c:swp2507calwaism2001cwa ch 3 final...section 3.1 limits 177 10. (a) (i) by reading the graph, as x...

Chapter 3

THE DERIVATIVE

3.1 Limits

1. Since limx!2¡

f(x) does not equal limx!2+

f(x), limx!2

f(x) does not exist. The answer is c.

2. Since limx!2¡

f(x) = limx!2+

f(x) = ¡1;

limx!2

f(x) = ¡1. The answer is a.

3. Since limx!4¡

f(x) = limx!4+

f(x) = 6,

limx!4 f(x) = 6. The answer is b.

4. Since limx!1¡

f(x) = limx!1+

f(x) = ¡1;

limx!1 f(x) = ¡1. The answer is b.

5. (a) By reading the graph, as x gets closer to 3from the left or right, f(x) gets closer to 3.

limx!3 f(x) = 3

(b) By reading the graph, as x gets closer to 0from the left or right, f(x) gets closer to 1.

limx!0 f(x) = 1

6. (a) By reading the graph, as x gets closer to 2from the left or right, F (x) gets closer to 4.

limx!2

F (x) = 4

(b) By reading the graph, as x gets closer to ¡1from left or right, F (x) gets closer to 4.

limx!¡1 F (x) = 4

7. (a) By reading the graph, as x gets closer to 0from the left or right, f(x) gets closer to 0.

limx!0 f(x) = 0

(b) By reading the graph, as x gets closer to 2from the left, f(x) gets closer to ¡2, but as x getscloser to 2 from the right, f(x) gets closer to 1.

limx!2 f(x) does not exist.

8. (a) By reading the graph, as x gets closer to 3from the left or right, g(x) gets closer to 2.

limx!3

g(x) = 2

(b) By reading the graph, as x gets closer to 5from the left, g(x) gets closer to ¡2, but as x getscloser to 5 from the right, g(x) gets closer to 1.

limx!5 g(x) does not exist.

9. (a) (i) By reading the graph, as x gets closer to¡2 from the left, f(x) gets closer to ¡1:

limx!¡2¡

f(x) = ¡1

(ii) By reading the graph, as x gets closer to¡2 from the right, f(x) gets closer to ¡1

2 :

limx!¡2+

f(x) = ¡12

(iii) Since limx!¡2¡

f(x) = ¡1 and limx!¡2+

f(x)

= ¡12 ; lim

x!¡2 f(x) does not exist.

(iv) f(¡2) does not exist since there is nopoint on the graph with an x-coordinateof ¡2:

(b) (i) By reading the graph, as x gets closer to¡1 from the left, f(x) gets closer to ¡1

2 :

limx!¡1¡

f(x) = ¡12

(ii) By reading the graph, as x gets closer to¡1 from the right, f(x) gets closer to ¡1

2 :

limx!¡1+

f(x) = ¡12

(iii) Since limx!¡1¡

f(x) = ¡12 and

limx!¡1+

f(x) = ¡12 ; lim

x!¡1f(x) = ¡1

2 :

(iv) f(¡1) = ¡12 since

¡¡1; ¡12

¢is a point

of the graph.176

Section 3.1 Limits 177

10. (a) (i) By reading the graph, as x gets closerto 1 from the left, f(x) gets closer to 1.

limx!1¡

f(x) = 1

(ii) By reading the graph, as x gets closer to 1from the right, f(x) gets closer to 1.

limx!1+

f(x) = 1

(iii) Since limx!1¡

f(x) = 1 and limx!1+

f(x) = 1;

limx!1

f(x) = 1:

(iv) f(1) = 2 since (1; 2) is part of the graph.

(b) (i) By reading the graph, as x gets closerto 2 from the left, f(x) gets closer to 0.

limx!2¡

f(x) = 0

(ii) By reading the graph, as x gets closer to 2from the right, f(x) gets closer to 0.

limx!2+

f(x) = 0

(iii) Since limx!2¡

f(x) = 0 and limx!2+

f(x) = 0;

limx!2 f(x) = 0:

(iv) f(2) = 0 since (2; 0) is point of the graph.

11. By reading the graph, as x moves furtherto the right, f(x) gets closer to 3. Therefore,limx!1 f(x) = 3:

12. By reading the graph, as x moves further to theleft, g(x) gets larger and larger. Therefore,lim

x!¡1 g(x) =1:

13. limx!2 F (x) in Exercise 6 exists because

limx!2¡

F (x) = 4 and limx!2+

F (x) = 4:

limx!¡2 f(x) in Exercise 9 does not exist since

limx!¡2¡

f(x) = ¡1; but limx!¡2+

f(x) = ¡12 :

15. From the table, as x approaches 1 from the left orthe right, f(x) approaches 4.

limx!1 f(x) = 4

16. f(x) = 2x2 ¡ 4x+ 7; …nd limx!1 f(x):

Substitute 0.9 for x in the expression at the rightto get f(0:9) = 5:02:Continue substituting to complete the table.

x 0:9 0:99 0:999

f(x) 5:02 5:0002 5:000002

x 1:001 1:01 1:1

f(x) 5:000002 5:0002 5:02

As x approaches 1 from the left or the right, f(x)approaches 5.

limx!1

f(x) = 5

17. k(x) =x3 ¡ 2x¡ 4x¡ 2 ; …nd lim

x!2 k(x):

x 1:9 1:99 1:999

k(x) 9:41 9:9401 9:9941

x 2:001 2:01 2:1

k(x) 10:006 10:0601 10:61

As x approaches 2 from the left or the right, k(x)approaches 10.

limx!2 k(x) = 10

18. f(x) =2x3 + 3x2 ¡ 4x¡ 5

x+ 1; …nd lim

x!¡1f(x):

x ¡1:1 ¡1:01 ¡1:001f(x) ¡3:68 ¡3:969 ¡3:996x ¡0:999 ¡0:99 ¡0:9f(x) ¡4:002 ¡4:02 ¡4:28As x approaches ¡1 from the left or the right,f(x) approaches ¡4:

limx!¡1 f(x) = ¡4

19. h(x) =px¡ 2x¡ 1 ; …nd limx!1 h(x):

x 0:9 0:99 0:999

h(x) 10:51317 100:50126 1000:50013

x 1:001 1:01 1:1

h(x) ¡999:50012 ¡99:50124 ¡9:51191limx!1¡

= ¡1limx!1+

= ¡1Thus, lim

x!1 h(x) does not exist.

178 Chapter 3 THE DERIVATIVE

20. f(x) =px¡ 3x¡ 3 ; …nd limx!3

f(x):

x 2:9 2:99 2:999

f(x) 12:9706 127:0838 1268:237

x 3:001 3:01 3:1

f(x) ¡1267:66 ¡125:506 ¡12:6795limx!3¡

f(x) =1

limx!3+

f(x) = ¡1

Thus, limx!3 f(x) does not exist.

21. limx!4

[f(x)¡ g(x)] = limx!4 f(x) ¡ lim

x!4 g(x)

= 9¡ 27 = ¡18

22. limx!4 [(g(x) ¢ f(x)]

=hlimx!4

g(x)i¢hlimx!4

f(x)i

= 27 ¢ 9 = 243

23. limx!4

f(x)

g(x)=limx!4 f(x)

limx!4 g(x)

=9

27=1

3

24. limx!4

log3 f(x) = log3 limx!4 f(x)

= log3 9 = 2

25. limx!4

pf(x) = lim

x!4 [f(x)]1=2

= [limx!4

f(x)]1=2

= 91=2 = 3

26. limx!4

3pg(x) = lim

x!4[g(x)]1=3

=hlimx!4 g(x)

i1=3

= 271=3 = 3

27. limx!4

2f(x) = 2limx!4

f(x)

= 29

= 512

28. limx!4

[1 + f(x)]2 =hlimx!4

(1 + f(x))i2

=hlimx!4 1+ lim

x!4f(x)

i2

= (1 + 9)2 = 102

= 100

29. limx!4

f(x) + g(x)

2g(x)

=limx!4

[f(x) + g(x)]

limx!4

2g(x)

=limx!4

f(x)+ limx!4

g(x)

2 limx!4

g(x)

=9 + 27

2(27)=36

54=2

3

30. limx!4

5g(x) + 2

1¡ f(x) =limx!4 [5g(x) + 2]

limx!4 [1¡ f(x)]

=5 limx!4

g(x)+ lim 2x!4

limx!4 1¡ lim

x!4 f(x)

=5 ¢ 27 + 21¡ 9

= ¡1378

31. limx!3

x2 ¡ 9x¡ 3 = lim

x!3(x¡ 3)(x+ 3)

x¡ 3= limx!3 (x+ 3)

= limx!3 x + lim

x!3 3

= 3 + 3

= 6

32. limx!¡2

x2 ¡ 4x+ 2

= limx!¡2

(x+ 2)(x¡ 2)(x+ 2)

= limx!¡2 (x¡ 2)

= ¡2¡ 2 = ¡4

33. limx!1

5x2 ¡ 7x+ 2x2 ¡ 1 = lim

x!1(5x¡ 2)(x¡ 1)(x+ 1)(x¡ 1)

= limx!1

5x¡ 2x+ 1

=5¡ 22

=3

2


34. limx!¡3

x2 ¡ 9x2 + x¡ 6 = lim

x!¡3(x¡ 3)(x+ 3)(x¡ 2)(x+ 3)

= limx!¡3

x¡ 3x¡ 2

=¡3¡ 3¡3¡ 2

=¡6¡5

=6

5

35. limx!¡2

x2 ¡ x¡ 6x+ 2

= limx!¡2

(x¡ 3)(x+ 2)x+ 2

= limx!¡2 (x¡ 3)

= limx!¡2 x + lim

x!¡2 (¡3)

= ¡2¡ 3= ¡5

36. limx!5

x2 ¡ 3x¡ 10x¡ 5 = lim

x!5(x¡ 5)(x+ 2)

(x¡ 5)= limx!5

(x+ 2)

= 5 + 2 = 7

37. limx!0

1x+3 ¡ 1

3

x

= limx!0

μ1

x+ 3¡ 13

¶μ1

x

¶

= limx!0

·3

3(x+ 3)¡ x+ 3

3(x+ 3)

¸μ1

x

¶

= limx!0

3¡ x¡ 33(x+ 3)(x)

= limx!0

¡x3(x+ 3)x

= limx!0

¡13(x+ 3)

=¡1

3(0 + 3)

= ¡19

38. limx!0

¡1x+2 ¡ 1

2

x

= limx!0

μ ¡1x+ 2

¡ 12

¶μ1

x

¶

= limx!0

· ¡22(x+ 2)

+x+ 2

2(x+ 2)

¸μ1

x

¶

= limx!0

¡2 + x+ 22(x+ 2)(x)

= limx!0

x

2x(x+ 2)

= limx!0

1

2(x+ 2)

=1

2(0 + 2)

=1

4

39. limx!25

px¡ 5x¡ 25

= limx!25

px¡ 5x¡ 25 ¢

px+ 5px+ 5

= limx!25

x¡ 25(x¡ 25)(px+ 5)

= limx!25

1px+ 5

=1p25 + 5

=1

10

40. limx!36

px¡ 6x¡ 36

= limx!36

px¡ 6x¡ 36 ¢

px+ 6px+ 6

= limx!36

(x¡ 36)(x¡ 36)(px+ 6)

= limx!36

1px+ 6

=1p36 + 6

=1

6 + 6

=1

12


41. limh!0

(x+ h)2 ¡ x2h

= limh!0

x2 + 2hx+ h2 ¡ x2h

= limh!0

2hx+ h2

h

= limh!0

h(2x+ h)

h

= limh!0

(2x+ h)

= 2x+ 0 = 2x

42. limh!0

(x+ h)3 ¡ x3h

= limh!0

(x3 + 3x2 + 3xh2 + h3)¡ x3h

= limh!0

3x2h+ 3xh2 + h3

h

= limh!0

h(3x2 + 3xh+ h2)

h

= limh!0

(3x2 + 3xh+ h2)

= 3x2 + 3x(0) + (0)2

= 3x2

43. limx!1

3x

7x¡ 1 = limx!1

3xx

7xx ¡ 1

x

= limx!1

3

7¡ 1x

=3

7¡ 0 =3

7

44. limx!¡1

8x+ 2

4x¡ 5 = limx!¡1

8xx +

2x

4xx ¡ 5

x

= limx!¡1

8 + 2x

4¡ 5x

=8+ 0

4¡ 0= 2

45. limx!¡1

¡x2 + 2x2x2 ¡ 2x+ 1

= limx!¡1

3x2

x2 +2xx2

2x2

x2 ¡ 2xx2 +

1x2

= limx!¡1

3 + 2x

2¡ 2x +

1x2

=3¡ 0

2 + 0 + 0=3

2

46. limx!1

x2 + 2x¡ 53x2 + 2

= limx!1

x2

x2 +2xx2 ¡ 5

x2

3x2

x2 +2x2

= limx!1

1 + 2x ¡ 5

x2

3 + 2x2

=1+ 0¡ 03 + 0

=1

3

47. limx!1

3x3 + 2x¡ 12x4 ¡ 3x3 ¡ 2

= limx!1

3x3

x4 +2xx4 ¡ 1

x4

2x4

x4 ¡ 3x3

x4 ¡ 2x4

= limx!1

3x +

2x3 ¡ 1

x4

2¡ 3x ¡ 2

x4

=0+ 0¡ 02¡ 0¡ 0 = 0

48. limx!1

2x2 ¡ 13x4 + 2

= limx!1

2x2

x4 ¡ 1x4

3x4

x4 +2x4

= limx!1

2x2 ¡ 1

x4

3 + 2x4

=0¡ 03 + 0

=0

3= 0

49. limx!1

2x3 ¡ x¡ 36x2 ¡ x¡ 1

= limx!1

2x3

x3 ¡ xx3 ¡ 3

x3

6x2

x3 ¡ xx3 ¡ 1

x3

= limx!1

2¡ 1x2 ¡ 3

x3

6x ¡ 1

x2 ¡ 1x3

=2¡ 0¡ 00¡ 0¡ 0 =

2

0=1

Therefore limx!1

2x3 ¡ x¡ 36x2 ¡ x¡ 1 does not exist.

50. limx!1

x4 ¡ x3 ¡ 3x7x2 + 9

= limx!1

x4

x4 ¡ x3

x4 ¡ 3xx4

7x2

x4 +9x4

= limx!1

1¡ 1x ¡ 3

x3

7x2 +

9x4

=1¡ 0¡ 00 + 0

=1

0=1

Division by 0 is unde…ned, so this limit does notexist.


51. limx!1

2x2 ¡ 7x49x2 + 5x¡ 6 = lim

x!1

2x2

x2 ¡ 7x4

x2

9x2

x2 +5xx2 ¡ 6

x2

= limx!1

2¡ 7x29 + 5

x ¡ 6x2

The denominator approaches 9, while the numer-ator becomes a negative number that is larger andlarger in magnitude, so

limx!1

2x2 ¡ 7x49x2 + 5x¡ 6 = ¡1:

52. limx!1

¡5x3 ¡ 4x2 + 86x2 + 3x+ 2

= limx!1

¡5x3x2 ¡ 4x2

x2 +8x2

6x2

x2 +3xx2 +

2x2

= limx!1

¡5x¡ 4 + 8x2

6 + 3x +

2x2

The denominator approaches 6, while the numer-ator becomes a negative number that is larger andlarger in magnitude, so

limx!1

¡5x3 ¡ 4x2 + 86x2 + 3x+ 2

= ¡1:

53. Find limx!3 f(x), where f(x) =

x2¡9x¡3 :

x 2.9 2.99 2.999 3.001 3.01 3.1f(x) 5.9 5.99 5.999 6.001 6.01 6.1

limx!3 f(x) = lim

x!3x2 ¡ 9x¡ 3 = 6:

54. Find limx!¡2 g(x), where g(x) =

x2¡4x+2 :

x ¡2:1 ¡2:01 ¡2:001 ¡1:999 ¡1:99 ¡1:9g(x) ¡4:1 ¡4:01 ¡4:001 ¡3:999 ¡3:99 ¡3:9

limx!¡2

g(x) = limx!¡2

x2 ¡ 4x+ 2

= ¡4:

55. Find limx!1

f(x), where f(x) = 5x2¡7x+2x2¡1 :

x 0.9 0.99 0.999 1.001 1.01 1.1f(x) 1.316 1.482 1.498 1.502 1.517 1.667

limx!1 f(x) = lim

x!15x2 ¡ 7x+ 2x2 ¡ 1 = 1:5 =

3

2:

56. Find limx!3 g(x), where g(x) =

x2¡9x2+x¡6 :

x ¡3:1 ¡3:01 ¡3:001g(x) 1:196 1:1996 1:19996

x ¡2:999 ¡2:99 ¡2:9g(x) 1:20004 1.2004 1:204

limx!¡3 g(x) = lim

x!¡3x2 ¡ 9

x2 + x¡ 6 = 1:2 =6

5:

57. (a) limx!¡2

3x(x+2)3 does not exist since

limx!¡2+

3x(x+2)3 = ¡1 and lim

x!¡2¡3x

(x+2)3 =1:

(b) Since (x + 2)3 = 0 when x = ¡2; x = ¡2 isthe vertical asymptote of the graph of F (x):

(c) The two answers are related. Since x = ¡2is a vertical asymptote, we know that lim

x!¡2 F (x)

does not exist.

58. G(x) =¡6

(x¡ 4)2

(a) limx!4

G(x) = ¡1; since by looking at thegraph of G(x); we see that as x gets closer to 4from either the right or the left, g(x) gets smaller.

(b) Since (x¡ 4)2 = 0 when x = 4; x = 4 is thevertical asymptote of the graph of G(x):

(c) The two answers are related. Since x = 4 is avertical asymptote, we know the lim

x!4G(x) does

not exist.

61. (a) limx!¡1 ex = 0 since, as the graph goes further

to the left, ex gets closer to 0.

(b) The graph of ex has a horizontal asymptoteat y = 0 since lim

x!¡1 ex = 0:

62. (a) y = xe¡x

From the graph, it appears that

limx!1 xe¡x = 0:

x 1 10 50

y 0:37 0:00045 9:64£ 10¡21

(b) y = x2e¡x


limx!1 x2e¡x = 0:

x 1 10 50

y 0:37 0:0045 4:82£ 10¡19

(c) limx!1 xne¡x = 0: As n gets larger, y is still

very small but larger than smaller values of n:


63. (a) limx!0+

ln x = ¡1 since, as the graph gets

closer to x = 0; the value of ln x get smaller.

(b) The graph of y = ln x has a vertical asymp-tote at x = 0 since lim

x!0+lnx = ¡1:

64. (a) y = x ln x


limx!0+

x ln x = 0:

x 1 0:5 0:1 0:01 0:001

y 0 ¡0:347 ¡0:230 ¡0:0461 ¡0:0069(b) y = x(ln x)2

From the graph it appears that

limx!0+

x(ln x)2 = 0:

x 1 0:5 0:1 0:01 0:001

y 0 0:240 0:53 0:212 0:048

(c) limx!0+

x(ln x)n = 0: As n gets closer to 0, jyjis still very small but bigger than smaller valuesof n:

67. limx!1

x4 + 4x3 ¡ 9x2 + 7x¡ 3x¡ 1

(a)

x 1:01 1:001 1:0001 0:99 0:999 0:9999

f(x) 5:0908 5:009 5:0009 4:9108 4:991 4:9991

As x!1¡ and as x!1+; we see that f(x)!5:(b) Graph

y =x4 + 4x3 ¡ 9x2 + 7x¡ 3

x¡ 1on a graphing calculator. One suitable choice forthe viewing window is [¡6; 6] by [¡10; 40] withXscl = 1, Yscl = 10.

Because x¡ 1 = 0 when x = 1; we know that thefunction is unde…ned at this x-value. The graphdoes not show an asymptote at x = 1: This indi-cates that the rational expression that de…nes thisfunction is not written in lowest terms, and thatthe graph should have an open circle to show a“hole” in the graph at x = 1: The graphing calcu-lator doesn’t show the hole, but if we try to …ndthe value of the function at x = 1; we see that itis unde…ned. (Using the TABLE feature on a

TI-83, we see that for x = 1; the y-value is listedas “ERROR.”)By viewing the function near x = 1 and using theZOOM feature, we see that as x gets close to 1from the left or the right, y gets close to 5, sug-gesting that

limx!1

x4 + 4x3 ¡ 9x2 + 7x¡ 3x¡ 1 = 5:

68. limx!2

x4 + x¡ 18x2 ¡ 4

(a)

x 2.01 2.001 2.0001 1.99 1.999 1.9999f(x) 8.29 8.25 8.25 8.21 8.25 8.25

(b) Graph

y =x4 + x¡ 18x2 ¡ 4 :

One suitable choice for the viewing window is [¡5; 5]by [0; 20]: Because x2 ¡ 4 = 0 when x = ¡2 orx = 2; we know that the function is unde…ned atthese two x-values. The graph shows an asymp-tote at x = ¡2: There should be open circle toshow a “hole” in the graph at x = 2: The graph-ing calculator doesn’t show the hole, but if we tryto …nd the value of the function at x = 2; we seethat it is unde…ned. (Using the TABLE featureon a TI-83, we see that for x = 2; the y-value islisted as “ERROR.”)By viewing the function near x = 2 and using theZOOM feature, we verify that the required limitis 8.25. (We may not be able to get this valueexactly.)

69. limx!¡1

x1=3 + 1

x+ 1

(a)x ¡1:01 ¡1:001 ¡1:0001f(x) 0:33223 0:33322 0:33332

x ¡0:99 ¡0:999 ¡0:9999f(x) 0:33445 0:33344 0:33334

We see that as x!¡1¡ and as x!¡1+;f(x)!0:3333 or 13 :

(b) Graph

y =x1=3 + 1

x+ 1:

One suitable choice for the viewing window is [¡5; 5]by [¡2; 2]:


Because x+1 = 0 when x = ¡1; we know that thefunction is unde…ned at this x-value. The graphdoes not show an asymptote at x = ¡1: This indi-cates that the rational expression that de…ned thisfunction is not written lowest terms, and that thegraph should have an open circle to show a “hole”in the graph at x = ¡1: The graphing calculatordoesn’t show the hole, but if we try to …nd thevalue of the function at x = ¡1; we see that it isunde…ned. (Using the TABLE feature on a TI-83,we see that for x = ¡1; the y-value is listed as“ERROR.”)By viewing the function near x = ¡1 and usingthe ZOOM feature, we see that as x gets close to¡1 from the left or right, y gets close to 0.3333,suggesting that

limx!¡1

x1=3 + 1

x+ 1= 0:3333 or

1

3:

70. limx!4

x3=2 ¡ 8x+ x1=2 ¡ 6

(a)x 4:1 4:01 4:001 4:0001

f(x) 2:4179 2:4018 2:4002 2:4

x 3:9 3:99 3:999 3:9999

f(x) 2:3819 2:3982 2:3998 2:4

(b) Graph

y =x3=2 ¡ 8

x+ x1=2 ¡ 6 :This function is unde…ned at x = 4 because thisvalue would make the denominator equal to 0.However, by viewing the function near x = 4 andusing the ZOOM feature, we verify that the re-quired limit is 2.4.

71. limx!1

p9x2 + 5

2x

Graph the functions on a graphing calculator. Agood choice for the viewing window is [¡10; 10] by[¡5; 5]:

(a) The graph appears to have horizontal asymp-totes at y = §1:5:We see that as x!1; y!1:5;

so we determine that

limx!1

p9x2 + 5

2x= 1:5:

(b) As x!1;p9x2 + 5!

p9x2 = 3 jxj ;

and p9x2 + 5

2x! 3 jxj

2x:

Since x > 0; jxj = x; so3 jxj2x

=3x

2x=3

2:

Thus,

limx!1

p9x2 + 5

2x=3

2or 1:5:

72. limx!¡1

p9x2 + 5

2x

Graph this function on a graphing calculator. Agood choice for the viewing window is [¡10; 10] by[¡5; 5]:

(a) The graph appears to have horizontal asymp-totes at y = §1:5: We see that as x!¡1;y!¡1:5; so we determine that

limx!¡1

p9x2 + 5

2x= ¡1:5:

(b) As x!¡1;p9x2 + 5!

p9x2 = 3 jxj ;

and p9x2 + 5

2x! 3 jxj

2x:

Since x < 0; jxj = ¡x; so3 jxj2x

=3(¡x)2x

= ¡32:


Thus,

limx!¡1

p9x2 + 5

2x= ¡3

2or ¡ 1:5:

73. limx!¡1

p36x2 + 2x+ 7

3x


(a) The graph appears to have horizontal asymp-totes at y = §2:We see that as x!¡1; y!¡2;so we determine that

limx!¡1

p36x2 + 2x+ 7

3x= ¡2:

(b) As x!¡1;p36x2 + 2x+ 7!

p36x2 = 6 jxj

andp36x2 + 2x+ 7

3x! 6 jxj

3x:

Since x < 0; jxj = ¡x; so

6 jxj3x

=6(¡x)3x

= ¡2:

Thus,

limx!¡1

p36x2 + 2x+ 7

3x= ¡2:

74. limx!1

p36x2 + 2x+ 7

3x


(a) The graph appears to have horizontal asymp-totes at y = §2: We see that as x!1; y!2; sowe determine that

limx!1

p36x2 + 2x+ 7

3x= 2:

(b) As x!1;p36x2 + 2x+ 7!

p36x2 = 6 jxj ;

and p36x2 + 2x+ 7

3x! 6 jxj

3x:

Since x > 0; jxj = x; so6 jxj3x

=6x

3x= 2:

Thus,

limx!1

p36x2 + 2x+ 7

3x= 2:

75. limx!1

¡1 + 5x1=3 + 2x5=3

¢3x5

Graph this function on a graphing calculator. Agood choice for the viewing window is [¡20; 20] by[0; 20] with Xscl = 5, Yscl = 5.


(a) The graph appears to have a horizontal as-ymptote at y = 8: We see that as x!1; y! 8;

so we determine that

limx!1

¡1 + 5x1=3 + 2x5=3

¢3x5

= 8:

(b) As x!1; the highest power term dominatesin the numerator, so

(1 + 5x1=3 + 2x5=3)3! (2x5=3)3 = 23x5= 8x5:

and ¡1 + 5x1=3 + 2x5=3

¢3x5

! 8x5

x5= 8:

Thus,

limx!1

¡1 + 5x1=3 + 2x5=3

¢3x5

= 8:

76. limx!¡1

¡1 + 5x1=3 + 2x5=3

¢3x5

Graph this function on a graphing calculator. Agood choice for the viewing window is [¡60; 60] by[0; 20] with Xscl = 10, Yscl = 10.

(a) The graph appears to have a horizontal as-ymptote at y = 8: We see that as x!¡1; y!8;so we determine that

limx!¡1

¡1 + 5x1=3 + 2x5=3

¢3x5

= 8:

(b) As x! ¡1; the highest power term domi-nates in the numerator, so³1 + 5x1=3 + 2x5=3

´3! (2x5=3) = 23x5 = 8x5;

and ¡1 + 5x1=3 + 2x5=3

¢3x5

! 8x5

x5= 8:

Thus,

limx!¡1

¡1 + 5x1=3 + 2x5=3

¢3x5

= 8:

78. (a) limx!12 G(t)

As t approaches 12 from either direction, the valueof G(t) for the corresponding point on the graphapproaches 3.Thus, lim

x!12G(t) = 3 which represents 3 million

gallons.

(b) limx!16+

G(t) = 1:5

limx!16¡

G(t) = 2

Since limx!16+

G(t) 6= limx!16¡

G(t); limx!16

G(t) does

not exist.

(c) G(16) is the value of function G(t) whent = 16: This value occurs at the solid dot onthe graph G(16) = 2 which represents 2 milliongallons.

(d) The tipping point occurs at the break in thegraph, when t = 16 months.

79. (a) limx!98 T (x) = 7:25 cents because T (x) is con-

stant at 7.25 cents as x approaches 98 from theleft or the right.

(b) limx!02¡

T (x) = 7 cents because as x approaches

02 from the left, T (x) is constant at 7 cents.

(c) limx!02+

T (x) = 7:25 cents because as x ap-

proaches 02 from the right, T (x) is constant at7.25 cents.

(d) limx!02

T (x) does not exist because

limx!02¡

T (x) 6= limx!02+

T (x):

(e) T (02) = 7:25 cents since (02; 7:25) is a pointon the graph.

80. C(x) = 15,000+ 6x

C(x) =C(x)

x=15,000+ 6x

x=15,000x

+ 6

limx!1 C(x) = lim

x!115,000x

+ 6 = 0 + 6 = 6

This means that the average cost approaches $6 asthe number of tapes produced becomes very large.


81. C(x) =C(x)

x

=0:0738x+ 111:83

x

= 0:0738 +111:83

x

limx!1 C(x) = 0:0738

The average cost approaches $0.0738 per mile asthe number of miles becomes very large.

82. P (s) =63s

s+ 8

lims!1

63s

s+ 8= lims!1

63ss

ss +

8s

= lims!1

63

1 + 8s

=63

1 + 0

= 63

The number of items of work a new employee pro-duces gets closer and closer to 63 as the numberof days of training increases.

83. limn!1

·R

·1¡ (1 + i)¡n

i

¸¸

=R

ilimn!1 [1¡ (1 + i)

¡n]

=R

i

hlimn!1 1¡ lim

n!1 (1 + i)¡ni

=R

i[1¡ 0] = R

i

84. limn!1

·R

i¡ g·1¡

μ1 + g

1 + i

¶n¸¸

=R

i¡ g limn!1

·1¡

μ1 + g

1 + i

¶n¸

=R

i¡ g·limn!1 1¡ lim

n!1

μ1 + g

1 + i

¶n¸

assuming i > g,

=R

i¡ g [1¡ 0]

=R

i¡ g85. (a) N(65) = 71:8e¡8:96e

(¡0:0685(65))

¼ 64:68To the nearest whole number, this species of alli-gator has approximately 65 teeth after 65 days ofincubation by this formula.

(b) Since limt!1 (¡8:96e

¡0:0685t) = ¡8:96 ¢0 = 0, itfollows that

limt!1 71:8e

¡8:96e(¡0:0685t) = 71:8e0

= 71:8 ¢ 1= 71:8

So, to the nearest whole number, limt!1 N(t) ¼ 72.

Therefore, by this model a newborn alligator ofthis species will have about 72 teeth.

86. (a) D(t) = 155(1¡ e¡0:0133t)D(20) = 155(1¡ e¡0:0133(20))

= 155(1¡ e¡0:266)¼ 36:2

The depth of the sediment layer deposited belowthe bottom of the lake in 1970 was 36.2 cm.

(b) limt!1 D(t) = lim

t!1 155(1¡ e¡0:0133t)

= 155 limt!1 (1¡ e

¡0:0133t)

= 155( limt!1 1¡ lim

t!1 e¡0:0133t)

= 155(1)¡ 155 limt!1 e

¡0:0133t

= 155¡ 155e limt!1¡0:0133t

limt!1 ¡0:0133t = ¡1As t!1;

155¡ 155e limt!1¡0:0133t ! 155¡ 155(0) = 155:

Thus,limt!1 D(t) = 155:

Going back in time (t is years before 1990), thedepth of the sediment approaches 155 cm.

87. A(h) =0:17h

h2 + 2

limx!1 A(h) = lim

x!10:17h

h2 + 2

= limx!1

0:17hh2

h2

h2 +2h2

= limx!1

0:17h

1 + 2h2

=0

1 + 0= 0

This means that the concentration of the drug inthe bloodstream approaches 0 as the number ofhours after injection increases.

Section 3.2 Continuity 187

88. (a) p2 =1

2+

μ0:7¡ 1

2

¶[1¡ 2(0:2)]2 = 0:572

(b) p4 =1

2+

μ0:7¡ 1

2

¶[1¡ 2(0:2)]4 = 0:526

(c) p8 =1

2+

μ0:7¡ 1

2

¶[1¡ 2(0:2)]8 = 0:503

(d) limn!1 pn = lim

n!1

·1

2+

μp0 ¡ 1

2

¶(1¡ 2p)n

¸

=1

2+ lim

n!1

μp0 ¡ 1

2

¶(1¡ 2p)n

=1

2+

μp0 ¡ 1

2

¶limn!1 (1¡ 2p)n

=1

2+

μp0 ¡ 1

2

¶¢ 0

=1

2

The number in parts (a), (b), and (c) represent theprobability that the legislator will vote yes on thesecond, fourth, and eighth votes. In (d), as thenumber of roll calls increases, the probability getsclose to 0.5, but is never less than 0.5.

3.2 Continuity

1. Discontinuous at x = ¡1(a) lim

x!¡1¡f(x) =

1

2

(b) limx!¡1+

f(x) =1

2

(c) limx!¡1 f(x) =

1

2(since (a) and (b) have the

same answers)

(d) f(¡1) does not exist.(e) f(¡1) does not exist.

2. Discontinuous at x = ¡1(a) lim

x!¡1¡f(x) = 2

(b) limx!¡1+ f(x) = 4

(c) limx!¡1 f(x) does not exist (since parts (a) and

(b) have di¤erent answers).

(d) f(¡1) = 2(e) lim

x!¡1f(x) does not exist.

3. Discontinuous at x = 1

(a) limx!1¡

f(x) = ¡2

(b) limx!1+

f(x) = ¡2

(c) limx!1 f(x) = ¡2 (since (a) and (b) have

the same answers)

(d) f(1) = 2

(e) limx!1 f(x) 6= f(1)

4. Discontinuous at x = ¡2 and x = 3

(a) limx!¡2¡

f(x) = ¡1 limx!3¡

f(x) = ¡1

(b) limx!¡2+

f(x) = ¡1 limx!3+

f(x) = ¡1

(c) limx!¡2 f(x) = ¡1 (since parts (a) and (b)

have the same answer)limx!3

f(x) = ¡1 (since parts (a) and (b)have the same answer)

(d) f(¡2) = 1 f(3) = 1

(e) limx!¡2 f(x) 6= f(¡2) lim

x!3 f(x) 6= f(¡3)

5. Discontinuous at x = ¡5 and x = 0

(a) limx!¡5¡

f(x) =1 (limit does not exist)

= limx!0¡

f(x) = 0

(b) limx!¡5+

f(x) = ¡1 (limit does not exist)

= limx!0+

f(x) = 0

(c) limx!¡5 f(x) does not exist, since the

answers to (a) and (b) are di¤erent.

limx!0

f(x) = 0; since the answers to

(a) and (b) are the same.

(d) f(¡5) does not exist. f(0) does not exist.

(e) f(¡5) does not exist and limx!¡5 f(x) does

not exist. f(0) does not exist.


6. Discontinuous at x = 0 and x = 2(a) lim

x!0¡f(x) = ¡1 (limit does not exist)

limx!2¡

f(x) = ¡2

(b) limx!0+


limx!2+ f(x) = ¡2

(c) limx!0


limx!2 f(x) = ¡2 (since parts (a) and (b)

have the same answer)

(d) f(0) does not exist.

f(2) does not exist.

(e) f(0) does not exist and limx!0 f(x) does

not exist.f(2) does not exist.

7. f(x) =5 + x

x(x¡ 2)f(x) is discontinuous at x = 0 and x = 2 since thedenominator equals 0 at these two values.limx!0 f(x) does not exist since lim

x!0¡f(x) = 1

and limx!0+ f(x) = ¡1.

limx!2 f(x) does not exist since lim

x!2¡f(x) = ¡1

and limx!2+

f(x) =1:

8. f(x) =¡2x

(2x+ 1)(3x+ 6)

f(x) is discontinuous at x = ¡12 and x = ¡2 since

the denominator equals 0 at these two values.

limx!¡2 f(x) does not exist since

limx!¡2¡

= +1 and limx!¡2+

f(x) = ¡1:

limx!¡12

f(x) does not exist since

limx!¡12

¡f(x) = ¡1 and lim

x!¡12+f(x) = +1:

9. f(x) =x2 ¡ 4x¡ 2

f(x) is discontinuous at x = 2 since the denomi-nator equals zero at that value.

Since

x2 ¡ 4x¡ 2 =

(x+ 2)(x¡ 2)x¡ 2 = x+ 2;

limx!2 f(x) = 2 + 2 = 4:

10. f(x) =x2 ¡ 25x+ 5

f(x) is discontinuous at x = ¡5 since the denom-inator equals zero at that value.

Since

x2 ¡ 25x+ 5

=(x+ 5)(x¡ 5)

x+ 5= x¡ 5;

limx!¡5 f(x) = ¡5¡ 5 = ¡10:

11. p(x) = x2 ¡ 4x+ 11Since p(x) is a polynomial function, it is continu-ous everywhere and thus discontinuous nowhere.

12. q(x) = ¡3x3 + 2x2 ¡ 4x+ 1Since q(x) is a polynomial function, it is continu-ous everywhere and thus discontinuous nowhere.

13. p(x) =jx+ 2jx+ 2

p(x) is discontinuous at x = ¡2 since the denom-inator is unde…ned at that value.Since lim

x!¡2¡p(x) = ¡1 and lim

x!¡2+p(x) = 1;

limx!¡2

p(x) does not exist.

14. r(x) =j5¡ xjx¡ 5

r(x) is discontinuous at x = 5 since the denomi-nator is unde…ned at that value.Since lim

x!5+r(x) = 1 and lim

x!5¡r(x) = ¡1;

limx!5

r(x) does not exist.

15. k(x) = epx¡1

The function is unde…ned for x < 1; so the func-tion is discontinuous for a < 1: The limit as xapproaches any a < 1 does not exist because thefunction is unde…ned for x < 1:

16. j(x) = e1=x

j(x) is discontinuous at x = 0 since the functionis unde…ned there.limx!0

j(x) does not exist since limx!0 j(x) = 0 and

limx!0

j(x) =1:


17. As x approaches 0 from the left or the right,¯̄̄xx¡1

¯̄̄

approaches 0 and r(x) = ln¯̄̄xx¡1

¯̄̄goes to ¡1. So

limx!0

r(x) does not exist. As x approaches 1 from

the left or the right,¯̄̄xx¡1

¯̄̄goes to 1 and so does

r(x) = ln¯̄̄xx¡1

¯̄̄. So lim

x!1r(x) does not exist.

18. As x approaches ¡2 from the left or the right,¯̄̄x+2x¡3

¯̄̄approaches 0 and r(x) = ln

¯̄̄x+2x¡3

¯̄̄goes to

¡1. So limx!¡2

r(x) does not exist. As x ap-

proaches 3 from the left or the right,¯̄̄x+2x¡3

¯̄̄goes

to 1 and so does r(x) = ln¯̄̄x+2x¡3

¯̄̄. So lim

x!3r(x)

does not exist.

19. f(x) =

8<:1 if x < 2

x+ 3 if 2 · x · 47 if x > 4

(a)

(b) f(x) is discontinuous at x = 2:

(c) limx!2¡

f(x) = 1 limx!2+

f(x) = 5

20. f(x) =

8<:x¡ 1 if x < 1

0 if 1 · x · 4x¡ 2 if x > 4

(a)

(b) f(x) is discontinuous at x = 4:

(c) limx!4¡

f(x) = 0; limx!4+

f(x) = 2

21. g(x) =

8<:11 if x < ¡1x2 + 2 if ¡1 · x · 311 if x > 3

(a)

(b) g(x) is discontinuous at x = ¡1:(c) lim

x!¡1¡g(x) = 11

limx!¡1+

g(x) = (¡1)2 + 2 = 3

22. g(x) =

8<:0 if x < 0

x2 ¡ 5x if 0 · x · 55 if x > 5

(a)

(b) g(x) is discontinuous at x = 5:

(c) limx!5¡

g(x) = 52 ¡ 5(5)= 0

limx!5+

g(x) = 5

23. h(x) =½4x+ 4 if x · 0x2 ¡ 4x+ 4 if x > 0

(a)

(b) There are no points of discontinuity.


24. h(x) =½x2 + x¡ 12 if x · 13¡ x if x > 1

(a)

(b) h(x) is discontinuous at x = 1:

(c) limx!1¡

h(x) = 12 + 1¡ 12= ¡10

limx!1+

h(x) = 3¡ 1= 2

25. Find k so that kx2 = x+ k for x = 2:

k(2)2 = 2 + k

4k = 2 + k

3k = 2

k =2

3

26. Find k so that x3 + k = kx¡ 5 for x = 3:33 + k = 3k ¡ 527 + k = 3k ¡ 5

32 = 2k

16 = k

27.2x2 ¡ x¡ 15

x¡ 3 =(2x+ 5)(x¡ 3)

x¡ 3 = 2x+ 5

Find k so that 2x+ 5 = kx¡ 1 for x = 3:2(3) + 5 = k(3)¡ 16 + 5 = 3k ¡ 111 = 3k ¡ 112 = 3k

4 = k

28.3x2 + 2x¡ 8

x+ 2=(3x¡ 4)(x+ 2)

x+ 2= 3x¡ 4

Find k so that 3x¡ 4 = 3x+ k for x = ¡2:

3(¡2)¡ 4 = 3(¡2) + k¡6¡ 4 = ¡6 + k¡10 = ¡6 + k¡4 = k

31. f(x) =x2 + x+ 2

x3 ¡ 0:9x2 + 4:14x¡ 5:4 =P (x)

Q(x)

(a) Graph

Y1 =P (x)

Q(x)=

x2 + x+ 2

x3 ¡ 0:9x2 + 4:14x¡ 5:4

on a graphing calculator. A good choice for theviewing window is [¡3; 3] by [¡10; 10]:

The graph has a vertical asymptote at x = 1:2;

which indicates that f is discontinuous at x = 1:2:

(b) Graph

Y2 = Q(x) = x3 ¡ :09x2 + 4:14x¡ 5:4

using the same viewing window.

We see that this graph has one x-intercept, 1.2.This indicates that 1.2 is the only real solution ofthe equation Q(x) = 0:

This result veri…es our answer from part (a) be-cause a rational function of the form

f(x) =P (x)

Q(x)

will be discontinuous wherever Q(x) = 0:


32. f(x) =x2 + 3x¡ 2

x3 ¡ 0:9x2 + 4:14x+ 5:4 =P (x)

Q(x)

(a) Graph

Y1 =P (x)

Q(x)=

x2 + 3x¡ 2x3 ¡ 0:9x2 + 4:14x+ 5:4

on a graphing calculator. A good choice for theviewing window is [¡3; 3] by [¡10; 10]:

The graph has a vertical asymptote at x ¼ ¡0:9:It is di¢cult to read this value accurately from thegraph.

(b) Graph

Y2 = Q(x) = x3 ¡ 0:9x2 + 4:14x+ 5:4

using the same viewing window.

We see that this graph has one x-intercept, ¼¡0:926: This indicates that ¡0:926 is the only realsolution of the equation Q(x) = 0:

A rational function of the form

f(x) =P (x)

Q(x)

will be discontinuous wherever Q(x) = 0; so wesee that f is discontinuous at x ¼ ¡0:926:The result in part (b) is consistent with the re-sult in part (a), but the result in part (b) is moreaccurate.

33. g(x) =x+ 4

x2 + 2x¡ 8=

x+ 4

(x¡ 2)(x+ 4)=

1

x¡ 2 , x 6= ¡4

If g(x) is de…ned so that g(¡4) = 1

¡4¡ 2 = ¡1

6,

then the function becomes continuous at ¡4. Itcannot be made continuous at 2. The correct an-swer is (a).

34. (a) limx!6 P (x)

As x approaches 6 from the left or the right, thevalue of P (x) for the corresponding point on thegraph approaches 500.

Thus, limx!6 P (x) = $500:

(b) limx!10¡

P (x) = $1500

because, as x approaches 10 from the left, P (x)approaches $1500.

(c) limx!10+

P (x) = $1000 because, as x approaches

10 from the right, P (x) approaches $1000.

(d) Since limx!10+

P (x) 6= limx!10¡

P (x);

limx!10 P (x) does not exist.

(e) From the graph, the function is discontinuousat x = 10: This may be the result of a change ofshifts.

(f) From the graph, the second shift will be aspro…table as the …rst shift when 15 units are pro-duced.

35. In dollars,

C(x) = 4x if 0 < x · 150C(x) = 3x if 150 < x · 400C(x) = 2:5x if 400 < x:

(a) C(130) = 4(130) = $520

(b) C(150) = 4(150) = $600

(c) C(210) = 3(210) = $630

(d) C(400) = 3(400) = $1200

(e) C(500) = 2:5(500) = $1250

(f) C is discontinuous at x = 150 and x = 400

because those represent points of price change.


36. In dollars,

F (x) =

½1:25x if 0 < x · 1001:00x if x > 100:

(a) F (80) = 1:25(80) = $100

(b) F (150) = 1:00(150) = $150

(c) F (100) = 1:25(100) = $125

(d) F is discontinuous at x = 100.

37. In dollars,

C(t) = 36t if 0 < t · 5C(t) = 36(5) = 180 if t = 6 or t = 7C(t) = 180 + 36(t¡ 7) if 7 < t · 12:The average cost per day is

A(t) =C(t)

t:

(a) A(4) =36(4)

4= $36

(b) A(5) =36(5)

5= $36

(c) A(6) =180

6= $30

(d) A(7) =180

7¼ $25:71

(e) A(8) =180 + 36(8¡ 7)

8

=216

8= $27

(f) limt!5¡

A(t) = 36 because as t approaches 5 from

the left, A(t) approaches 36 (think of the graph fort = 1; 2; :::; 5):

(g) limt!5+

A(t) = 30 because as t approaches 5

from the right, A(t) approaches 30. 1, 2, 3, 4, 7,8, 9, 10

(h) A is discontinuous at t = 11; and becausethe average cost will di¤er for each di¤erent rentallength.

38. (a) limx!3¡

C(x) = $0:84 + $0:96 + $0:95 = $2:75

(b) limx!3+

C(x) = $0:84+$0:96+2($0:95) = $3:70

(c) limx!3 C(x) does not exist since lim

x!3¡C(x) =

$2:75 and limx!3+

C(x) = $3:70:

(d) C(3) = $0:84 + $0:96 + $0:95 = $2:75

(e)The function is discontinuous at all whole num-bers where an additional daily charge kicks in,that is 1, 2, 3, 4, 5, 6, and 7.

(f)

39. (a) Since t = 0 weeks the woman weighs 120 lbs.and at t = 40 weeks she weighs 147 lbs., graphthe line beginning at coordinate (0; 120) and end-ing at (40; 147), with closed circles at these points.Since immediately after giving birth, she loses 14lbs. and continues to lose 13 more lbs. over thefollowing 20 weeks, graph the line between thepoints (40; 133) and (60; 120) with an open circleat (40; 133) and a closed circle at (60; 120):

(b) From the graph, we see that

limt!40¡

w(t) = 147 6= 133= lim

t!40+w(t);

where w(t) is the weight in pounds t weeks afterconception. Therefore, w is discontinuous at t =40:

40. W (t) =½48 + 3:64t+0:6363t2+0:00963t3; 1 · t · 28;¡1,004 + 65:8t; 28 < t · 56

(a) W (25) = 48 + 3:64(25) + 0:6363(25)2

+ 0:00963(25)3

¼ 687:156A male broiler at 25 days weighs about 687 grams.

Section 3.3 Rates of Change 193

(b) W (t) is not a continuous function.At t = 28

limt!28¡

W (t)

= limt!28¡

48 + 3:64t+ 0:6363t2 + 0:00963t3

= 48 + 3:64(28) + 0:6363(28)2 + 0:00963(28)3

¼ 860:18and

limt!28¡

W (t) 6= limt!28+

(¡1004 + 65:8t)= ¡1004 + 65:8(28)= 838:4

so

limt!28¡

W (t) 6= limt!28+

W (t)

Thus W (t) is discontinuous.

(c)

3.3 Rates of Change

1. y = x2 + 2x = f(x) between x = 1 and x = 3

Average rate of change

=f(3)¡ f(1)3¡ 1

=15¡ 32

= 6

2. y = ¡4x2 ¡ 6 = f(x) between x = 2 and x = 6Average rate of change

=f(6)¡ f(2)6¡ 2

=(¡150)¡ (¡22)

6¡ 2

=¡150 + 22

4

=¡1284

= ¡32

3. y = ¡3x3 + 2x2 ¡ 4x+ 1 = f(x)between x = ¡2 and x = 1

Average rate of change =f(1)¡ f(¡2)1¡ (¡2)

=(¡4)¡ (41)1¡ (¡2)

=¡453

= ¡15

4. y = 2x3 ¡ 4x2 + 6x = f(x) between x = ¡1and x = 4


=f(4)¡ f(¡1)4¡ (¡1)

=88¡ (¡12)

5=100

5

= 20

5. y =px = f(x) between x = 1 and x = 4


=f(4)¡ f(1)4¡ 1

=2¡ 13

=1

3

6. y =p3x¡ 2 = f(x) between x = 1 and x = 2

Average rate of change =f(2)¡ f(1)2¡ 1

=2¡ 12¡ 1

=1

1= 1

7. y = ex = f(x) between x = ¡2 and x = 0


=f(0)¡ f(¡2)0¡ (¡2)

=1¡ e¡22

¼ 0:4323


8. y = lnx = f(x) between x = 2 and x = 4

Average rate of change =f(4)¡ f(2)4¡ 2

=ln 4¡ ln 2

2

¼ 0:3466

9. limh!0

s(6 + h)¡ s(6)h

= limh!0

(6+h)2+5(6+h)+2¡[62+5(6)+ 2]h

= limh!0

h2 + 17h+ 68¡ 68h

= limh!0

h2 + 17h

h

= limh!0

h(h+ 17)

h= limh!0

(h+ 17) = 17

The instantaneous velocity at t = 6 is 17.

10. s(t) = t2 + 5t+ 2

limh!0

s(1 + h)¡ s(1)h

= limh!0

[(1 + h)2 + 5(1 + h) + 2]¡[(1)2 + 5(1) + 2]h

= limh!0

[1 + 2h+ h2 + 5+ 5h+ 2]¡[1 + 5 + 2]h

= limh!0

8 + 7h+ h2 ¡ 8h

= limh!0

7h+ h2

h

= limh!0

h(7 + h)

h

= limh!0

(7 + h) = 7


11. s(t) = 5t2 ¡ 2t¡ 7

limh!0

s(2 + h)¡ s(2)h

= limh!0

[5(2 + h)2 ¡ 2(2 + h)¡ 7]¡[5(2)2 ¡ 2(2)¡7]h

= limh!0

[20 + 20h+ 5h2 ¡ 4¡ 2h¡ 7]¡[20¡4¡7]h

= limh!0

9 + 18h+ 5h2 ¡ 9h

= limh!0

18h+ 5h2

h

= limh!0

h(18 + 5h)

h

= limh!0

(18 + 5h) = 18


12. s(t) = 5t2 ¡ 2t¡ 7

limh!0

s(3 + h)¡ s(3)h

= limh!0

[5(3 + h)2 ¡ 2(3 + h)¡ 7]¡ [5(3)2 ¡ 2(3)¡ 7]h

= limh!0

[45 + 30h+ 5h2 ¡ 6¡ 2h¡ 7]¡ [45¡ 6¡ 7]h

= limh!0

32 + 28h+ 5h2 ¡ 32h

= limh!0

28h+ 5h2

h

= limh!0

h(28 + 5h)

h

= limh!0

(28 + 5h) = 28


13. s(t) = t3 + 2t+ 9

limh!0

s(1 + h)¡ s(1)h

= limh!0

[(1+h)3+2(1+h)+9]¡[(1)3+2(1)+9]h

= limh!0

[1+3h+3h2+h3+2+2h+9]¡[1+2+9]h

= limh!0

h3 + 3h2 + 5h+ 12¡ 12h

= limh!0

h3 + 3h2 + 5h

h= limh!0

h(h2 + 3h+ 5)

h

= limh!0

(h2 + 3h+ 5) = 5


14. s(t) = t3 + 2t+ 9

limh!0

s(4 + h)¡ s(4)h

= limh!0

(4+h)3+2(4+h)+9¡[43+2(4)+9]h

= limh!0

43+3(42)h+3(4h2)+h3+8+2h+9¡(43+8+9)h

= limh!0

h3 + 12h2 + 48h+ 2h

h

= limh!0

h3 + 12h2 + 50h

h

= limh!0

h(h2 + 12h+ 50)

h

= limh!0

(h2 + 12h+ 50) = 50



15. f(x) = x2 + 2x at x = 0

limh!0

f(0 + h)¡ f(0)h

= limh!0

(0 + h)2 + 2(0 + h)¡ [02 + 2(0)]h

= limh!0

h2 + 2h

h= limh!0

h(h+ 2)

h

= limh!0

h+ 2 = 2

The instantaneous rate of change at x = 0 is 2.

16. s(t) = ¡4t2 ¡ 6 at t = 2

limh!0

s(2 + h)¡ s(2)h

= limh!0

¡4(2 + h)2 ¡ 6¡ [¡4(2)2 ¡ 6]h

= limh!0

¡4(4 + 4h+ h2)¡ 6 + 16 + 6h

= limh!0

¡16h¡ 4h2h

= limh!0

h(¡16¡ 4h)h

= limh!0

(¡16¡ 4h)= ¡16

The instantaneous rate of change at t = 2 is ¡16:17. g(t) = 1¡ t2 at t = ¡1

limh!0

g(¡1 + h)¡ g(¡1)h

= limh!0

1¡ (¡1 + h)2 ¡ [1¡ (¡1)2]h

= limh!0

1¡ (1¡ 2h+ h2)¡ 1 + 1h

= limh!0

2h¡ h2h

= limh!0

h(2¡ h)h

= limh!0

(2¡ h) = 2The instantaneous rate of change at t = ¡1 is 2.

18. F (x) = x2 + 2 at x = 0

limh!0

F (0 + h)¡ F (0)h

= limh!0

(0 + h)2 + 2¡ [02 + 2]h

= limh!0

h2

h

= limh!0

h

= 0

The instantaneous rate of change at x = 0 is 0.

19. f(x) = xx at x = 2

h

0.01f(2 + 0:01)¡ f(2)

0:01

=2:012:01 ¡ 22

0:01

= 6:84

0.001f(2 + 0:001)¡ f(2)

0:001

=2:0012:001 ¡ 22

0:001

= 6:779

0.0001f(2 + 0:0001)¡ f(2)

0:0001

=2:00012:0001 ¡ 22

0:0001

= 6:773

0.00001f(2 + 0:00001)¡ f(2)

0:00001

=2:000012:00001 ¡ 22

0:00001

= 6:7727

0.000001f(2 + 0:000001)¡ f(2)

0:000001

=2:0000012:000001 ¡ 22

0:000001

= 6:7726

The instantaneous rate of change at x = 2 is6.7726.


20. f(x) = xx at x = 3

h

0:01f(3 + 0:01)¡ f(3)

0:01

=3:013:01 ¡ 33

0:01

= 57:3072

0:001f(3 + 0:001)¡ f(3)

0:001

=3:0013:001 ¡ 33

0:001

= 56:7265

0:00001f(3 + 0:00001)¡ f(3)

0:00001

=3:000013:00001 ¡ 33

0:00001

= 56:6632

0:000001f(3 + 0:000001)¡ f(3)

0:000001

=3:0000013:000001 ¡ 33

0:000001

= 56:6626

0:0000001f(3 + 0:0000001)¡ f(3)

0:0000001

=3:00000013:0000001 ¡ 33

0:0000001

= 56:6625


21. f(x) = xln x at x = 2

h

0.01f(2 + 0:01)¡ f(2)

0:01

=2:01ln 2:01 ¡ 2ln 2

0:01

= 1:1258

h

0.001f(2 + 0:001)¡ f(2)

0:001

=2:001ln 2:001 ¡ 22ln 2

0:001

= 1:1212

0.0001f(2 + 0:0001)¡ f(2)

0:0001

=2:0001ln 2:0001 ¡ 2ln 2

0:0001

= 1:1207

0.00001f(2 + 0:00001)¡ f(2)

0:00001

=2:00001ln 2:00001 ¡ 2ln 2

0:00001

= 1:1207


22. f(x) = xln x at x = 3

h

0:01f(3 + 0:01)¡ f(3)

0:01

=3:01ln 3:01 ¡ 3ln 3

0:01

= 2:4573

0:001f(3 + 0:001)¡ f(3)

0:001

=3:001ln 3:001 ¡ 3ln 3

0:001

= 2:4495

0:00001f(3 + 0:00001)¡ f(3)

0:00001

=3:00001ln 3:00001 ¡ 3ln 3

0:00001

= 2:4486



24. If the instantaneous rate of change of f(x) withrespect to x is positive when x = 1; the functionwould be increasing.

25. Let B(t) = the amount in the Medicare TrustFund for year t.

(a) B(1994) = 152B(1998) = 125

Average change in fund

=125¡ 1521998¡ 1994

=¡274

= ¡6:75

On average, the amount in the fund decreased ap-proximately $6.75 billion per year from 1994 ¡1998.

(b) B(1998) = 125B(2010) = 300


=300¡ 1252010¡ 1998

=175

12

¼ 14:58

On average, the amount in the fund increases ap-proximately $14.58 billion per year from 1998 ¡2010.

(c) B(1990) = 125B(1998) = 125


=125¡ 1251998¡ 1990

=0

8

= 0

On average, the amount in the fund changes ap-proximately $0 per year from 1990 to 1998.

26. (a) p(1998) = 16:9p(2000) = 22:8

Slope =p(2000)¡ p(1998)2000¡ 1998

=22:8¡ 16:9

2

=5:9

2= 2:95

The rate of change is 2:95 which means that thepercentage of sales consisting of imports increasedan average of 2.95% per year from 1998 to 2000.

(b) p(2002) = 27:5p(2004) = 28:6

Slope =p(2004)¡ p(2002)2004¡ 2002

=28:6¡ 27:5

2

=1:1

2= 0:55

The rate of change is 0:55; which means that thepercentage of sales consisting of imports increasedan average of 0.55% per year from 2002 to 2004.

(c) No

27. (a) From June 2005 to December 2005 is 6 months.During that time, the number of single familyhousing starts went from 1,724,000 to 1,633,000—a drop of 91,000. The average monthly rate ofchange is

¡91,0006

= ¡15,167 starts per month

(b) FromDecember 2005 to June 2006 is 6 months.During that time, the number of single familyhousing starts went from 1,633,000 to 1,486,000—a drop of 147,000. The average monthly rate ofchange is

¡147,0006



(c) From June 2005 to June 2006 is 12 months.During that time, the number of single familyhousing starts went from 1,724,000 to 1,486,000—a drop of 238,000. The average monthly rate ofchange is

¡238,00012


(d)¡15,167 + (¡ 24,500)

2= ¡19,833 (allowing

for rounding)

They are equal. This will not be true for all timeperiods (only for time periods of equal length).

28. (a) s(4) = 10

s(1) = 1


=s(4)¡ s(1)4¡ 1

=10¡ 14¡ 1

=9

3

which is $3000 per year.

(b) s(7) = 10

s(4) = 10


=s(7)¡ s(4)7¡ 4

=10¡ 107¡ 4 =

0

3= 0

which is $0 per year.

(c) s(12) = 1

s(7) = 10


=s(12)¡ s(7)12¡ 7

=1¡ 1012¡ 7 = ¡1:8

which is ¡$1800 per year.(d) Sales increase during the …rst four years, thenstay constant until year 7, then decrease.

(e) Many answers are possible; usually this trendoccurs with high-tech products or products thatmirror trends, such as CDs.

29. P (x) = 2x2 ¡ 5x+ 6(a) P (4) = 18

P (2) = 4

Average rate of change of pro…t

=P (4)¡ P (2)

4¡ 2=18¡ 42

=14

2= 7;

which is $700 per item.

(b) P (3) = 9

P (2) = 4

Average rate of change of pro…t

=P (3)¡ P (2)

3¡ 2=9¡ 41

= 5


(c) limh!0

P (2 + h)¡ P (2)h

= limh!0

2(2 + h)2 ¡ 5(2 + h) + 6¡ 4h

= limh!0

8 + 8h+ 2h2 ¡ 10¡ 5h+ 2h

= limh!0

2h2 + 3h

h

= limh!0

h(2h+ 3)

h= limh!0

(2h+ 3) = 3;


(d) limh!0

P (4 + h)¡ P (4)h

= limh!0

2(4 + h)2 ¡ 5(4 + h) + 6¡ 18h

= limh!0

32 + 16h+ 2h2 ¡ 20¡ 5h¡ 12h

= limh!0

2h2 + 11h

h

= limh!0

h(2h+ 11)

h

= limh!0

2h+ 11 = 11;



30. R = 10x¡ 0:002x2

(a) Average rate of change =R(1001)¡R(1000)

1001¡ 1000 =8005:998¡ 8000

1= 5:998

The average rate of change is $5998.

(b) Marginal revenue = limh!0

R(1000 + h)¡R(1000)h

= limh!0

[10(1000 + h)¡ 0:002(1000 + h)2]¡ [10(1000)¡ 0:002(1000)2]h

= limh!0

[10,000 + 10h¡ 0:002(1,000,000 + 2000h+ h2]¡ 8000h

= limh!0

10,000 + 10h¡ 2000¡ 4h¡ 0:002h2 ¡ 8000h

= limh!0

6h¡ 0:002h2h

= limh!0

h(6¡ 0:002h)h

= limh!0

(6¡ 0:002h) = 6The marginal revenue is $6000. This is approximately the revenue generated by the 1000th unit produced.

(c) Additional revenue = R(1001)¡R(1000) = [10(1001)¡ 0:002(1001)2]¡ [10(1000)¡ 0:002(1000)2]= 8005:998¡ 8000

The additional revenue is $5998.

(d) The answers to parts (a) and (c) are the same and are approximately equal to the answer to part (b).

31. N(p) = 80¡ 5p2; 1 · p · 4(a) Average rate of change of demand is

N(3)¡N(2)3¡ 2 =

35¡ 601

= ¡25 boxes per dollar.(b) Instantaneous rate of change when p is 2 is

limh!0

N(2 + h)¡N(2)h

= limh!0

80¡ 5(2 + h)2 ¡ [80¡ 5(2)2]h

= limh!0

80¡ 20¡ 20h¡ 5h2 ¡ (80¡ 20)h

= limh!0

¡5h2 ¡ 20hh

= ¡20 boxes per dollar. Around the $2 point, a $1 price increase (say, from $1.50 to $2.50) causes a drop indemand of about 20 boxes.

(c) Instantaneous rate of change when p is 3 is

limh!0

80¡ 5(3 + h)2 ¡ [80¡ 5(3)2]h

= limh!0

80¡ 45¡ 30h¡ 5h2 ¡ 80 + 45h

= limh!0

¡30h¡ 5h2h

= ¡30 boxes per dollar.(d) As the price increases, the demand decreases; this is an expected change.


32. p(t) = t2 + t

(a) p(1) = 12 + 1 = 2p(4) = 42 + 4 = 20

Average rate of change =p(4)¡ p(1)4¡ 1 =

20¡ 23

= 6

The average rate of change is 6% per day.

(b) limh!0

p(3 + h)¡ p(3)h

= limh!0

(3 + h)2 + (3 + h)¡ (32 + 3)h

= limh!0

9 + 6h+ h2 + 3+ h¡ 12h

= limh!0

h2 + 7h

h= lim

h!0h(h+ 7)

h= lim

h!0(h+ 7) = 7

The instantaneous rate of change is 7% per day. The number of people newly infected on day 3 is about 7% ofthe total population.

33. Let P (t) = world population estimated in billions for year t.

(a) P (1990) = 5:3

If replacement-level fertility is reached in 2010, P (2050) = 8:6.

Average rate of change =P (2050)¡ P (1990)

2050¡ 1990=8:6¡ 5:360

= 0:055

On average, the population will increase 55 million per year.If replacement-level fertility is reached in 2030, P (2050) = 9:2.


2050¡ 1990=9:2¡ 5:360

= 0:065

On average, the population will increase 65 million per year.If replacement-level fertility is reached in 2050, P (2050) = 9:8.


2050¡ 1990=9:8¡ 5:360

= 0:075

On average, the population will increase 75 million per year.The projection for replacement-level fertility by 2010 predicts the smallest rate of change in world population.

(b) If replacement-level fertility is reached in 2010

P (2090) = 9:3

P (2130) = 9:6


2130¡ 2090=9:6¡ 9:340

= 0:0075


On average, the population will increase 7.5 million per year.If replacement-level fertility is reached in 2030,

P (2090) = 10:3

P (2130) = 10:6


2130¡ 2090=10:6¡ 10:3

40

= 0:0075

On average, the population will increase 7.5 million per year.If replacement-level fertility is reached in 2050,

P (2090) = 11:35

P (2130) = 11:75


2130¡ 2090=11:75¡ 11:35

40

= 0:01

On average, the population will increase 10 million per year.From 2090¡ 2130 the three projections show almost the same rate of change in world population.

34. (a) P (2) = 5; P (1) = 3


2¡ 1 =5¡ 32¡ 1 =

2

1= 2

From 1 min to 2 min, the population of bacteria increases, on the average, 2 million per min.

(b) P (3) = 4:2; P (2) = 5


3¡ 2 =4:2¡ 53¡ 2 =

¡:81= ¡:8

From 2 min to 3 min, the population of bacteria decreases, on the average, ¡:8 million or 800,000 per min.

(c) P (3) = 4:2; P (4) = 2


=P (4)¡ P (3)

4¡ 3 =2¡ 4:24¡ 3 =

¡2:21

= ¡2:2From 3 min to 4 min, the population of bacteria decreases, on the average, ¡2:2 million per min.(d) P (4) = 2; P (5) = 1


5¡ 4 =1¡ 25¡ 4 =

¡11= ¡1

From 4 min to 5 min, the population decreases, on the average, ¡1 million per min.(e) The population increased up to 2 min after the bactericide was introduced, but decreased after 2 min.

(f) The rate of decrease of the population slows down at about 3 min.The graph becomes less and less steep after that point.


35. L(t) = ¡0:01t2 + 0:788t¡ 7:048

(a)L(28)¡ L(22)28¡ 22 =

7:176¡ 5:4486

= 0:288

The average rate of growth during weeks 22 through 28 is 0.288 mm per week.

(b) limh!0

L(t+ h)¡ L(t)h

= limh!0

L(22 + h)¡ L(22)h

= limh!0

[¡0:01(22+h)2+0:788(22+h)¡7:048]¡5:448h

= limh!0

¡0:01(h2+44h+484)+17:336+0:788h¡12:496h

= limh!0

¡0:01h2 + 0:348hh

= limh!0

(¡0:01h+ 0:348)

= 0:348

The instantaneous rate of growth at exactly 22 weeks is 0.348 mm per week.

(c)

36. (a)

(b) The average rate of change during the …rst hour is

F (1)¡ F (0)1¡ 0 ¼ 81:51

kilojoules per hour per hour.

(c) Store F (t) in a function menus of a graphing calculator.Store Y1(1+X)¡Y1(1)

X as Y2 in the function menu, where Y1 represents F (t). Substitute small values for X in Y2perhaps with use of a table feature of the graphing calculator. As X is allowed to get smaller, Y2 approaches18.81 kilojoules per hour per hour.

(d) Through use of a MAX/feature program of a graphing calculator, the maximum point seen in part (a) isestimated to occur at approximately t = 1:3 hours.


37. (a) The average rate of change of M(t) on theinterval [105; 115] is

M(115)¡M(105)115¡ 105 =

0:8

10= 0:08

kilograms per day.

(b) Calculate limh!0

M(105 + h)¡M(105)h

M(105 + h) = 27:5 + 0:3(105 + h)

¡ 0:001(105 + h)2= 27:5 + 31:5 + 0:3h

¡ (11:025 + 0:21h+ 0:001h2)= 47:975 + 0:09h¡ 0:001h2

M(105) = 47:975

So, the instantaneous rate of change of M(t) att = 105 is

limh!0

μ47:975 + 0:09h¡ 0:001h2 ¡ 47:975

h

¶

= limh!0

μ0:09h¡ 0:001h2

h

¶

= limh!0

(0:09¡ 0:001h)= 0:09 kilograms per day.

(c)

38. Let I(t) represent immigration (in thousands) inyear t:

(a)I(2000)¡ I(1996)2000¡ 1996 =

1552¡ 11714

= 95:25

The average rate of change is 95,250 immigrantsper year.

(b)I(2004)¡ I(2000)2004¡ 2000 =

1226¡ 15524

= ¡81:5The average rate of change is ¡81,500 immigrantsper year.

(c)I(2004)¡ I(1996)2004¡ 1996 =

1226¡ 11718

= 6:875

The average rate of change is 6875 immigrants peryear.

(d)95,250 + (¡ 81,500)

2= 6875

They are equal. This will not be true for all timeperiods (only for time periods of equal length).

39. (a) Let D(t) represent percent of kids who haveused drugs by grade 8 in year t:

D(2001)¡D(1998)2001¡ 1998 =

26:8¡ 293

¼ ¡0:73The average rate of change from 1998 to 2001 is¡0:73 percent per year.D(2005)¡D(2002)

2005¡ 2002 =21:4¡ 24:5

3

¼ ¡1:03The average rate of change from 2002 to 2005 is¡1:03 percent per year.(b) Let D(t) represent percent of kids who haveused drugs by grade 10 in year t:

D(2001)¡D(1998)2001¡ 1998 =

45:6¡ 44:93

¼ 0:23The average rate of change from 1998 to 2001 is0:23 percent per year.

D(2005)¡D(2002)2005¡ 2002 =

38:2¡ 44:63

¼ ¡2:13The average rate of change from 2002 to 2005 is¡2:13 percent per year.(c) Let D(t) represent percent of kids who haveused drugs by grade 12 in year t:

D(2001)¡D(1998)2001¡ 1998 =

53:9¡ 54:13

¼ ¡0:07The average rate of change from 1998 to 2001 is¡0:07 percent per year.D(2005)¡D(2002)

2005¡ 2002 =50:4¡ 53

3

¼ ¡0:87The average rate of change from 2002 to 2005 is¡0:87 percent per year.


40. (a)T (3000)¡ T (1000)3000¡ 1000 =

23¡ 152000

=8

2000

=4

1000

From 1000 to 3000 ft, the temperature changesabout 4± per 1000 ft; the temperature rises (onthe average).

(b)T (5000)¡ T (1000)5000¡ 1000 =

22¡ 154000

=7

4000

=1:75

1000

From 1000 to 5000 ft, the temperature changesabout 1.75± per 1000 ft; the temperature rises(on the average).

(c)T (9000)¡ T (3000)9000¡ 3000 =

15¡ 236000

=¡86000

=¡43

1000

From 3000 to 9000 ft, the temperature changesabout ¡4

3± per 1000 ft; the temperature falls (on

the average).

(d) T (9000)¡ T (1000)9000¡ 1000 =

15¡ 158000

= 0

From 1000 to 9000 ft, the temperature changesabout 0± per 1000 ft; the temperature stays con-stant (on the average).

(e) The temperature is highest at 3000 ft and low-est at 1000 ft. If 7000 ft is changed to 10,000 ft,the lowest temperature would be at 10,000 ft.

(f) The temperature at 9000 ft is the same as1000 ft.

41. (a)s(2)¡ s(0)2¡ 0 =

10¡ 02

= 5 ft/sec

(b)s(4)¡ s(2)4¡ 2 =

14¡ 102

= 2 ft/sec

(c)s(6)¡ s(4)6¡ 4 =

20¡ 142

= 3 ft/sec

(d)s(8)¡ s(6)8¡ 6 =

30¡ 202

= 5 ft/sec

(e) (i)f(x0 + h)¡ f(x0 ¡ h)

2h

=f(4 + 2)¡ f(4¡ 2)

(2)(2)

=f(6)¡ f(2)

4

=20¡ 104

=10

4= 2:5 ft/sec

(ii)2 + 3

2= 2:5 ft/sec

(f) (i)f(x0 + h)¡ f(x0 ¡ h)

2h

=f(6 + 2)¡ f(6¡ 2)

(2)(2)

=f(8)¡ f(4)

4

=30¡ 144

=16

4= 4 ft/sec

(ii)3 + 5

2= 4 ft/sec

42. (a) Average rate of change from 0.5 to 1:

f(1)¡ f(0:5)1¡ 0:5 =

55¡ 300:5

= 50 mph

Average rate of change from 1 to 1.5:

f(1:5)¡ f(1)1:5¡ 1 =

80¡ 550:5

= 50 mph

Estimate of instantaneous velocity is

50 + 50

2= 50 mph.

(b) Average rate of change from 1.5 to 2:

f(2)¡ f(1:5)2¡ 1:5 =

104¡ 800:5

= 48 mph

Average rate of change from 2 to 2.5

f(2:5)¡ f(2)2:5¡ 2 =

124¡ 1040:5

= 40 mph

Estimate of instantaneous velocity is

48 + 40

2= 44 mph.

Section 3.4 De…nition of the Derivative 205

43. s(t) = t2 + 5t+ 2

(a) Average velocity

=s(6)¡ s(4)6¡ 4

=68¡ 386¡ 4

=30

2= 15 ft/sec

(b) Average velocity

=s(5)¡ s(4)5¡ 4

=52¡ 385¡ 4

14

1= 14 ft/sec

(c) limh!0

s(4 + h)¡ s(4)h

= limh!0

(4 + h)2 + 5(4 + h) + 2¡ 38h

= limh!0

16 + 8h+ h2 + 20 + 5h+ 2¡ 38h

= limh!0

h2 + 13h

h

= limh!0

h(h+ 13)

h

= limh!0

(h+ 13)

= 13 ft/sec

3.4 De…nition of the Derivative

1. (a) f(x) = 5 is a horizontal line and has slope 0;the derivative is 0.

(b) f(x) = x has slope 1; the derivative is 1.

(c) f(x) = ¡x has slope of ¡1; the derivative is¡1:

(d) x = 3 is vertical and has unde…ned slope; thederivative does not exist.

(e) y =mx+ b has slope m; the derivative is m:

2. (a)

The line tangent to g(x) = 3px at x = 0 is a

vertical line. Since the slope of a vertical line isunde…ned, g0(0) does not exist.

3. f(x) = x2¡1x+2 is not di¤erentiable when x+ 2 = 0

or x = ¡2 because the function is unde…ned anda vertical asymptote occurs there.

4. If the rate of change of f(x) is zero when x = a,the tangent line at that point must have a slope ofzero. Thus, the tangent line is horizontal at thatpoint.

5. Using the points (5; 3) and (6; 5); we have

m =5¡ 36¡ 5 =

2

1

= 2:

6. Using the points (2; 2) and (¡2; 6); we have

m =6¡ 2¡2¡ 2 =

4

¡4 = ¡1:

7. Using the points (¡2; 2) and (2; 3); we have

m =3¡ 2

2¡ (¡2)=1

4:

8. Using the points (3;¡1) and (¡2; 3); we have

m =3¡ (¡1)¡2¡ 3 =

4

¡5 = ¡4

5:

9. Using the points (¡3;¡3) and (0;¡3); we have

m =¡3¡ (¡3)0¡ 3 =

0

¡3= 0:

10. The line tangent to the curve at (4; 2) is a verticalline. A vertical line has unde…ned slope.


11. f(x) = ¡4x2 + 9x+ 2Step 1 f(x+ h)

= ¡4(x+ h)2 + 9(x+ h) + 2= ¡4(x2 + 2xh+ h2) + 9x+ 9h+ 2= ¡4x2 ¡ 8xh¡ 4h2 + 9x+ 9h+ 2

Step 2 f(x+ h)¡ f(x)= ¡4x2 ¡ 8xh¡ 4h2 + 9x+ 9h+ 2¡ (¡4x2 + 9x+ 2)

= ¡8xh¡ 4h2 + 9h= h(¡8x¡ 4h+ 9)

Step 3f(x+ h)¡ f(x)

h

=h(¡8x¡ 4h+ 9)

h

= ¡8x¡ 4h+ 9

Step 4 f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

(¡8x¡ 4h+ 9)

= ¡8x+ 9f 0(¡2) = ¡8(¡2) + 9 = 25f 0(0) = ¡8(0) + 9 = 9f 0(3) = ¡8(3) + 9 = ¡15

12. f(x) = 6x2 ¡ 5x¡ 1Step 1 f(x+ h)

= 6(x+ h)2 ¡ 5(x+ h)¡ 1= 6(x2 + 2xh+ h2)¡ 5x¡ 5h¡ 1= 6x2 + 12xh+ 6h2 ¡ 5x¡ 5h¡ 1

Step 2 f(x+ h)¡ f(x)= 6x2 + 12xh+ 6h2 ¡ 5x¡ 5h¡ 1¡ 6x2 + 5x+ 1= 6h2 + 12xh¡ 5h= h(6h+ 12x¡ 5)

Step 3f(x+ h)¡ f(x)

h=h(6h+ 12x¡ 5)

h

= 6h+ 12x¡ 5

Step 4 f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

(6h+ 12x¡ 5)= 12x¡ 5

f 0(¡2) = 12(¡2)¡ 5 = ¡29f 0(0) = 12(0)¡ 5 = ¡5f 0(3) = 12(3)¡ 5 = 31

13. f(x) =12

x

f(x+ h) =12

x+ h

f(x+ h)¡ f(x) = 12

x+ h¡ 12x

=12x¡ 12(x+ h)

x(x+ h)

=12x¡ 12x¡ 12h

x(x+ h)

=¡12hx(x+ h)

f(x+ h)¡ f(x)h

=¡12h

hx(x+ h)

=¡12

x(x+ h)

=¡12

x2 + xh

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

¡12x2 + xh

=¡12x2

f 0(¡2) = ¡12(¡2)2 =

¡124

= ¡3

f 0(0) = ¡1202 which is unde…ned so f 0(0) does not

exist.

f 0(3) =¡1232

=¡129

= ¡43

14. f(x) =3

x

f(x+ h) =3

x+ h

f(x+ h)¡ f(x) = 3

x+ h¡ 3

x

=3x¡ 3(x+ h)x(x+ h)

=¡3h

x(x+ h)

f(x+ h)¡ f(x)h

=¡3h

hx(x+ h)

=¡3

x(x+ h)


f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

¡3x(x+ h)

=¡3x2

f 0(¡2) = ¡34

f 0(0) = ¡30which is unde…ned, so f 0(0) does not

exist.

f 0(3) = ¡ 332= ¡1

3

15. f(x) =px

Steps 1-3 are combined.

f(x+ h)¡ f(x)h

=

px+ h¡px

h

=

px+ h¡px

h¢px+ h+

pxp

x+ h+px

=x+ h¡ x

h(px+ h+

px)

=1p

x+ h+px

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

1px+ h+

px

=1

2px

f 0(¡2) = 1

2p¡2 which is unde…ned so f

0(¡2)does not exist.

f 0(0) =1

2p0=1

0which is unde…ned so f 0(0) does

not exist.

f 0(3) =1

2p3

16. f(x) = ¡3pxSteps 1-3 are combined.

f(x+ h)¡ f(x)h

=¡3px+ h+ 3px

h

Rationalize the numerator.

=¡3px+ h+ 3px

h¢ ¡3

px+ h¡ 3px

¡3px+ h¡ 3px

=9(x+ h)¡ 9x

h(¡3px+ h¡ 3px)

=9x+ 9h¡ 9x

h(¡3px+ h¡ 3px)

=9

¡3px+ h¡ 3px

=3

¡px+ h¡px

f 0(x) = limh!0

3

¡px+ h¡px=

3

¡px¡px =3

¡2px

f 0(¡2) = 3

¡2p¡2 which is unde…ned so f0(¡2)

does not exist.

f 0(0) =3

¡2p0 =3

0which is unde…ned so

f 0(0) does not exist.

f 0(3) =3

¡2p3 = ¡3

2p3

17. f(x) = 2x3 + 5


f(x+ h)¡ f(x)h

=2(x+ h)3 + 5¡ (2x3 + 5)

h

=2(x3 + 3x2h+ 3xh2 + h3) + 5¡ 2x3 ¡ 5

h

=2x3 + 6x2h+ 6xh2 + 2h3 + 5¡ 2x3 ¡ 5

h

=6x2h+ 6xh2 + 2h3

h

=h(6x2 + 6xh+ 2h2)

h

= 6x2 + 6xh+ 2h2


f 0(x) = limh!0

(6x2 + 6xh+ 2h2)

= 6x2

f 0(¡2) = 6(¡2)2 = 24f 0(0) = 6(0)2 = 0f 0(3) = 6(3)2 = 54

18. f(x) = 4x3 ¡ 3


f(x+ h)¡ f(x)h

=4(x+ h)3 ¡ 3¡ (4x3 ¡ 3)

h

=4(x3 + 3x2h+ 3xh2 + h3)¡ 3¡ 4x3 + 3

h

=4x3 + 12x2h+ 12xh2 + 4h3 ¡ 3¡ 4x3 + 3

h

=12x2h+ 12xh2 + 4h3

h

=h(12x2 + 12xh+ 4h2)

h

= 12x2 + 12xh+ 4h2

f 0(x) = limh!0

(12x2 + 12xh+ 4h2)

= 12x2

f 0(¡2) = 12(¡2)2 = 48f 0(0) = 12(0)2 = 0f 0(3) = 12(3)2 = 108

19. (a) f(x) = x2 + 2x;x = 3; x = 5

Slope of secant line =f(5)¡ f(3)5¡ 3

=(5)2 + 2(5)¡ [(3)2 + 2(3)]

2

=35¡ 152

= 10

Now use m = 10 and (3; f(3)) = (3; 15) in thepoint-slope form.

y ¡ 15 = 10(x¡ 3)y ¡ 15 = 10x¡ 30

y = 10x¡ 30 + 15y = 10x¡ 15

(b) f(x) = x2 + 2x; x = 3

f(x+ h)¡ f(x)h

=[(x+ h)2 + 2(x+ h)]¡ (x2 + 2x)

h

=(x2 + 2hx+ h2 + 2x+ 2h)¡ (x2 + 2x)

h

=2hx+ h2 + 2h

h= 2x+ h+ 2

f 0(x) = limh!0

(2x+ h+ 2) = 2x+ 2

f 0(3) = 2(3) + 2 = 8 is the slope of the tangentline at x = 3:Use m = 8 and (3; 15) in the point-slope form.

y ¡ 15 = 8(x¡ 3)y = 8x¡ 9

20. (a) f(x) = 6¡ x2; x = ¡1; x = 3

Slope of secant line =f(3)¡ f(¡1)3¡ (¡1)

=6¡ (3)2 ¡ [6¡ (¡1)2]

4

=¡3¡ 54

= ¡2Now use m = ¡2 and (¡1; f(¡1)) = (¡1; 5) inthe point-slope form.

y ¡ 5 = ¡2[x¡ (¡1)]y ¡ 5 = ¡2x¡ 2

y = ¡2x¡ 2 + 5y = ¡2x+ 3

(b) f(x) = 6¡ x2; x = ¡1f(x+ h)¡ f(x)

h

=[6¡ (x+ h)2]¡ [6¡ (x)2]

h

=[6¡ (x2 + 2xh+ h2)]¡ [6¡ x2]

h

=6¡ x2 ¡ 2xh¡ h2 ¡ 6 + x2

h

=¡2xh¡ h2

h=h(¡2x¡ h)

h= ¡2x¡ h

f 0(x) = limh!0

(¡2x¡ h) = ¡2x


f 0(¡1) = ¡2(¡1) = 2 is the slope of the tangentline at x = ¡1: Use m = 2 and (¡1; 5) in thepoint-slope form.

y ¡ 5 = 2(x+ 1)y ¡ 5 = 2x+ 2

y = 2x+ 7

21. (a) f(x) = 5x ;x = 2; x = 5


=55 ¡ 5

2

3

=1¡ 5

2

3

= ¡12

Now use m = ¡12 and (5; f(5)) = (5; 1) in the

point-slope form.

y ¡ 1 = ¡12[x¡ 5]

y ¡ 1 = ¡12x+

5

2

y = ¡12x+

5

2+ 1

y = ¡12x+

7

2

(b) f(x) =5

x; x = 2

f(x+ h)¡ f(x)h

=5

x+h ¡ 5x

h

=

5x¡5(x+h)(x+h)x

h

=5x¡ 5x¡ 5hh(x+ h)(x)

=¡5h

h(x+ h)x

=¡5

(x+ h)x

f 0(x) = limh!0

¡5(x+ h)(x)

= ¡ 5

x2

f 0(2) = ¡522 = ¡5

4 is the slope of the tangent lineat x = 2:

Now use m = ¡54 and

¡2; 52

¢in the point-slope

form.

y ¡ 52= ¡5

4(x¡ 2)

y ¡ 52= ¡5

4x+

10

4

y = ¡54x+ 5

5x+ 4y = 20

22. (a) f(x) = ¡ 3x+1 ;x = 1; x = 5


=¡ 3(5+1) ¡

h¡ 3(1+1)

i4

=¡12 +

32

4

=1

4

Now use m = 14 and (1; f(1)) = (1;¡3

2) in thepoint-slope form.

y ¡μ¡32

¶=1

4(x¡ 1)

y +3

2=1

4x¡ 1

4

y =1

4x¡ 1

4¡ 32

y =1

4x¡ 7

4

(b) f(x) =¡3x+ 1

; x = 1

f(x+ h)¡ f(x)h

=

¡3(x+ h) + 1

¡ ¡3x+ 1

h

=

¡3(x+ 1) + 3(x+ h+ 1)(x+ 1)(x+ h+ 1)

h

=

¡3x¡ 3 + 3x+ 3h+ 3(x+ 1)(x+ h+ 1)

h

=3h

h(x+ 1)(x+ h+ 1)

f 0(x) = limh!0

3

(x+ 1)(x+ h+ 1)

=3

(x+ 1)2


f 0(1) = 3(1+1)2 =

34 is the slope of the tangent line

at x = 1: Use m = 34 and

¡1;¡3

2

¢in the point-

slope form.

y ¡μ¡32

¶=3

4(x¡ 1)

y +3

2=3

4x¡ 3

4

y =3

4x¡ 9

4

23. (a) f(x) = 4px; x = 9; x = 16


=4p16¡ 4p97

=16¡ 127

=4

7

Now use m = 47 and (9; f(9)) = (9; 12) in the

point-slope form.

y ¡ 12 = 4

7(x¡ 9)

y ¡ 12 = 4

7x¡ 36

7

y =4

7x¡ 36

7+ 12

y =4

7x+

48

7

(b) f(x) = 4px; x = 9

f(x+ h)¡ f(x)h

=4px+ h¡ 4px

h¢ 4px+ h+ 4

px

4px+ h+ 4

px

=16(x+ h)¡ 16xh(4px+ h+ 4

px)

f 0(x) = limh!0

16(x+ h)¡ 16xh(4px+ h+ 4

px)

= limh!0

16h

h(4px+ h+ 4

px)

= limh!0

4

(px+ h+

px)=

4

2px

=2px

f 0(9) = 2p9= 2

3 is the slope of the tangent line atx = 9:

Use m = 23 and (9; 12) in the point-slope form.

y ¡ 12 = 2

3(x¡ 9)

y =2

3x+ 6

3y = 2x+ 18

24. (a) f(x) =px; x = 25; x = 36


=

p36¡p2511

=6¡ 511

=1

11

Now use m = 111 and (25; f(25)) = (25; 5) in the

point-slope form.

y ¡ 5 = 1

11(x¡ 25)

y ¡ 5 = 1

11x¡ 25

11

y =1

11x¡ 25

11+ 5

y =1

11x+

30

11

(b) f(x) =px; x = 25

f(x+ h)¡ f(x)h

=

px+ h¡px

h

=

px+ h¡px

h¢px+ h+

pxp

x+ h+px

=x+ h¡ x

h(px+ h+

px)

=h

h(px+ h+

px)=

1px+ h+

px

f 0(x) =limh!0

1px+ h+

px=

1

2px

f 0(25) =1

2p25=

1

2 ¢ 5 =1

10

Use m = 110 and (25; 5) in the point-slope form.

y ¡ 5 = 1

10(x¡ 25)

y ¡ 5 = 1

10x¡ 25

10

y =1

10x+

5

2


25. f(x) = ¡4x2 + 11xf(x+ h)¡ f(x)

h

=¡4(x+ h)2 + 11(x+ h)¡ (¡4x2 + 11x)

h

=¡8xh¡ 4h2 + 11h

h

f 0(x) = limh!0

(¡8x¡ 4h+ 11)= ¡8x+ 11

f 0(2) = ¡8(2) + 11 = ¡5f 0(16) = ¡8(16) + 11 = ¡117f 0(¡3) = ¡8(¡3) + 11 = 35

26. f(x) = 6x2 ¡ 4xf(x+ h)¡ f(x)

h

=6(x+ h)2 ¡ 4(x+ h)¡ (6x2 ¡ 4x)

h

=12xh+ 6h2 ¡ 4h

h

= 12x+ 6h¡ 4

f 0(x) = limh!0

(12x+ 6h¡ 4)

= 12x¡ 4f 0(2) = 12(2)¡ 4 = 24¡ 4 = 20f 0(16) = 12(16)¡ 4 = 192¡ 4 = 188f 0(¡3) = 12(¡3)¡ 4 = ¡36¡ 4 = ¡40

27. f(x) = ex

f(x+ h)¡ f(x)h

=ex+h ¡ ex

h

f 0(x) = limh!0

ex+h ¡ exh

f 0(2) ¼ 7:3891; f 0(16) ¼ 8; 886; 111; f 0(¡3) ¼ 0:0498

28. f(x) = ln jxjf(x+ h)¡ f(x)

h

=ln jx+ hj ¡ ln jxj

h

f 0(x) = limh!0

ln jx+ hj ¡ ln jxjh

f 0(2) = 0:5f 0(16) = 0:0625f 0(¡3) = ¡0:3

29. f(x) = ¡ 2x

f(x+ h)¡ f(x)h

=

¡2x+ h

¡μ¡2x

¶

h

=

¡2x+ 2(x+ h)(x+ h)x

h

=2h

h(x+ h)x

=2

(x+ h)x

f 0(x) = limh!0

2

(x+ h)x

=2

x2

f 0(2) =2

22=1

2

f 0(16) =2

162

=2

2561

128:

f 0(¡3) = 2

(¡3)2

=2

9

30. f(x) =6

x

f(x+ h)¡ f(x)h

=6

x+h ¡ 6x

h

=6x¡ 6(x+ h)hx(x+ h)

=¡6

x(x+ h)

f 0(x) = limh!0

¡6x(x+ h)

=¡6x2

f 0(2) =¡6(2)2

=¡64= ¡3

2

f 0(16) =¡6162

=¡6256

= ¡ 3

128

f 0(¡3) = ¡6(¡3)2 =

¡69= ¡2

3


31. f(x) =px

f(x+ h)¡ f(x)h

=

px+ h¡px

h¢px+ h+

pxp

x+ h+px

=(x+ h)¡ x

h(px+ h+

px)

=h

h(px+ h+

px)

=1p

x+ h+px

f 0(x) = limh!0

1px+ h+

px=

1

2px

f 0(2) =1

2p2

f 0(16) =1

2p16=1

8

f 0(¡3) = 1

2p¡3 is not a real number, so

f 0(¡3) does not exist.

32. f(x) = ¡3pxf(x+ h)¡ f(x)

h

=¡3px+ h+ 3px

h

=¡3px+ h+ 3px

h¢ ¡3

px+ h¡ 3px

¡3px+ h¡ 3px

=9(x+ h)¡ 9x

h(¡3px+ h¡ 3px)=

9

¡3px+ h¡ 3px=

3

¡px+ h¡px

f 0(x) = limh!0

3

¡px+ h¡px=

3

¡px¡px = ¡3

2px

f 0(2) = ¡ 3

2p2

f 0(16) =¡32p16= ¡3

8

f 0(¡3) = ¡32p¡3 ; which is not a real number, so

f 0(¡3) does not exist.

33. At x = 0; the graph of f(x) has a sharp point.Therefore, there is no derivative for x = 0:

34. No derivative exists at x = ¡6 because the func-tion is not de…ned at x = ¡6:

35. For x = ¡3 and x = 0; the tangent to the graph off(x) is vertical. For x = ¡1; there is a gap in thegraph f(x): For x = 2; the function f(x) does notexist. For x = 3 and x = 5; the graph f(x) hassharp points. Therefore, no derivative exists forx = ¡3; x = ¡1; x = 0; x = 2; x = 3; and x = 5:

36. For x = ¡5 and x = 0, the function f(x) is notde…ned. For x = ¡3 and x = 2 the graph of f(x)has sharp points. For x = 4; the tangent to thegraph is vertical. Therefore, no derivative existsfor x = ¡5; x = ¡3; x = 0; x = 2; or x = 4:

37. (a) The rate of change of f(x) is positive whenf(x) is increasing, that is, on (a; 0) and (b; c):

(b) The rate of change of f(x) is negative whenf(x) is decreasing, that is, on (0; b):

(c) The rate of change is zero when the tangentto the graph is horizontal, that is, at x = 0 andx = b:

38. The zeros of graph (b) correspond to the turningpoints of graph (a), the points where the derivativeis zero. Graph (a) gives the distance, while graph(b) gives the velocity.

39. The zeros of graph (b) correspond to the turningpoints of graph (a), the points where the derivativeis zero. Graph (a) gives the distance, while graph(b) gives the velocity.


40. f(x) = xx; a = 2

(a) h

0:01f(2 + 0:01)¡ f(2)

0:01

=2:012:01 ¡ 22

0:01

= 6:84

0:001

f(2 + 0:001)¡ f(2)0:001

=2:0012:001 ¡ 22

0:001

= 6:779

0:0001

f(2 + 0:0001)¡ f(2)0:0001

=2:00012:0001 ¡ 22

0:0001

= 6:773

0:00001

f(2 + :00001)¡ f(2)0:00001

=2:000012:00001 ¡ 22

0:00001

= 6:7727

0:000001

f(2 + 0:000001)¡ f(2)0:000001

=2:0000012:000001 ¡ 22

0:000001

= 6:7726

It appears that f 0(2) = 6:7726:

(b) Graph the function on a graphing calculatorand move the cursor to an x-value near x = 2: Agood choice for the initial viewing window is [0; 3]by [0; 10]:Now zoom in on the function several times. Eachtime you zoom in, the graph will look less likea curve and more like a straight line. When thegraph appears to be a straight line, use the TRACEfeature to select two points on the graph, andrecord their coordinates. Use these two points tocompute the slope. The result will be very close tothe most accurate value found in part (a), whichis 6.7726.

41. f(x) = xx; a = 3

(a) h

0:01f(3 + 0:01)¡ f(3)

0:01

=3:013:01 ¡ 33

0:01

= 57:3072

0:001f(3 + 0:001)¡ f(3)

0:001

=3:0013:001 ¡ 33

0:001

= 56:7265

0:00001f(3 + :00001)¡ f(3)

0:00001

=3:000013:00001 ¡ 33

0:00001

= 56:6632

0:000001f(3 + 0:000001)¡ f(3)

0:000001

=3:0000013:000001 ¡ 33

0:000001

= 56:6626

0:0000001f(3 + 0:0000001)¡ f(3)

0:0000001

=3:00000013:0000001 ¡ 33

0:0000001

= 56:6625


(b) Graph the function on a graphing calculatorand move the cursor to an x-value near x = 3: Agood choice for the initial viewing window is [0; 4]by [0; 60] with Xscl = 1, Yscl = 10.

Now zoom in on the function several times. Eachtime you zoom in, the graph will look less likea curve and more like a straight line. Use theTRACE feature to select two points on the graph,and record their coordinates. Use these two pointsto compute the slope. The result will be close tothe most accurate value found in part (a), whichis 56.6625.

Note: In this exercise, the method used in part(a) gives more accurate results than the methodused in part (b).


42. f(x) = x1=x; a = 2

(a) h

0.01f(2 + :01)¡ f(2)

0:01

=2:011=2:01 ¡ 21=2

0:01

= 0:1071

0.001

f(2 + 0:001)¡ f(2)0:001

=2:0011=2:001 ¡ 21=2

0:001

= 0:1084

0.0001

f(2 + 0:0001)¡ f(2)0:0001

=2:00011=2:0001 ¡ 21=2

0:0001

= 0:1085

0.00001

f(2 + 0:00001)¡ f(2)0:00001

=2:000011=2:00001 ¡ 21=2

0:00001

= 0:1085

0.000001

f(2 + 0:000001)¡ f(2)0:000001

=2:0000011=2:000001 ¡ 22

0:000001

= 0:1085


(b) Graph this function on a graphing calculatorand move the cursor to an x-value near x = 2: Agood choice for the initial viewing window is [0; 5]by [0; 3]:Follow the procedure outlined in the solution forExercise 38, part (b). The …nal result will be closeto the value found in part (a) of this exercise,which is 0.1085.

43. f(x) = x1=x; a = 3

(a) h

0:01f(3 + 0:01)¡ f(3)

0:01

=3:011=3:01 ¡ 31=3

0:01

= ¡0:0160

0.001f(3 + 0:001)¡ f(3)

0:001

=3:0011=3:001 ¡ 31=3

0:001

= ¡0:0158

0.0001f(3 + 0:0001)¡ f(3)

0:0001

=3:00011=3:0001 ¡ 31=3

0:0001

= ¡0:0158

It appears that f 0(3) = ¡0:0158:

(b) Graph the function on a graphing calculatorand move the cursor to an x-value near x = 3: Agood choice for the initial viewing window is [0; 5]by [0; 3]:

Follow the procedure outlined in the solution forExercise 39, part (b). Note that near x = 3; thegraph is very close to a horizontal line, so we ex-pect that it slope will be close to 0. The …nalresult will be close to the value found in part (a)of this exercise, which is ¡0:0158:

44. For Column A, let h = 0:01f(x) = ln jxj

Graph y =ln jx+ 0:01j ¡ ln jxj

0:01.

f(x) = ex


Graph y =ex+0:01 ¡ ex

0:01.

f(x) = x3

Graph y =(x+ 0:01)3 ¡ x3

0:01.

Column BGraph y = ex

Graph y = 3x2

Graph y =1

x

We observe that the graph of

y =ln jx+ 0:01j ¡ ln jxj

0:01

is very similar to the graph of

y =1

x,

the graph of

y =ex+0:01 ¡ ex

0:01


y = ex, and

the graph of

y =(x+ 0:01)3 ¡ x3

0:01


y = 3x2.

Thus the derivative of lnx is 1x , the derivative of

ex is ex, and the derivative of x3 is 3x2.

46. (a) f(x) = ¡4x2 + 11xf(x+ h) = ¡4(x+ h)2 + 11(x+ h)

= ¡4(x2 + 2xh+ h2) + 11(x+ h)= ¡4x2 ¡ 8xh¡ 4h2 + 11x+ 11h

f(x+ h)¡ f(x)= ¡4x2 ¡ 8xh¡ 4h2 + 11x+ 11h+ 4x2 ¡ 11x= ¡8xh¡ 4h2 + 11h

f(x+ h)¡ f(x)h

=¡8xh¡ 4h2 + 11h

h

= ¡8x+ 4h+ 11

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

(¡8x+ 4h+ 11)= ¡8x+ 11

f 0(3) = ¡8(3) + 11 = ¡13


f(3 + 0:1)¡ f(3)0:1

=f(3:1)¡ f(3)

0:1

=¡4(3:1)2 + 11(3:1)¡ (¡4(3)2 + 11(3))

0:1

=¡4(9:61) + 11(3:1) + 4(9)¡ 11(3)

0:1

=¡38:44 + 34:1 + 36¡ 33

0:1

=1:34

0:1= ¡13:4

f(3 + 0:1)¡ f(3¡ 0:1)2(0:1)

=f(3:1)¡ f(2:9)

0:2

=¡4(3:1)2 + 11(3:1)¡ (¡4(2:9)2 + 11(2:9))

0:2

=¡4(9:61) + 11(3:1) + 4(8:41)¡ 11(2:9)

0:2

=¡38:44 + 34:1 + 33:64¡ 31:9

0:2=¡2:60:2

= ¡13(b) f 0(3) = ¡8(3) + 11 = ¡13f(3 + 0:01)¡ f(3)

0:01

=f(3:01)¡ f(3)

0:01

=¡4(3:01)2 + 11(3:01)¡ (¡4(3)2 + 11(3))

0:01

=¡4(9:0601) + 11(3:01) + 4(9)¡ 11(3)

0:01

=¡36:2404 + 33:11 + 36¡ 33

0:01

=¡0:13040:01

= ¡13:04f(3 + 0:01)¡ f(3¡ 0:01)

2(0:01)

=f(3:01)¡ f(2:99)

0:02

=¡4(3:01)2 + 11(3:01)¡ (¡4(2:99)2 + 11(2:99))

0:02

=¡4(9:0601) + 11(3:01) + 4(8:9401)¡ 11(2:99)

0:02

=¡36:2404 + 33:11 + 35:7604¡ 32:89

0:02

=¡0:260:02

= ¡13

(c) f(x) =¡2x

f(x+ h) =¡2x+ h

f(x+ h)¡ f(x) = ¡2x+ h

¡μ¡2x

¶

=¡2x+ h

+2

x

=¡2x+ 2(x+ h)x(x+ h)

=2h

x(x+ h)

f(x+ h)¡ f(x)h

=

2hx(x+h)

h

=2h

x(x+ h)¢ 1h

=2

x(x+ h)

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

2

x(x+ h)=2

x2

f 0(3) =2

32=2

9¼ 0:222222

f(3 + 0:1)¡ f(3)0:1

=f(3:1)¡ f(3)

0:1

=¡23:1 ¡ ¡2

3

0:1

¼ :215054f(3 + 0:1)¡ f(3¡ 0:1)

2(0:1)=f(3:1)¡ f(2:9)

0:2

=¡23:1 ¡ ¡2

2:9

0:2

¼ 0:222469

(d) f 0(3) =2

32=2

9¼ 0:222222

f(3 + :01)¡ f(3)0:01

=f(3:01)¡ f(3)

0:01

=¡23:01 ¡ ¡2

3

0:01

¼ 0:221484f(3 + 0:01)¡ f(3¡ 0:01)

2(0:01)=f(3:01)¡ f(2:99)

0:02

=¡23:01 ¡ ¡2

2:99

0:02

¼ 0:222225


(e) f(x) =px

f(x+ h)¡ f(x)h

=

px+ h¡px

h¢px+ h+

pxp

x+ h+px

=(x+ h)¡ x

h(px+ h+

px)

=h

h(px+ h+

px)

=1

h(px+ h+

px)

f 0(x) = limh!0

1

h(px+ h+

px)=

1

2px

f 0(3) =1

2p3¼ 0:288675

f(3 + 0:1)¡ f(3)0:1

=f(3:1)¡ f(3)

0:1

=

p3:1¡p30:1

¼ 0:286309f(3 + 0:1)¡ f(3¡ 0:1)

2(0:1)=f(3:1)¡ f(2:9)

0:2

=

p3:1¡p2:90:2

¼ 0:288715

(f) f 0(3) =1

2p3¼ 0:288675

f(3 + 0:01)¡ f(3)0:01

=f(3:01)¡ f(3)

0:01

=

p3:01¡p30:01

¼ 0:288435f(3 + 0:01)¡ f(3¡ 0:01)

2(0:01)=f(3:01)¡ f(2:99)

0:02

=

p3:01¡p2:990:02

¼ 0:288676

47. D(p) = ¡2p2 ¡ 4p+ 300D is demand; p is price.

(a)Given thatD0(p) = ¡4p¡4; the rate of changeof demand with respect to price is ¡4p ¡ 4; thederivative of the function D(p):

(b) D0(10) = ¡4(10)¡ 4= ¡44

The demand is decreasing at the rate of about 44items for each increase in price of $1.

48. P (x) = 1000 + 32x¡ 2x2

(a) $8000 is 8 thousands, so x = 8:

P 0(8) = 32¡ 4(8) = 32¡ 32 = 0

No, the …rm should not increase production, sincethe marginal pro…t is 0.

(b) $6000, x = 6

P 0(6) = 32¡ 4(6) = 32¡ 24 = 8

Yes, the …rst should increase production, since themarginal pro…t is positive.

(c) $12,000, x = 12

P 0(12) = 32¡ 4(12)= 32¡ 48 = ¡16

No, because the marginal pro…t is negative.

(d) $20,000, x = 20

P 0(20) = 32¡ 4(20)= 32¡ 80 = ¡48

No, because the marginal pro…t is negative.

49. R(x) = 20x¡ x2

500

(a) R0(x) = 20¡ 1

250x

At y = 1000;

R0(1000) = 20¡ 1

250(1000)

= $16 per table.

(b) The marginal revenue for the 1001st table isapproximately R0(1000): From (a), this is about$16.

(c) The actual revenue is

R(1001)¡R(1000) = 20(1001)¡ 10012

500

¡·20(1000)¡ 1000

2

500

¸

= 18; 015:998¡ 18; 000= $15:998 or $16.


(d) The marginal revenue gives a good approxi-mation of the actual revenue from the sale of the1001st table.

50. C(x) = ¡0:00375x2 + 1:5x+ 1000; 0 · x · 180(a) The marginal cost is given by

C0(x) = ¡0:0075x+ 1:5; 0 · x · 180(b) C0(100) = ¡0:0075(100) + 1:5 = 0:75This represents the fact that the cost of producingthe next (101st) taco is approximately $0.75.(c) The exact cost to produce the 101st taco isgiven by

C(101)¡C(100)= [¡0:00375(101)2 + 1:5(101) + 1000]¡ [¡0:00375(100)2 + 1:5(100) + 1000]= (¡38:25375 + 151:5 + 1000)¡ (37:5 + 150 + 1000)= 0:74625

That is, the exact cost of producing the 101st tacois $0.74625.

(d) The exact cost of producing the 101st tacois $0.00375 less than the approximate cost. Theyare very close.

(e) C(x) = ax2 + bx+ c

C 0(x) = 2ax+ b

j[C(x+ 1)¡C(x)]¡C 0(x)j=¯̄[a(x+ 1)2 + b(x+ 1) + c]

¡[ax2 + bx+ c]¡ (2ax+ b)¯̄=¯̄ax2 + 2ax+ a+ bx+ b+ c

¡ax2 ¡ bx¡ c¡ 2ax¡ b¯̄= jaj

(f) C(x) = ax2 + bx+ c

C0(x) = 2ax+ bC(x+ 1)¡C(x)= [a(x+ 1)2 + b(x+ 1) + c]

¡ [ax2 + bx+ c]= ax2 + 2ax+ a+ bx+ b+ c¡ ax2 ¡ bx¡ c= 2ax+ a+ b

C 0μx+

1

2

¶= 2a

μx+

1

2

¶+ b

= 2ax+ a+ b

Thus, C(x+ 1)¡C(x) = C 0μx+

1

2

¶:

51. (a) f(x) = ¡0:0142x4 + 0:6698x3 ¡ 6:113x2+ 84:05x+ 203:9

f(10) = 961

f(20) = 2526

f(30) = 3806

(b) Y1 = ¡0:0142x4 + 0:6698x3 ¡ 6:113x2+ 84:05x+ 203:9

nDeriv(Y1; x; 10) ¼ 106nDeriv(Y1; x; 20) ¼ 189nDeriv(Y1; x; 30) ¼ ¡8nDeriv(Y1; x; 35) ¼ ¡318

52. (a) From the graph, Vmp is just about at the turn-ing point of the curve. Thus, the slope of thetangent line is approximately zero. The power ex-penditure is not changing.

(b) From the graph, the slope of the tangent lineat Vmr is approximately 0.1. The power expendedis increasing 0.1 unit per unit increase in speed.

(c) The slope of the tangent line at Vopt is a bitgreater than that at Vmr; about 0.12. The powerexpended increases 0.12 units for each unit in-crease in speed.

(d) The power level …rst decreases to Vmp; thenincreases at greater rates.

(e) Vmr is the point which produces the smallestslope of a line.

53. The derivative at (2; 4000) can be approximatedby the slope of the line through (0; 2000) and (2; 4000).The derivative is approximately

4000¡ 20002¡ 0 =

2000

2= 1000.

Thus the shell…sh population is increasing at arate of 1000 shell…sh per unit time.The derivative at about (10; 10,300) can be ap-proximated by the slope of the line through (10; 10,300)and (13; 12,000). The derivative is approximately

12,000¡ 10,30013¡ 10 =

1700

3¼ 570.

The shell…sh population is increasing at a rateof about 583 shell…sh per unit time. The deriv-ative at about (13; 11,250) can be approximatedby the slope of the line through (13; 11,250) and(16; 12,000). The derivative is approximately

12,000¡ 11,25016¡ 13 =

750

3= 250.


The shell…sh population is increasing at a rate of250 shell…sh per unit time.

54. I(t) = 27 + 72t¡ 1:5t2

(a)

I 0(t) = limh!0

I(t+ h)¡ I(t)h

= limh!0

27+72t+72h¡1:5t2¡3th¡1:5h2¡27¡72t+1:5t2h

= limh!0

72h¡ 3th¡ 1:5h2h

= limh!0

72¡ 3t¡ 1:5h= 72¡ 3t

I 0(5) = 72¡ 3(5)= 72¡ 15= 57

The rate of change of the intake of food 5 minutesinto a meal is 57 grams per minute.

(b) I 0(24) ?= 0

72¡ 3(24) ?= 0

0 = 0

24 minutes after the meal starts the rate of foodconsumption is 0.

(c) After 24 minutes the rate of food consumptionis negative according to the function where a rateof zero is more accurate. A logical range for thisfunction is

0 · t · 24:

55. (a) Set M(v) = 150 and solve for v:

0:0312443v2 ¡ 101:39v + 82,264 = 1500:0312443v2 ¡ 101:39v + 82,114 = 0

Solve using the quadratic formula.Let D equal the discriminant.

D = b2 ¡ 4ac= (¡101:39)2 ¡ 4(0:0312443)(82,114)¼ 17:55

v =101:39§pD2(0:0312443)

v ¼ 1690 meter per second orv ¼ 1560 meters per second.Since the functions is de…ned only for v ¸ 1620;the only solution is 1690 meters per second.

(b) Calculate limh!0

M(1700 + h)¡M(1700)h

M(1700+ h)= 0:0312443(1700 + h)2 ¡ 101:39(1700 + h)+ 82,264= 90,296.027 + 106.23062h+ 0:0312443h2

¡ 172,363¡ 101:39h+ 82,264= 0:01312443h2 + 4:84062h+ 197:027

M(1700) = 197:027, so the derivative of M(v) atv = 1700 is

limh!0

μ0:0312443h2+4:84062h+197:027¡197:027

h

¶

= limh!0

μ0:0312443h2+4:84062h

h

¶

= limh!0

(0:0312443h+ 4:84062)

= 4:84062

¼ 4:84 days per meter per secondThe increase in velocity for this cheese from1700 m/s to 1701 m/s indicates that the approxi-mate age of the cheese has increased by 4.84 days.

56. The slope of the tangent line to the graph at the…rst point is found by …nding two points on thetangent line.

(x1; y1) = (1000; 13:5)

(x2; y2) = (0; 18:5)

m =18:5¡ 13:50¡ 1000 =

5

¡1000 = ¡0:005

At the second point, we have

(x1; y1) = (1000; 13:5)

(x2; y2) = (2000; 21:5):

m =21:5¡ 13:52000¡ 1000

=8

1000

= 0:008

At the third point, we have

(x1; y1) = (5000; 20)

(x2; y2) = (3000; 22:5):

m =22:5¡ 203000¡ 5000

=2:5

¡2000= ¡0:00125


At 500 ft, the temperature decreases 0.005± perfoot. At about 1500 ft, the temperature increases0.008± per foot. At 5000 ft, the temperature de-creases 0.00125± per foot.

57. (a) The derivative does not exist at the two “cor-ners” or “sharp points.” The x-values of thesepoints are 0.75 and 3.

(b) To …nd T 0(0:5); calculate the slope of the linesegment with positive slope, since this portion ofthe graph includes the value x = 0:5 (marked as12 hr on the graph.) To …nd this slope, use thepoints (0; 100) (the starting point) and (0:75; 875)(the beginning point of the cleaning cycle).

m = T 0(0:5) =875¡ 1000:75¡ 0

=775

0:75

¼ 1033

The oven temperature is increasing at 1033± perhour.

(c) To …nd T 0(2); …nd the slope of the line seg-ment containing the value x = 2: This segment ishorizontal, so

m = T 0(2) = 0:

The oven temperature is not changing.

(d) To …nd T 0(3:5); calculate the slope of the linesegment with negative slope, since this portion ofthe graph includes the value x = 3:5: To …nd thisslope, use the points (3; 875) (the end of the clean-ing cycle) and (3:75; 100) (the stopping point).

m = T 0(3:5) =100¡ 8753:75¡ 3

=¡7750:75

¼ ¡1033

The oven temperature is decreasing at 1033± perhour.

58. The slope of the graph at x = 24 can be estimatedusing the points (24; 360) and (33; 395).

slope =395¡ 36033¡ 24 =

35

9= 389

Thus, the derivative for a 24 ounce bat is about 4ft per oz which means that the distance the balltravels is increasing 4 feet per ounce.The slope of the graph at x = 51 can be estimatedusing the points (60; 380) and (51; 390).

slope =390¡ 38051¡ 60 =

10

¡9 = ¡10

9¼ ¡1:1

Thus, the derivative for a 51 ounce bat is about¡1:1 ft per oz which means that the distance theball travels is decreasing 1.1 feet per ounce.

59. The velocities are equal at approximately t = 0:13:The acceleration for the hands is approximately0 mph per sec. The acceleration for the bat isapproximately 640 mph per sec.

3.5 Graphical Di¤erentiation3. Since the x-intercepts of the graph of f 0 occurwhenever the graph of f has a horizontal tangentline, Y1 is the derivative of Y2. Notice that Y1 has2 x-intercepts; each occurs at an x-value wherethe tangent line to Y2 is horizontal.Note also that Y1 is positive whenever Y2 is in-creasing, and that Y1 is negative whenever Y2 isdecreasing.

4. Since the x-intercepts of the graph f 0 occur when-ever the graph of f has a horizontal tangent line,Y2 is the derivative of Y1. Notice that Y2 has 2x-intercepts; each occurs at an x-value where thetangent line to Y1 is horizontal.Note also that Y2 is negative whenever Y1 is de-creasing, and Y2 is positive whenever Y1 is increas-ing.

5. Since the x-intercepts of the graph of f 0 occurwhenever the graph of f has a horizontal tangentline, Y2 is the derivative of Y1. Notice that Y2 has1 x-intercept which occurs at the x-value wherethe tangent line to Y1 is horizontal. Also noticethat the range on which Y1 is increasing, Y2 is pos-itive and the range on which it is decreasing, Y2is negative.

Section 3.5 Graphical Di¤erentiation 221

6. Since the x-intercepts of the graph f 0 occur when-ever the graph of f has a horizontal tangent line,Y1 is the derivative of Y2. Notice that Y1 has 4x-intercepts; each occurs at an x-value where thetangent line to Y2 is horizontal.Note also that Y1 is negative whenever Y2 is de-creasing and Y1 is positive whenever Y2 is increas-ing.

7. To graph f 0; observe the intervals where the slopesof tangent lines are positive and where they arenegative to determine where the derivative is pos-itive and where it is negative. Also, whenever fhas a horizontal tangent, f 0 will be 0, so the graphof f 0 will have an x-intercept. The x-values of thethree turning point on the graph of f become thethree x-intercepts of the graph of f:

Estimate the magnitude of the slope at severalpoints by drawing tangents to the graph of f:

8. To graph f 0, observe the intervals where the slopesof lines are positive and where they are negativeto determine where the derivative is positive andwhere it is negative. Also, whenever f has a hor-izontal tangent, f 0 will be 0.Estimate the magnitude of the slope at severalpoints by drawing tangents to the graph of f:

9. On the interval (¡1;¡2); the graph of f is a hori-zontal line, so its slope is 0. Thus, on this interval,the graph of f 0 is y = 0 on (¡1;¡2): On the in-terval (¡2; 0); the graph of f is a straight line, soits slope is constant. To …nd this slope, use thepoints (¡2; 2) and (0; 0):

m =2¡ 0¡2¡ 0 =

2

¡2 = ¡1

On the interval (0; 1); the slope is also constant.To …nd this slope, use the points (0; 0) and (1; 1):

m =1¡ 01¡ 0 = 1

On the interval (1;1); the graph is again a hor-izontal line, so m = 0: The graph of f 0 will bemade up of portions of the y-axis and the linesy = ¡1 and y = 1:Because the graph of f has “sharp points” or “cor-ners” at x = ¡2; x = 0; and x = 1; we know thatf 0(¡2); f 0(0), and f 0(1) do not exist. We showthis on the graph of f 0 by using open circles atthe endpoints of the portions of the graph.

10. On the interval (¡1;¡3); the graph of f is astraight line, so its slope is constant. To …nd thisslope, use the points (¡6;¡3) and (¡3; 0):

m =0¡ (¡3)¡3¡ (¡6) =

3

3= 1

On the interval (3;1); the slope of f is also con-stant. To …nd this slope, use the points (3; 0) and(6;¡3):

m =¡3¡ 06¡ 3 =

¡33= ¡1

Thus, we have f 0(x) = 1 on (¡1;¡3) and f 0(x) =¡1 on (3;1): Because the graph of f has sharppoints at x = ¡3 and x = 3; we know that f 0(¡3)and f 0(3) do not exist. We show this on the graphof f 0(x) by using open circles.We also observe that the slopes of tangent linesare negative on (¡3; 0); that the graph has a hori-zontal tangent at x = 0, and that the slopes oftangent lines are positive on (0; 3): Thus, f 0 isnegative on (¡1; 3); 0 at x = 0; and positive on(3;1): Furthermore, by drawing tangents, we seethat on (¡3; 3); the value of f 0 increases from ¡1to 1.


11. On the interval (¡1;¡2); the graph of f is astraight line, so its slope is constant. To …nd thisslope, use the points (¡4; 2) and (¡2; 0):

m =0¡ 2

¡2¡ (¡4) =¡22= ¡1

On the interval (2;1); the slope of f is also con-stant. To …nd this slope, use the points (2; 0) and(3; 2):

m =2¡ 03¡ 2 =

2

1= 2

Thus, we have f 0(x) = ¡1 on (¡1;¡2) andf 0(x) = 2 on (2;1):Because f is discontinuous at x = ¡2 and x = 2;we know that f 0(¡2) and f 0(2) do not exist, whichwe indicate with open circles at (¡2;¡1) and (2; 2)on the graph of f 0:

On the interval (¡2; 2); all tangent lines have pos-itive slopes, so the graph of f 0 will be above they-axis. Notice that the slope of f (and thus they-value of f 0) decreases on (¡2; 0) and increaseson (0; 2); with a minimum value on this intervalof 1 at x = 0:

12. On the interval (¡1; 0); the graph of f is a hori-zontal line, so its slope is 0. Thus, the graph of f 0

is y = 0 (the x-axis) on (¡1; 0):Since f is discontinuous at x = 0; f 0(0) does notexist. Thus, the graph of f 0 has an open circle atx = 0:

On the interval (0;1); the slopes are positive butdecreasing at a slower rate as x gets larger. There-fore, the value of f 0 will be positive but decreas-ing on this interval. This value approaches 0, butnever becomes 0.

13. We observe that the slopes of tangent lines arepositive on the interval (¡1; 0) and negative onthe interval (0; 1); so the value of f 0 will be pos-itive on (¡1, 0) and negative on (0,1): Since fis unde…ned at x = 0; f 0(0) does not exist.

Notice that the graph of f becomes very ‡at whenjxj!1: The value of f approaches 0 and also theslope approaches 0. Thus, y = 0 (the x-axis) is ahorizontal asymptote for both the graph of f andthe graph of f 0:

As x! 0¡ and x! 0+; the graph of f gets verysteep, so jf 0(x)j!1: Thus, x = 0 (the y-axis) isa vertical asymptote for both the graph of f andthe graph of f 0:

14. The graph of f is a step function. (This is thegreatest integer function, f(x) = [[x]]:) The graphis made up of an in…nite series of horizontal linesegments. Thus, the derivative will be 0 every-where it is de…ned. However, since f is discontin-uous wherever x is an integer, f 0(x) does not existat any integer.The graph of f 0 is an in…nite set of line segmentson the x-axis (where y = f 0(x) = 0) separated byopen circles wherever x is an integer.

Section 3.5 Graphical Di¤erentiation 223

15. The slope of f(x) is unde…ned at x = ¡2;¡1; 0; 1;and 2; and the graph approaches vertical (unboundedslope) as x approaches those values. Accordingly,the graph of f 0(x) has vertical asymptotes atx = ¡2;¡1; 0; 1; and 2. f(x) has turning points(zero slope) at x = ¡1:5;¡0:5; 0:5, and 1.5, so thegraph of f 0(x) crosses the x-axis at those values.Elsewhere, the graph of f 0(x) is negative wheref(x) is decreasing and positive where f(x) is in-creasing.

16. The slope of f(x) is unde…ned at x = ¡1 and 1.The graph approaches vertical (unbounded slope)as x approaches ¡1 from the left and 1 from theright. Accordingly, the graph of f 0(x) has verticalasymptotes for x = ¡1 approached from the leftand for x = 1 approached from the right. f(x)has a turning point (zero slope) near x = 1:7; sothe graph of f 0(x) crosses the x-axis near x = 1:7:Elsewhere, the graph of f 0(x) is negative wheref(x) is decreasing and positive where f(x) is in-creasing. Between x = ¡1 and x = 1, the slopeof f(x) is 0.5; therefore, f 0(x) = 0:5 between ¡1and 1.

17. The graph rises steadily, with varying degrees ofsteepness. The graph is steepest around 1976 andnearly ‡at around 1950 and 1980. Accordingly,the rate of change is always positive, with a maxi-mum value around 1976 and values near zero around1950 and 1980.

18. (a) The curve slants downward up to about age4.25, where it turns and begins to rise. There is aslight decline in steepness between ages 10 and 18.Correspondingly, the graph of the rate of changelies below the horizontal axis to the left of 4.25years and above the horizontal axis to the rightof that point. The graph of the rate of change isdeclining between ages 10 and 18.

(b) The curve slants downward up to about age5.75, where it turns and begins to rise. There is aslight decline in steepness between ages 13 and 20.Correspondingly, the graph of the rate of changelies below the horizontal axis to the left of 5.75years and above the horizontal axis to the rightof that point. The graph of the rate of change isdeclining between ages 13 and 20.

19. The growth rate of the function y = f(x) is givenby the derivative of this function y0 = f(x): Weuse the graph of f to sketch the graph of f 0:First, notice as x increase, y increases throughoutthe domain of f; but at a slower and slower rate.The slope of f is positive but always decreasing,and approaches 0 as t gets large. Thus, y0 will al-ways be positive and decreasing. It will approachbut never reach 0.

To plot point on the graph of f 0; we need to es-timate the slope of f at several points. From thegraph of f; we obtain the values given in the fol-lowing table.

t y0

2 100010 700

13 250


Use these points to sketch the graph.

20. Let P (V ) represent the power corresponding to agiven value of the tern’s speed, V:

The rate of change of power as a function of timeis given by the derivative of this function, P 0(V ):We use the graph of P to sketch the graph of P 0:

First, we observe that the graph of P has one turn-ing point, at V = Vmp ¼ 5:5: At this value of V;the graph has a horizontal tangent, so the graphof V 0 has an x-intercept at this value of V:

Since the slopes of tangent lines are negative onthe interval (0; Vmp) and positive when V > Vmp;the value of V 0 is negative on (0; Vmp) and positivewhen V > Vmp:

Use the tangents drawn on the graph and addi-tional tangent as needed to estimate the slope atseveral points on the graph of V to improve theaccuracy of the graph of V 0:

21.

About 9 cm; about 2.6 cm less per year

22.

Chapter 3 Review Exercises4. The derivative can be used to …nd the instanta-neous rate of change at a point on a function, andthe slope of a tangent line at a point on a function.

5. (a) limx!¡3¡

= 4

(b) limx!¡3+

= 4

(c) limx!¡3= 4 (since parts (a) and (b) have the

same answer)

(d) f(¡3) = 4; since (¡3; 4) is a point of thegraph.

6. (a) limx!¡1¡

g(x) = ¡2

(b) limx!¡1+

g(x) = 2

(c) limx!¡1 g(x) does not exist since parts (a) and

(b) have di¤erent answers.

(d) g(¡1) = ¡2; since (¡1;¡2) is a point on thegraph.

7. (a) limx!4¡

f(x) =1

(b) limx!4+

f(x) = ¡1

(c) limx!4

f(x) does not exist since the answers to

parts (a) and (b) are di¤erent.

(d) f(4) does not exist since the graph has nopoint with an x-value of 4.

8. (a) limx!2¡

h(x) = 1

(b) limx!2+

h(x) = 1

(c) limx!2

h(x) = 1

(d) h(2) does not exists since the graph has nopoint with an x-value of 2.

Chapter 3 Review Exercises 225

9. limx!¡1 g(x) =1 since the y-value gets very large

as the x-value gets very small.

10. limx!1 f(x) = ¡3 since the line y = ¡3 is a hori-zontal asymptote for the graph.

11. limx!6

2x+ 7

x+ 3=2(6) + 7

6 + 3

=19

9

12. Let f(x) =2x+ 5

x+ 3:

x ¡3:1 ¡3:01 ¡3:001f(x) 12 102 1002

x ¡2:9 ¡2:99 ¡2:999f(x) ¡8 ¡98 ¡998

As x approaches ¡3 from the left, f(x) gets in…-nitely larger. As x approaches ¡3 from the right,f(x) gets in…nitely smaller. Therefore, lim

x!¡32x+5x+3

does not exist.

13. limx!4

x2 ¡ 16x¡ 4 = lim

x!4(x¡ 4)(x+ 4)

x¡ 4= limx!4

(x+ 4)

= 4 + 4

= 8

14. limx!2

x2 + 3x¡ 10x¡ 2 = lim

x!2(x+ 5)(x¡ 2)

x¡ 2= limx!2 (x+ 5) = 2 + 5 = 7

15. limx!¡4

2x2 + 3x¡ 20x+ 4

= limx!¡4

(2x¡ 5)(x+ 4)x+ 4

= limx!¡4 (2x¡ 5)

= 2(¡4)¡ 5= ¡13

16. limx!3

3x2 ¡ 2x¡ 21x¡ 3

= limx!3

(3x+ 7)(x¡ 3)x¡ 3

= limx!3 (3x+ 7) = 9 + 7 = 16

17. limx!9

px¡ 3x¡ 9 = lim

x!9

px¡ 3x¡ 9 ¢

px+ 3px+ 3

= limx!9

x¡ 9(x¡ 9)(px+ 3)

= limx!9

1px+ 3

=1p9 + 3

=1

6

18. limx!16

px¡ 4x¡ 16

= limx!16

px¡ 4x¡ 16 ¢

px+ 4px+ 4

= limx!16

x¡ 16(x¡ 16)(px+ 4)

= limx!16

1px+ 4

=1p16 + 4

=1

4 + 4=1

8

19. limx!1

2x2 + 5

5x2 ¡ 1 = limx!1

2x2

x2 +5x2

5x2

x2 ¡ 1x2

= limx!1

2 + 5x2

5¡ 1x2

=2+ 0

5¡ 0=2

5

20. limx!1

x2 + 6x+ 8

x3 + 2x+ 1= limx!1

x2

x3 +6xx3 +

8x3

x3

x3 +2xx3 +

1x3

= limx!1

1x +

6x2 +

8x3

1 + 2x2 +

1x3

=0+ 0 + 0

1 + 0 + 0= 0

21. limx!¡1

μ3

8+3

x¡ 6

x2

¶

= limx!¡1

3

8+ lim

x!¡13

x¡ lim

x!¡16

x2

=3

8+ 0¡ 0

=3

8


22. limx!¡1

μ9

x4+10

x2¡ 6¶

= limx!¡1

9

x4+ lim

x!¡110

x2¡ lim

x!¡1 6

= 0 + 0¡ 6 = ¡623. As shown on the graph, f(x) is discontinuous at

x2 and x4:

24. As shown on the graph, f(x) is discontinuous atx1 and x4:

25. f(x) is discontinuous at x = 0 and x = ¡13 since

that is where the denominator of f(x) equals 0.f(0) and f

¡¡13

¢do not exist.

limx!0 f(x) does not exist since lim

x!0+f(x) = ¡1;

but limx!0¡

f(x) = 1: limx!¡13

f(x) does not exist

since limx!¡13

¡= ¡1; but lim

x!¡13+f(x) =1:

26. f(x) =7¡ 3x

(1¡ x)(3 + x)The function is discontinuous at x = ¡3 and x = 1because those values make the denominator of thefraction equal to zero.limx!¡3 f(x) does not exist since lim

x!¡3¡f(x) = ¡1

and limx!¡3+

f(x) =1:limx!1 f(x) does not exist since lim

x!1¡f(x) =1 and

limx!1+

f(x) = ¡1:f(¡3) and f(1) do not exist since there is no pointof the graph that has an x-value of ¡3 or 1.

27. f(x) is discontinuous at x = ¡5 since that is wherethe denominator of f(x) equals 0.f(¡5) does not exist.limx!¡5 f(x) does not exist since lim

x!¡5¡f(x) =1;

but limx!¡5+

f(x) = ¡1:

28. f(x) =x2 ¡ 9x+ 3

The function is discontinuous at x = ¡3 since thisvalue makes the denominator of the fraction equalto zero.

limx!¡3

x2 ¡ 9x+ 3

= limx!¡3

(x+ 3)(x¡ 3)x+ 3

= limx!¡3 (x¡ 3)

= ¡3¡ 3 = ¡6f(¡3) does not exist since there is no point on thegraph with an x-value of ¡3:

29. f(x) = x2+3x¡ 4 is continuous everywhere sincef is a polynomial function.

30. f(x) = 2x2 ¡ 5x ¡ 3 has no points of disconti-nuity since it is a polynomial function, which iscontinuous everywhere.

31. (a)

(b) The graph is discontinuous at x = 1:

(c) limx!1¡

f(x) = 0; limx!1+

f(x) = 2

32. f(x) =

8<:2 if x < 0

¡x2 + x+ 2 if 0 · x · 21 if x > 2

(a)

(b) The graph is discontinuous at x = 2:

(c) limx!2¡

f(x) = ¡4 + 2 + 2 = 0limx!2+

f(x) = 1

33. f(x) =x4 + 2x3 + 2x2 ¡ 10x+ 5

x2 ¡ 1(a) Find the values of f(x) when x is close to 1.

x y

1:1 2:6005

1:01 2:06

1:001 2:006

1:0001 2:0006

0:99 1:94

0:999 1:994

0:9999 1:9994

It appears that limx!1 f(x) = 2:


(b) Graph

y =x4 + 2x3 + 2x2 ¡ 10x+ 5

x2 ¡ 1on a graphing calculator. One suitable choice forthe viewing window is [¡2; 6] by [¡10; 10]:Because x2 ¡ 1 = 0 when x = ¡1 or x = 1; thisfunction is discontinuous at these two x-values.The graph shows a vertical asymptote at x = ¡1but not at x = 1: The graph should have an opencircle to show a “hole” in the graph at x = 1: Thegraphing calculator doesn’t show the hole, but try-ing to …nd the value of the function of x = 1 willshow that this value is unde…ned.

By viewing the function near x = 1 and using theZOOM feature, we see that as x gets close to 1from the left or the right, y gets close to 2, sug-gesting that

limx!1

x4 + 2x3 + 2x2 ¡ 10x+ 5x2 ¡ 1 = 2:

34. f(x) =x4 + 3x3 + 7x2 + 11x+ 2

x3 + 2x2 ¡ 3x¡ 6(a) Find values of f(x) when x is close to ¡2:

x f(x)

¡2:01 ¡12:62¡2:001 ¡12:96¡2:0001 ¡13¡1:99 ¡13:41¡1:999 ¡13:04¡1:9999 ¡13

It appears that limx!¡2 f(x) = ¡13:

(b) Graph

y =x4 + 3x3 + 7x2 + 11x+ 2

x3 + 2x2 ¡ 3x¡ 6on a graphing calculator. One suitable choice forthe viewing window is [¡5; 5] by [¡10; 10]:By viewing the function near x = ¡2; we see thatas x gets close to ¡2 from the left on the right, ygets close to ¡13; suggesting that

limx!¡2

x4 + 3x3 + 7x2 + 11x+ 2

x3 + 2x2 ¡ 3x¡ 6 = ¡13:

35. y = 6x3 + 2 = f(x); from x = 1 to x = 4

f(4) = 6(4)3 + 2 = 386

f(1) = 6(1)3 + 2 = 8

Average rate of change:

=386¡ 84¡ 1 =

378

3= 126

y0 = 18x

Instantaneous rate of change at x = 1:

f 0(1) = 18(1) = 18

36. y = ¡2x3 ¡ 3x2 + 8 = f(x)f(6) = ¡2(6)3 ¡ 3(6)2 + 8 = ¡532

f(¡2) = ¡2(¡2)3 ¡ 3(¡2)2 + 8 = 12Average rate of change:

=f(6)¡ f(¡2)6¡ (¡2)

=¡532¡ 126 + 2

=¡5448

= ¡68

y0 = ¡6x2 ¡ 6x

Instantaneous rate of change at x = ¡2 :f 0(¡2) = ¡6(¡2)2 ¡ 6(¡2) = ¡6(4) + 12 = ¡12

37. y =¡63x¡ 5 = f(x); from x = 4 to x = 9

f(9) =¡6

3(9)¡ 5 =¡622

= ¡ 3

11

f(4) =¡6

3(4)¡ 5 = ¡6

7


=¡311 ¡

¡¡67

¢9¡ 4

=¡21+6677

5

=45

5(77)=9

77

y0 =(3x¡ 5)(0)¡ (¡6)(3)

(3x¡ 5)2

=18

(3x¡ 5)2Instantaneous rate of change at x = 4:

f 0(4) =18

(3 ¢ 4¡ 5)2 =18

72=18

49


38. y =x+ 4

x¡ 1 = f(x)

f(5) =5 + 4

5¡ 1 =9

4

f(2) =2 + 4

2¡ 1 = 6


=94 ¡ 65¡ 2 =

¡154

3= ¡5

4

y0 =(x¡ 1)(1)¡ (x+ 4)(1)

(x¡ 1)2

=x¡ 1¡ x¡ 4(x¡ 1)2

=¡5

(x¡ 1)2

Instantaneous rate of change at x = 2 :

f 0(2) =¡5

(2¡ 1)2 =¡51= ¡5

39. (a) f(x) = 3x2 ¡ 5x+ 7; x = 2; x = 4

Slope of secant line

=f(4)¡ f(2)4¡ 2

=[3(4)2 ¡ 5(4) + 7]¡ [3(2)2 ¡ 5(2) + 7]

2

=35¡ 92

= 13

Now use m = 13 and (2; f(2)) = (2; 9) in thepoint-slope form.

y ¡ 9 = 13(x¡ 2)y ¡ 9 = 13x¡ 26

y = 13x¡ 26 + 9y = 13x¡ 17

(b) f(x) = 3x2 ¡ 5x+ 7; x = 2f(x+ h)¡ f(x)

h

=[3(x+ h)2 ¡ 5(x+ h) + 7]¡ [3x2 ¡ 5x+ 7]

h

=3x2 + 6xh+ 3h2 ¡ 5x¡ 5h+ 7¡ 3x2 + 5x¡ 7

h

=6xh+ 3h2 ¡ 5h

h

= 6x+ 3h¡ 5

f 0(x) = limh!0

6x+ 3h¡ 5= 6x¡ 5

f 0(2) = 6(2)¡ 5= 7

Now use m = 7 and (2; f(2)) = (2; 9) in the point-slope form.

y ¡ 9 = 7(x¡ 2)y ¡ 9 = 7x¡ 14

y = 7x¡ 14 + 9y = 7x¡ 5

40. (a) f(x) =1

x; x =

1

2; x = 3

Slope of secant line =f(3)¡ f ¡12¢

3¡ 12

=13 ¡ 23¡ 1

2

=2¡ 1218¡ 3

=¡1015

= ¡23

Now use m = ¡23 and

¡12 ; f

¡12

¢¢=¡12 ; 2¢in the

point-slope form.

y ¡ 2 = ¡23

μx¡ 1

2

¶

y ¡ 2 = ¡23x+

1

3

y = ¡23x+

1

3+ 2

y = ¡23x+

7

3


(b) f(x) =1

x; x =

1

2

f(x+ h)¡ f(x)h

=1

x+h ¡ 1x

h

=x¡ (x+ h)xh(x+ h)

= ¡ h

xh(x+ h)

= ¡ 1

x(x+ h)

f 0(x) = limh!0

¡ 1

x(x+ h)

= ¡ 1

x2

f 0μ1

2

¶= ¡ 1¡

12

¢2= ¡4

Now use m = ¡4 and ¡12 ; f ¡12¢¢ = ¡12 ; 2¢in the

point-slope form.

y ¡ 2 = ¡4μx¡ 1

2

¶

y ¡ 2 = ¡4x+ 2y = ¡4x+ 2+ 2y = ¡4x+ 4

41. (a) f(x) =12

x¡ 1 ; x = 3; x = 7


=127¡1 ¡ 12

3¡14

=2¡ 64

= ¡1

Now use m = ¡1 and (3; f(x)) = (3; 6) in thepoint-slope form.

y ¡ 6 = ¡1(x¡ 3)y ¡ 6 = ¡x+ 3

y = ¡x+ 3+ 6y = ¡x+ 9

(b) f(x) =12

x¡ 1 ; x = 3

f(x+ h)¡ f(x)h

=12

x+h¡1 ¡ 12x¡1

h

=12(x¡ 1)¡ 12(x+ h¡ 1)h(x¡ 1)(x+ h¡ 1)

=¡12h

h(x¡ 1)(x+ h¡ 1)= ¡ 12

(x¡ 1)(x+ h¡ 1)

f 0(x) = limh!0

¡ 12

(x¡ 1)(x+ h¡ 1)= ¡ 12

(x¡ 1)2

f 0(3) = ¡ 12

(3¡ 1)2= ¡3

Now use m = ¡3 and (3; f(x)) = (3; 6) in thepoint-slope form.

y ¡ 6 = ¡3(x¡ 3)y ¡ 6 = ¡3x+ 9

y = ¡3x+ 9+ 6y = ¡3x+ 15

42. f(x) = 2px¡ 1; x = 5; x = 10


=2p10¡ 1¡ 2p5¡ 1

5

=2(3)¡ 2(2)

5

=2

5

Now usem = 25 and (5; f(x)) = (5; 4) in the point-

slope form.

y ¡ 4 = 2

5(x¡ 5)

y ¡ 4 = 2

5x¡ 2

y =2

5x¡ 2 + 4

y =2

5x+ 2


(b) f(x) = 2px¡ 1; x = 5

f(x+ h)¡ f(x)h

=2px+ h¡ 1¡ 2px¡ 1

h

=2(px+ h¡ 1¡px¡ 1)(px+ h¡ 1 +px¡ 1)

h(px+ h¡ 1 +px¡ 1)

=2(x+ h¡ 1¡ x+ 1)

h(px+ h¡ 1 +px¡ 1)

=2h

h(px+ h¡ 1 +px¡ 1)

=2p

x+ h¡ 1 +px¡ 1

f 0(x) = limh!0

2px+ h¡ 1 +px¡ 1

=2

2px¡ 1

=1px¡ 1

f 0(5) =1p5¡ 1

=1

2

Now use m = 12 and (5; f(x)) = (5; 4) in the point-slope form.

y ¡ 4 = 1

2(x¡ 5)

y ¡ 4 = 1

2x¡ 5

2

y =1

2x¡ 5

2+ 4

y =1

2x+

3

2

43. y = 4x2 + 3x¡ 2 = f(x)

y0 = limh!0

f(x+ h)¡ f(x)h

= limh!0

[4(x+ h)2 + 3(x+ h)¡ 2]¡ [4x2 + 3x¡ 2]h

= limh!0

4(x2 + 2xh+ h2) + 3x+ 3h¡ 2¡ 4x2 ¡ 3x+ 2h

= limh!0

4x2 + 8xh+ 4h2 + 3x+ 3h¡ 2¡ 4x2 ¡ 3x+ 2h

= limh!0

8xh+ 4h2 + 3h

h

= limh!0

h(8x+ 4h+ 3)

h

= limh!0

(8x+ 4h+ 3)

= 8x+ 3


44. y = 5x2 ¡ 6x+ 7 = f(x)

y0 = limh!0

f(x+ h)¡ f(x)h

= limh!0

[5(x + h)2 ¡ 6(x + h) + 7] ¡ [5x2 ¡ 6x + 7]

h

= limh!0

5(x2 + 2xh + h2) ¡ 6x ¡ 6h + 7 ¡ 5x2 + 6x

h

= limh!0

5x2 + 10xh + 5h2 ¡ 6x ¡ 6h + 7 ¡ 5x2 + 6x + 7

h

= limh!0

10xh+ 5h2 ¡ 6hh

= limh!0

h(10x+ 5h¡ 6)h

= limh!0

(10x+ 5h¡ 6)= 10x¡ 6

45. f(x) = (ln x)x; x0 = 3

(a)

h

0.01f(3 + 0:01)¡ f(3)

0:01

=(ln 3:01)3:01 ¡ (ln 3)3

0:01

= 1:3385

0.001

f(3 + 0:001)¡ f(3)0:001

=(ln 3:001)3:001 ¡ (ln 3)3

0:001

= 1:3323

0.0001

f(3 + 0:0001)¡ f(3)0:0001

=(ln 3:0001)3:0001 ¡ (ln 3)3

0:0001

= 1:3317

0.00001

f(3 + 0:00001)¡ f(3)0:00001

=(ln 3:00001)3:00001 ¡ (ln 3)3

0:00001

= 1:3317

(b) Using a graphing calculator will con…rm this result.


46. f(x) = xln x; x0 = 2

(a) h

0:01f(2 + 0:01)¡ f(2)

0:01

=2:01ln 2:01 ¡ 2ln 2

0:01

= 1:1258

0:001f(2 + 0:001)¡ f(2)

0:001

=2:001ln 2:001 ¡ 2ln 2

0:001

= 1:1212

0:0001f(2 + 0:0001)¡ f(2)

0:0001

=2:0001ln 2:0001 ¡ 2ln 2

0:0001

= 1:1207

0:00001f(2 + 0:00001)¡ f(2)

0:00001

=2:00001ln 2:00001 ¡ 2ln 2

0:00001

= 1:1207

It appears that limx!2 f(x) = 1:1207:

(b) Graph the function on a graphing calculator and move the cursor to an x-value near x = 2: A good choicefor the viewing window is [0; 10] by [0; 10]:

Zoom in on the function until the graph looks like a straight line. Use the TRACE feature to select two pointson the graph, and use these points to compute the slope. The result will be close to the most accurate valuefound in part (a), which is 1.1207.

47. On the interval (¡1; 0); the graph of f is a straight line, so its slope is constant. To …nd this slope, use thepoints (¡2; 2) and (0; 0):

m =0¡ 2

0¡ (¡2) =¡22= ¡1

Thus, the value of f 0 will be ¡1 on this interval.The graph of f has a sharp point at 0, so f 0(0) does not exist. To show this, we use an open circle on the graphof f 0 at (0;¡1):We also observe that the slope of f is positive but decreasing from x = 0 to about x = 1; and then negativefrom there on. As x!1; f(x)!0 and also f 0(x) = 0:


Use this information to complete the graph of f 0:

48. On the intervals (¡1; 0) and (0;1); the slope ofany tangent line will be positive, so the derivativewill be positive. Thus, the graph of f 0 will lieabove the y-axis. The slope of f and thus the valueof f 0 approaches 0 when x!¡1 and x!1 andapproaches some particular but unknown positivevalue > 1 when x!0¡ and x!0+:

Because f is discontinuous at x = 0; we knowthat f 0(0) does not exist, which we indicate withan open circle at x = 0 on the graph of f 0:

49. limx!1

cf(x)¡ dg(x)f(x)¡ g(x) =

limx!1 [cf(x)¡ dg(x)]limx!1 [f(x)¡ g(x)]

=limx!1 [cf(x)]¡ lim

x!1 [dg(x)]

limx!1 [f(x)]¡ lim

x!1 [g(x)

=c limx!1 [f(x)]¡ d limx!1 [g(x)]

limx!1 [f(x)]¡ lim

x!1 [g(x)]

=c ¢ c¡ d ¢ dc¡ d

=(c+ d)(c¡ d)

c¡ d= c+ d

The answer is (e).

50. Notice how the slope of the graph changes withtime.The slope is negative as it approaches 1980, then isslightly positive until about 1985, then it is nega-tive until about 1994, and then it remains positive.Use this information to sketch a graph of the rateof change of earnings for this period.

From the given graph, the earnings are approx-imately $10/hr. The slope of the graph at thatpoint appears to be approximately $10 hr

5 years or $2/hr/yr.

51. R(x) = 5000 + 16x¡ 3x2

(a) R0(x) = 16¡ 6x(b) Since x is in hundreds of dollars, $1000 corre-sponds to x = 10:

R0(10) = 16¡ 6(10)= 16¡ 60 = ¡44

An increase of $100 spent on advertising when ad-vertising expenditures are $1000 will result in therevenue decreasing by $44.

52. C(x) =½1:50x for 0 < x · 1251:35x for x > 125

(a) C(100) = 1:50(100) = $150

(b) C(125) = 1:50(125) = $187:50

(c) C(140) = 1:35(140) = $189

(d)

(e) By reading the graph, C(x) is discontinuousat x = $125:


The average cost per pound is given by

C(x) =C(x)

x:

C(x) =

½1:50 for 0 < x · 1251:35 for x > 125

(f) C(100) = $1:50

(g) C(125) = $1:50

(h) C(140) = $1:35

The marginal cost is given by

C(x) =

½1:50 for 0 < x · 1251:35 for x > 125:

(i) C0(100) = 1:50; the 101st pound will cost$1.50.

(j) C 0(140) = 1:35; the 141st pound will cost$1.35.

53. P (x) = 15x+ 25x2

(a) P (6) = 15(6) + 25(6)2

= 90 + 900 = 990

P (7) = 15(7) + 25(7)2

= 105 + 1225 = 1330


=P (7)¡ P (6)

7¡ 6 =1330¡ 990

1

= 340 cents or $3.40

(b) P (6) = 990

P (6:5) = 15(6:5) + 25(6:5)2

= 97:5 + 1056:25

= 1153:75


=P (6:5)¡ P (6)

6:5¡ 6

=1153:75¡ 990

0:5

= 327:5 cents or $3.28

(c) P (6) = 990

P (6:1) = 15(6:1) + 25(6:1)2

= 91:5 + 930:25

= 1021:75


=P (6:1)¡ P (6)

6:1¡ 6

=1021:75¡ 990

0:1

= 317:5 cents or $3.18

(d) P 0(x) = 15 + 50xP 0(6) = 15 + 50(6)

= 15 + 300

= 315 cents or $3.15

(e) P 0(20) = 15 + 50(20)= 1015 cents or $10.15

(f) P 0(30) = 15 + 50(30)= 1515 cents or $15.15

(g) The domain of x is [0;1) since pounds cannotbe measured with negative numbers.

(h) Since P 0(x) = 15 + 50x gives the marginalpro…t, and x ¸ 0; P 0(x) can never be negative.

(i) P (x) =P (x)

x

=15x+ 25x2

x

= 15 + 25x

(j) P0(x) = 25

(k) The marginal average pro…t cannot changesince P

0(x) is constant. The pro…t per pound

never changes, no matter now many pounds aresold.


54. (b) The value of x for which the average cost issmallest is x = 7:5: This can be found by drawinga line from the origin to any point of C(x): Atx = 7:5; you will get a line with the smallest slope.

(c) The marginal cost equals the average cost atthe point where the average cost is smallest.

55. (a) limx!29,300¡

T (x) = (29,300)(0.15)

= $4395

(b) limx!29,300+

T (x) = 4350 + (0:27)(29,300 ¡ 29,300)

= $4350

(c) limx!29,300

T (x) does not exist since parts (a)

and (b) have di¤erent answers.

(d)

(e) The graph is discontinuous at x = 29,300.

(f) For 0 · x · 29,300;

A(x) =T (x)

x=0:15x

x= 0:15:

For x > 29,300,

A(x) =T (x)

x

=4350 + (0:27)(x¡ 29,300)

x

=0:27x¡ 3561

x

= 0:27¡ 3561x:

(g) limx!29,300¡

A(x) = 0:15

(h) limx!29,300+

A(x) = 0:27¡ 3561

29,300= 0:14846

(i) limx!29,300 A(x) does not exist since parts (g)

and (h) have di¤erent answers.

(j) limx!1 A(x) = 0:27¡ 0 = 0:27

(k)

56. (a) The slope of the tangent line at x = 100 is0.02. This means that the risk of heart attack isgoing up at a rate of 0.02 per 1000 people for eachincrease in the blood cholesterol of 1 mg/dL.

(b) The slope of the tangent line at x = 200 is0.15. This means that the risk of heart attack isgoing up at a rate of 0.15 per 1000 people for eachincrease in the blood cholesterol of 1 mg/dL.

(c)C(300)¡C(100)

300¡ 100 =20¡ 7200

=13

200= 0:065

The average rate of change in the risk is 0.065 per1000 people per mg/dL of cholesterol.

57. V (t) = ¡t2 + 6t¡ 4

(a)

(b) The x-intercepts of the parabola are 0.8 and5.2, so a reasonable domain would be [0:8; 5:2];which represents the time period from 0.8 to 5.2weeks.


(c) The number of cases reaches a maximum at the vertex;

x =¡b2a

=¡6¡2 = 3

V (3) = ¡32 + 6(3)¡ 4 = 5The vertex of the parabola is (3; 5): This represents a maximum at 3 weeks of 500 cases.

(d) The rate of change function isV 0(t) = ¡2t+ 6:

(e) The rate of change in the number of cases at the maximum is

V 0(3) = ¡2(3) + 6 = 0:

(f) The sign of the rate of change up to the maximum is + because the function is increasing. The sign of therate of change after the maximum is ¡ because the function is decreasing.

58. (a) The rate can be estimated by estimating the slope of the tangent line to the curve at the given time.Answers will vary depending on the points used to determine the slope of the tangent line.

(i) The tangent line to the curve at 17:37 can be found by using the endpoints at (17:35.5, 0) and (17:40, 400).The rate the whale is descending at 17:37 is about

400¡ 017:40¡ 17:35.5 =

400

4:5¼ 90 meters per minute.

(ii) The tangent line to the curve at 17:39 can be found by using the endpoints at (17:36.2, 0) and(17:40.8, 400). The rate the whale is descending at 17:39 is about

400¡ 017:40:8¡ 17:36.2 =

400

4:6¼ 85 meters per minute.

(b) The whale appears to have 5 distinct rates at which it is descending.

Interval Rate (meters per minute)

17:35¡17:35.3 f(17:35:3)¡f(17:35)17:35:3¡17:35 =

10¡00:3

¼ 13

17:36¡17:37 f(17:37)¡f(17:36)17:37¡17:36 =

130¡401

= 90

17:37:3¡17:37.7 f(17:37:7)¡f(17:37:3)17:37:7¡17:37:3 =

150¡1500:4

= 0

17:38:3¡17:39.5 f(17:39.5)¡f(17:38:3)17:39:5¡17:38:3 =

300¡2001:2

¼ 83

17:40¡17:41 f(17:41)¡f(17:40)17:41¡17:40 =

400¡3331

= 67


Making smooth transitions between each interval, we get

59. (a) The curve slants downward up to about age 4, where it turns and begins to rise. There is a slight decline insteepness between ages 10.5 and 18. Correspondingly, the graph of the rate of change lies below the horizontalaxis to the left of 4 years and above the horizontal axis to the right of that point. The graph of the rate ofchange is declining between ages 10.5 and 18.

(b) The curve slants downward up to about age 5.25, where it turns and begins to rise. There is a slight declinein steepness between ages 11 and 20. Correspondingly, the graph of the rate of change lies below the horizontalaxis to the left of 5.25 years and above the horizontal axis to the right of that point. The graph of the rate ofchange is declining between ages 11 and 20.

60.

For a 10-year old girl, the remaining growth is about 14 cm and the rate of change is about ¡2:75 cm per year.61. (a) The slope of the tangent line at x = 100 is 1. This means that the ball is rising 1 ft for each foot it travels

horizontally.


(b) The slope of the tangent line at x = 200 is ¡2:7: This means that the ball is dropping 2.7 ft for each footit travels horizontally.

62. (a) The graph is discontinuous nowhere.

(b) The graph is not di¤erentiable where the graph makes a sudden change, namely at x = 50;x = 130; x = 230; and x = 770:

(c)

c:swp2507calwaism2001cwa ch 3 final...section 3.1 limits 177 10. (a) (i) by reading the graph, as x...

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