0

2

280

1 1

0

0

0

1 1

1

1 0 2 0 0

0 1

3

5

680 4 + 1 0

450 (1 3) 3

960

1 0 0 0 20 1 0 0

1 0 0 300 0 0 1 150

→

1

1 0

→

1

0

→

0

0

0 2

1 2

1 0 0 0 0 0

0

0 1

2

1 0 2 0 0 0 1

1 0 0 0

20

0 1 0 0 300 0 0 0 1 150

1 5

1 1

280 1 2 0

450 3 2

960 4 2 2

→

280 1 2 3 0

150

2 + 2 3

960 4 5 3

3 4 →

→

1 0 0 0

20

0 1 0 0 300 0 0 1 0 210

0 0 1 0 210 0 0 0 1 150 0 0 1 0 210 Solution: There were originally 20 good reviews, 300 bad ones, 210 mediocre ones, and 150 blank ones. 15. Unknowns: = the number of empty seats on United, = the number of empty seats on American, = the number of empty seats on Southwest We are given the following information: (1) United, American, and Southwest flew a total of 210 empty seats: + + = 210.

Solutions Section 4.3 (2) The total cost of these seats was $108,900. (Note that the figures in the table are for 1,000 miles, but the trip was 3,000 miles): 3(181) + 3(197) + 3(134) = 108,900. That is, 543 + 591 + 402 = 108,900. (3) United had three times as many empty seats as American: = 3 , or 3 = 0 We use the Pivot and Gauss-Jordan Tool on the Website: →

→

→ Solution: United: 120; American: 40; Southwest: 50

16. Unknowns: = the number of empty seats on United, = the number of empty seats on JetBlue, = the number of empty seats on Southwest We are given the following information: (1) United, JetBlue, and Southwest flew a total of 300 empty seats: + + = 300. (2) The total cost of these seats was $86,500. (Note that the figures in the table are for one mile, but the trip was 2,000 miles): 2(181) + 2(197) + 2(134) = 86,500. That is, 362 + 246 + 268 = 86,500. (3) JetBlue had three times as many empty seats as Southwest: = 3 , or 3 = 0. We use the Pivot and Gauss-Jordan Tool on the Website: →

→

→ Solution: United: 100; JetBlue: 150; Southwest: 50

17. Unknowns: = amount invested in SHPIX, = amount invested in RYURX, = amount invested in RYCWX + + = 9,000. = , or = 0. 0.17 + 0.16 = 1,170.

The total investment was $9,000: You invested an equal amount in RYURX and RYCWX: YTD loss from the first two funds was $1,170: Solving: 1 1 0

1

0.17 0.16 1 0

1 1

1

1

1

1 0

9000 0

1170 100 3

9000 0

0 1 17 36000 3 + 2

0

1

1

1

1

→ 0 1

1

→

1 2

1

9000 0

117000 3 17 1

→

0

0 0 18 36000 (1 18) 3

→

17 16 1 0

0

2

9000

1 0

2

0 1

9000

1

0 0

1

0

2000

1 0 0 5000 0 1 0 2000

1 + 2 3 2 3

Solutions Section 4.3

1 0

0

→ 0 1

5000

0

2000

0 0 1 2000 3

→

= 5,000, = 2,000, = 2,000

0 0 1 2000 Solution: You invested $5,000 in SHPIX, $2,000 in RYURX, $2,000 in RYCWX.

18. Unknowns: = amount invested in SHPIX, = amount invested in RYURX, = amount invested in RYCWX The total investment was $7,000: You invested an equal amount in SHPIX and RYURX: Total loss for the year was $1,410:

Solving: 1 0

0

2

1

2

1

1 1

1

1

1

0

7000 0

0.17 0.16 0.25 1410 100 3

→

22000 2 3 2

0

1

7000

1 7000 8

0

1

7000

0

0

2000

2 1 + 2

1 3

2

1

1

1

1

+ + = 7,000. = , or = 0. 0.17 + 0.16 + 0.25 = 1,410. 1

0

7000 0

17 16 25 141000 3 17 1

0

1

7000

→ 0 2 1 7000 2

0

0

2 1

0

17

51000 (1 17) 3

4000

→

→

(1 2) 1

0 2 1 7000 2 + 3 → 0 2 0 4000 (1 2) 2 → 0 1

0

1

3000

0

0

1

1 0 0 2000 → 0 1 0 2000 2 0 1 0 2000

3000

= 2,000, = 2,000, = 3,000

0 0 1 3000 0 0 1 3000 Solution: You invested $2,000 in each of SHPIX and RYURX and $3,000 in RYCWX.

19. Unknowns: = the number of shares of WSR, = the number of shares of TKOMY, = the number of shares of STX The total investment was $7,850. Investment in WSR = shares @ $10 = 10 Investment in TKOMY = shares @ $53 = 53 Investment in STX = shares @ $84 = 84 Thus, 10 + 53 + 84 = 7,850. You expected to earn $272 in dividends: WSR dividend = 4% of 10 invested = 0.04(10 ) = 0.4 TKOMY dividend = 4% of 53 invested = 0.04(53 ) = 2.12 STX dividend = 3% of 84 invested = 0.03(84 ) = 2.52 Thus, 0.4 + 2.12 + 2.52 = 272. You purchased a total of 200 shares: + + = 200. We therefore have the following system: + + = 200 10 + 53 + 84 = 7,850 0.4 + 2.12 + 2.52 = 272. Solving:

1

10

1

53

1

84

0.4 2.12 2.52 1

1

200

7850

200 43 1 2

1

0

43

74 5850

0

43

74

0

43

1 0

0

0 1

→

0 0

5850 2 + 74 3 →

1

100 50

50

0 0 1 50 3

1

200

10 53 63 6800 3 10 1

31 2750 1 31 3

0

1

→ 10 53 84 7850 2 10 1 →

53 4800 3 2

43

0

272 25 3

1

Solutions Section 4.3

43 0 0

0

43

43 0

0 0

1 0 0 100

→ 0 1 0

50

0 0 1

31 74

2750

5850

0

21 1050 (1 21) 3

43 0