grosse pointe public schools · chapter 5 graphs and the derivative 5.1 increasing and decreasing...

Chapter 5

GRAPHS AND THE DERIVATIVE

5.1 Increasing and DecreasingFunctions

1. By reading the graph, f is

(a) increasing on (1;1) and(b) decreasing on (¡1; 1):


(a) increasing on (¡1; 4) and(b) decreasing on (4;1):

3. By reading the graph, g is

(a) increasing on (¡1;¡2) and(b) decreasing on (¡2;1):

4. By reading the graph, g is

(a) increasing on (3;1) and(b) decreasing on (¡1; 3):

5. By reading the graph, h is

(a) increasing on (¡1;¡4) and(¡2;1) and

(b) decreasing on (¡4;¡2):6. By reading the graph, h is

(a) increasing on (1; 5) and

(b) decreasing on (¡1; 1) and (5;1):7. By reading the graph, f is

(a) increasing on (¡7;¡4) and(¡2;1) and

(b) decreasing on (¡1;¡7) and(¡4;¡2):


(a) increasing on (¡3; 0) and (3;1) and(b) decreasing on (¡1;¡3) and (0; 3):

9. (a) Since the graph of the function is positive forx < ¡1 and x > 3, the intervals where f(x) isincreasing are (¡1;¡1) and (3;1):

(b) Since the graph of the function is negative for¡1 < x < 3, the interval where f(x) is decreasingis (¡1; 3).

10. (a) Since the graph of the function is positive for3 < x < 5; the intervals where f(x) is increasingis (3; 5):

(b) Since the graph of the function is negativefor x < 3 and x > 5, the interval where f(x) isdecreasing are (¡1; 3) and (5;1).

11. (a) Since the graph of the function is positivefor x < ¡8;¡6 < x < ¡2:5 and x > ¡1:5;the intervals where f(x) is increasing are(¡1;¡8); (¡6;¡2:5); and (¡1:5;1):(b) Since the graph of the function is negativefor ¡8 < x < ¡6 and ¡2:5 < x < ¡1:5, the in-tervals where f(x) is decreasing are (¡8;¡6) and(¡2:5;¡1:5):

12. (a) Since the graph of the function is positive forx < ¡3;¡3 < x < 3, and x > 3, the intervalswhere f(x) is increasing are (¡1;¡3); (¡3; 3), and(3;1):(b) Since the graph of the function is nevernegative, there are no intervals where f(x)is decreasing.

13. y = 2:3 + 3:4x¡ 1:2x2

(a) y0 = 3:4¡ 2:4xy 0 is zero when

3:4¡ 2:4x = 0

x =3:4

2:4=17

12

and there are no values of x where y0 does notexist, so the only critical number is x = 17

12 :

297

298 Chapter 5 GRAPHS AND THE DERIVATIVE

Test a point in each interval.

When x = 0; y0 = 3:4¡ 2:4(0) = 3:4 > 0:When x = 2; y0 = 3:4¡ 2:4(2) = ¡1:4 < 0:(b) The function is increasing on

¡¡1; 1712¢ :(c) The function is decreasing on

¡1712 ;1

¢:

14. y = 1:1¡ 0:3x¡ 0:3x2

(a) y0 = ¡0:3¡ 0:6xy0 is zero when

¡0:3¡ 0:6x = 0

x = ¡0:30:6

= ¡12

and there are no values of x where y0 does notexist, so the only critical number is x = ¡1

2 :


When x = ¡1; y0 = ¡0:3¡ 0:6(¡1) = 0:3 > 0:When x = 0; y0 = ¡0:3¡ 0:6(0) = ¡0:3 < 0:(b) The function is increasing on

¡¡1;¡12

¢:

(c) The function is decreasing on¡¡1

2 ;1¢:

15. f(x) =2

3x3 ¡ x2 ¡ 24x¡ 4

(a) f 0(x) = 2x2 ¡ 2x¡ 24= 2(x2 ¡ x¡ 12)= 2(x+ 3)(x¡ 4)

f 0(x) is zero when x = ¡3 or x = 4; so the criticalnumbers are ¡3 and 4:


f 0(¡4) = 16 > 0f 0(0) = ¡24 < 0f 0(5) = 16 > 0

(b) f is increasing on (¡1;¡3) and (4;1):(c) f is decreasing on (¡3; 4):

16. f(x) =2

3x3 ¡ x2 ¡ 4x+ 2

(a) f 0(x) = 2x2 ¡ 2x¡ 4= 2(x2 ¡ x¡ 2)= 2(x+ 1)(x¡ 2)

f 0(x) is zero when x = ¡1 or x = 2; so the criticalnumbers are ¡1 and 2:


f 0(¡2) = 8 > 0f 0(0) = ¡4 < 0f 0(3) = 8 > 0

(b) f is increasing on (¡1;¡1) and (2;1):(c) f is decreasing on (¡1; 2):

17. f(x) = 4x3 ¡ 15x2 ¡ 72x+ 5(a) f 0(x) = 12x2 ¡ 30x¡ 72

= 6(2x2 ¡ 5x¡ 12)= 6(2x+ 3)(x¡ 4)

f 0(x) is zero when x = ¡32 or x = 4; so the critical

numbers are ¡32 and 4:

f 0(¡2) = 36 > 0f 0(0) = ¡72 < 0f 0(5) = 78 > 0

(b) f is increasing on¡¡1;¡3

2

¢and (4;1):

(c) f is decreasing on¡¡3

2 ; 4¢:

18. f(x) = 4x3 ¡ 9x2 ¡ 30x+ 6(a) f 0(x) = 12x2 ¡ 18x¡ 30

= 6(2x2 ¡ 3x¡ 5)= 6(2x¡ 5)(x+ 1)

f 0(x) is zero when x = 52 or x = ¡1; so the critical

numbers are 52 and ¡1:

Section 5.1 Increasing and Decreasing Functions 299


f(¡2) = 54 > 0f(0) = ¡30 < 0f(3) = 24 > 0

(b) f is increasing on (¡1;¡1) and ¡52 ;1¢ :(c) f is decreasing on

¡¡1; 52¢ :19. f(x) = x4 + 4x3 + 4x2 + 1

(a) f 0(x) = 4x3 + 12x2 + 8x= 4x(x2 + 3x+ 2)

= 4x(x+ 2)(x+ 1)

f 0(x) is zero when x = 0; x = ¡2; or x = ¡1; sothe critical numbers are 0; ¡2; and ¡1:


f 0(¡3) = ¡12(¡1)(¡2) = ¡24 < 0f 0(¡1:5) = ¡6(:5)(¡:5) = 1:5 > 0f 0(¡:5) = ¡2(1:5)(:5) = ¡1:5 < 0f 0(1) = 4(3)(2) = 24 > 0

(b) f is increasing on (¡2;¡1) and (0;1):(c) f is decreasing on (¡1;¡2) and (¡1; 0):

20. f(x) = 3x4 + 8x3 ¡ 18x2 + 5(a) f 0(x) = 12x3 + 24x2 ¡ 36x

= 12x(x2 + 2x¡ 3)= 12x(x+ 3)(x¡ 1)

f 0(x) is zero when x = 0; x = ¡3; or x = 1; sothe critical numbers are 0;¡3; and 1:


f 0(¡4) = ¡240 < 0f 0(¡1) = 48 > 0

f 0μ1

2

¶= ¡21

2< 0

f 0(2) = 120 > 0

(b) f is increasing on (¡3; 0) and (1;1):(c) f is decreasing on (¡1;¡3) and (0; 1):

21. y = ¡3x+ 6

(a) y0 = ¡3 < 0

There are no critical numbers since y0 is never 0and always exists.

(b) Since y0 is always negative, the function is in-creasing on no interval.

(c) y0 is always negative, so the function is de-creasing everywhere, or on the interval (¡1;1):

22. y = 6x¡ 9

(a) y0 = 6 > 0

There are no critical numbers since y0 can neverbe 0 and always exists.

(b) Since y0 is always positive, the function is in-creasing everywhere, or on the interval (¡1;1):

(c) y0 is never negative, so the function is decreas-ing on no interval.

23. f(x) =x+ 2

x+ 1

(a) f 0(x) =(x+ 1)(1)¡ (x+ 2)(1)

(x+ 1)2

=¡1

(x+ 1)2

The derivative is never 0, but it fails to exist atx = ¡1: Since ¡1 is not in the domain of f; how-ever, ¡1 is not a critical number.

f 0(¡2) = ¡1 < 0f 0(0) = ¡1 < 0

(b) f is increasing on no interval.

(c) f is decreasing everywhere that it is de…ned,on (¡1;¡1) and on (¡1;1):


24. f(x) =x+ 3

x¡ 4(a) f 0(x) =

(1)(x¡ 4)¡ (1)(x+ 3)(x¡ 4)2

=x¡ 4¡ x¡ 3(x¡ 4)2 =

¡7(x¡ 4)2

f 0 is never 0, but it fails to exist when x = 4: Since4 is not in the domain of f; 4 is not a critical num-ber. Thus, there are no critical numbers. How-ever, the line x = 4 is an asymptote of the graph,so the function might change direction from oneside of the asymptote to the other.

f 0(0) = ¡ 7

16< 0

f 0(5) = ¡7 < 0(b) f 0(x) is always negative, so f(x) is increasingon no interval.

(c) f 0(x) is always negative, so f(x) is decreas-ing everywhere that is de…ned. Since f(x) is notde…ned at x = 4; these intervals are (¡1; 4) and(4;1):

25. y =px2 + 1

= (x2 + 1)1=2

(a) y0 =1

2(x2 + 1)¡1=2(2x)

= x(x2 + 1)¡1=2

=xpx2 + 1

y0 = 0 when x = 0:Since y does not fail to exist for any x; and sincey0 = 0 when x = 0; 0 is the only critical number.

y0(1) =1p2> 0

y0(¡1) = ¡1p2< 0

(b) y is increasing on (0;1):(c) y is decreasing on (¡1; 0):

26. y = xp9¡ x2 = x(9¡ x2)1=2

(a) Use the product rule.

y0 = (1)(9¡ x2)1=2

+1

2(9¡ x2)¡1=2(¡2x)(x)

= (9¡ x2)1=2 ¡ x2(9¡ x2)¡1=2= (9¡ x2)¡1=2(9¡ x2 ¡ x2)= (9¡ x2)¡1=2(9¡ 2x2)

=9¡ 2x2p9¡ x2

Critical numbers occur when y0 = 0 or when y0

fails to exist.y0 = 0 when

9¡ 2x2 = 0

x =§3p2=§3p22

:

y0 fails to exist when

9¡ x2 = 0x = §3:

Thus, the critical numbers are §3p22 and §3:

These four values determine three intervals sincef(x) is de…ned only on [¡3; 3]: Note that §3

p22 ¼

§2:12:

f 0(¡2:5) = ¡2:11 < 0f 0(0) = 3 > 0

f 0(2:5) = ¡2:11 < 0

(b) f is increasing on³¡3

p22 ;

3p22

´:

(c) f is decreasing on³¡3;¡3

p22

´and

³3p22 ; 3

´:

27. f(x) = x2=3

(a) f 0(x) =2

3x¡1=3 =

2

3x1=3

f 0(x) is never zero, but fails to exist when x = 0;so 0 is the only critical number.


f 0(¡1) = ¡23< 0

f 0(1) =2

3> 0

(b) f is increasing on (0;1):(c) f is decreasing on (¡1; 0):

28. f(x) = (x+ 1)4=5

(a) f 0(x) =4

5(x+ 1)¡1=5 =

4

5(x+ 1)1=5

f 0(x) is never zero, but fails to exist when x = ¡1;so the critical number is ¡1:

f 0(¡2) = ¡45< 0

f 0(0) =4

5> 0

(b) f is increasing on (¡1;1):(c) f is decreasing on (¡1;¡1):

29. y = x¡ 4 ln (3x¡ 9)

(a) y0 = 1¡ 12

3x¡ 9 = 1¡4

x¡ 3=x¡ 7x¡ 3

y0 is zero when x = 7: The derivative does notexist at x = 3; but note that the domain of f is(3;1):Thus, the only critical number is 7.

Choose values in the intervals (3; 7) and (7;1):

f 0(4) = ¡3 < 0

f(8) =1

5> 0

(b) The function is increasing on (7;1):(c) The function is decreasing on (3; 7):

30. f(x) = ln5x2 + 4

x2 + 1= ln(5x2 + 4)¡ ln(x2 + 1)

(a) f 0(x) =10x

5x2 + 4¡ 2x

x2 + 1

=10x(x2 + 1)

(5x2 + 4)(x2 + 1)¡ 2x(5x2 + 4)

(5x2 + 4)(x2 + 1)

=10x3 + 10x¡ 10x3 ¡ 8x(5x2 + 4)(x2 + 1)

=2x

(5x2 + 4)(x2 + 1)

f 0(x) is zero when x = 0 and there are no valuesof x where f 0(x) does not exist, so the criticalnumber is 0.


f 0(¡1) = 2(¡1)(5(¡1)2 + 4)((¡1)2 + 1) = ¡

1

9< 0

f 0(1) =2(1)

(5(1)2 + 4)((1)2 + 1)=1

9> 0

(b) The function is increasing on (0;1):(c) The function is decreasing on (¡1; 0):

31. f(x) = xe¡3x

(a) f 0(x) = e¡3x + x(¡3e¡3x)= (1¡ 3x)e¡3x

=1¡ 3xe3x

f 0(x) is zero when x = 13 and there are no values

of x where f 0(x) does not exist, so the criticalnumber is 13 :


f 0(0) =1¡ 3(0)e3(0)

= 1 > 0

f 0(1) =1¡ 3(1)e3(1)

= ¡ 2e3< 0


(b) The function is increasing on¡¡1; 13¢ :

(c) The function is decreasing on¡13 ;1

¢:

32. y = xex2¡3x

(a) y0 = xex2¡3x(2x¡ 3) + ex2¡3x(1)

= ex2¡3x[x(2x¡ 3) + 1]

= ex2¡3x(2x2 ¡ 3x+ 1)

= ex2¡3x(2x¡ 1)(x¡ 1)

y0 is zero when x = 12 or x = 1; so the critical

numbers are 12 and 1.


f 0(0) = 1(¡1)(¡1) = 1 > 0

f 0(0:6) = e¡1:44(0:2)(¡0:4) = ¡0:08e1:44

< 0

f 0(2) = e¡2(3)(1) =3

e2> 0

(b) The function is increasing on¡¡1; 12¢ and

(1;1):(c) The function is decreasing on

¡12 ; 1¢:

33. f(x) = x22¡x

(a) f 0(x) = x2[ln 2(2¡x)(¡1)] + (2¡x)2x= 2¡x(¡x2 ln 2 + 2x)

=x(2¡ x ln 2)

2x

f 0(x) is zero when x = 0 or x = 2ln 2 and there

are no values of x where f 0(x) does not exist. Thecritical numbers are 0 and 2

ln 2 :


f 0(¡1) = (¡1)(2¡ (¡1) ln 2)2¡1

= ¡2(2 + ln 2) < 0

f 0(1) =(1)(2¡ (1) ln 2)

21=2¡ ln 22

> 0

f 0(3) =(3)(2¡ (3) ln 2)

23=3(2¡ 3 ln 2)

8< 0

(b) The function is increasing on¡0; 2

ln 2

¢:

(c) The function is decreasing on (¡1; 0)and

¡2ln 2 ;1

¢:

34. f(x) = x2¡x2

(a) f 0(x) = x[ln 2(2¡x2

)(¡2x)] + (2¡x2)= 2¡x

2

(¡2x2 ln 2 + 1)

=1¡ 2x2 ln 2

2x

f 0(x) = 0 when

1¡ 2x2 ln 2 = 0

x2 =1

2 ln 2

x = § 1p2 ln 2

and there are no values of x where f 0(x) does notexist. The critical numbers are § 1p

2 ln 2:


f 0(¡1) = 1¡ 2(¡1)2 ln 22¡1

= 2(1¡ 2 ln 2) < 0

f 0(0) =1¡ 2(0)2 ln 2

20= 1 > 0

f 0(1) =1¡ 2(1)2 ln 2

21=1¡ 2 ln 2

2< 0

(b)The function is increasing on³¡ 1p

2 ln 2; 1p

2 ln 2

´:

(c) The function is decreasing on (¡1;¡ 1p2 ln 2

)

and³

1p2 ln 2

;1´:

35. y = x2=3 ¡ x5=3

(a) y0 =2

3x¡1=3 ¡ 5

3x2=3 =

2¡ 5x3x1=3

y0 = 0 when x = 25 . The derivative does not exist

at x = 0. So the critical numbers are 0 and 25 .



y0(¡1) = 7

¡3 < 0

y0μ1

5

¶=

1

3¡15

¢1=3 = 51=3

3> 0

y0(1) =¡33= ¡1 < 0

(b) y is increasing on (0; 25).

(c) y is decreasing on (¡1; 0) and (25 ;1).36. y = x1=3 + x4=3

(a) y0 =1

3x¡2=3 +

4

3x1=3 =

1+ 4x

3x2=3

y0 = 0 when x = ¡14 . The derivative does not

exist at x = 0. So the critical numbers are ¡14

and 0.

Test a number in each interval.

y0(¡1) = ¡1 < 0

y0μ¡18

¶=2

3> 0

y0(1) =5

3> 0

(b) The derivative is positive on¡¡1

4 ; 0¢and (0;1)

and the function is de…ned at x = 0. Thus, y isincreasing on

¡¡14 ,1

¢.

(c) y is decreasing on¡¡1;¡1

4

¢.

38. f(x) = ax2 + bx+ c; a > 0f 0(x) = 2ax+ b

Let f 0(x) = 0 to …nd the critical number.

2ax+ b = 0

2ax = ¡b

x =¡b2a

Choose a value in the interval¡¡1; ¡b2a ¢ : Since

a > 0;

¡b2a¡ 1

2a=¡b¡ 12a

<¡b2a:

f 0μ¡b¡ 1

2a

¶= 2a

μ¡b¡ 12a

¶+ b

= ¡1 < 0

Choose a value in the interval¡¡b2a ;1

¢: Since

a > 0;

¡b2a+1

2a=¡b+ 12a

>¡b2a:

f 0μ¡b+ 1

2a

¶= 1 > 0

f(x) is increasing on¡¡b2a ;1

¢and decreasing on¡¡1; ¡b2a ¢ :

This tells us that the curve opens upward and x =¡b2a is the x-coordinate of the vertex.

f

μ¡b2a

¶= a

μ¡b2a

¶2+ b

μ¡b2a

¶+ c

=ab2

4a2¡ b2

2a+ c

=b2

4a¡ 2b

2

4a+4ac

4a

=4ac¡ b24a

The vertex is³¡b2a ;

4ac¡b24a

´or³¡ b2a ;

4ac¡b24a

´:

39. f(x) = ax2 + bx+ c; a < 0f 0(x) = 2ax+ b

Let f 0(x) = 0 to …nd the critical number.

2ax+ b = 0

2ax = ¡b

x =¡b2a

Choose a value in the interval¡¡1;¡¡b

2a

¢: Since

a < 0;

¡b2a¡ ¡12a

=¡b+ 12a

<¡b2a:

f 0μ¡b+ 1

2a

¶= 2a

μ¡b+ 12a

¶+ b

= 1 > 0

Choose a value in the interval¡¡b2a ;1

¢: Since

a < 0;

¡b2a¡ ¡12a

=¡b¡ 12a

<¡b2a:

f 0μ¡b¡ 1

2a

¶= 2a

μ¡b¡ 12a

¶+ b

= ¡1 < 0

f 0 is increasing on¡¡1; ¡b2a ¢ and decreasing on¡¡b

2a ;1¢:


This tells us that the curve opens downward andx = ¡b

2a is the x-coordinate of the vertex.

f

μ¡b2a

¶= a

μ¡b2a

¶2+ b

μ¡b2a

¶+ c

=ab2

4a2¡ b2

2a+ c

=b2

4a¡ 2b

2

4a+4ac

4a

=4ac¡ b24a

The vertex is³¡b2a ;

4ac¡b24a

´or³¡ b2a ;

4ac¡b24a

´:

40. f(x) = ex

f 0(x) = ex > 0

f(x) = ex is increasing on (¡1;1):f(x) = ex is decreasing nowhere.Since f 0(x) is always positive, it never equals zero.Therefore, the tangent line is horizontal nowhere.

41. f(x) = ln x

f 0(x) =1

x

f 0(x) is unde…ned at x = 0: f 0(x) never equalszero. Note that f(x) has a domain of (0;1):Pick a value in the interval (0;1):

f 0(2) =1

2> 0

f(x) is increasing on (0;1):f(x) is never decreasing.Since f(x) never equals zero, the tangent line ishorizontal nowhere.

42. (a) The critical numbers are ¡3 and 4: The aver-age is

¡3 + 42

= 0:5:

(b) Use a graphing calculator to …nd the zeros ofthe function f(x) = 2

3x3 ¡ x2 ¡ 24x¡ 4:

The roots are ¡5:2005;¡0:1680; and 6:8685: Theaverage is

¡5:2005 + (¡0:1680) + 6:86853

= 0:5:

(c) The answers are the same.

(d) The critical numbers are ¡32 and 4. The av-

erage is¡32 + 4

2= 1:25:

(e) Use a graphing calculator to …nd the zeros ofthe function f(x) = 4x3¡15x2¡72x+5: The rootsare ¡2:8112; 0:0685; and 6.4927. The average is

¡2:8112 + (0:0685) + 6:49273

= 1:25:

(f) The answers are the same.

43. f(x) = e0:001x ¡ lnxf 0(x) = 0:001e0:001x ¡ 1

x

Note that f(x) is only de…ned for x > 0: Use agraphing calculator to plot f 0(x) for x > 0:

(a) f 0(x) > 0 about (567;1); so f(x) is increasingabout (567;1):(b) f 0(x) < 0 about (0; 567), so f(x) is decreasingabout (0; 567):

44. f(x) = ln(x2 + 1)¡ x0:3

f 0(x) =2x

x2 + 1¡ 0:3x¡0:7

Note that f(x) is only de…ned for x > 0: Use agraphing calculator to plot f 0(x) for x > 0:

(a) f 0(x) > 0 about (0; 558), so f(x) is increasingabout (558;1):(b) f 0(x) < 0 on (558;1); so f(x) is decreasingon (558;1):

45. H(r) =300

1 + 0:03r2= 300(1 + 0:03r2)¡1

H 0(r) = 300[¡1(1 + 0:03r2)¡2(0:06r)]

=¡18r

(1 + 0:03r2)2

Since r is a mortgage rate (in percent), it is alwayspositive. Thus, H 0(r) is always negative.

(a) H is increasing on nowhere.

(b) H is decreasing on (0;1):46. C(x) = x3 ¡ 2x2 + 8x+ 50

C 0(x) = 3x2 ¡ 4x+ 8Since C(x) is a polynomial function, the only crit-ical numbers are found by solving C0(x) = 0:

3x2 ¡ 4x+ 8 = 0

x =4§p16¡ 96

6

=4§p¡80

6


Sincep¡80 is not a real number, there are no crit-

ical points. The function either always increasesor always decreases.Test any point, say x = 0:

C0(0) = 8 > 0

(a) C(x) is decreasing nowhere.

(b) C(x) is increasing everywhere.

47. C(x) = 0:32x2 ¡ 0:00004x3R(x) = 0:848x2 ¡ 0:0002x3P (x) = R(x)¡C(x)

= (0:848x2 ¡ 0:0002x3)¡ (0:32x2 ¡ 0:00004x3)= 0:528x2 ¡ 0:00016x3

P 0(x) = 1:056x¡ 0:00048x2

1:056x¡ 0:00048x2 = 0x(1:056¡ 0:00048x) = 0

x = 0 or x = 2200

Choose x = 1000 and x = 3000 as test points.

P 0(1000) = 1:056(1000)¡ 0:00048(1000)2 = 576P 0(3000) = 1:056(3000)¡ 0:00048(3000)2

= ¡1152

The function is increasing on (0; 2200).

48. P (x) = ¡(x¡ 4)ex ¡ 4 (0 < x · 3:9)P 0(x) = ¡(x¡ 4)ex + ex(¡1)

= ¡ex[(x¡ 4) + 1]= ¡ex(x¡ 3)

P 0(x) = 0 when x = 3:

P 0(1) = ¡e1(¡2) = 2e > 0

P 0(3:5) = ¡e3:5(0:5) = ¡0:5e3=5

< 0

(a) The pro…t is increasing on (0; 3); which repre-sents production between 0 and 300 players.

(b) The pro…t is decreasing on (3, 3.9), which rep-resents production between 300 and 390 players.

49. (a) These curves are graphs of functions since theyall pass the vertical line test.

(b) The graph for particulates increases from Aprilto July; it decreases from July to November; it isconstant from January to April and November toDecember.

(c) All graphs are constant from January to Apriland November to December. When the tempera-ture is low, as it is during these months, air pol-lution is greatly reduced.

50. P (t) =10 ln(0:19t+ 1)

0:19t+ 1

P 0(t) = 10

Ã(0:19t+1) ¢ 1

0:19t+1 ¢0:19¡0:19 ln(0:19t+ 1)(0:19t+ 1)2

!

= 10

μ0:19¡ 0:19 ln(0:19t+ 1)

(0:19t+ 1)2

¶

= 1:9

μ1¡ ln(0:19t+ 1)(0:19t+ 1)2

¶

Note that P (t) is de…ned when 0:19t+1 > 0, whichimplies that t > ¡1

0:19 ¼ ¡5:26, so the domain of Pis ftjt > ¡100

19 ¼ ¡5:26g.Find critical numbers. Set numerator equal to 0.

1¡ ln(0:19t+ 1) = 0ln(0:19t+ 1) = 1

0:19t+ 1 = e

t =e¡ 10:19

¼ 9:04

So, P 0(t) = 0 when t = 10019 (e¡ 1) ¼ 9:04:

Set denominator equal to 0.

(0:19t+ 1)2 = 0

0:19t+ 1 = 0

t =¡10:19

¼ ¡5:26

Since ¡10019 ¼ ¡5:26 is not in the domain of P ,

10019 (e¡ 1) is the only critical number.


Test a number in each interval:

P 0(0) = 1:9μ1¡ ln(1)12

¶= 1:9 > 0

P 0(10) = 1:9μ1¡ ln(2:9)2:92

¶< 0

since ln (2:9) > ln(e) = 1:Therefore, P increases on

¡¡10019 ;

10019 (e¡ 1)

¢and

decreases on¡10019 (e¡ 1);1

¢, so the number of

people infected will start to decline after day10019 (e¡ 1) ¼ 9:04, or day 9.

51. A(x) = 0:003631x3 ¡ 0:03746x2+ 0:1012x+ 0:009

A0(x) = 0:010893x2 ¡ 0:07492x+ 0:1012Solve for A0(x) = 0:

x ¼ 1:85 or x ¼ 5:03Choose x = 1 and x = 4 as test points.

A0(1) = 0:010893(1)2 ¡ 0:07492(1)+ 0:1012

= 0:037173

A0(4) = 0:010893(4)2 ¡ 0:07492(4)+ 0:1012

= ¡0:024192(a) The function is increasing on (0; 1:85):

(b) The function is decreasing on (1:85; 5):

52. K(x) =4x

3x2 + 27

K0(x) =(3x2 + 27)4¡ 4x(6x)

(3x2 + 27)2

=108¡ 12x2(3x2 + 27)2

K0(x) is zero when

108¡ 12x2 = 012(9¡ x2) = 0

(3¡ x)(3 + x) = 0x = 3 or x = ¡3:

However, in this problem x represents time,which cannot be negative, so 3 is the onlycritical number.

K0(0) =108

272> 0

K0(4) =¡84

(48 + 27)2< 0

(a) K(x) is increasing on (0; 3):(Note: x must be at least 0.)

(c) K(x) is decreasing on (3;1):

53. K(t) =5t

t2 + 1

K0(t) =5(t2 + 1)¡ 2t(5t)

(t2 + 1)2

=5t2 + 5¡ 10t2(t2 + 1)2

=5¡ 5t2(t2 + 1)2

K0(t) = 0 when

5¡ 5t2(t2 + 1)2

= 0

5¡ 5t2 = 05t2 = 5

t = §1:

Since t is the time after a drug is administered, thefunction applies only for [0;1); so we discard t =¡1: Then 1 divides the interval into two intervals.

K0(0:5) = 2:4 > 0K0(2) = ¡0:6 < 0(a) K is increasing on (0; 1):

(b) K is decreasing on (1;1):

54. D(p) = 0:000002p3 ¡ 0:0008p2 + 0:1141p+ 16:683D0(p) = 0:000006p2 ¡ 0:0016p+ 0:1141By the quadratic formula, there are no real num-ber solutions for p when D0(p) = 0:Choose a value in the interval (55; 130):

D0(60) = 0:0397 > 0

The function is increasing on the entire interval(55; 130); so it is decreasing nowhere.

Section 5.2 Relative Extrema 307

55. (a) F (t) = ¡10:28 + 175:9te¡t=1:3

F 0(t) = (175:9)(e¡t=1:3)

+ (175:9:9t)

μ¡ 1

1:3e¡t=1:3

¶

= (175:9)(e¡t=1:3)μ1¡ t

1:3

¶

¼ 175:9e¡t=1:3(1¡ 0:769t)(b) F 0(t) is equal to 0 at t = 1:3. Therefore, 1.3 isa critical number. Since the domain is (0;1), testvalues in the intervals from (0; 1:3) and (1:3;1):

F 0(1) ¼ 18:83 > 0 and F 0(2) ¼ ¡20:32 < 0F 0(t) is increasing on (0; 1:3) and decreasing on(1:3;1):

56. W1(t) = 619(1¡ 0:905e¡0:002t)1:2386W

01(t) = 1:388e

¡0:002t(1¡ 0:905e¡0:002t)0:2386 > 0Since W

01(t) > 0 and t is a positive number, this

function is increasing on (0;1):

57. f(x) =1p2¼e¡x

2=2

f 0(x) =1p2¼e¡x

2=2(¡x)

=¡xp2¼e¡x

2=2

f 0(x) = 0 when x = 0:Choose a value from each of the intervals (¡1; 0)and (0;1):

f 0(¡1) = 1p2¼e¡1=2 > 0

f 0(1) =¡1p2¼e¡1=2 < 0

The function is increasing on (¡1; 0) and decreas-ing on (0;1):

58. (a) The stockpiles of both countries are increasingon about (1945; 1966) and (1970; 1974):

(b) The stockpiles of both countries are decreasing onabout (1986; 2002):

59. As shown on the graph,

(a) horsepower increases with engine speed on(1500; 6250);

(b) horsepower decreases with engine speed on(6250; 7200);

(c) torque increases with engine speed on(1500; 2500) and (3500; 4400);

(d) torque decreases with engine speed on(3000; 3500) and (6000; 7200):

60. (a) As x increases, f(x) decreases, so f isdecreasing for these x-values. Therefore,f 0(x) is negative.

(b) mpglb , or miles per gallon per pound.

5.2 Relative Extrema

1. As shown on the graph, the relative minimum of¡4 occurs when x = 1:

2. As shown on the graph, the relative maximum of1 occurs when x = 4:

3. As shown on the graph, the relative maximum of3 occurs when x = ¡2:

4. As shown on the graph, the relative maximum of¡4 occurs when x = 3:

5. As shown on the graph, the relative maximum of3 occurs when x = ¡4 and the relative minimumof 1 occurs when x = ¡2:

6. As shown on the graph, the relative minimum of¡6 occurs when x = 1 and the relative maximumof 2 occurs when x = 5:

7. As shown on the graph, the relative maximum of3 occurs when x = ¡4; the relative minimum of¡2 occurs when x = ¡7 and x = ¡2

8. As shown on the graph, the relative maximum of4 occurs when x = 0; the relative minimum of 0occurs when x = ¡3 and x = 3:

9. Since the graph of the function is zero at x = ¡1and x = 3, the critical numbers are ¡1 and 3.Since the graph of the function is positive on(¡1;¡1) and negative on (¡1; 3); there is a rela-tive maximum at ¡1: Since the graph of the func-tion is negative on (¡1; 3) and positive on (3;1);there is a relative minimum at 3.


10. Since the graph of the function is zero at x = 3

and x = 5, the critical numbers are 3 and 5.

Since the graph of the function is negative on(¡1; 3) and positive on (3; 5); there is a relativeminimum at 3: Since the graph of the function ispositive on (3; 5) and negative on (5;1); there isa relative maximum at 5.

11. Since the graph of the function is zero at x =¡8; x = ¡6; x = ¡2:5 and x = ¡1:5; the criticalnumbers are ¡8;¡6;¡2:5; and ¡1:5:

Since the graph of the function is positive on(¡1;¡8) and negative on (¡8;¡6), there is a rel-ative maximum at ¡8. Since the graph of thefunction is on (¡8;¡6) and positive on (¡6;¡2:5),there is a relative minimum at¡6. Since the graphof the function is positive on (¡6;¡2:5) and nega-tive on (¡2:5;¡1:5), there is a relative maximumat ¡2:5. Since the graph of the function is neg-ative on (¡2:5;¡1:5) and positive on (¡1:5;1),there is a relative minimum at ¡1:5:

12. Since the graph of the function is zero at x = ¡3and x = 3, the critical numbers are ¡3 and 3.

Since the graph of the function is positive on(¡1;¡3); (¡3; 3), and (3;1), there are no rela-tive extrema.

13. f(x) = x2 ¡ 10x+ 33f 0(x) = 2x¡ 10f 0(x) is zero when x = 5:

f 0(0) = ¡10 < 0f 0(6) = 2 > 0

f is decreasing on (¡1; 5) and increasing on (5;1).Thus, a relative minimum occurs at x = 5:

f(5) = 8

Relative minimum of 8 at 5

14. f(x) = x2 + 8x+ 5

f 0(x) = 2x+ 8

f 0(x) is zero when x = ¡4:

f 0(¡5) = ¡2 < 0f 0(0) = 8 > 0

f is decreasing on (¡1;¡4) and increasing on(¡4;1). Thus, a relative minimum occurs at x =¡4:

f(¡4) = ¡11

Relative minimum of ¡11 at ¡4

15. f(x) = x3 + 6x2 + 9x¡ 8f 0(x) = 3x2 + 12x+ 9 = 3(x2 + 4x+ 3)

= 3(x+ 3)(x+ 1)

f 0(x) is zero when x = ¡1 or x = ¡3:

f 0(¡4) = 9 > 0f 0(¡2) = ¡3 < 0f 0(0) = 9 > 0

Thus, f is increasing on (¡1;¡3); decreasing on(¡3;¡1); and increasing on (¡1;1):f has a relative maximum at ¡3 and a relativeminimum at ¡1:

f(¡3) = ¡8f(¡1) = ¡12

Relative maximum of ¡8 at ¡3; relative minimumof ¡12 at ¡1


16. f(x) = x3 + 3x2 ¡ 24x+ 2f 0(x) = 3x2 + 6x¡ 24

= 3(x2 + 2x¡ 8)= 3(x+ 4)(x¡ 2)

f 0(x) is zero when x = ¡4 and x = 2:

f 0(¡5) = 21 > 0f 0(0) = ¡24 < 0f 0(3) = 21 > 0

f is increasing on (¡1;¡4) and decreasing on(¡4; 2): Thus, a relative maximum occurs at x =¡4: f(x) is decreasing on (¡4; 2) and increasing on(2;1): Thus, a relative minimum occurs at x = 2:f(¡4) = (¡4)3 + 3(¡4)2 ¡ 24(¡4) + 2

= 82

f(2) = (2)3 + 3(2)2 ¡ 24(2) + 2= ¡26

Relative maximum of 82 at ¡4; relative minimumof ¡26 at 2

17. f(x) = ¡43x3 ¡ 21

2x2 ¡ 5x+ 8

f 0(x) = ¡4x2 ¡ 21x¡ 5= (¡4x¡ 1)(x+ 5)

f 0(x) is zero when x = ¡5, or x = ¡14 :

f 0(¡6) = ¡23 < 0f 0(¡4) = 15 > 0f 0(0) = ¡5 < 0

f is decreasing on (¡1;¡5); increasing on ¡¡5;¡14

¢;

and decreasing on¡¡1

4 ;1¢: f has a relative min-

imum at ¡5 and a relative maximum at ¡14 :

f(¡5) = ¡3776

f

μ¡14

¶=827

96

Relative maximum of 82796 at ¡1

4 ; relative mini-mum of ¡377

6 at ¡5

18. f(x) = ¡23x3 ¡ 1

2x2 + 3x¡ 4

f 0(x) = ¡2x2 ¡ x+ 3= ¡(2x2 + x¡ 3)= ¡(2x+ 3)(x¡ 1)

f 0(x) is zero when x = ¡32 or x = 1:

f 0(¡2) = ¡3 < 0f 0(0) = 3 > 0f 0(2) = ¡7 < 0

f is decreasing on¡¡1;¡3

2

¢and increasing on¡¡3

2 ; 1¢: Thus, a relative minimum occurs at x =

¡32 : f(x) is increasing on

¡¡32 ; 1¢and decreasing

on (1;1): Thus, a relative maximum occurs atx = 1:

f

μ¡32

¶= ¡59

8

f(1) = ¡136

Relative maximum of ¡136 at 1; relative minimum

of ¡598 at ¡3

2

19. f(x) = x4 ¡ 18x2 ¡ 4f 0(x) = 4x3 ¡ 36x

= 4x(x2 ¡ 9)= 4x(x+ 3)(x¡ 3)

f 0(x) is zero when x = 0 or x = ¡3 or x = 3:

f 0(¡4) = 4(¡4)3 ¡ 36(¡4) = ¡112 < 0f 0(¡1) = ¡4 + 36 = 32 > 0f 0(1) = 4¡ 36 = ¡32 < 0f 0(4) = 4(4)3 ¡ 36(4) = 112 > 0

f is decreasing on (¡1;¡3) and (0; 3); f is in-creasing on (¡3; 0) and (3;1):

f(¡3) = ¡85f(0) = ¡4f(3) = ¡85

Relative maximum of ¡4 at 0; relative minimumof ¡85 at 3 and ¡3


20. f(x) = x4 ¡ 8x2 + 9f 0(x) = 4x3 ¡ 16x

= 4x(x2 ¡ 4)= 4x(x+ 2)(x¡ 2)

f 0(x) is zero when x = 0 or x = ¡2 or x = 2:

f 0(¡3) = ¡60 < 0f 0(¡1) = 12 > 0f 0(1) = ¡12 < 0f(3) = 60 > 0

f is increasing on (¡2; 0) and (2;1); f is decreas-ing on (¡1;¡2) and (0; 2):

f(¡2) = ¡7f(0) = 9

f(2) = ¡7

Relative maximum of 9 at 0; relative minimum of¡7 at ¡2 and 2

21. f(x) = 3¡ (8 + 3x)2=3

f 0(x) = ¡23(8 + 3x)¡1=3(3)

= ¡ 2

(8 + 3x)1=3

Critical number:

8 + 3x = 0

x = ¡83

f 0(¡3) = 2 > 0f 0(0) = ¡1 < 0

f is increasing on¡¡1;¡8

3

¢and decreasing on¡¡8

3 ;1¢:

f

μ¡83

¶= 3

Relative maximum of 3 at ¡83

22. f(x) =(5¡ 9x)2=3

7+ 1

f 0(x) =μ2

3

¶(5¡ 9x)¡1=3

7(¡9)

= ¡ 6

7(5¡ 9x)1=3Critical number:

5¡ 9x = 0

x =5

9

f 0(0) ¼ ¡0:5013 < 0f 0(1) ¼ 0:5400 > 0

f is decreasing on¡¡1; 59¢ and increasing on ¡59 ;1¢ :f

μ5

9

¶= 1

Relative minimum of 1 at 59

23. f(x) = 2x+ 3x2=3

f 0(x) = 2 + 2x¡1=3

= 2 +23px

Find the critical numbers.

f 0(x) = 0 when

2 +23px= 0

23px= ¡2

13px= ¡1

x = (¡1)3x = ¡1:

f 0(x) does not exist when

3px = 0

x = 0:


f 0(¡2) = 2 + 23p¡2 ¼ 0:41 > 0

f 0μ¡12

¶= 2 +

2

3

q¡12

= 2 +2 3p2

¡1 ¼ ¡0:52 < 0

f 0(1) = 2 +23p1= 4 > 0

f is increasing on (¡1;¡1) and (0;1):f is decreasing on (¡1; 0):

f(¡1) = 2(¡1) + 3(¡1)2=3 = 1f(0) = 0

Relative maximum of 1 at ¡1; relative minimumof 0 at 0

24. f(x) = 3x5=3 ¡ 15x2=3f 0(x) = 5x2=3 ¡ 10x¡1=3

= 5x¡1=3(x¡ 2)

=5(x¡ 2)x1=3

Find the critical numbers:

5(x¡ 2) = 0 x1=3 = 0

x = 2 x = 0

f 0(¡1) = 15 > 0f 0(1) = ¡5 < 0f 0(3) = 3:47 > 0

f is increasing on (¡1; 0) and (2;1):f is decreasing on (0; 2):

f(x) = 3x2=3(x¡ 5)f(0) = 3 ¢ 0(0¡ 5) = 0f(2) = 3 ¢ 22=3(2¡ 5)

= ¡9 ¢ 22=3 ¼ ¡14:287

Relative maximum of 0 at 0; relative minimum of¡9 ¢ 22=3 ¼ ¡14:287 at 2

25. f(x) = x¡ 1

x

f 0(x) = 1 + 1x2 is never zero, but fails to exist at

x = 0:

Since f(x) also fails to exist at x = 0; there areno critical numbers and no relative extrema.

26. f(x) = x2 +1

x

f 0(x) = 2x¡ 1

x2

=2x3 ¡ 1x2

Find the critical number:

2x3 ¡ 1 = 0

x =13p2=

3p4

2¼ 0:79

Note that both f(x) and f 0(x) do not exist atx = 0; so 0 is not a critical number.

f 0(¡1) = ¡3 < 0f 0(1) = 1 > 0

f(x) is decreasing on³¡1; 3p4

2

´and increasing

on³

3p42 ;1

´:

f

Ã3p4

2

!=

Ã3p4

2

!2+

Ã3p4

2

!¡1

=

Ã3p4

2

!¡1 24Ã 3p4

2

!3+ 1

35

=23p4

μ4

8+ 1

¶=

23p4

μ3

2

¶¢

3p2

3p2

=3 3p2

2¼ 1:890

Relative minimum of 33p22 ¼ 1:890 at 3p4

2


27. f(x) =x2 ¡ 2x+ 1x¡ 3

f 0(x) =(x¡ 3)(2x¡ 2)¡ (x2 ¡ 2x+ 1)(1)

(x¡ 3)2

=x2 ¡ 6x+ 5(x¡ 3)2


x2 ¡ 6x+ 5 = 0(x¡ 5)(x¡ 1) = 0x = 5 or x = 1

Note that f(x) and f 0(x) do not exist at x = 3; sothe only critical numbers are 1 and 5.

f 0(0) =5

9> 0

f 0(2) = ¡3 < 0

f 0(6) =5

9> 0

f(x) is increasing on (¡1; 1) and (5;1):f(x) is decreasing on (1; 5).

f(1) = 0

f(5) = 8

Relative maximum of 0 at 1; relative minimum of8 at 5

28. f(x) =x2 ¡ 6x+ 9x+ 2

f 0(x) =(2x¡ 6)(x+ 2)¡ (1)(x2 ¡ 6x+ 9)

(x+ 2)2

=x2 + 4x¡ 21(x+ 2)2

=(x+ 7)(x¡ 3)(x+ 2)2


(x+ 7)(x¡ 3) = 0x = ¡7 or x = 3

Note that f(x) and f 0(x) do not exist at x = ¡2,so the only critical numbers are ¡7 and 3.

f 0(¡8) = 11

36> 0

f 0(¡3) = ¡24 < 0

f 0(0) = ¡214< 0

f 0(4) =11

36> 0

f is increasing on (¡1;¡7) and (3;1):f is decreasing on (¡7;¡2) and (¡2; 3).

f(¡7) = ¡20f(3) = 0

Relative maximum of ¡20 at ¡7; relative mini-mum of 0 at 3

29. f(x) = x2ex ¡ 3f 0(x) = x2ex + 2xex

= xex(x+ 2)

f 0(x) is zero at x = 0 and x = ¡2:

f 0(¡3) = 3e¡3 = 3

e3> 0

f 0(¡1) = ¡e¡1 = ¡1e< 0

f 0(1) = 3e1 > 0

f is increasing on (¡1;¡2) and (0;1):f is decreasing on (¡2; 0):

f(0) = 0 ¢ e0 ¡ 3 = ¡3f(¡2) = (¡2)2e¡2 ¡ 3

=4

e2¡ 3

¼ ¡2:46

Relative minimum of ¡3 at 0; relative maximumof ¡2:46 at ¡2


30. f(x) = 3xex + 2f 0(x) = 3xex + 3ex

= 3ex(x+ 1)


3ex = 0 or x+ 1 = 0

ex = 0 or x = ¡1

ex is always positive, so the only critical numberis ¡1:

f 0(¡2) = 3e¡2(¡2 + 1)= ¡3e¡2 < 0

f 0(0) = 3e0(0 + 1)= 3 > 0

f is increasing on (¡1;1) and decreasing on (¡1;¡1):

f(¡1) = 3(¡1)e¡1 + 2

=¡3e+ 2 ¼ 0:9

Relative minimum of 0.90 at ¡1

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

f 0(x) = 2 +1

x=2x+ 1

x

f 0(x) is zero at x = ¡12 : The domain of f(x) is

(0;1): Therefore f 0(x) is never zero in the domainof f(x):f 0(1) = 3 > 0: Since f(x) is always increasing, fhas no relative extrema.

32. f(x) =x2

ln x

f 0(x) =(ln x) 2x¡ x2 ¡ 1x¢

(ln x)2=2x ln x¡ x(ln x)2

=x(2 ln x¡ 1)(ln x)2


x = 0 or 2 ln x¡ 1 = 0 or ln x = 0

2 ln x = 1 x = 1

ln x =1

2x = e1=2

=pe

¼ 1:65

Since the domain of y = ln x is (0;1); 0 is not acritical number. Since ln 1 = 0; we see that 1 isalso not in the domain of f: Thus

pe ¼ 1:65 is the

only critical number.

f 0(1:5) =1:5(2 ln 1:5¡ 1)

(ln 1:5)2< 0

f 0(2) =2(2 ln 2¡ 1)(ln 2)2

> 0

f(x) is increasing on (1:65;1) and decreasing on(1; 1:65):

f(1:65) =(1:65)2

ln (1:65)¼ 5:44

Relative minimum of 5.44 at 1.65

33. f(x) =2x

x

f 0(x) =(x) ln 2(2x)¡ 2x(1)

x2

=2x(x ln 2¡ 1)

x2


x ln 2¡ 1 = 0 or x2 = 0

x =1

ln 2x = 0

Since f is not de…ned for x = 0; 0 is not a crit-ical number. x = 1

ln 2 ¼ 1:44 is the only criticalnumber.

f 0(1) ¼ ¡0:6137 < 0f 0(2) ¼ 0:3863 > 0

f is decreasing on¡0; 1ln 2

¢and increasing on

¡1ln 2 ;1

¢:

f

μ1

ln 2

¶=21= ln 2

1ln 2

= e ln 2

Relative minimum of e ln 2 at1

ln 2


34. f(x) = x+ 8¡x

f 0(x) = 1 + (ln 8)8¡x(¡1)

= 1¡ ln 88x


1¡ ln 88x

= 0

1 =ln 8

8x

8x = ln 8

x ln 8 = ln(ln 8)

x =ln(ln 8)

ln 8¼ 0:35

f 0(0) ¼ ¡1:0794 < 0f 0(1) ¼ 0:7401 > 0

f is decreasing on³¡1; ln(ln 8)ln 8

´and increasing

on³ln(ln 8)ln 8 ;1

´:

f

μln(ln 8)

ln 8

¶=ln(ln 8)

ln 8+ 8¡ ln(ln 8)= ln 8

=ln(ln 8)

ln 8+

1

ln 8

=ln(ln 8) + 1

ln 8

Relative minimum of ln(ln 8)+1ln 8 at ln(ln 8)ln 8

35. y = ¡2x2 + 12x¡ 5y0 = ¡4x+ 12= ¡4(x¡ 3)

The vertex occurs when y0 = 0 or when

x¡ 3 = 0x = 3

When x = 3;

y = ¡2(3)2 + 12(3)¡ 5 = 13

The vertex is (3; 13):

36. y = ax2 + bx+ cy0 = 2ax+ b

The vertex occurs when y0 = 0:

2ax+ b = 0

x =¡b2a

a

μ¡b2a

¶2+ b

μ¡b2a

¶+ c =

ab2

4a2¡ b2

2a+ c

=b2

4a¡ 2b

2

4a+4ac

4a

=4ac¡ b24a

:

The vertex is at³¡b2a ;

4ac¡b24a

´:

37. f(x) = x5 ¡ x4 + 4x3 ¡ 30x2 + 5x+ 6

f 0(x) = 5x4 ¡ 4x3 + 12x2 ¡ 60x+ 5

Graph f 0 on a graphing calculator. A suitablechoice for the viewing window is [¡4; 4] by [¡50; 50];Yscl = 10.

Use the calculator to estimate the x-intercepts ofthis graph. These numbers are the solutions of theequation f 0(x) = 0 and thus the critical numbersfor f: Rounded to three decimal places, these x-values are 0.085 and 2.161.

Examine the graph of f 0 near x = 0:085 andx = 2:161: Observe that f 0(x) > 0 to the left ofx = 0:085 and f 0(x) < 0 to the right of x = 0:085:Also observe that f 0(x) < 0 to the left of x = 2:161and f 0(x) > 0 to the right of x = 2:161: The …rstderivative test allows us to conclude that f hasa relative maximum at x = 0:085 and a relativeminimum at x = 2:161:

f(0:085) ¼ 6:211f(2:161) ¼ ¡57:607

Relative maximum of 6.211 at 0.085; relative min-imum of ¡57:607 at 2.161.


38. f(x) = ¡x5 ¡ x4 + 2x3 ¡ 25x2 + 9x+ 12

f 0(x) = ¡5x4 ¡ 4x3 + 6x2 ¡ 50x+ 9

Graph f 0 on a graphing calculator. A suitablechoice for the viewing window is [¡4; 4] by [¡100; 100];Yscl = 20.Use the calculator to estimate the x-intercepts ofthis graph. These numbers are the solutions of theequation f 0(x) = 0 and thus the critical numbersfor f: Rounded to three decimal places, these x-values are 0.183 and ¡2:703:Examine the graph of f 0 near x = ¡2:703 andx = 0:183: Observe that f 0(x) < 0 to the leftof x = ¡2:703 and f 0(x) > 0 to the right ofx = 2:703: Also observe that f 0(x) > 0 to theleft of x = 0:183 and f 0(x) < 0 to the rightof x = 0:183: The …rst derivative test allows usto conclude that f has a relative minimum atx = ¡2:703 and a relative maximum at x = 0:183:

f(0:183) ¼ 12:821f(¡2:703) ¼ ¡143:572

Relative maximum of 12.821 at 0.183; relative min-imum of ¡143:572 at ¡2:703

39. f(x) = 2 jx+ 1j+ 4 jx¡ 5j ¡ 20

Graph this function in the window [¡10; 10] by[¡15; 30], Yscl = 5.

The graph shows that f has no relative maxima,but there is a relative minimum at x = 5:(Note that the graph has a sharp point at (5;¡8),indicating that f 0(5) does not exist.)

40. (a) When graphing g(x) in the standard window,no graph seems to appear.

(b) g(x) =1

x12¡ 2

μ1000

x

¶6

g0(x) =¡12x13

¡ 12μ1000

x

¶5μ¡1000x2

¶

=¡12x13

+1:2£ 1019

x7

=¡12x13

+(1:2£ 1019)x6

x13

=¡12 + (1:2£ 1019)x6

x13


¡12 + (1:2£ 1019)x6 = 0 or x13 = 0

(1:2£ 1019)x6 = 12 x = 0

x6 =12

1:2£ 1019

x6 =10

1019= 10¡18

x =6p10¡18

= §0:001 or x = 0

g0(¡1) = ¡12 + (1:2£ 1019)(¡1)6(¡1)13 < 0

g0(¡0:0001) = ¡12 + (1:2£ 1019)(¡0:0001)6(¡0:0001)13 > 0

g0(0:0001) =¡12 + (1:2£ 1019)(0:0001)6

(0:0001)13< 0

g0(1) =¡12 + (1:2£ 1019)16

113> 0

g(0:001) = ¡10¡36; g(¡0:001) = ¡1036Minimum of ¡1036 at x = §0:001

41. C(q) = 80 + 18q; p = 70¡ 2qP (q) = R(q)¡C(q) = pq ¡ C(q)

= (70¡ 2q)q ¡ (80 + 18q)= ¡2q2 + 52q ¡ 80

(a) Since the graph of P is a parabola that opensdownward, we know that its vertex is a maximumpoint. To …nd the q-value of this point, we …ndthe critical number.

P 0(q) = ¡4q + 52P 0(q) = 0 when

¡4q + 52 = 04q = 52

q = 13


The number of units that produce maximum pro…tis 13.

(b) If q = 13;

p = 70¡ 2(13)= 44

The price that produces maximum pro…t is $44.

(c) P (13) = ¡2(13)2 + 52(13)¡ 80 = 258The maximum pro…t is $258.

42. C(q) = 25q + 5000; p = 90¡ 0:02qP (q) = R(q)¡C(q) = pq ¡C(q)

= (90¡ 0:02q)q ¡ (25q + 5000)= ¡0:02q2 + 65q ¡ 5000

(a) Since the graph of P is a parabola that opensdownward, we know that its vertex is a maximumpoint. To …nd the q-value of this point, we …ndthe critical number.

P 0(q) = ¡0:04q + 65P 0(q) = 0 when

¡0:04q + 65 = 00:04q = 65

q = 1625

The number of units that produces maximum pro…tis 1625.

(b) If q = 1625;

p = 90¡ 0:02(1625)= 57:5

The price per unit that produces maximum pro…tis $57.50.

(c) P (13) = ¡0:02(1625)2 + 65(1625)¡ 5000= 47,812.5

The maximum pro…t is $47,812.50.

43. C(q) = 100 + 20qe¡0:01q; p = 40e¡0:01q

P (q) = R(q)¡C(q) = pq ¡C(q)= (40e¡0:01q)q ¡ (100 + 20qe¡0:01q)= 20qe¡0:01q ¡ 100

(a) P 0(q) = 20e¡0:01q + 20qe¡0:01q(¡0:01)= (20¡ 0:2q)e¡0:01q

Solve P 0(q) = 0:

(20¡ 0:2q)e¡0:01q = 020¡ 0:2q = 0

q = 100

Since e¡0:01q > 0 for all values of q, the sign ofP 0(q) is the same as the sign of 20 ¡ 0:2q: Forq < 100; P 0(q) > 0; for q > 100; P 0(q) < 0: There-fore, the number of units that produces maximumpro…t is 100.

(b) If q = 100;

p = 40e¡0:01(100)

= 40e¡1

¼ 14:72The price per unit that produces maximum pro…tis $14.72.

(c) P (100) = 20(100)e¡0:01(100) ¡ 100= 2000e¡1 ¡ 100¼ 635:76

The maximum pro…t is $635.76.

44. C(q) = 21:047q + 3; p = 50¡ 5 ln(q + 10)P (q) = R(q)¡C(q) = pq ¡C(q)

= [50¡ 5 ln(q + 10)]q ¡ (21:047q + 3)= 28:953q ¡ 3¡ 5q ln(q + 10)

(a) P 0(q) = 28:953¡ 5 ln(q + 10)¡ 5qq+10

Use a graphing calculator to …nd the critical pointsfor q > 0: The critical point is 120. For q <120; P 0(q) > 0; for q > 120; P 0(q) < 0. There-fore, the number of units that produces maximumpro…t is 120.

(b) If q = 120;

p = 50¡ 5 ln(120 + 10)¼ 25:66

The price per unit that produces maximum pro…tis $25.66.

(c) P (120) = 28:953(120)¡ 3¡ 5(120) ln(120 + 10)¼ 550:84

The maximum pro…t is $550.84.

45. P (t) = ¡0:01432t3 + 0:3976t2 ¡ 2:257t+ 23:41P 0(t) = ¡0:04296t2 + 0:7952t¡ 2:257Solve for P 0(t) = 0 :

t ¼ 3:5001 or t ¼ 15:0101Test points t = 1; t = 4; t = 16:

P 0(1) ¼ ¡1:5048P 0(4) ¼ 0:2364P 0(16) ¼ ¡0:5316


P is decreasing on (0; 3:5) and (15; 18) and in-creasing on (3:5; 15):

P (0) = 23:41

P (3:5) ¼ 19:767P (15) ¼ 30:685P (18) ¼ 28:092

Relative maximum of 23,410 megawatts at mid-night; relative minimum of 19,767 megawatts at3:30 a.m.; relative maximum of 30,685 at 3:00p.m.; relative minimum of 28,092 at 6:00 p.m.

46. P (x) = ln(¡x3 + 3x2 + 72x+ 1), for x in [0; 10]:

(a) P 0(x) =¡3x2 + 6x+ 72

¡x3 + 3x2 + 72x+ 1=

¡3(x¡ 6)(x+ 4)¡x3 + 3x2 + 72x+ 1

P 0(x) = 0 when x = 6 or x = ¡4, but 4 is notin the domain of [0; 10]. P 0(x) fails to exist forx ¼ ¡0:14 and x ¼ 10:123, neither of which are inthe domain of [0; 10]. Thus, the critical number is6.Use the …rst derivative test to verify that x = 6

gives a maximum pro…t.

P 0(5) =27

311> 0

P 0(7) = ¡ 11

103< 0

The maximum pro…t results when 6 units are sold.

(b) P (6) ¼ 5:784The maximum pro…t is about $5784.

47. p = D(q) = 200e¡0:1q

R(q) = pq

= 200qe¡0:1q

R0(q) = 200qe¡0:1q(¡0:1) + 200e¡0:1q= 20e¡0:1q(10¡ q)

R0(q) = 0 when q = 10 , the only critical number.Use the …rst derivative test to verify that q = 10gives the maximum revenue.

R0(9) = 20e¡0:9 > 0R0(11) = ¡20e¡1:1 < 0

The maximum revenue results when q = 10

p = D(10) = 200e ¼ 73:58, or when telephones are

sold at $73.58.

48. p = D(q) = 500qe¡0:0016q2

R(q) = pq = (500qe¡0:0016q2

)q

= 500q2e¡0:0016q2

R0(q) = 500e¡0:0016q2

(2q) + 500q2e¡0:0016q2(¡0:0016¢2q)

= (1000q ¡ 1:6q3)e¡0:0016q2

Since e¡:0016q2

> 0 for all values of q,

R0(q) = 0(1000q ¡ 1:6q3)e¡0:0016q2 = 0

1000q ¡ 1:6q3 = 01:6q3 = 1000q

q2 = 625

q = 25

If q < 25, R0(q) > 0; if q > 25, R0(q) < 0. So phas a maximum value at q = 25:

p = D(25) = 500(25)e¡0:0016(25)2 ¼ 4600

Maximum revenue occurs when the price is about$4600 and 25 computer systems are sold.

49. C(x) = 0:002x3 = 9x+ 6912

C(x) =C(x)

x= 0:002x2 + 9+

6912

x

C0(x) = 0:004x¡ 6912

x2

C0(x) = 0 when

0:004x¡ 6912x2

= 0

0:004x3 = 6912

x3 = 1,728,000x = 120

A product level of 120 units will produce the min-imum average cost per unit.

50. a(t) = 0:008t3 ¡ 0:288t2 + 2:304t+ 7a0(t) = 0:024t2 ¡ 0:576t+ 2:304Set a0 = 0 and use the quadratic formula to solvefor t.

0:024t2 ¡ 0:576t+ 2:304 = 00:024(t2 ¡ 24t+ 96) = 0

t2 ¡ 24t+ 96 = 0t ¼ 5:07 or t ¼ 18:93

t = 5:07 = 5 hours + 0:07¢60 minutes correspondsto 5:04 P.M.t = 18:93 = 18 hours + 0.93 ¢ 60 minutes corre-sponds to 6:56 A.M.


51. (a) M(t) = 6:281t0:242e¡0:025t

M 0(t) = (1:520002t¡0:758)(e¡0:025t)+ (6:281t0:242)(¡0:025e¡0:025t)

= e¡0:025t(1:520002t¡0:758 ¡ 0:157025t0:242)M 0(t) = 0 when

1:520002t¡0:758 ¡ 0:157025t0:242 = 0t = 9:68

Let t = 9:68 in M(t):

M(9:68) = 6:281(9:68)0:242e¡0:025(9:68)

¼ 8:54 kgThe maximum daily consumption is 8.54 kg andit occurs at 9.68 weeks.

(b) M(t) = atbe¡ct

M 0(t) = (batb¡1)(e¡ct) + (atb)(¡ce¡ct)= ae¡ct(btb¡1 ¡ ctb)

M 0(t) = 0 when

btb¡1 ¡ ctb = 0btb¡1 = ctb

b

c= t

Let t =b

cin M(t):

M

μb

c

¶= a

μb

c

¶be¡c³bc

´

= a

μb

c

¶be¡b

The maximum daily consumption is a(b=c)be¡b kgand it occurs at b=c weeks.

52. M(t) = 369(0:93)t(t)0:36

M 0(t) = (369)(0:93)t ln(0:93)(t0:36)

+ 369(0:93)t(0:36)(t)¡0:64

= (369t0:36)(0:93t ln 0:93) +132:84(0:93)t

t0:64

M 0(t) = 0 when t ¼ 4:96:Verify that t ¼ 4:96 gives a maximum.

M 0(4) > 0M 0(5) < 0

Find M(4:96)

M(4:96) = 369(0:93)4:96(4:96)0:36 ¼ 458:22The female moose reaches a maximum weight ofabout 458.22 kilograms at about 4.96 years.

53. F (t) = ¡10:28 + 175:9te¡t=1:3

F 0(t) = (175:9t)μ¡ 1

1:3

¶e¡t=1:3 + (175:9)(e¡t=1:3)

= e¡t=1:3μ¡175:9t

1:3+ 175:9

¶F 0(t) = 0 when

¡175:9t1:3

+ 175:9 = 0

t = 1:3

At 1.3 hours, the termal e¤ect of the food is max-imized.

54. D(x) = ¡x4 + 8x3 + 80x2D0(x) = ¡4x3 + 24x2 + 160x

= ¡4x(x2 ¡ 6x¡ 40)= ¡4x(x+ 4)(x¡ 10)

D0(x) = 0 when x = 0; x = ¡4; or x = 10:Disregard the nonpositive values.Verify that x = 10 gives a maximum.

D0(9) = 468 > 0D0(11) = ¡660 < 0

The speaker should aim for a degree of discrepancyof 10.

55. R(t) =20t

t2 + 100

R0(t) =20(t2 + 100)¡ 20t(2t)

(t2 + 100)2=2000¡ 20t2(t2 + 100)2

R0(t) = 0 when

2000¡ 20t2 = 0¡20t2 = ¡2000

t2 = 100

t = §10:Disregard the negative value.Use the …rst derivative test to verify that t = 10gives a maximum rating.

R0(9) = 0:0116 > 0R0(11) = ¡0:0086 < 0

The …lm should be 10 minutes long.

56. s(t) = ¡16t2 + 64t+ 3s0(t) = ¡32t+ 64(a) When s0(t) = 0;

0 = ¡32t+ 6432t = 64

t = 2:

Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 319

Verify that t = 2 gives a maximum.

s0(1) = 32 > 0s0(3) = ¡32 < 0

Now …nd the height when t = 2:

s(2) = ¡16(2)2 + 64(2) + 3= 67

Therefore, the maximum height is 67 ft.

(b) The cork remains in the air as long as s(t) > 0:So, solve the equation s(t) = 0 using the quadraticformula.

s(t) = 0

¡16t2 + 64t+ 3 = 0

t =¡64§p642 ¡ 4(¡16)(3)

2(¡16)

t =8§p67

4

t ¼ ¡0:05; 4:05The cork remains in the air for about 4.05 seconds.

5.3 Higher Derivatives, Concavity,and the Second Derivative Test

1. f(x) = 5x3 ¡ 7x2 + 4x+ 3f 0(x) = 15x2 ¡ 14x+ 4f 00(x) = 30x¡ 14f 00(0) = 30(0)¡ 14 = ¡14f 00(2) = 30(2)¡ 14 = 46

2. f(x) = 4x3 + 5x2 + 6x¡ 7f 0(x) = 12x2 + 10x+ 6f 00(x) = 24x+ 10f 00(0) = 24(0) + 10 = 10f 00(2) = 24(2) + 10 = 58

3. f(x) = 4x4 ¡ 3x3 ¡ 2x2 + 6f 0(x) = 16x3 ¡ 9x2 ¡ 4xf 00(x) = 48x2 ¡ 18x¡ 4f 00(0) = 48(0)2 ¡ 18(0)¡ 4 = ¡4f 00(2) = 48(2)2 ¡ 18(2)¡ 4 = 152

4. f(x) = ¡x4 + 7x3 ¡ x2

2f 0(x) = ¡4x3 + 21x2 ¡ xf 00(x) = ¡12x2 + 42x¡ 1f 00(0) = ¡12(0)2 + 42(0)¡ 1 = ¡1f 00(2) = ¡12(2)2 + 42(2)¡ 2 = 35

5. f(x) = 3x2 ¡ 4x+ 8f 0(x) = 6x¡ 4f 00(x) = 6

f 00(0) = 6f 00(2) = 6

6. f(x) = 8x2 + 6x+ 5f 0(x) = 16x+ 6f 00(x) = 16f 00(0) = 16f 00(2) = 16

7. f(x) =x2

1 + x

f 0(x) =(1 + x)(2x)¡ x2(1)

(1 + x)2

=2x+ x2

(1 + x)2

f 00(x) =(1 + x)2(2 + 2x)¡ (2x+ x2)(2)(1 + x)

(1 + x)4

=(1 + x)(2 + 2x)¡ (2x+ x2)(2)

(1 + x)3

=2

(1 + x)3

f 00(0) = 2

f 00(2) =2

27

8. f(x) =¡x1¡ x2

f 0(x) =(1¡ x2)(¡1)¡ (¡x)(¡2x)

(1¡ x2)2

=¡1 + x2 ¡ 2x2(1¡ x2)2

=¡1¡ x2(1¡ x2)2

f 00(x) =(1¡x2)2(¡2x)¡(¡1¡x2)(2)(1¡x2)(¡2x)

(1¡ x2)4

=(1¡ x2)[¡2x(1¡ x2) + 4x(¡1¡ x2)]

(1¡ x2)4

=¡2x3 ¡ 6x(1¡ x2)3

=¡2x(x2 + 3)(1¡ x2)3


f 00(0) =¡2(0)(02 + 3)(1¡ 02)3 = 0

f 00(2) =¡2(2)(4 + 3)(1¡ 22)3

=¡28¡27

=28

27

9. f(x) =px2 + 4 = (x2 + 4)1=2

f 0(x) =1

2(x2 + 4)¡1=2 ¢ 2x

=x

(x2 + 4)1=2

f 00(x) =(x2 + 4)1=2(1)¡ x £12(x2 + 4)¡1=2¤ 2x

x2 + 4

=(x2 + 4)1=2 ¡ x2

(x2+4)1=2

x2 + 4

=(x2 + 4)¡ x2(x2 + 4)3=2

=4

(x2 + 4)3=2

f 00(0) =4

(02 + 4)3=2

=4

43=2=4

8=1

2

f 00(2) =4

(22 + 4)3=2

=4

83=2=

4

16p2=

1

4p2

10. f(x) =p2x2 + 9

= (2x2 + 9)1=2

f 0(x) =1

2(2x2 + 9)¡1=2 ¢ 4x

=2x

(2x2 + 9)1=2

f 00(x) =(2x2 + 9)1=2(2)¡ 2x £12(2x2 + 9)¡1=2¤ 4x

2x2 + 9

=2(2x2 + 9)1=2 ¡ 4x2

(2x2+9)1=2

2x2 + 9

=2(2x2 + 9)¡ 4x2(2x2 + 9)3=2

=18

(2x2 + 9)3=2

f 00(0) =18

[2(0)2 + 9]3=2

=18

93=2=18

27=2

3

f 00(2) =18

[2(2)2 + 9]3=2

=18

173=2=

18

17p17

11. f(x) = 32x3=4

f 0(x) = 24x¡1=4

f 00(x) = ¡6x¡5=4 = ¡ 6

x5=4

f 00(0) does not exist.

f 00(2) = ¡ 6

25=4

= ¡ 3

21=4

12. f(x) = ¡6x1=3

f 0(x) = ¡2x¡2=3

f 00(x) =4

3x¡5=3 =

4

3x5=3

f 00(0) does not exist.

f 00(2) =4

3(25=3)=21=3

3

13. f(x) = 5e¡x2

f 0(x) = 5e¡x2

(¡2x) = ¡10xe¡x2f 00(x) = ¡10xe¡x2(¡2x) + e¡x2(¡10)

= 20x2e¡x2 ¡ 10e¡x2

f 00(0) = 20(02)e¡02 ¡ 10e

= 0¡ 10 = ¡10f 00(2) = 20(22)e¡(2

2) ¡ 10e¡(22)= 80e¡4 ¡ 10e¡4 = 70e¡4¼ 1:282

14. f(x) = 0:5ex2

f 0(x) = (0:5)(2x)ex2

= xex2

f 00(x) = xex2

(2x) + ex2

(1) = 2x2ex2

+ ex2

or ex2

(2x2 + 1)

f 00(0) = e02

[2(0)2 + 1] = 1(1) = 1

f 00(2) = e22

[2(2)2 + 1] = e4(9) = 9e4

¼ 491:4


15. f(x) =ln x

4x

f 0(x) =4x¡1x

¢¡ (ln x)(4)(4x)2

=4¡ 4 ln x16x2

=1¡ ln x4x2

f 00(x) =4x2

¡¡ 1x

¢¡ (1¡ ln x)8x16x4

=¡4x¡ 8x+ 8x ln x

16x4=¡12x+ 8x ln x

16x4

=4x(¡3 + 2 ln x)

16x4=¡3 + 2 ln x

4x3

f 00(0) does not exist because ln 0 is unde…ned.

f 00(2) =¡3 + 2 ln 24(2)3

=¡3 + 2 ln 2

32¼ 0:050

16. f(x) = ln x+1

x

f 0(x) =1

x¡ 1

x2

f 00(x) =¡1x2+2

x3

=¡xx3

+2

x3=2¡ xx3

f 00(0) does not exist since thedenominator is 0.

f 00(2) =2¡ 223

=0

8= 0

17. f(x) = 7x4 + 6x3 + 5x2 + 4x+ 3

f 0(x) = 28x3 + 18x2 + 10x+ 4f 00(x) = 84x2 + 36x+ 10f 000(x) = 168x+ 36f(4)(x) = 168

18. f(x) = ¡2x4 + 7x3 + 4x2 + xf 0(x) = ¡8x3 + 21x2 + 8x+ 1f 00(x) = ¡24x2 + 42x+ 8f 000(x) = ¡48x+ 42f(4)(x) = ¡48

19. f(x) = 5x5 ¡ 3x4 + 2x3 + 7x2 + 4f 0(x) = 25x4 ¡ 12x3 + 6x2 + 14xf 00(x) = 100x3 ¡ 36x2 + 12x+ 14f 000(x) = 300x2 ¡ 72x+ 12f(4)(x) = 600x¡ 72

20. f(x) = 2x5 + 3x4 ¡ 5x3 + 9x¡ 2f 0(x) = 10x4 + 12x3 ¡ 15x2 + 9f 00(x) = 40x3 + 36x2 ¡ 30xf 000(x) = 120x2 + 72x¡ 30f (4)(x) = 240x+ 72

21. f(x) =x¡ 1x+ 2

f 0(x) =(x+ 2)¡ (x¡ 1)

(x+ 2)2=

3

(x+ 2)2

f 00(x) =¡3(2)(x+ 2)(x+ 2)4

=¡6

(x+ 2)3

f 000(x) =(¡6)(¡3)(x+ 2)2

(x+ 2)6

= 18(x+ 2)¡4 or18

(x+ 2)4

f (4)(x) =¡18(4)(x+ 2)3

(x+ 2)8

= ¡72(x+ 2)¡5 or¡72

(x+ 2)5

22. f(x) =x+ 1

x

f 0(x) =(1)(x)¡ 1(x+ 1)

x2

= ¡ 1

x2= ¡x¡2

f 00(x) = 2x¡3 or2

x3

f 000(x) = ¡6x¡4 or ¡ 6

x4

f (4)(x) = 24x¡5 or24

x5

23. f(x) =3x

x¡ 2f 0(x) =

(x¡ 2)(3)¡ 3x(1)(x¡ 2)2 =

¡6(x¡ 2)2

f 00(x) =¡6(¡2)(x¡ 2)(x¡ 2)4 =

12

(x¡ 2)3

f 000(x) =¡12(3)(x¡ 2)2

(x¡ 2)6 = ¡36(x¡ 2)¡4

or¡36

(x¡ 2)4

f (4)(x) =¡36(¡4)(x¡ 2)3

(x¡ 2)8 = 144(x¡ 2)¡5

or144

(x¡ 2)5


24. f(x) =x

2x+ 1

f 0(x) =(1)(2x+ 1)¡ (2)(x)

(2x+ 1)2

=1

(2x+ 1)2= (2x+ 1)¡2

f 00(x) = ¡2(2x+ 1)¡3(2)= ¡4(2x+ 1)¡3

f 000(x) = 12(2x+ 1)¡4(2)= 24(2x+ 1)¡4

or24

(2x+ 1)4

f(4)(x) = ¡96(2x+ 1)¡5(2)= ¡192(2x+ 1)¡5

or¡192

(2x+ 1)5

25. f(x) = ln x

(a) f 0(x) =1

x= x¡1

f 00(x) = ¡x¡2 = ¡1x2

f 000(x) = 2x¡3 =2

x3

f (4)(x) = ¡6x¡4 = ¡6x4

f (5)(x) = 24x¡5 =24

x5

(b) f (n)(x) =(¡1)n¡1(n¡ 1)!

xn

26. f(x) = ex

f 0(x) = ex

f 00(x) = ex

f 000(x) = ex

f(n)(x) = ex

27. Concave upward on (2;1)Concave downward on (¡1; 2)In‡ection point at (2; 3)

28. Concave upward on (¡1; 3)Concave downward on (3;1)In‡ection point at (3; 7)

29. Concave upward on (¡1;¡1) and (8;1)Concave downward on (¡1; 8)In‡ection point at (¡1; 7) and (8; 6)

30. Concave upward on (¡2; 6)Concave downward on (¡1;¡2) and (6;1)In‡ection point at (¡2;¡4) and (6;¡1)

31. Concave upward on (2;1)Concave downward on (¡1; 2)No points of in‡ection

32. Concave upward on (¡1; 0)Concave downward on (0;1)No in‡ection points

33. f(x) = x2 + 10x¡ 9f 0(x) = 2x+ 10f 00(x) = 2 > 0 for all x:

Always concave upwardNo in‡ection points

34. f(x) = 8¡ 6x¡ x2f 0(x) = ¡6¡ 2xf 00(x) = ¡2 < 0 for all x:Always concave downwardNo in‡ection points

35. f(x) = ¡2x3 + 9x2 + 168x¡ 3f 0(x) = ¡6x2 + 18x+ 168f 00(x) = ¡12x+ 18f 00(x) = ¡12x+ 18 > 0

¡6(2x¡ 3) > 02x¡ 3 < 0

x <3

2:

when

Concave upward on¡¡1; 32¢

f 00(x) = ¡12x+ 18 < 0 when¡6(2x¡ 3) < 0

2x¡ 3 > 0

x >3

2:

Concave downward on¡32 ;1

¢f 00(x) = ¡12x+ 18 = 0 when

¡6(2x+ 3) = 02x+ 3 = 0

x =3

2:

f

μ3

2

¶=525

2

In‡ection point at¡32 ;

5252

¢


36. f(x) = ¡x3 ¡ 12x2 ¡ 45x+ 2f 0(x) = ¡3x2 ¡ 24x¡ 45f 00(x) = ¡6x¡ 24f 00(x) = ¡6x¡ 24 > 0 when

¡6(x+ 4) > 0x+ 4 < 0

x < ¡4:

Concave upward on (¡1;¡4)

f 00(x) = ¡6x¡ 24 < 0 when¡6(x+ 4) < 0

x+ 4 > 0

x > ¡4:

Concave downward on (¡4;1)

f 00(x) = ¡6x¡ 24 = 0 when¡6(x+ 4) = 0

x = ¡4:f(¡4) = 54

In‡ection point at (¡4; 54)

37. f(x) =3

x¡ 5f 0(x) =

¡3(x¡ 5)2

f 00(x) =¡3(¡2)(x¡ 5)(x¡ 5)4 =

6

(x¡ 5)3

f 00(x) =6

(x¡ 5)3 > 0 when(x¡ 5)3 > 0x¡ 5 > 0

x > 5:

Concave upward on (5;1)

f 00(x) =6

(x¡ 5)3 < 0 when

(x¡ 5)3 < 0x¡ 5 < 0

x < 5:

Concave downward on (¡1; 5)

f 00(x) 6= 0 for any value for x; it does not existwhen x = 5: There is a change of concavity there,but no in‡ection point since f(5) does not exist.

38. f(x) =¡2x+ 1

= ¡2(x+ 1)¡1

f 0(x) = 2(x+ 1)¡2

f 00(x) = ¡4(x+ 1)¡3 = ¡4(x+ 1)3

f 00(x) =¡4

(x+ 1)3> 0 when

x+ 1 < 0

x < ¡1:Concave upward on (¡1;¡1)

f 00(x) =¡4

(x+ 1)3< 0 when

x+ 1 > 0

x > ¡1:Concave downward on (¡1;1)f 00(x) 6= 0 for any value of x; it does not exist whenx = ¡1: There is a change of concavity there, butno in‡ection point since f(¡1) does not exist.

39. f(x) = x(x+ 5)2

f 0(x) = x(2)(x+ 5) + (x+ 5)2

= (x+ 5)(2x+ x+ 5)

= (x+ 5)(3x+ 5)

f 00(x) = (x+ 5)(3) + (3x+ 5)= 3x+ 15 + 3x+ 5 = 6x+ 20

f 00(x) = 6x+ 20 > 0 when2(3x+ 10) > 0

3x > ¡10

x > ¡103:

Concave upward on¡¡10

3 ;1¢

f 00(x) = 6x+ 20 < 0 when2(3x+ 10) < 0

3x < ¡10

x < ¡103:

Concave downward on¡¡1;¡10

3

¢f

μ¡103

¶= ¡10

3

μ¡103+ 5

¶2

=¡103

μ¡10 + 153

¶2

= ¡103¢ 259= ¡250

27

In‡ection point at¡¡10

3 ;¡25027

¢


40. f(x) = ¡x(x¡ 3)2f 0(x) = ¡1(x¡ 3)2 + 2(x¡ 3)(¡x)

= ¡(x¡ 3)2 ¡ 2x2 + 6xf 00(x) = ¡2(x¡ 3)¡ 4x+ 6

= ¡2x+ 6¡ 4x+ 6= ¡6x+ 12

f 00(x) = ¡6x+ 12 > 0 when¡6(x¡ 2) > 0

x¡ 2 < 0x < 2:

Concave upward on (¡1; 2)

f 00(x) = ¡6x+ 12 < 0 when¡6(x¡ 2) < 0

x¡ 2 > 0x > 2:

Concave downward on (2;1)

f 00(x) = ¡6x+ 12 = 0 when x = 2:f(2) = ¡2

In‡ection point at (2;¡2)

41. f(x) = 18x¡ 18e¡xf 0(x) = 18¡ 18e¡x(¡1) = 18 + 18e¡xf 00(x) = 18e¡x(¡1) = ¡18e¡x

f 00(x) = ¡18e¡x < 0 for all x

f(x) is never concave upward and always concavedownward. There are no points of in‡ection since¡18e¡x is never equal to 0.

42. f(x) = 2e¡x2

f 0(x) = 2e¡x2

(¡2x) = ¡4xe¡x2f 00(x) = ¡4xe¡x2(¡2x) + e¡x2(¡4)

= ¡4e¡x2(¡2x2 + 1)

f 00(x) = 0 when ¡2x2 + 1 = 01 = 2x2

1

2= x2

§p2

2= x

Check the sign of f 00(x) in each of the intervals

determined by x =p22 and x = ¡

p22 using test

points.

f 00(¡1) = ¡4e¡(¡1)2 [¡2(¡1)2 + 1]

=¡4e(¡1) = 4

e> 0

f 00(0) = ¡4e¡02 [¡2(0)2 + 1]= ¡4(1) = ¡4 < 0

f 00(1) = ¡4e¡12 [¡2(1)2 + 1]

=¡4e(¡1) = 4

e> 0

Concave upward on³¡1;¡

p22

´and

³p22 ;1

´;

concave downward on³¡p22 ;

p22

´:

f³§p22

´= 2e¡1=2 =

2pe

In‡ection points at³¡p22 ;

2pe

´and

³p22 ;

2pe

´

43. f(x) = x8=3 ¡ 4x5=3

f 0(x) =8

3x5=3 ¡ 20

3x2=3

f 00(x) =40

9x2=3 ¡ 40

9x¡1=3 =

40(x¡ 1)9x1=3

f 00(x) = 0 when x = 1f 00(x) fails to exist when x = 0Note that both f(x) and f 0(x) exist at x = 0.Check the sign of f 00(x) in the three intervals de-termined by x = 0 and x = 1 using test points.

f 00(¡1) = 40(¡2)9(¡1) =

80

9> 0

f 00μ1

8

¶=40¡¡7

8

¢9¡12

¢ = ¡709< 0

f 00(8) =40(7)

9(2)=140

9> 0

Concave upward on (¡1; 0) and (1;1); concavedownward on (0; 1)

f(0) = (0)8=3 ¡ 4(0)5=3 = 0f(1) = (1)8=3 ¡ 4(1)5=3 = ¡3

In‡ection points at (0; 0) and (1;¡3)


44. f(x) = x7=3 + 56x4=3

f 0(x) =7

3x4=3 +

224

3x1=3

f 00(x) =28

9x1=3 +

224

9x¡2=3

=28(x+ 8)

9x2=3

f 00(x) = 0 when x = ¡8f 00(x) fails to exist when x = 0Note that both f(x) and f 0(x) exist at x = 0:Check the sign of f 00(x) in three intervals deter-mined by x = ¡8 and x = 0 using test points.

f 00(¡27) = 28(¡19)9(9)

= ¡53281

< 0

f 00(¡1) = 28(7)

9(1)=196

9> 0

f 00(1) =28(9)

9(1)= 28 > 0

Concave upward on (¡8;1); concave downwardon (¡1;¡8)

f(¡8) = (¡8)7=3 + 56(¡8)4=3 = ¡128 + 896= 768

In‡ection point at (¡8; 768)45. f(x) = ln(x2 + 1)

f 0(x) =2x

x2 + 1

f 00(x) =(x2 + 1)(2)¡ (2x)(2x)

(x2 + 1)2

=¡2x2 + 2(x2 + 1)2

f 00(x) =¡2x2 + 2(x2 + 1)2

> 0 when

¡2x2 + 2 > 0¡2x2 > ¡2x2 < 1

¡1 < x < 1Concave upward on (¡1; 1)

f 00(x) =¡2x2 + 2(x2 + 1)2

< 0 when

¡2x2 + 2 < 0¡2x2 < ¡2x2 > 1

x > 1 or x < ¡1

Concave downward on (¡1;¡1) and (1;1)

f(1) = ln[(1)2 + 1] = ln 2

f(¡1) = ln[(¡1)2 + 1] = ln 2

In‡ection points at (¡1; ln 2) and (1; ln 2)

46. f(x) = x2 + 8 ln jx+ 1j

f 0(x) = 2x+8

x+ 1

f 00(x) = 2¡ 8

(x+ 1)2

f 00(x) = 2¡ 8

(x+ 1)2> 0 when

2¡ 8

(x+ 1)2> 0

2 >8

(x+ 1)2

(x+ 1)2 > 4

x+ 1 > 2 or x+ 1 < ¡2x > 1 or x < ¡3

Concave upward on (¡1;¡3) and (1;1)

f 00(x) = 2¡ 8

(x+ 1)2< 0 when

2¡ 8

(x+ 1)2< 0

2 <8

(x+ 1)2

(x+ 1)2 < 4

¡2 < x+ 1 < 2¡3 < x < 1

Note that f(x) is not de…ned at x = ¡1.Concave downward on (¡3;¡1) and (¡1; 1)

f(¡3) = (¡3)2 + 8 ln j¡3 + 1j = 9 + 8 ln 2f(1) = (1)2 + 8 ln j1 + 1j = 1 + 8 ln 2

In‡ection points at (¡3; 9 + 8 ln 2) and(1; 1 + 8 ln 2)


47. f(x) = x2 log jxj

f 0(x) = 2x log jxj+ x2μ

1

x ln 10

¶

= 2x log jxj+ x

ln 10

f 00(x) = 2 log jxj+ 2xμ

1

x ln 10

¶+

1

ln 10

= 2 log jxj+ 3

ln 10

f 00(x) = 2 log jxj+ 3

ln 10> 0 when

2 log jxj+ 3

ln 10> 0

2 log jxj > ¡ 3

ln 10

log jxj > ¡ 3

2 ln 10

ln jxjln 10

> ¡ 3

2 ln 10

ln jxj > ¡32

jxj > e¡3=2x > e¡3=2 or x < ¡e¡3=2

Concave upward on (¡1;¡e¡3=2) and (e¡3=2;1)

f 00(x) = 2 log jxj+ 3

ln 10< 0 when

2 log jxj+ 3

ln 10< 0

2 log jxj < ¡ 3

ln 10

log jxj < ¡ 3

2 ln 10

ln jxjln 10

< ¡ 3

2 ln 10

ln jxj < ¡32

jxj < e¡3=2¡e¡3=2 < x < e¡3=2

Note that f(x) is not de…ned at x = 0:

Concave downward on (¡e¡3=2; 0) and (0; e¡3=2):f(¡e¡3=2) = (¡e¡3=2)2 log ¯̄¡e¡3=2¯̄

= e¡3 log e¡3=2 = ¡ 3e¡3

2 ln 10

f(e¡3=2) = (e¡3=2)2 log¯̄e¡3=2

¯̄= e¡3 log e¡3=2 = ¡ 3e¡3

2 ln 10

In‡ection points at³¡e¡3=2;¡ 3e¡3

2 ln 10

ánd

³e¡3=2;¡ 3e¡3

2 ln 10

´48. f(x) = 5¡x

2

f 0(x) = (ln 5)5¡x2 ¢ (¡2x)

= ¡(2x ln 5)5¡x2

f 00(x) = ¡2 ln 5 ¢ 5¡x2 + (¡2 ln 5)x ¢ (ln 5)(5¡x2)(¡2x)= (2 ln 5)5¡x

2

[2 ln 5(x2)¡ 1]f 00(x) = (2 ln 5)5¡x

2

[2 ln 5(x2)¡ 1] > 0 when2 ln 5(x2)¡ 1 > 0

2 ln 5(x2) > 1

x2 >1

2 ln 5

x >1p2 ln 5

or x < ¡ 1p2 ln 5

Concave upward on³¡1;¡ 1p

2 ln 5

ánd

³1p2 ln 5

;1´

f 00(x) = (2 ln 5)5¡x2

[2 ln 5(x2)¡ 1] < 0 when2 ln 5(x2)¡ 1 < 0

2 ln 5(x2) < 1

x2 <1

2 ln 5

¡ 1p2 ln 5

< x <1p2 ln 5

Concave downward on³¡ 1p

2 ln 5; 1p

2 ln 5

´

f

μ¡ 1p

2 ln 5

¶= 5¡(¡1=

p2 ln 5)2 = 5¡1=(2 ln 5) = e¡1=2

f

μ1p2 ln 5

¶= 5¡(1=

p2 ln 5)2 = 5¡1=(2 ln 5) = e¡1=2

In‡ection points at³¡ 1p

2 ln 5; e¡1=2

ánd³

1p2 ln 5

; e¡1=2´

49. Since the graph of f 0(x) is increasing on (¡1; 0)and (4;1); the function is concave upward on(¡1; 0) and (4;1): Since the graph of f 0(x) isdecreasing on (0; 4); the function is concave down-ward on (0; 4). The in‡ection points are at 0 and4.

50. Since the graph of f 0(x) is increasing on (0; 5);the function is concave upward on (0; 5): Sincethe graph of f 0(x) is decreasing on (¡1; 0) and(5;1); the function is concave downward on (¡1; 0)and (5;1). The in‡ection points are at 0 and 5.


51. Since the graph of f 0(x) is increasing on (¡7; 3)and (12;1), the function is concave upward on(¡7; 3) and (12;1). Since the graph of f 0(x) is de-creasing on (¡1;¡7) and (3; 12), the function isconcave downward on (¡1;¡7) and (3; 12). Thein‡ection points are at ¡7; 3, and 12.

52. Since the graph of f 0(x) is increasing on (¡1;¡5)and (1; 9), the function is concave upward on (¡1;¡5)and (1; 9). Since the graph of f 0(x) is decreas-ing on (¡5; 1) and (9;1); the function is concavedownward on (¡5; 1) and (9;1). The in‡ectionpoints are at ¡5; 1; and 9.

53. Choose f(x) = xk, where 1 < k < 2.

If k = 43 , then

f 0(x) =4

3x1=3 f 00(x) =

4

9x¡2=3 =

4

9x2=3

Critical number: 0Since f 0(x) is negative when x < 0 and positivewhen x > 0, f(x) = x4=3 has a relative minimumat x = 0.

If k = 53 ; then

f 0(x) =5

3x2=3 f 00(x) =

10

9x¡1=3 =

10

9x1=3

f 00(x) is never 0, and does not exist when x = 0;so, the only candidate for an in‡ection point is atx = 0.Since f 00(x) is negative when x < 0 and positivewhen x > 0, f(x) = x5=3 has an in‡ection pointat x = 0:

54. (a)

(b) Both f(x) and g(x) are concave down on (¡1; 0)and concave up on (0;1). Thus, they both havea point of in‡ection at (0; 0).

(c) f(x) = x7=3

f 0(x) =7

3x4=3

f 00(x) =28

9x1=3

f 00(0) = 0

g(x) = x5=3

g0(x) =5

3x2=3

g00(x) =10

9x¡1=3

g00(0) is unde…ned

(d) No.

55. (a) The slope of the tangent line to f(x) = ex

as x ! ¡1 is close to 0 since the tangent lineis almost horizontal, and a horizontal line has aslope of 0.

(b) The slope of the tangent line to f(x) = ex

as x ! 0 is close to 1 since the …rst derivativerepresents the slope of the tangent line, f 0(x) =ex; and e0 = 1:

56. The slope of the tangent line to f(x) = ln x asx!1 approaches 0. The slope of the tangent lineto the graph of f(x) = ln x as x! 0 approaches1: For example, the slope of the tangent line atx = 0:00001 is 100,000.

57. f(x) = ¡x2 ¡ 10x¡ 25f 0(x) = ¡2x¡ 10

= ¡2(x+ 5) = 0

Critical number: ¡5f 00(x) = ¡2 < 0 for all x:

The curve is concave downward, which means arelative maximum occurs at x = ¡5:

58. f(x) = x2 ¡ 12x+ 36f 0(x) = 2x¡ 12

f 0(x) = 0 when

2x¡ 12 = 0x = 6:

Critical number: 6

f 00(x) = 2 > 0 for all x:

The curve is concave upward, which means a rel-ative minimum occurs at x = 6:


59. f(x) = 3x3 ¡ 3x2 + 1f 0(x) = 9x2 ¡ 6x

= 3x(3x¡ 2) = 0Critical numbers: 0 and 2

3

f 00(x) = 18x¡ 6f 00(0) = ¡6 < 0; which means that a relative max-imum occurs at x = 0:

f 00¡23

¢= 6 > 0; which means that a relative min-

imum occurs at x = 23 :

60. f(x) = 2x3 ¡ 4x2 + 2f 0(x) = 6x2 ¡ 8xf 0(x) = 0 when

6x2 ¡ 8x = 02x(3x¡ 4) = 0x = 0 or x =

4

3:

Critical numbers: 0, 43

f 00(x) = 12x¡ 8f 00(0) = ¡8 < 0; which means that a relative max-imum occurs at x = 0:f 00¡43

¢= 8 > 0; which means that a relative min-

imum occurs at x = 43 :

61. f(x) = (x+ 3)4

f 0(x) = 4(x+ 3)3 = 0

Critical number: x = ¡3f 00(x) = 12(x+ 3)2

f 00(¡3) = 12(¡3 + 3)2 = 0The second derivative test fails.Use the …rst derivative test.

f 0(¡4) = 4(¡4 + 3)2= 4(¡1)3 = ¡4 < 0

This indicates that f is decreasing on (¡1;¡3):

f 0(0) = 4(0 + 3)3

= 4(3)3 = 108 > 0

This indicates that f is increasing on (¡3;1):A relative minimum occurs at ¡3:

62. f(x) = x3

f 0(x) = 3x2

f 0(x) = 0 when

3x2 = 0

x = 0:

Critical number: 0

f 00(x) = 6xf 00(0) = 0

The second derivative test fails.

Use the …rst derivative test.

f 0(¡1) = 3(¡1)2 = 3 > 0This indicates that f is increasing on (¡1; 0):

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

This indicates that f is increasing on (0;1):Neither a relative maximum nor relative minimumoccurs at x = 0:

63. f(x) = x7=3 + x4=3

f 0(x) =7

3x4=3 +

4

3x1=3

f 0(x) = 0 when

7

3x4=3 +

4

3x1=3 = 0

x1=3

3(7x+ 4) = 0

x = 0 or x = ¡47:

Critical numbers: ¡47 ; 0

f 00(x) =28

9x1=3 +

4

9x¡2=3

f 00μ¡47

¶=28

9

μ¡47

¶1=3+4

9

μ¡47

¶¡2=3¼ ¡1:9363

Relative maximum occurs at ¡47 .

f 00(0) does not exist, so the second derivative testfails.



f 0μ¡12

¶=7

3

μ¡12

¶4=3+4

3

μ¡12

¶1=3¼ ¡0:1323

This indicates that f is decreasing on¡¡4

7 ; 0¢:

f 0(1) =7

3(1)4=3 +

4

3(1)1=3 =

11

3

This indicates that f is increasing on (0;1):Relative minimum occurs at 0.

64. f(x) = x8=3 + x5=3

f 0(x) =8

3x5=3 +

5

3x2=3

f 0(x) = 0 when

8

3x5=3 +

5

3x2=3 = 0

x2=3

3(8x+ 5) = 0

x = 0 or x = ¡58:

Critical numbers: ¡58 ; 0

f 00(x) =40

9x2=3 +

10

9x¡1=3

f 00μ¡58

¶=40

9

μ¡58

¶2=3+10

9

μ¡58

¶¡1=3¼ 1:9493

Relative minimum occurs at ¡58 :

f 00(0) does not exist, so the second derivative testfails.


f 0μ¡12

¶=8

3

μ¡12

¶5=3+5

3

μ¡12

¶2=3¼ 0:2100

This indicates that f is increasing on¡¡5

8 ; 0¢:

f 0(1) =8

3(1)5=3 +

5

3(1)2=3 =

13

3

This indicates that f is increasing on (0;1):Neither a relative minimum or maximum occursat 0.

65. There are many examples. The easiest is f(x) =px: This graph is increasing and concave down-

ward.f 0(x) = 1

2x¡1=2 = 1

2px

f 0(0) does not exist, while f 0(x) > 0 for all x > 0:(Note that the domain of f is [0;1).)As x increases, the value of f 0(x) decreases, butremains positive. It approaches zero, but neverbecomes zero or negative.

66. f 0(x) = x3 ¡ 6x2 + 7x+ 4f 00(x) = 3x2 ¡ 12x+ 7Graph f 0 and f 00 in the window [¡5; 5] by [¡5; 15];Xscl = 0.5.Graph of f 0:

Graph of f 00:

(a) f has relative extrema where f 0(x) = 0: Usethe graph to approximate the x-intercepts of thegraph of f 0. These numbers are the solutions ofthe equation f 0(x) = 0: We …nd that the criticalnumbers of f are about ¡0:4; 2:4; and 4.0.


By either looking at the graph of f 0 and applyingthe …rst derivative test or by looking at the graphof f 00 and applying the second derivative test, wesee that f has relative minima at about ¡0:4 and4.0 and a relative maximum at about 2.4.

(b) Examine the graph of f 0 to determine the in-tervals where the graph lies above and below thex-axis. We see that f 0(x) > 0 on about (¡0:4; 2:4)and (4:0;1); indicating that f is increasing onthe same intervals. We also see that f 0(x) < 0 onabout (¡1;¡0:4) and (2:4; 4:0); indicating thatf is decreasing on the same intervals.

(c) Examine the graph of f 00. We see that thisgraph has two x-intercepts, so there are two x-values where f 00(x) = 0: These x-values are about0.7 and 3.3. Because the sign of f 00 changes atthese two values, we see that the x-values of thein‡ection points of the graph of f are about 0:7and 3.3.

(d)We observe from the graph of f 00 that f 00(x) >0 on about (¡1; 0:7) and (3:3;1); so f is concaveupward on the same intervals.Likewise, we observe that f 00(x) < 0 on about(0.7, 3.3), so f is concave downward on the sameinterval.

67. f 0(x) = 10x2(x¡ 1)(5x¡ 3)= 10x2(5x2 ¡ 8x+ 3)= 50x4 ¡ 80x3 + 30x2

f 00(x) = 200x3 ¡ 240x2 + 60x= 20x(10x2 ¡ 12x+ 3)

Graph f 0 in the window [¡1; 1:5] by [¡2; 2];Xscl = 0.1.

This window does not give a good view of thegraph of f 00; so we graph f 00 in the window [¡1; 1:5]by [¡20; 20]; Xscl = 0.1. Yscl = 5.

(a) The critical numbers of f are the x-interceptsof the graph of f 0: (Note that there are no val-ues where f 0(x) does not exist.) From the graphor by examining the factored expression for f 0;wesee that the critical numbers of f are 0, 0.6, and1.

By either looking at the graph of f 0 and applyingthe …rst derivative test or by looking at the graphof f 00 and applying the second derivative test, wesee that f has a relative minimum at 1 and a rel-ative maximum at 0.6.(At x = 0; the second derivative test fails sincef 00(0) = 0; and the …rst derivative does not changesign, so there is no relative extremum at 0.)

(b) Examine the graph of f 0 to determine the in-tervals where the graph lies above and below thex-axis. We see that f 0(x) ¸ 0 on (¡1; 0:6); f 0(x) <0 on (0:6; 1); and f 0(x) > 0 on (1;1): Therefore,f is increasing on (¡1; 0:6) and (1;1) and de-creasing on (0:6; 1):

(c) Examine the graph of f 00: We see that thisgraph has three x-intercepts, so there are threevalues where f 00(x) = 0: These x-values are 0,about 0.36, and about 0.85. Because the sign off 00 and thus the concavity of f changes at thesethree values, we see that the x-values of the in-‡ection points of the graph of f are 0, about 0.36,and about 0.85.

(d)We observe from the graph of f 00 that f 00(x) >0 on (0; 0:36) and (0:85;1); so f is concave up-ward on the same intervals. Likewise, f 00(x) < 0on (¡1; 0) and (0:36; 0:85); so f is concave down-ward on the same intervals.


68. f 0(x) =1¡ x2(x2 + 1)2

f 00(x) =(x2+1)2(¡2x)¡(1¡x2)(2)(x2+1)2x

(x2 + 1)4

=¡2x(x2 + 1)[(x2 + 1) + 2(1¡ x2)]

(x2 + 1)4

=¡2x(3¡ x2)(x2 + 1)3

Graph f 0 and f 00 in the window [¡3; 3] by [¡1:5; 1; 5];Xscl = 0.2.Graph of f 0:

Graph of f 00:

(a) The critical numbers of f are the x-interceptsof the graph of f 0. (Note that there are no valueswhere f 0 does not exist.) We see from the graphthat these x-values are ¡1 and 1.By either looking at the graph of f 0 and applyingthe …rst derivative test or by looking at the graphof f 00 and applying the second derivative test, wesee that f has a relative minimum at ¡1 and arelative maximum at 1:

(b) Examine the graph of f 0 to determine theintervals where the graph lies above and belowthe x-axis. We see that f 0(x) > 0 on (¡1; 1);indicating that f is increasing on the same inter-val. We also see that f 0(x) < 0 on (¡1;¡1) and(1;1); indicating that f is decreasing on the sameintervals.

(c) Examine the graph of f 00: We see that thisgraph has three x-intercepts, so there are threevalues where f 00(x) = 0: These x-values are about¡1:7; 0, and about 1.7. Because the sign of f 00and thus the concavity of f changes at these threevalues, we see that the x-values of the in‡ectionpoints of the graph of f are about ¡1:7; 0; andabout 1.7.

(d)We observe from the graph of f 00 that f 00(x) >0 on about (¡1:7; 0) and (1:7;1); so f is concaveupward on the same intervals.Likewise, we observe that f 00(x) < 0 on about(¡1;¡1:7) and (0; 1:7); so f is concave downwardon the same intervals.

69. f 0(x) = x2 + x lnx

f 00(x) = 2x+ (x)μ1

x

¶+ (1)(lnx)

= 2x+ 1+ lnx

Graph f 0 and f 00 in the window [0; 1] by [¡2; 3];Xscl = 0.1, Yscl = 1.

Graph of f 0:

Graph of f 00:

(a) The critical number of f is the x-intercept ofthe graph of f 0. Using the graph, we …nd a crit-ical number of f is about 0.5671. By looking atthe graph of f 00 and applying the second deriva-tive test, we see f has a minimum at 0.5671.


(b) Examine the graph of f 0 to determine the in-tervals where the graph lies above and below thex-axis. We see that f 0(x) > 0 on about (0:5671;1),indicating that f is increasing on about (0:5671;1).We also see that f 0(x) < 0 on about (0; 0:5671), in-dicating that f is decreasing on about (0; 0:5671):

(c) Examine the graph of f 00. We see that thegraph has one x-intercept, so there is one x-valuewhere f 00(x) = 0. This value is about 0.2315. Be-cause the sign of f 00 changes at this value, we seethat x-value of the in‡ection point of the graph off is about 0.2315.

(d) We observe from the graph f 00 that f 00 > 0

on about (0:2315;1), so f is concave upward onabout (0:2315;1). Likewise, we observe from thegraph f 00 that f 00 < 0 on about (0; 0:2315), so f isconcave downward on about (0; 0:2315).

70. (a) The left side of the graph changes from con-cave upward to concave downward at the in‡ec-tion point between voice recognition software andSmart phones. The rate of growth of sales beginsto decline at the in‡ection point.

(b) Telephone answering machines are closest tothe right-hand in‡ection point. This in‡ectionpoint indicates that the rate of decline of salesis beginning to slow.

71. f(t) = 1:5207t4 ¡ 19:166t3 + 62:91t2+ 6:0726t+ 1026

f 0(t) = 6:0828t3 ¡ 57:498t2 + 125:82t+ 6:0726f 00(t) = 18:2484t2 ¡ 114:996t+ 125:82Solve f 00(x) = 0 to …nd the in‡ection points.

18:2484t2 ¡ 114:996t+ 125:82 = 0

t =114:996§p(¡114:996)2 ¡ 4(18:2484)(125:82)

2(18:2484)

t ¼ 1:409 or t ¼ 4:892

f 0(1:409) = 6:0828(1:409)3 ¡ 57:498(1:409)2+ 125:82(1:409) + 6:0726

¼ 86:218f 0(4:892) = 6:0828(4:892)3 ¡ 57:498(4:892)2

+ 125:82(4:892) + 6:0726

¼ ¡42:303Rents were increasing most rapidly when t ¼ 1:409or about mid 1999.

72. R(x) = 10,000¡ x3 + 42x2 + 800x;0 · x · 20R0(x) = ¡3x2 + 84x+ 800R00(x) = ¡6x+ 84A point of diminishing returns occurs at a pointof in‡ection, or where R00(x) = 0:

¡6x+ 84 = 0x = 14

TestR00(x) to determine whether concavity changesat x = 14:

R00(12) = 12 > 0R00(16) = ¡12 < 0

R(x) is concave upward on (0; 14) and concavedownward on (14; 20):

R(14) = 10,000¡ (14)3 + 42(14)2 + 800(14)= 26,688

The point of diminishing returns is (14; 26,688):

73. R(x) =4

27(¡x3 + 66x2 + 1050x¡ 400)

0 · x · 25R0(x) =

4

27(¡3x2 + 132x+ 1050)

R00(x) =4

27(¡6x+ 132)

A point of diminishing returns occurs at a pointof in‡ection, or where R00(x) = 0:

4

27(¡6x+ 132) = 0

¡6x+ 132 = 06x = 132

x = 22

TestR00(x) to determine whether concavity changesat x = 22:

R00(20) =4

27(¡6 ¢ 20 + 132) = 16

9> 0

R00(24) =4

27(¡6 ¢ 24 + 132) = ¡16

9< 0

R(x) is concave upward on (0; 22) and concavedownward on (22; 25):

R(22) =4

27[¡(22)3+66(22)2+1060(22)¡400]

¼ 6517:9The point of diminishing returns is (22; 6517:9):


74. R(x) = ¡0:3x3 + x2 + 11:4x; 0 · x · 6R0(x) = ¡0:9x2 + 2x+ 11:4R00(x) = ¡1:8x+ 2A point of diminishing returns occurs at a pointof in‡ection, or where R00(x) = 0:

¡1:8x+ 2 = 02 = 1:8x

1:11 ¼ x

TestR00(x) to determine whether concavity changesat x = 1:11:

R00(2) = ¡1:8(2) + 2 = ¡3:6 + 2= ¡1:8 < 0

R00(1) = ¡1:8(1) + 2 = 0:2 > 0

R(x) is concave upward on (0; 1:11) and concavedownward on (1:11; 6):

R(1:11) = ¡0:3(1:11)3 + (1:11)2 + 11:4(1:11)= 13:476 ¼ 13:5

The point of diminishing returns is (1:11; 13:5):

75. R(x) = ¡0:6x3 + 3:7x2 + 5x; 0 · x · 6R0(x) = ¡1:8x2 + 7:4x+ 5R00(x) = ¡3:6x+ 7:4A point of diminishing returns occurs at a pointof in‡ection or where R00(x) = 0:

¡3:6x+ 7:4 = 0¡3:6x = ¡7:4

x =¡7:4¡3:6 ¼ 2:06

TestR00(x) to determine whether concavity changesat x = 2:05:

R00(2) = ¡3:6(2) + 7:4= ¡7:2 + 7:4 = 0:2 > 0

R00(3) = ¡3:6(3) + 7:4= ¡10:8 + 7:4 = ¡3:4 < 0

R(x) is concave upward on (0; 2:06) and concavedownward on (2:06; 6):

R(2:06) = ¡0:6(2:06)3 + 3:7(2:06)2 + 5(2:06)¼ 20:8

The point of diminishing returns is (2:06; 20:8):

76. I(M) =¡U 00(M)U 0(M)

For U(M) =pM =M1=2;

U 0(M) =1

2M¡1=2

U 00(M) = ¡14M¡3=2:

Then

I(M) =¡ ¡¡1

4M¡3=2¢

12M

¡1=2

=1

2M¡1

=1

2M:

For U(M) =M2=3;

U 0(M) =2

3M¡1=3

U 00(M) = ¡29M¡4=3:

Then

I(M) =¡ ¡¡2

9M¡4=3¢

23M

¡1=3

=1

3M¡1

=1

3M:

U(M) =pM indicates a greater risk aversion be-

cause 12M > 1

3M for 0 <M < 1:

77. Let D(q) represent the demand function.The revenue function, R(q); is R(q) = qD(q):The marginal revenue is given by

R0(q) = qD0(q) +D(q)(1)= qD0(q) +D(q):

R00(q) = qD00(q) +D0(q)(1) +D0(q)= qD00(q) + 2D0(q)

gives the rate of decline of marginal revenue.D0(q) gives the rate of decline of price.If marginal revenue declines more quickly thanprice,

qD00(q) + 2D0(q)¡D0(q) < 0or qD00(q) +D0(q) < 0:


78. (a) f0 represents initial population (t = 0):

(b) (a; f(a)) is the point where the graph changesconcavity or the in‡ection point.

(c) fM is the maximum carrying capacity.

79. (a) R(t) = t2(t¡ 18) + 96t+ 1000; 0 < t < 8= t3 ¡ 18t2 + 96t+ 1000

R 0(t) = 3t2 ¡ 36t+ 96Set R0(t) = 0:

3t2 ¡ 36t+ 96 = 0t2 ¡ 12t+ 32 = 0(t¡ 8)(t¡ 4) = 0

t = 8 or t = 4

8 is not in the domain of R(t):R00(t) = 6t¡ 36R00(4) = ¡12 < 0 implies that R(t) is maximizedat t = 4, so the population is maximized at 4hours.

(b) R(4) = 16(¡14) + 96(4) + 1000= ¡224 + 384 + 1000= 1160

The maximum population is 1160 million.

80. The chemicals are accumulating, but at a slowerrate. Therefore, c(t) is increasing and concavedownward. Thus, c0(t) > 0 and c00(t) < 0:

81. K(x) =3x

x2 + 4

(a) K0(x) =3(x2 + 4)¡ (2x)(3x)

(x2 + 4)2

=¡3x2 + 12(x2 + 4)2

= 0

¡3x2 + 12 = 0x2 = 4

x = 2 or x = ¡2For this application, the domain of K is [0;1); sothe only critical number is 2.

K00(x) =(x2+4)2(¡6x)¡(¡3x2+12)(2)(x2+4)(2x)

(x2 + 4)4

=¡6x(x2 + 4)¡ 4x(¡3x2 + 12)

(x2 + 4)3

=6x3 ¡ 72x(x2 + 4)3

K00(2) = ¡96512 = ¡ 3

16 < 0 implies that K(x) ismaximized at x = 2.

Thus, the concentration is a maximum after 2hours.

(b) K(2) =3(2)

(2)2 + 4=3

4

The maximum concentration is 34%:

82. K(x) =4x

3x2 + 27

(a) K0(x) =(3x2 + 27)(4)¡ 4x(6x)

(3x2 + 27)2

=¡12x2 + 108(3x2 + 27)2

Set K0(x) = 0:

¡12(x2 ¡ 9) = 0x = ¡3 or x = 3

For this application, the domain of K is [0;1); sothe only critical number is 3.To determine whether a relative maximum or min-imum occurs at x = 3; we …nd K00(3) and use thesecond derivative test.To …nd K00(x); apply the quotient rule to …nd thederivative of K0(t): The numerator of K00(x) willbe

(3x2 + 27)2(¡24x)¡ (¡12x2 + 108)(2)(3x2 + 27)(6x)= (3x2 + 27)2(¡24x)¡ (12x)(¡12x2 + 108)(3x2 + 27):

The denominator of K00 is (3x2 + 27)4; so

K00(3) =(54)2(¡72)¡ (36)(0)(54)

(54)4

= ¡ 72542

< 0:

This indicates that the concentration is a maxi-mum after 3 hours.

(b) K(3) =4(3)

3(3)2 + 27

=2

9

The maximum concentration is 29%:


83. G(t) =10,000

1 + 49e¡0:1t

G0(t) =(1 + 49e¡0:1t)(0)¡ (10,000)(¡4:9e¡0:1t)

(1 + 49e¡0:1t)2

=49,000e¡0:1t

(1 + 49e¡0:1t)2

To …nd G00(t); apply the quotient rule to …nd thederivative of G0(t):The numerator of G00(t) will be

(1 + 49e¡0:1t)2(¡4900e¡0:1t)¡ (49,000e¡0:1t)(2)(1 + 49e¡0:1t)(¡4:9e¡0:1t)= (1 + 49e¡0:1t)(¡4900e¡0:1t)¢ [(1 + 49e¡0:1t)¡ 20(4:9e¡0:1t)]= (¡4900e¡0:1t)[1 + 49e¡0:1t ¡ 98e¡0:1t]= (¡4900e¡0:1t)(1¡ 49e¡0:1t):

Thus,

G00(t) =(¡4900e¡0:1t)(1¡ 49e¡0:1t)

(1 + 49e¡0:1t)4:

G00(t) = 0 when ¡4900e¡0:1t = 0 or1¡ 49e¡0:1t = 0:¡4900e¡0:1t < 0; and thus never equals zero.

1¡ 49e¡0:1t = 01 = 49e¡0:1t

1

49= e¡0:1t

ln

μ1

49

¶= ¡0:1t

ln 1¡ ln 49 = ¡0:1t¡ ln 49 = ¡0:1tln 49 = 0:1t

ln 72 = 0:1t

2 ln 7 = 0:1t

20 ln 7 = t

38:9182 ¼ t

The point of in‡ection is (38:9182; 5000):

84. G(t) =5200

1 + 12e¡0:52t

G0(t) =(1 + 12e¡0:52t)(0)¡ 5200(¡6:24e¡0:52t)

(1 + 12e¡0:52t)2

=32,448e¡0:52t

(1 + 12e¡0:52t)2

To …nd G00(t); use the quotient rule to …nd thederivative of G0(t): The numerator of G00(t) willbe

(1 + 12e¡0:52t)2(¡16,873e¡0:52t)¡ (32,448e¡0:52t)(2)(1 + 12e¡0:52t)(¡6:24e¡0:52t)= ¡16,873e¡0:52t(1 + 12e¡0:52t)¢ [(1 + 12e¡0:52t)¡ 24e¡0:52t]= ¡16,873e¡0:52t(1 + 12e¡0:52t)(1¡ 12e¡0:52t):

Thus,

G00(t) =¡16,873e¡0:52t(1 + 12e¡0:52t)(1¡ 12e¡0:52t)

(1 + 12e¡0:52t)4

=¡16,873e¡0:52t(1¡ 12e¡0:52t)

(1 + 12e¡0:52t)3:

G00(t) = 0 when 1¡ 12e¡0:52t = 0:

1 = 12e¡0:52t

1

12= e¡0:52t

ln1

12= ¡0:52t

ln 1¡ ln 12 = ¡0:52t¡ ln 12 = ¡0:52t

1:9231 ln 12 = t ¼ 4:779The point of in‡ection is (4:779; 2600):

85. L(t) = Be¡ce¡kt

L0(t) = Be¡ce¡kt(¡ce¡kt)0

= Be¡ce¡kt[¡ce¡kt(¡kt)0]

= Bcke¡ce¡kt¡kt

L00(t) = Bcke¡ce¡kt¡kt(¡ce¡kt ¡ kt)0

= Bcke¡ce¡kt¡kt[¡ce¡kt(¡kt)0 ¡ k]

= Bcke¡ce¡kt¡kt(cke¡kt ¡ k)

= Bck2e¡ce¡kt¡kt(ce¡kt ¡ 1)

L00(t) = 0 when ce¡kt ¡ 1 = 0

ce¡kt ¡ 1 = 0c

ekt= 1

ekt = c

kt = ln c

t =ln c

k

Letting c = 7:267963 and k = 0:670840

t =ln 7:267963

0:670840¼ 2:96 years

Verify that there is a point of in‡ection att = ln c

k ¼ 2:96. For

L00(t) = Bck2e¡ce¡kt¡kt(ce¡kt ¡ 1);


we only need to test the factor ce¡kt ¡ 1 on theintervals determined by t ¼ 2:96 since the otherfactors are always positive.L00(1) has the same sign as

7:267963e¡0:670840(1) ¡ 1 ¼ 2:72 > 0:L00(3) has the same sign as

7:267963e¡0:670840(3) ¡ 1 ¼ ¡0:029 < 0:Therefore L, is concave up on

¡0; ln ck ¼ 2:96¢ and

concave down on¡ln ck ;1

¢, so there is a point of

in‡ection at t = ln ck ¼ 2:96 years.

This signi…es the time when the rate of growthbegins to slow down since L changes from concaveup to concave down at this in‡ection point.

86. N(t) = ec(1¡e¡kt)

N 0(t) = ec(1¡e¡kt)[c(1¡ e¡kt)]0

= ec(1¡e¡kt)(¡ce¡kt)(¡kt)0

= ckec¡ce¡kt¡kt

N 00(t) = ckec¡ce¡kt¡kt(c¡ ce¡kt ¡ kt)0

= ckec¡ce¡kt¡kt[¡ce¡kt(¡kt)0 ¡ k]

= ckec¡ce¡kt¡kt(cke¡kt ¡ k)

= ck2ec¡ce¡kt¡kt(ce¡kt ¡ 1)

N 00(t) = 0 when ce¡kt ¡ 1 = 0ce¡kt ¡ 1 = 0

c

ekt= 1

ekt = c

kt = ln c

t =ln c

k

Letting c = 27:3 and k = 0:011

t =ln 27:3

0:011¼ 301

Verify that there is a point of in‡ection at t =ln 27:30:011 . For N 00(t) = ck2ec¡ce

¡kt¡kt(ce¡kt ¡ 1);we only need to test the factor ce¡kt ¡ 1 on theintervals determined by t = ln 27:3

0:011 ¼ 301 since theother factors are always positive.N 00(200) has the same sign as

27:3e¡0:011(200) ¡ 1 = 27:3e¡2:2 ¡ 1¼ 2:02 > 0:

N 00(400) has the same sign as

27:3e¡0:011(400) ¡ 1 = 27:3e¡4:4 ¡ 1¼ ¡0:66 < 0:

Therefore, N is concave up on¡0; ln 27:30:011

¢and N

is concave down on¡ln 27:30:011 ;1

¢, so there is a point

of in‡ection at t = ln 27:30:011 ¼ 301 days.

This signi…es the time when the rate of growthbegins to slow down since L changes from concaveup to concave down at this in‡ection point.

87. v(x) = ¡35:98 + 12:09x¡ 0:4450x2v0(x) = 12:09¡ 0:89xv00(x) = ¡0:89

Since ¡0:89 < 0, the function is always concavedown.

88. N(t) = 71:8e¡8:96e¡0:0685t

N 0(t) = 71:8(¡8:96)(¡0:0685)e¡0:0685t(e¡8:96e¡0:0685t)= (44:067968e¡0:0685t)(e¡8:96e

¡0:0685t)

N 00(t) = (44:067968)(0:61376e¡0:0685t

¡ 0:0685)e(¡8:96e¡0:0685t¡0:0685t)

N 00(t) = 0 when 0:61376e¡0:0685t ¡ 0:0685 = 00:61376e¡0:0685t ¡ 0:0685 = 0

e¡0:0685t ¼ 0:11162¡0:0685t ¼ ln 0:11162

t ¼ 32:01

N(t) has an in‡ection point at (32:01; N(32:01))or (32:01; 26:41):

89. Since the rate of violent crimes is decreasing but ata slower rate than in previous years, we know thatf 0(t) < 0 but f 00(t) > 0: Note that since f 0(t) < 0;f is decreasing, and since f 00(t) > 0; the graph off is concave upward.

90. V (x) = 12x(100¡ x)= 1200x¡ 12x2

V 0(x) = 1200¡ 24x

Set V 0(x) = 0:

1200¡ 24x = 0x = 50

V 00(x) = ¡24V 00(50) = ¡24

Thus, a value of 50 will produce a maximumrate of reaction.

Section 5.4 Curve Sketching 337

91. s(t) = ¡16t2v(t) = s0(t) = ¡32t(a) v(3) = ¡32(3) = ¡96 ft/sec(b) v(5) = ¡32(5) = ¡160 ft/sec(c) v(8) = ¡32(8) = ¡256 ft/sec(d) a(t) = v0(t) = s00(t)

= ¡32 ft/sec2

92. s(t) = ¡16t2 + 140t+ 37s0(t) = ¡32t+ 140s0(t) = 0 when ¡32t+ 140 = 0

140 = 32t

4:375 = t:

s0(1) = ¡32(1) + 140 > 0s0(5) = ¡32(5) + 140 < 0

s(4:375) = 343:25

(a) Maximum height of 343.25 ft is reached 4.375sec after the ball is thrown.

(b) The ball hits ground when s(t) = 0:

¡140§p1402 ¡ 4(¡16)(37)¡32 ¼ 9

s0(9) = ¡32(9) + 140 = ¡148The ball hits ground in about 9 seconds with aspeed of about ¡148 ft/sec:

93. s(t) = 256t¡ 16t2v(t) = s0(t) = 256¡ 32ta(t) = v0(t) = s00(t) = ¡32To …nd when the maximum height occurs, sets0(t) = 0:

256¡ 32t = 0t = 8

Find the maximum height.

s(8) = 256(8)¡ 16(82)= 1024

The maximum height of the ball is 1024 ft.The ball hits the ground when s = 0:

256t¡ 16t2 = 016t(16¡ t) = 0t = 0 (initial moment)t = 16 (…nal moment)

The ball hits the ground 16 seconds after beingthrown.

94. s(t) = 1:5t2 + 4t

(a) s(10) = 1:5(102) + 4(10)= 150 + 40 = 190

The car will move 190 ft in 10 seconds.

(b) v(t) = s0(t) = 3t+ 4

s0(5) = 3 ¢ 5 + 4 = 19s0(10) = 3 ¢ 10 + 4 = 34

The velocity at 5 sec is 19 ft/sec, and the velocityat 10 sec is 34 ft/sec.

(c) The car stops when v(t) = 0; but v(t) > 0 forall t ¸ 0:(d) a(t) = v0(t) = s00(t) = 3

The acceleration at 5 sec is 3 ft/sec2; and the acel-eration at 10 sec is also 3 ft/sec2.

(e) As t increases, the velocity increases, but theacceleration is constant.

95. The car was moving most rapidly when t ¼ 6; be-cause acceleration was positive on (0; 6) and neg-ative after t = 6; so velocity was a maximum att = 6:

5.4 Curve Sketching

1. Graph y = x ln jxj on a graphing calculator. Asuitable choice for the viewing window is [¡1; 1]by [¡1; 1]; Xscl = 0.1, Yscl = 0.1.

The calculator shows no y-value when x = 0 be-cause 0 is not in the domain of this function. How-ever, we see from the graph that

limx!0¡

x ln jxj = 0

andlimx!0+

x ln jxj = 0:


Thus,

limx!0

x ln jxj = 0:

3. f(x) = ¡2x3 ¡ 9x2 + 108x¡ 10

Domain is (¡1;1):

f(¡x) = ¡2(¡x)3 ¡ 9(¡x)2 + 108(¡x)¡ 10= 2x3 ¡ 9x2 ¡ 108x¡ 10

No symmetry

f 0(x) = ¡6x2 ¡ 18x+ 108= ¡6(x2 + 3x¡ 18)= ¡6(x+ 6)(x¡ 3)

f 0(x) = 0 when x = ¡6 or x = 3:Critical numbers: ¡6 and 3Critical points: (¡6;¡550) and (3; 179)

f 00(x) = ¡12x¡ 18f 00(¡6) = 54 > 0f 00(3) = ¡54 < 0

Relative maximum at 3, relative minimum at ¡6Increasing on (¡6; 3)Decreasing on (¡1;¡6) and (3;1)

f 00(x) = ¡12x¡ 18 = 0¡6(2x+ 3) = 0

x = ¡32

Point of in‡ection at (¡1:5;¡185:5)Concave upward on (¡1;¡1:5)Concave downward on (¡1:5;1)y-intercept:y = ¡2(0)3 ¡ 9(0)2 + 108(0)¡ 10 = ¡10

4. f(x) = x3 ¡ 152x2 ¡ 18x¡ 1

Domain is (¡1;1):

f(¡x) = (¡x)3 ¡ 152(¡x)2 ¡ 18(¡x)¡ 1

= ¡x3 ¡ 152x2 + 18x¡ 1

No symmetry

f 0(x) = 3x2 ¡ 15x¡ 18= 3(x2 ¡ 5x¡ 6)= 3(x¡ 6)(x+ 1) = 0

Critical numbers: 6 and ¡1Critical points: (6;¡163) and (¡1; 8:5)f 00(x) = 6x¡ 15f 00(6) = 21 > 0

f 00(¡1) = ¡21 < 0Relative maximum at x = ¡1; relative minimumat x = 6Increasing on (¡1;¡1) and (6;1)Decreasing on (¡1; 6)

f 00(x) = 6x¡ 15 = 0

x =5

2

Point of in‡ection at¡52 ;¡77:25

¢Concave upward on

¡52 ;1

¢Concave downward on

¡¡1; 52¢y-intercept:

y = (0)3 ¡ 152(0)2 ¡ 18(0)¡ 1 = ¡1


5. f(x) = ¡3x3 + 6x2 ¡ 4x¡ 1Domain is (¡1;1):f(¡x) = ¡3(¡x)3 + 6(¡x)2 ¡ 4(¡x)¡ 1

= 3x3 + 6x2 + 4x¡ 1No symmetry

f 0(x) = ¡9x2 + 12x¡ 4= ¡(3x¡ 2)2

(3x¡ 2)2 = 0

x =2

3

Critical number: 23

f¡23

¢= ¡3 ¡23¢3 + 6 ¡23¢2 ¡ 4 ¡23¢¡ 1 = ¡17

9

Critical point:¡23 ;¡17

9

¢f 0(0) = ¡9(0)2 + 12(0)¡ 4 = ¡4 < 0f 0(1) = ¡9(1)2 + 12(1)¡ 4 = ¡1 < 0No relative extremum at

¡23 ;¡17

9

¢Decreasing on (¡1;1)f 00(x) = ¡18x+ 12

= ¡6(3x¡ 2)3x¡ 2 = 0

x =2

3

Point of in‡ection at¡23 ;¡17

9

¢f 00(0) = ¡18(0) + 12 = 12 > 0f 00(1) = ¡18(1) + 12 = ¡6 < 0Concave upward on

¡¡1; 23¢Concave downward on

¡23 ;1

¢Point of in‡ection at

¡23 ;¡17

9

¢y-intercept: y = ¡3(0)3 + 6(0)2 ¡ 4(0)¡ 1 = ¡1

6. f(x) = x3 ¡ 6x2 + 12x¡ 11Domain is (¡1;1):f(¡x) = (¡x)3 ¡ 6(¡x)2 + 12(¡x)¡ 11

= ¡x3 ¡ 6x2 ¡ 12x¡ 11No symmetry

f 0(x) = 3x2 ¡ 12x+ 12 = 3(x2 ¡ 4x+ 4)= 3(x¡ 2)2

Critical number: 2Critical point: (2;¡3)f 00(x) = 6x¡ 12f 00(0) = 6(0)¡ 12 = ¡12 < 0f 00(3) = 6(3)¡ 12 = 6 > 0No relative extremaIncreasing on (¡1;1)Point of in‡ection at (2;¡3)Concave upward on (2;1); concave downward on(¡1; 2)y-intercept:

y = 03 ¡ 6(0)2 + 12(0)¡ 11 = ¡11

7. f(x) = x4 ¡ 24x2 + 80Domain is (¡1;1):f(¡x) = (¡x)4 ¡ 24(¡x)2 + 80

= x4 ¡ 24x2 + 80 = f(x)The graph is symmetric about the y-axis.

f 0(x) = 4x3 ¡ 48x

4x3 ¡ 48x = 04x(x2 ¡ 12) = 0

4x(x¡ 2p3)(x+ 2p3) = 0

Critical numbers: ¡2p3; 0; and 2p3Critical points: (¡2p3;¡64); (0; 80); and (2p3;¡64)


f 00(x) = 12x2 ¡ 48f 00(¡2p3) = 12(¡2p3)2 ¡ 48 = 96 > 0

f 00(0) = 12(0)2 ¡ 48 = ¡48 < 0f 00(2

p3) = 12(2

p3)2 ¡ 48 = 96 > 0

Relative maximum at 0, relative minima at ¡2p3and 2

p3

Increasing on (¡2p3; 0) and (2p3;1)Decreasing on (¡1;¡2p3) and (0; 2p3)12x2 ¡ 48 = 012(x2 ¡ 4) = 0

x = §2Points of in‡ection at (¡2; 0) and (2; 0)Concave upward on (¡1;¡2) and (2;1)Concave downward on (¡2; 2)x-intercepts: 0 = x4 ¡ 24x2 + 80Let u = x2.

u2 ¡ 24u+ 80 = 0(u¡ 4)(u¡ 20) = 0u = 4 or u = 20

x = §2 or x = §2p5y-intercept: y = (0)4 ¡ 24(0)2 + 80 = 80

8. f(x) = ¡x4 + 6x2

Domain is (¡1;1):f(¡x) = ¡(¡x)4 + 6(¡x)2 = ¡x4 + 6x2 = f(x)The graph is symmetric about the y-axis.

f 0(x) = ¡4x3 + 12x¡4x3 + 12x = 0¡4x(x2 ¡ 3) = 0

¡4x(x¡p3)(x+p3) = 0Critical numbers:¡p3; 0; and p3Critical points: (¡p3; 9); (0; 0); and (p3; 9)f 00(x) = ¡12x2 + 12

f 00(¡p3) = ¡12(p3)2 + 12 = ¡24 < 0f 00(0) = ¡12(0)2 + 12 = 12 > 0

f 00(p3) = ¡12(p3)2 + 12 = ¡24 < 0

Relative minimum at 0, relative maxima at ¡p3and

p3

Increasing on (¡1;¡p3) and (0;p3)Decreasing on (¡p3; 0) and (p3;1)

¡12x2 + 12 = 0¡12(x2 ¡ 1) = 0

x = §1Points of in‡ection at (¡1; 5) and (1; 5)Concave upward on (¡1; 1)Concave downward on (¡1;¡1) and (1;1)x-intercepts: ¡x4 + 6x2 = 0

¡x2(x2 ¡ 6) = 0x = 0 or x = §p6

y-intercept: y = ¡(0)4 + 6(0)2 = 0

9. f(x) = x4 ¡ 4x3

Domain is (¡1;1):

f(¡x) = (¡x)4¡4(¡x)3 = x4+4x3 6= f(x) or ¡f(x)

The graph is not symmetric about the y-axis orthe origin.

f 0(x) = 4x3 ¡ 12x2

4x3 ¡ 12x2 = 04x2(x¡ 3) = 0Critical numbers: 0 and 3Critical points: (0; 0) and (3;¡27)f 00(x) = 12x2 ¡ 24x

f 00(0) = 12(0)2 ¡ 24(0) = 0f 00(3) = 12(3)2 ¡ 24(3) = 36 > 0


Second derivative test fails for 0. Use …rst deriva-tive test.

f 0(¡1) = 4(¡1)3 ¡ 12(¡1)2 = ¡16 < 0f 0(1) = 4(1)3 ¡ 12(1)2 = ¡8 < 0

Neither a relative minimum nor maximum at 0Relative minimum at 3Increasing on (3;1)Decreasing on (¡1; 3)12x2 ¡ 24x = 012x(x¡ 2) = 0

x = 0 or x = 2

Points of in‡ection at (0; 0) and (2;¡16)Concave upward on (¡1; 0) and (2;1)Concave downward on (0; 2)

x-intercepts: x4 ¡ 4x3 = 0x3(x¡ 4) = 0

x = 0 or x = 4

y-intercept: y = (0)4 ¡ 4(0)3 = 0

10. f(x) = x5 ¡ 15x3Domain is (¡1;1):f(¡x) = (¡x)5 ¡ 15(¡x)3

= ¡x5 + 15x3 = ¡f(x)The graph is symmetric about the origin.

f 0(x) = 5x4 ¡ 45x2 = 05x2(x2 ¡ 9) = 0

5x2(x+ 3)(x¡ 3) = 0Critical numbers: 0; ¡3; and 3Critical points: (0; 0); (¡3; 162); and (3;¡162)f 00(x) = 20x3 ¡ 90x = 10x(x2 ¡ 9)f 00(0) = 0

f 00(¡3) = ¡270 < 0f 00(3) = 270 > 0

Relative maximum at ¡3Relative minimum at 3No relative extremum at 0

Increasing on (¡1;¡3) and (3;1)Decreasing on (¡3; 3)

f 00(x) = 20x3 ¡ 90x = 010x(2x2 ¡ 9) = 0

x = 0 or x = § 3p2

Points of in‡ection at (0; 0);³¡ 3p

2; 100:23

´; and³

3p2;¡100:23

Ćoncave upward on

³¡ 3p

2; 0ánd

³3p2;1´

Concave downward on³¡1;¡ 3p

2

ánd

³0; 3p

2

´x-intercepts: 0 = x5 ¡ 15x3

0 = x3(x2 ¡ 15)x = 0; x = §p15

y-intercept: y = 05 ¡ 15(0)3 = 0

11. f(x) = 2x+10

x

= 2x+ 10x¡1

Since f(x) does not exist when x = 0, the domainis (¡1; 0) [ (0;1):f(¡x) = 2(¡x) + 10(¡x)¡1

= ¡(2x+ 10x¡1)= ¡f(x)

The graph is symmetric about the origin.

f 0(x) = 2¡ 10x¡2

2¡ 10x2= 0

2(x2 ¡ 5)x2

= 0

x = §p5Critical numbers: ¡p5 and p5Critical points: (¡p5;¡4p5) and (p5; 4p5)


Test a point in the intervals (¡1;¡p5);(¡p5; 0); (0;p5); and (p5;1):

f 0(¡3) = 2¡ 10(¡3)¡2 = 8

9> 0

f 0(¡1) = 2¡ 10(¡1)¡2 = ¡8 < 0f 0(1) = 2¡ 10(1)¡2 = ¡8 < 0

f 0(3) = 2¡ 10(3)¡2 = 8

9> 0

Relative maximum at ¡p5Relative minimum at

p5

Increasing on (¡1;¡p5) and (p5;1)Decreasing on (¡p5; 0) and (0;p5)(Recall that f(x) does not exist at x = 0:)

f 00(x) = 20x¡3 =20

x3

f 00(x) =20

x3is never equal to zero.

There are no in‡ection points.Test a point in the intervals (¡1; 0) and (0;1):

f 00(¡1) = 20

(¡1)3 = ¡20 < 0

f 00(1) =20

(1)3= 20 > 0

Concave upward on (0;1)Concave downward on (¡1; 0)f(x) is never zero, so there are no x-intercepts.f(x) does not exist for x = 0; so there is no y-intercept.

Vertical asymptote at x = 0y = 2x is an oblique asymptote.

f(¡x) = 2(¡x) + 10(¡x)¡1= ¡(2x+ 10x¡1)= ¡f(x)

12. f(x) = 16x+1

x2

= 16x+ x¡2

Since f(x) does not exist when x = 0, the domainis (¡1; ) [ (0;1):f(¡x) = 16(¡x) + (¡x)¡2 = ¡16x+ x¡2


f 0(x) = 16¡ 2x¡3 = 16¡ 2

x3

16¡ 2

x3= 0

2(8x3 ¡ 1)x3

= 0

x =1

2

Critical number: 12

Critical point:¡12 ; 12

¢Test a point in the intervals (¡1; 0); ¡0; 12¢, and¡12 ;1

¢:

f 0(¡1) = 16¡ 2(¡1)¡3 = 18 > 0

f 0μ1

3

¶= 16¡ 2

μ1

3

¶¡3= ¡38 < 0

f 0(1) = 16¡ 2(1)¡3 = 14 > 0Relative minimum at 12

Increasing on (¡1; 0) and (12 ;1)Decreasing on (0; 12)

f 00(x) = 6x¡4 =6

x4

f 00(x) =6

x4> 0 and is never zero.

There are no in‡ection points.Concave upward on (1; 0) and (0;1)x-intercept: 16x+ x¡2 = 0

16x3 + 1

x2= 0

x = ¡ 1

2 3p2

f(x) does not exist for x = 0, so there is no x-intercept.

Vertical asymptote at x = 0


y = 16x is an oblique asymptote.

13. f(x) =¡x+ 4x+ 2

Since f(x) does not exist when x = ¡2, the do-main is (¡1;¡2) [ (¡2;1).

f(¡x) = ¡(¡x) + 4(¡x) + 2 =

x+ 4

¡x+ 2The graph is not symmetric about the y-axis orthe origin.

f 0(x) =(x+ 2)(¡1)¡ (¡x+ 4)(1)

(x+ 2)2

=¡6

(x+ 2)2

f 0(x) < 0 and is never zero. f 0(x) fails to exist forx = ¡2:No critical numbers; no relative extremaDecreasing on (¡1;¡2) and (¡2;1)

f 00(x) =12

(x+ 2)3

f 00(x) fails to exist for x = ¡2:No points of in‡ectionTest a point in the intervals (¡1;¡2) and (¡2;1):f 00(¡3) = ¡12 < 0f 00(¡1) = 12 > 0Concave upward on (¡2;1)Concave downward on (¡1;¡2)x-intercept:

¡x+ 4x+ 2

= 0

x = 4

y-intercept: y =¡0 + 40 + 2

= 2

Vertical asymptote at x = ¡2

Horizontal asymptote at y = ¡1

14. f(x) =3x

x¡ 2Since f(x) does not exist when x = 2, the domainis (¡1; 2) [ (2;1):

f(¡x) = 3(¡x)(¡x)¡ 2 =

¡3x¡x¡ 2


f 0(x) =(x¡ 2)(3)¡ (3x)(1)

(x¡ 2)2

=¡6

(x¡ 2)2

f 0(x) < 0 is never zero. f 0(x) fails to exist forx = 2:

No critical numbers; no relative extremaDecreasing on (¡1; 2) and (2;1)

f 00(x) =12

(x¡ 2)3f 00(x) fails to exist for x = 2:

No points of in‡ection

Test a point in the intervals (¡1; 2) and (2;1):

f 00(1) = ¡12 < 0f 00(3) = 12 > 0

Concave upward on (2;1)Concave downward on (¡1; 2)

x-intercept:3x

x¡ 2 = 0

x = 0

y-intercept: y =3(0)

0¡ 2 = 0



Horizontal asymptote at y = 3

15. f(x) =1

x2 + 4x+ 3

=1

(x+ 3)(x+ 1)

Since f(x) does not exist when x = ¡3 and x = ¡1, the domain is (¡1;¡3) [ (¡3;¡1) [ (¡1;1):

f(¡x) = 1

(¡x)2 + 4(¡x) + 3 =1

x2 ¡ 4x+ 3

The graph is not symmetric about the y-axis or the origin.

f 0(x) =0¡ (2x+ 4)(x2 + 4x+ 3)2

=¡2(x+ 2)

[(x+ 3)(x+ 1)]2

Critical number: ¡2

Test a point in the intervals (¡1;¡3); (¡3;¡2); (¡2;¡1), and (¡1;1):

f 0(¡4) = ¡2(¡4 + 2)[(¡4 + 3)(¡4 + 1)]2 =

4

9> 0

f 0μ¡52

¶=

¡2 ¡¡52 + 2

¢£¡¡5

2 + 3¢ ¡¡5

2 + 1¢¤2 = 16

9> 0

f 0μ¡32

¶=

¡2 ¡¡32 + 2

¢£¡¡3

2 + 3¢ ¡¡3

2 + 1¢¤2 = ¡169 < 0

f 0(0) =¡2(0 + 2)

[(0 + 3)(0 + 1)]2= ¡4

9< 0

f(¡2) = 1

(¡2 + 3)(¡2 + 1) = ¡1

Relative maximum at (¡2;¡1)Increasing on (¡1;¡3) and (¡3;¡2)Decreasing on (¡2;¡1) and (¡1;1)


f 00(x) =(x2 + 4x+ 3)2(¡2)¡ (¡2x¡ 4)(2)(x2 + 4x+ 3)(2x+ 4)

(x2 + 4x+ 3)4

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

(x2 + 4x+ 3)4

=¡2(x2 + 4x+ 3¡ 4x2 ¡ 16x¡ 16)

(x2 + 4x+ 3)3

=¡2(¡3x2 ¡ 12x¡ 13)

(x2 + 4x+ 3)3

=2(3x2 + 12x+ 13)

[(x+ 3)(x+ 1)]3

Since 3x2 + 12x+ 13 = 0 has no real solutions, there are no x-values where f 00(x) = 0. f 00(x) does not existwhere x = ¡3 and x = ¡1: Since f(x) does not exist at these x-values, there are no points of in‡ection.

Test a point in the intervals (¡1;¡3); (¡3;¡1), and (¡1;1).

f 00(¡4) = 2[3(¡4)2 + 12(¡4) + 13][(¡4 + 3)(¡4 + 1)]3 =

26

27> 0

f 00(¡2) = 2[3(¡2)2 + 12(¡2) + 13][(¡2 + 3)(¡2 + 1)]3 = ¡2 < 0

f 00(0) =2[3(0)2 + 12(0) + 13]

[(0 + 3)(0 + 1)]3=26

27> 0

Concave upward on (¡1;¡3) and (¡1;1)

Concave downward on (¡3;¡1)

f(x) is never zero, so there are no x-intercepts.

y-intercept: y =1

(0 + 3)(0 + 1)=1

3

Vertical asymptotes where f(x) is unde…ned at x = ¡3 and x = ¡1:Horizontal asymptote at y = 0


16. f(x) =¡8

x2 ¡ 6x¡ 7 =¡8

(x+ 1)(x¡ 7)

Since f(x) does not exist when x = ¡1 and x = 7, the domain is (¡1;¡1) [ (¡1; 7) [ (7;1):

f(¡x) = 8

(¡x)2 ¡ 6(¡x)¡ 7 =8

x2 + 6x¡ 7The graph is not symmetric about the y-axis or the origin.

f 0(x) =0¡ (¡8)(2x¡ 6)(x2 ¡ 6x¡ 7)2

=16(x¡ 3)

[(x+ 1)(x¡ 7)]2

Critical number: 3

Test a point in the intervals (¡1;¡1); (¡1; 3); (3; 7); and (7;1):

f 0(¡2) = 16(¡2¡ 3)[(¡2 + 1)(¡2¡ 7)]2 = ¡

80

81< 0

f 0(0) =16(0¡ 3)

[(0 + 1)(0¡ 7)]2 = ¡48

49< 0

f 0(6) =16(6¡ 3)

[(6 + 1)(6¡ 7)]2 =48

49> 0

f 0(8) =16(8¡ 3)

[(8 + 1)(8¡ 7)]2 =80

81> 0

f(3) =¡8

(3 + 1)(3¡ 7) =1

2

Relative minimum at¡3; 12

¢Increasing on (3; 7) and (7;1)Decreasing on (¡1;¡1) and (¡1; 3)

f 00(x) =(x2 ¡ 6x¡ 7)2(16)¡ 16(x¡ 3)(2)(x2 ¡ 6x¡ 7)(2x¡ 6)

(x2 ¡ 6x¡ 7)4

=16(x2 ¡ 6x¡ 7)[(x2 ¡ 6x¡ 7)¡ 4(x¡ 3)(x¡ 3)]

(x2 ¡ 6x¡ 7)4

=16(x2 ¡ 6x¡ 7¡ 4x2 + 24x¡ 36)

(x2 ¡ 6x¡ 7)3

=16(¡3x2 + 18x¡ 43)(x2 ¡ 6x¡ 7)3

=¡16(3x2 ¡ 18x+ 43)[(x+ 1)(x¡ 7)]3

Since 3x2 ¡ 18x+ 43 = 0 has no real solutions, there are no x-values where f 00(x) = 0.f 00(x) does not exist where x = ¡1 and x = 7. Since f(x) does not exist at these x-values, there are no pointsof in‡ection.

Test a point in the intervals (¡1;¡1); (¡1; 7), and (7;1).


f 00(¡2) = ¡16[3(¡2)2 ¡ 18(¡2) + 43)[(¡2 + 1)(¡2¡ 7)]3 = ¡1456

729< 0

f 00(0) =¡16[3(0)2 ¡ 18(0) + 43)

[(0 + 1)(0¡ 7)]3 =688

343> 0

f 00(8) =¡16[3(8)2 ¡ 18(8) + 43)

[(8 + 1)(8¡ 7)]3 = ¡1456729

< 0

Concave upward on (¡1; 7)Concave downward on (¡1;¡1) and (7;1)f(x) is never zero, so there are no x-intercepts.

y-intercept: y =¡8

(0 + 1)(0¡ 7) =8

7

Vertical asymptotes where f(x) is unde…ned at x = ¡1 and x = 7:Horizontal asymptote at y = 0

17. f(x) =x

x2 + 1

Domain is (¡1;1)f(¡x) = ¡x

(¡x)2 + 1 = ¡x

x2 + 1= ¡f(x)


f 0(x) =(x2 + 1)(1)¡ x(2x)

(x2 + 1)2

=1¡ x2(x2 + 1)2

1¡ x2 = 0Critical numbers: 1 and ¡1Critical points:

¡1; 12

¢and

¡¡1;¡12

¢f 00(x) =

(x2+1)2(¡2x)¡(1¡x2)(2)(x2+1)(2x)(x2 + 1)4

=¡2x3 ¡ 2x¡ 4x+ 4x3

(x2 + 1)3=2x3 ¡ 6x(x2 + 1)3

f 00(1) = ¡12< 0

f 00(¡1) = 1

2> 0

Relative maximum at 1Relative minimum at ¡1Increasing on (¡1; 1)


Decreasing on (¡1;¡1) and (1;1)

f 00(x) =2x3 ¡ 6x(x2 + 1)3

= 0

2x3 ¡ 6x = 02x(x2 ¡ 3) = 0x = 0; x = §p3

In‡ection points at (0; 0);³p3;

p34

ánd

³¡p3;¡

p34

´

Concave upward on (¡p3; 0) and (p3;1)Concave downward on (¡1;¡p3) and (0;p3)

x-intercept: 0 =x

x2 + 1

0 = x

y-intercept: y =0

02 + 1= 0


18. f(x) =1

x2 + 4

Domain is (¡1;1):

f(¡x) = 1

(¡x)2 + 4 =1

x2 + 4= f(x)

The graph is symmetric about the y-axis.

f 0(x) =¡2x

(x2 + 4)2

Critical number: 0

Critical point:¡0; 14

¢

f 00(x) =(x2+4)2(¡2)¡(¡2x)(2)(x2+4)(2x)

(x2 + 4)4

=¡2(x2 + 4)[(x2 + 4) + (¡2x)(2x)]

(x2 + 4)4

=¡2(x2 + 4¡ 4x2)

(x2 + 4)3

=¡2(¡3x2 + 4)(x2 + 4)3

=2(3x2 ¡ 4)(x2 + 4)3

f 00(0) =2[3(0)2 ¡ 4][(0)2 + 4]3

= ¡18< 0

Relative maximum at¡0; 14

¢Increasing on (¡1; 0)Decreasing on (0;1)

f(0) =1

(0)2 + 4=1

4

f 00(x) = 0 when

2(3x2 ¡ 4) = 0

x2 =4

3

x = § 2p3

In‡ection points at³¡ 2p

3; 316

ánd

³2p3; 316

Ćoncave upward on

³¡1;¡ 2p

3

ánd

³2p3;1´

Concave downward on³¡ 2p

3; 2p

3

´f(x) is never zero, so there are no x-intercepts.

y-intercept: y =1

02 + 4=1

4



19. f(x) =1

x2 ¡ 9=

1

(x+ 3)(x¡ 3)Since f(x) does not exist when x = ¡3 and x = 3,the domain is (¡1;¡3) [ (¡3; 3) [ (3;1):

f(¡x) = 1

(¡x)2 ¡ 9 =1

x2 ¡ 9 = f(x)


f 0(x) =¡2x

(x2 ¡ 9)2Critical number: 0

Critical point:¡0;¡1

9

¢Test a point in the intervals (¡1;¡3);(¡3; 0); (0; 3), and (3;1):

f 0(¡4) = ¡2(¡4)[(¡4)2 ¡ 9]2 =

8

49> 0

f 0(¡1) = ¡2(¡4)[(¡1)2 ¡ 9]2 =

1

32> 0

f 0(1) =¡2(1)

[(1)2 ¡ 9]2 = ¡1

32< 0

f 0(4) =¡2(4)

[(4)2 ¡ 9]2 = ¡8

49< 0

Relative maximum at¡0;¡1

9

¢Increasing on (¡1;¡3) and (¡3; 0)Decreasing on (0; 3) and (3;1)

f 00(x) =(x2 ¡ 9)2(¡2)¡ (¡2x)(2)(x2 ¡ 9)(2x)

(x2 ¡ 9)4

=¡2(x2 ¡ 9)[(x2 ¡ 9) + (¡2x)(2x)]

(x2 + 4)4

=¡2(x2 ¡ 9¡ 4x2)

(x2 ¡ 9)3

=¡2(¡3x2 ¡ 9)(x2 ¡ 9)3

=6(x2 + 3)

[(x+ 3)(x¡ 3)]3Since x2 + 3 = 0 has no solutions, there are nox-values where f 00(x) = 0. f 00(x) does not ex-ist where x = ¡3 and x = 3. Since f(x) doesnot exist at these x-values, there are no points ofin‡ection.

Test a point in the intervals (¡1;¡3); (¡3; 3),and (3;1):

f 00(¡4) = 6[(¡4)2 + 3][(¡4 + 3)(¡4¡ 3)]3 =

114

343> 0

f 00(0) =6[(0)2 + 3]

[(0 + 3)(0¡ 3)]3 = ¡2

81< 0

f 00(4) =6[(4)2 + 3]

[(4 + 3)(4¡ 3)]3 =114

343> 0

Concave upward on (¡1;¡3) and (3;1)Concave downward on (¡3; 3)

f(x) is never zero, so there are no x-intercepts.

y-intercept: y =1

02 ¡ 9 = ¡1

9

Vertical asymptotes where f(x) is unde…ned atx = ¡3 and x = 3:Horizontal asymptote at y = 0

20. f(x) =¡2xx2 ¡ 4

=¡2x

(x+ 2)(x¡ 2)

Since f(x) does not exist when x = ¡2 and x = 2,the domain is (¡1;¡2) [ (¡2; 2) [ (2;1):

f(¡x) = ¡2(¡x)(¡x)2 ¡ 4 =

2x

x2 ¡ 4 = ¡f(x)


f 0(x) =(x2 ¡ 4)(¡2)¡ (¡2x)(2x)

(x2 ¡ 4)2

=2x2 + 8

(x2 ¡ 4)2

f 0(x) > 0 and is never zero.No critical points, no relative extremaIncreasing on (¡1;¡2); (¡2; 2) and (2;1)


f 00(x) =(x2 ¡ 4)2(4x)¡ (2x2 + 8)(2)(x2 ¡ 4)(2x)

(x2 ¡ 4)4

=4x(x2 ¡ 4)[(x2 ¡ 4)¡ (2x2 + 8)]

(x2 ¡ 4)4

=4x(¡x2 ¡ 12)(x2 ¡ 4)3

=¡4x(x2 + 12)(x2 ¡ 4)3

f 00(x) = 0 when

¡4x(x2 + 12) = 0x = 0

Test a point in the intervals(¡1;¡2); (¡2; 0); (0; 2),and (2;1).

f 00(¡3) = ¡4(¡3)[(¡3)2 + 12][(¡3)2 ¡ 4]3 =

252

125> 0

f 00(¡1) = ¡4(¡1)[(¡1)2 + 12][(¡1)2 ¡ 4]3 = ¡52

27< 0

f 00(1) =¡4(1)[(1)2 + 12][(1)2 ¡ 4]3 =

52

27> 0

f 00(3) =¡4(3)[(3)2 + 12][(3)2 ¡ 4]3 = ¡252

125< 0

In‡ection point at (0; 0)Concave upward on (¡1;¡2) and (0; 2)Concave downward on (¡2; 0) and (2;1)

x-intercept:¡2xx2 ¡ 4 = 0

¡2x = 0x = 0

y-intercept: y =¡2(0)02 ¡ 4 = 0

Vertical asymptotes where f(x) is unde…ned atx = ¡2 and x = 2:Horizontal asymptote at y = 0

21. f(x) = x ln jxjThe domain of this function is (¡1; 0) [ (0;1):f(¡x) = ¡x ln¡x j¡xj

= x ln j¡xj = ¡f(x)The graph is symmetric about the origin.

f 0(x) = x ¢ 1x+ ln jxj

= 1 + ln jxjf 0(x) = 0 when

0 = 1 + ln jxj¡1 = ln jxje¡1 = jxj

x = §1e¼ §0:37:

Critical numbers: §1e ¼ §0:37:

f 0(¡1) = 1 + ln j¡1j = 1 > 0f 0(¡0:1) = 1 + ln j¡0:1j ¼ ¡1:3 < 0f 0(0:1) = 1 + ln j0:1j ¼ ¡1:3 < 0f 0(1) = 1 + ln j1j = 1 > 0

f

μ1

e

¶=1

eln

¯̄̄¯1e¯̄̄¯ = ¡1e

f

μ¡1e

¶= ¡1

eln

¯̄̄¯¡1e

¯̄̄¯ = 1

e

Relative maximum of (¡1e ;

1e ); relative minimum

of ( 1e ;¡1e ):

Increasing on¡¡1;¡1

e

¢and

¡1e ;1

¢and decreas-

ing on¡¡1

e ; 0¢and

¡0; 1e

¢:

f 00(x) =1

x

f 00(¡1) = 1

¡1 = ¡1 < 0

f 00(1) =1

1= 1 > 0

Concave downward on (¡1; 0);Concave upward on (0;1):There is no y-intercept.

x-intercept: 0 = x ln jxjx = 0 or ln jxj = 0

jxj = e0 = 1x = §1


Since 0 is not in the domain, the only x-interceptsare ¡1 and 1.

22. f(x) = x¡ ln jxj

Domain is (¡1; 0) [ (0;1)

f(¡x) = ¡x¡ ln j¡xj= ¡x¡ ln jxj

The graph has no symmetry.

f 0(x) = 1¡ 1

x=x¡ 1x

Critical number: x = 1(Recall that f(x) does not exist at x = 0.)

f 0(¡1) = ¡1¡ 1¡1 = 2 > 0

f 0μ1

2

¶=

12 ¡ 112

= ¡1 < 0

f 0(2) =2¡ 12

=1

2> 0

f(1) = 1¡ ln j1j = 1

Relative minimum at (1; 1)Increasing on (¡1; 0) and (1;1)Decreasing on (0; 1)

f 00(x) =x(1)¡ (x¡ 1)(1)

x2=1

x2

No in‡ection pointf 00(x) is positive everywhere it is de…ned,Concave upward on (¡1; 0) and (0;1)There is no y-interceptx-intercept: x¡ ln jxj = 0

x = ln jxjx ¼ ¡0:567


23. f(x) =lnx

x

Note that the domain of this function is (0;1):

f(¡x) = ln(¡x)¡x does not exist when x ¸ 0; no

symmetry.

f 0(x) =x¡1x

¢¡ lnx(1)x2

=1¡ lnxx2

Critical numbers:

1¡ lnx = 01 = lnx

e1 = x

f(e) =ln e

e=1

e

Critical points:μe;1

e

¶

f 0(1) =1¡ ln 112

=1

1= 1 > 0

f 0(3) =1¡ ln 332

= ¡0:01 < 0

There is a relative maximum at¡e; 1e

¢:

The function is increasing on (0; e) and decreasingon (e;1):

f 00(x) =x2¡¡ 1

x

¢¡ (1¡ lnx)2xx4

=¡x¡ 2x(1¡ lnx)

x4

=¡x[1 + 2(1¡ lnx)]

x4

=¡(1 + 2¡ 2 lnx)

x3

=¡3 + 2 lnx

x3


f 00(x) = 0 when ¡3 + 2 lnx = 02 lnx = 3

lnx =3

2= 1:5

x = e1:5 ¼ 4:48:

f 00(1) =¡3 + 2 ln 1

13= ¡3 < 0

f 00(5) =¡3 + 2 ln 5

53¼ 0:0018 > 0

In‡ection point at¡e1:5; 1:5e1:5

¢ ¼ (4:48; 0:33)Concave downward on (0; e1:5); concave upwardon (e1:5;1)

f(e1:5) =ln e1:5

e1:5=1:5

e1:5

=3

2e1:5¼ 0:33

Since x 6= 0; there is no y-intercept.x-intercept: f(x) = 0 when ln x = 0

x = e0 = 1

Vertical asymptote at x = 0Horizontal asymptote at y = 0

24. f(x) =lnx2

x2

Domain is (¡1; 0) [ (0;1)

f(¡x) = ln(¡x)2(¡x)2

=lnx2

x2= f(x)


f 0(x) =x2 ¢ 1x2 ¢ 2x¡ (lnx2)2x

x4

=2x(1¡ lnx2)

x4=2(1¡ lnx2)

x3

f 0(x) = 0 when 1¡ lnx2 = 0lnx2 = 1

x2 = e

x = §pe:

Critical numbers: ¡pe;pe

Critical points:¡¡pe; 1e¢ ; ¡pe; 1e¢

f 00(x) =x3(¡2 ¢ 1x2 ¢ 2x)¡ 2(1¡ lnx2)3x2

x6

=x2[¡4¡ 6(1¡ lnx2)]

x6

=¡10 + 6 lnx2

x4

f 00(x)(§pe) = ¡10 + 6 ln ee2

=¡4e2< 0

There are relative maxima at¡§pe; 1e¢ :

Increasing on (¡1;¡pe) [ (0;pe)Decreasing on (¡pe; 0) [ (pe;1)

f 00(x) = 0when ¡10 + 6 lnx2 = 06 lnx2 = 10

lnx2 =5

3

x2 = e5=3

x = §pe5=3 = §e5=6

f 00(3) =¡10 + 6 ln 9

34= 0:0393 > 0

f 00(1) =¡10 + 6 ln 12

14= ¡10 < 0

f 00(¡1) = ¡10 + 6 ln(¡1)2(¡1)4 = ¡10 < 0

f 00(¡3) = ¡10 + 6 ln(¡3)2(¡3)4 = 0:0393 > 0

In‡ection points at¡§e5=6; 5

3e5=3

¢Concave upward on (¡1;¡e5=6) and (e5=6;1)Concave downward on (¡e5=6; 0) and (0; e5=6)

There is a vertical asymptote at x = 0:There is no y-intercept.There are x-intercepts when f(x) = 0:

ln x2 = 0

x2 = 1

x = §1:


25. f(x) = xe¡x

Domain is (¡1;1):

f(¡x) = ¡xex


f 0(x) = ¡xe¡x + e¡x= e¡x(1¡ x)

f 0(x) = 0 when e¡x(1¡ x) = 0x = 1

Critical numbers: 1

Critical points:¡1; 1e

¢f 0(0) = e¡0(1¡ 0) = 1 > 0

f 0(2) = e¡2(1¡ 2) = ¡1e2< 0

Relative maximum at¡1; 1e

¢Increasing on (¡1; 1); decreasing on (1;1)

f 00(x) = e¡x(¡1) + (1¡ x)(¡e¡x)= ¡e¡x(1 + 1¡ x)= ¡e¡x(2¡ x)

f 00 = 0 when ¡e¡x(2¡ x) = 0x = 2:

f 00(0) = ¡e¡0(2¡ 0) = ¡2 < 0

f 00(3) = ¡e¡3(2¡ 3) = 1

e3> 0

In‡ection point at¡2; 2e2

¢Concave downward on (¡1; 2); concave upwardon (2;1)

x-intercept: 0 = xe¡x

x = 0

y-intercept: y = 0 ¢ e¡0 = 0


26. f(x) = x2e¡x

Domain is (¡1;1)f(¡x) = (¡x)2ex

= x2ex


f 0(x) = x2e¡x(¡1) + e¡x(2x)= xe¡x(¡x+ 2)

Critical numbers x = 0 or ¡x+ 2 = 0x = 2:

f 0(¡1) = (¡1)e¡(¡1)[¡(¡1) + 2] = ¡3e < 0

f 0(1) = (1)e¡1(¡1 + 2) = 1

e> 0

f 0(3) = 3e¡3(¡3 + 2) = ¡3e3< 0

The function is decreasing on (¡1; 0) and (2;1)and increasing on (0; 2):

f(0) = 02e¡0 = 0

f(2) = 22e¡2 =4

e2

Relative minimum at (0; 0); relative maximum at¡2; 4e2

¢f 00(x) = xe¡x(¡1) + (¡x+ 2)(¡xe¡x + e¡x)

= ¡xe¡x + x2e¡x ¡ xe¡x ¡ 2xe¡x + 2e¡x= x2e¡x ¡ 4xe¡x + 2e¡x= e¡x(x2 ¡ 4x+ 2)

f 00(x) = 0 when

x2 ¡ 4x+ 2 = 0x =

4§p16¡ 4(1)(2)2

=4§p82

= 2§p2:


x = 2 +p2 ¼ 3:4 or 2¡p2 ¼ 0:6

f 00(0) = e¡0[02 ¡ 4(0) + 2] = 2 > 0

f 00(1) = e¡1(12 ¡ 4(1) + 2) = ¡1e< 0

f 00(4) = e¡4[42 ¡ 4(4) + 2] = 2

e4> 0

In‡ection points at (0:6; 0:19) and (3:4; 0:38)

Concave upward on (¡1; 0:6) and (3:4;1); down-ward on (0:6; 3:4)y-intercept: y = 02e¡0 = 0

0 is the only intercept.

27. f(x) = (x¡ 1)e¡x

Domain is (¡1;1)f(¡x) = (¡x¡ 1)ex


f 0(x) = ¡(x¡ 1)e¡x + e¡x(1)= e¡x[¡(x¡ 1) + 1]= e¡x(2¡ x)

f 0(x) = 0 when e¡x(2¡ x) = 0x = 2:

Critical number: 2

Critical point:¡2; 1e2

¢f 00(x) = ¡e¡x + (2¡ x)(¡e¡x)

= ¡e¡x[1 + (2¡ x)]= ¡e¡x(3¡ x)

f 00(2) = ¡e¡2(3¡ 2) = ¡1e2< 0

Relative maximum at¡2; 1e2

¢f 0(0) = e¡0(2¡ 0) = 2 > 0

f 0(3) = e¡3(2¡ 3) = ¡1e3< 0

Increasing on (¡1; 2); decreasing on (2;1):

f 00(x) = 0 when ¡e¡x(3¡ x) = 0x = 3:

f 00(0) = ¡e¡0(3¡ 0) = ¡3 < 0

f 00(4) = ¡e¡4(3¡ 4) = 1

e4> 0

In‡ection point at¡3; 2e3

¢Concave downward on (¡1; 3); concave upwardon (3;1)

f(3) = (3¡ 1)e¡3 = 2

e3

y-intercept: y = (0¡ 1)e¡0= (¡1)(1) = ¡1

x-intercept: 0 = (x¡ 1)e¡xx¡ 1 = 0

x = 1


28. f(x) = ex + e¡x

Domain is (¡1; )f(¡x) = e¡x + ex = f(x)The graph has symmetry about the y-axis.

f 0(x) = ex ¡ e¡x

Critical numbers:

ex ¡ e¡x = 0

ex ¡ 1

ex= 0

e2x ¡ 1ex

= 0

e2x ¡ 1 = 0e2x = 1

2x = ln 1

2x = 0

x = 0

The only critical number is 0.


The only critical point is (0; 2).

f 0(¡1) = e¡1 ¡ e¡(¡1) = 1

e¡ e ¼ ¡2:35 < 0

f 0(1) = e1 ¡ e¡1 ¼ 2:35 > 0

Relative minimum at (0; 2)Decreasing on (¡1; 0); increasing on (0;1)

f(x)00 = ex + e¡x =e2x + 1

ex

Since f(x)00 6= 0 (e2x+1 is always positive), thereis no in‡ection point.

f 00(0) = e0 + e¡0 = 2 > 0

The entire graph is concave upward on (¡1;1):y-intercept: y = e0 + e¡0 = 1 + 1 = 2

The y-intercept is 2.There is no x-intercept.

29. f(x) = x2=3 ¡ x5=3

Domain is (¡1;1):f(¡x) = x2=3 + x5=3


f 0(x) =2

3x¡1=3 ¡ 5

3x2=3

=2¡ 5x3x1=3

f 0(x) = 0 when 2¡ 5x = 0

Critical number: x =2

5

f

μ2

5

¶=

μ2

5

¶2=3¡μ2

5

¶5=3

=3 ¢ 22=355=3

¼ 0:326

Critical point: (0:4; 0:326)

f 00(x) =3x1=3(¡5)¡ (2¡ 5x)(3) ¡13¢x¡2=3

(3x1=3)2

=¡15x1=3 ¡ (2¡ 5x)x¡2=3

9x2=3

=¡15x¡ (2¡ 5x)

9x4=3

=¡10x¡ 29x4=3

f 00μ2

5

¶=¡10 ¡25¢¡ 29¡25

¢4=3¼ ¡2:262 < 0

Relative maximum at³25 ;

3¢22=355=3

´¼ (0:4; 0:326)

f 0(x) does not exist when x = 0

Since f 00(0) is unde…ned, use the …rst derivativetest.

f 0(¡1) = 2¡ 5(¡1)3(¡1)1=3 =

7

¡3 < 0

f 0μ1

8

¶=2¡ 5 ¡18¢3¡18

¢1=3 = 11

12> 0

f 0(1) =2¡ 53 ¢ 11=3 = ¡1 < 0

Relative minimum at (0; 0)

f increases on¡0; 25

¢.

f decreases on (¡1; 0) and ¡25 ;1¢ :f 00(x) = 0 when ¡10x¡ 2 = 0

x = ¡15

f 00(x) unde…ned when 9x4=3 = 0x = 0

f 00(¡1) = ¡10(¡1)¡ 29(¡1)4=3 =

8

9> 0

f 00μ¡18

¶=¡10 ¡¡1

8

¢¡ 29¡¡1

8

¢4=3 = ¡43< 0

f 00(1) =¡10(1)¡ 29(1)4=3

= ¡43< 0

Concave upward on¡¡1;¡1

5


¡¡15 ;1

¢In‡ection point at

¡¡15 ;

655=3

¢ ¼ (¡0:2; 0:410)


y-intercept: y = 02=3 ¡ 05=3 = 0x-intercept: 0 = x2=3 ¡ x5=3

= x2=3(1¡ x)x = 0 or x = 1

30. f(x) = x1=3 + x4=3

Domain is (¡1;1):


f 0(x) =1

3x¡2=3 +

4

3x1=3

=1+ 4x

3x2=3

f 0(x) = 0 when 1 + 4x = 0

x = ¡14

Critical number: x = ¡14 and x = 0

f

μ¡14

¶=

μ¡14

¶1=3+

μ¡14

¶4=3

= ¡ 3

44=3¼ ¡0:472

Critical points:¡¡1

4 ;¡0:472)¢, (0; 0)

f 00(x) =3x2=3(4)¡ (1 + 4x)3 ¡23¢x¡1=3

(3x2=3)2

=12x2=3 ¡ 2(1 + 4x)x¡1=3

9x4=3

=12x¡ 2(1 + 4x)

9x5=3

=4x¡ 29x5=3

f 00μ¡14

¶=4¡¡1

4

¢¡ 29¡¡1

4

¢5=3 = 45=3

3

¼ 3:360 > 0

Relative minimum at¡¡1

4 ;¡ 344=3

¢ ¼ (¡0:25;¡0:472)f 0(x) =

1 + 4x

3x2=3unde…ned at x = 0:

Test sign of f 0(x) on intervals de…ned by x = ¡14 ;

x = 0:

f 0(¡1) = ¡33= ¡1 < 0

f 0μ¡18

¶=1¡ 1

234

> 0

f 0(1) =5

3> 0

f increases on¡¡1

4 ;1¢, f decreases on

¡¡1;¡14

¢No extreme point at (0; f(0)) = (0; 0)

f 00(x) = 0 when 4x¡ 2 = 0

x =1

2

f 00(x) is unde…ned when 9x5=3 = 0x = 0

f 00(¡1) = 4(¡1)¡ 29(¡1)5=3 =

2

3> 0

f 00μ1

8

¶=4¡18

¢¡ 29¡18

¢5=3 = ¡163< 0

f 00(1) =4(1)¡ 29(1)5=3

=2

9> 0

In‡ection points at (0; 0) and¡12 ;

324=3

¢ ¼ (0:5; 1:191)Concave upward on (¡1; 0) and ¡12 ;1¢Concave downward on

¡0; 12

¢y-intercept: y = 01=3 + 04=3 = 0x-intercept: 0 = x1=3 + x4=3

= x1=3(1 + x)

x = 0 or x = ¡1


31. For Exercises 3, 7, and 9, the relative maxima orminima are outside the vertical window of ¡10 ·y · 10.

For Exercise 11, the default window shows only asmall portion of the graph.

For Exercise 15, the default window does not allowthe graph to properly display the vertical asymp-totes.

32. For Exercise 6, the y-intercept is outside the ver-tical window of ¡10 · y · 10:

For Exercises 4, 10, and 12, the relative max-ima or minima are outside the vertical windowof ¡10 · y · 10:


33. For Exercises 17, 19, 23, 25, and 27, the y-coordinateof the relative minimum, relative maximum, or in-‡ection point is so small, it may be hard to dis-tinguish.

34. For Exercises 18 and 24, the y-coordinate of therelative minimum, relative maximum, or in‡ectionpoint is so small, it may be hard to distinguish.


For Exercises 35¡39 other graphs are possible.

35. (a) indicates a smooth, continuous curve exceptwhere there is a vertical asymptote.

(b) indicates that the function decreases on bothsides of the asymptote, so there are no relativeextrema.

(c) gives the horizontal asymptote y = 2:

(d) and (e) indicate that concavity does not changeleft of the asymptote, but that the right portion ofthe graph changes concavity at x = 2 and x = 4:

There are in‡ection points at 2 and 4.

36. (a) indicates that the curve may not contain breaks.

(b) and (c) indicate relative minima at ¡6 and 3and a relative maximum at 1.

(d) and (e), when combined with (b) and (c) showthat concavity does not change between relativeextrema.

(f) gives the y-intercept.

37. (a) indicates that there can be no asymptotes,sharp “corners”, holes, or jumps. The graph mustbe one smooth curve.

(b) and (c) indicate relative maxima at ¡3 and4 and a relative minimum at 1.

(d) and (e) are consistent with (g).

(f) indicates turning points at the critical num-bers ¡3 and 4.



(b) and (c) indicate relative maxima at ¡2 and3 and a relative minimum at 0.

(d) shows that concavity does not change at 0.

(d) and (e) are consistent with (i).

(f) shows critical values.

(g) and (h) indicate that the function is not dif-ferentiable at 0, and is di¤erentiable everywhereelse.Thus, a sharp corner must exist at 0.

(i) indicates that concavity changes just once, at(5; 1):


(b) indicates that there is a sharp “corner” at 4.

(c) gives a point at (1; 5):

(d) shows critical numbers.

(e) and (f) indicate (combined with (c) and (d))a relative maximum at (1; 5), and (combined with(b)) a relative minimum at 4.

(g) is consistent with (b).

(h) indicates the curve is concave upward on (2; 3):

(i) indicates the curve is concave downward on(¡1; 2); (3; 4) and (4;1):

Chapter 5 Review Exercises

5. f(x) = x2 + 9x+ 8

f 0(x) = 2x+ 9

f 0(x) = 0 when x = ¡92 and f

0 exists everywhere.


Test an x-value in the intervals¡¡1;¡9

2

¢and¡¡9

2 ;1¢:

f 0(¡5) = ¡1 < 0f 0(¡4) = 1 > 0f is increasing on

¡¡92 ;1

¢and decreasing on¡¡1;¡9

2

¢:

6. f(x) = ¡2x2 + 7x+ 14f 0(x) = ¡4x+ 7f 0(x) = 0 when x = 7

4 and f0 exists everywhere.

Critical number: 74

Test an x-value in the intervals¡¡1; 74¢ and¡

74 ;1

¢:

f 0(0) = 7 > 0f 0(2) = ¡1 < 0f is increasing on

¡¡1; 74¢ and decreasing on ¡74 ;1¢ :7. f(x) = ¡x3 + 2x2 + 15x+ 16

f 0(x) = ¡3x2 + 4x+ 15= ¡(3x2 ¡ 4x¡ 15)= ¡(3x+ 5)(x¡ 3)

f 0(x) = 0 when x = ¡53 or x = 3 and f 0 exists

everywhere.

Critical numbers: ¡53 and 3

Test an x-value in the intervals¡¡1;¡5

3

¢,¡¡5

3 ; 3¢,

and (3;1):f 0(¡2) = ¡5 < 0f 0(0) = 15 > 0f 0(4) = ¡17 < 0

f is increasing on¡¡5

3 ; 3¢and decreasing on¡¡1;¡5

3

¢and (3;1):

Chapter 5 Review Exercises 359

8. f(x) = 4x3 + 8x2 ¡ 16x+ 11f 0(x) = 12x2 + 16x¡ 16

= 4(3x2 + 4x¡ 4)= 4(3x¡ 2)(x+ 2)

f 0(x) = 0 when x = 23 or x = ¡2 and f 0 exists

everywhere.


Test an x-value in the intervals (¡1;¡2) ; ¡¡2; 23¢,and (23 ;1):

f 0(¡3) = 44 > 0f 0(0) = ¡16 < 0f 0(1) = 12 > 0

f is increasing on (¡1;¡2) and (23 ;1) and de-creasing on (¡2; 23):

9. f(x) =16

9¡ 3xf 0(x) =

16(¡1)(¡3)(9¡ 3x)2 =

48

(9¡ 3x)2

f 0(x) > 0 for all x (x 6= 3), and f is not de…nedfor x = 3:

f is increasing on (¡1; 3) and (3;1) and neverdecreasing.

10. f(x) =15

2x+ 7

f 0(x) =15(¡1)(2)(2x+ 7)2

=¡30

(2x+ 7)2

f 0(x) < 0 for all x¡x 6= ¡7

2

¢, and f is not de…ned

for x = ¡72 .

f is never increasing, and decreasing on¡¡1;¡7

2

¢and

¡¡72 ;1

¢:

11. f(x) = ln¯̄x2 ¡ 1¯̄

f 0(x) =2x

x2 ¡ 1f is not de…ned for x = ¡1 and x = 1.

f 0(x) = 0 when x = 0.

Test an x-value in the intervals (¡1;¡1); (¡1; 0);(0; 1), and (1;1).

f 0(¡2) = ¡43< 0

f 0μ¡12

¶=4

3> 0

f 0μ1

2

¶= ¡4

3< 0

f 0(2) =4

3> 0

f is increasing on (¡1; 0) and (1;1) and decreas-ing on (¡1;¡1) and (0; 1).

12. f(x) = 8xe¡4x

f 0(x) = 8(e¡4x) + 8x(¡4e¡4x)= 8e¡4x ¡ 32xe¡4x= 8e¡4x(1¡ 4x)

f 0(x) = 0 when x = 14 .

Test an x-value in the intervals¡¡1; 14¢ and ¡14 ;1¢ :

f 0(0) = 8 > 0f 0(1) = ¡24e¡4 < 0f is increasing on

¡¡1; 14¢ and decreasing on ¡14 ;1¢ :13. f(x) = ¡x2 + 4x¡ 8

f 0(x) = ¡2x+ 4 = 0Critical number: x = 2

f 00(x) = ¡2 < 0 for all x; so f(2) is a relativemaximum.

f(2) = ¡4Relative maximum of ¡4 at 2

14. f(x) = x2 ¡ 6x+ 4f 0(x) = 2x¡ 6f 0(x) = 0 when x = 3:Critical number: 3

f 00(x) = 2 > 0 for all x; so f(3) is a relative mini-mum.

f(3) = ¡5Relative minimum of ¡5 at 3

15. f(x) = 2x2 ¡ 8x+ 1f 0(x) = 4x¡ 8 = 0Critical number: x = 2

f 00(x) = 4 > 0 for all x; so f(2) is a relative mini-mum.

f(2) = ¡7Relative minimum of ¡7 at 2


16. f(x) = ¡3x2 + 2x¡ 5f 0(x) = ¡6x+ 2

f 0(x) = 0 when x = 13 :

Critical number:1

3

Since f 00(x) = ¡6 < 0 for all x; f ¡13¢ is a relativemaximum.

f

μ1

3

¶= ¡14

3

Relative maximum of ¡143 at

13

17. f(x) = 2x3 + 3x2 ¡ 36x+ 20f 0(x) =6x2 + 6x¡ 36 = 0

6(x2 + x¡ 6) = 0(x+ 3)(x¡ 2) = 0


f 00(x) = 12x+ 6

f 00(¡3) = ¡30 < 0; so a maximum occursat x = ¡3:

f 00(2) = 30 > 0; so a minimum occursat x = 2:

f(¡3) = 101

f(2) = ¡24

Relative maximum of 101 at ¡3Relative minimum of ¡24 at 2

18. f(x) = 2x3 + 3x2 ¡ 12x+ 5f 0(x) = 6x2 + 6x¡ 12

= 6(x2 + x¡ 2)= 6(x+ 2)(x¡ 1)

f 0(x) = 0 when x = ¡2 or x = 1:Critical numbers: ¡2; 1

f 00(x) = 12x+ 6

f 00(¡2) = ¡18 < 0; so a maximum occurs at x =¡2:f 00(1) = 18 > 0; so a minimum occurs at x = 1:

f(¡2) = 25f(1) = ¡2

Relative maximum of 25 at ¡2Relative minimum of ¡2 at 1

19. f(x) =xex

x¡ 1f 0(x) =

(x¡ 1)(xex + ex)¡ xex(1)(x¡ 1)2

=x2ex + xex ¡ xex ¡ ex ¡ xex

(x¡ 1)2

=x2ex ¡ xex ¡ ex

(x¡ 1)2

=ex(x2 ¡ x¡ 1)(x¡ 1)2

f 0(x) is unde…ned at x = 1; but 1 is not in thedomain of f(x):f 0(x) = 0 when x2 ¡ x¡ 1 = 0

x =1§p1¡ 4(1)(¡1)

2

=1§p52

1 +p5

2¼ 1:618 or

1¡p52

= ¡0:618

Critical numbers are ¡0:618 and 1.618.

f 0(1:4) =e1:4(1:42 ¡ 1:4¡ 1)

(1:4¡ 1)2 ¼ ¡11:15 < 0

f 0(2) =e2(22 ¡ 2¡ 1)(2¡ 1)2 = e2 ¼ 7:39 > 0

f 0(¡1) = e¡1[(¡1)2 ¡ (¡1)¡ 1](¡1¡ 1)2 ¼ 0:09 > 0

f 0(0) =e0(02 ¡ 0¡ 1)(0¡ 1)2 = ¡1 < 0

There is a relative maximum at (¡0:618; 0:206)and a relative minimum at (1:618; 13:203):

20. y =ln(3x)

2x2

y0 =2x2 ¢ 1x ¡ (ln 3x) ¢ 4x

4x4

=2x(1¡ 2 ln 3x)

4x4

=1¡ 2 ln 3x2x3


y0 = 0 when

1¡ 2 ln 3x = 01 = 2 ln 3x1

2= ln 3x

e1=2 = 3x

e1=2

3= x

pe

3¼ 0:55

f 0(1) =1¡ 2 ln 3

2¼ ¡0:6 < 0

f 0μ1

3

¶=1¡ 2 ln 12¡13

¢3 =1

2 ¢ 127=27

2> 0

f

μpe

3

¶¼ 0:83

Relative maximum at³p

e3 ; 0:83

´or (0:55; 0:83)

21. f(x) = 3x4 ¡ 5x2 ¡ 11xf 0(x) = 12x3 ¡ 10x¡ 11f 00(x) = 36x2 ¡ 10f 00(1) = 36(1)2 ¡ 10 = 26

f 00(¡3) = 36(¡3)2 ¡ 10 = 314

22. f(x) = 9x3 +1

x= 9x3 + x¡1

f 0(x) = 27x2 ¡ x¡2

f 00(x) = 54x+ 2x¡3 = 54x+2

x3

f 00(1) = 54(1) +2

(1)3= 56

f 00(¡3) = 54(¡3) + 2

(¡3)3 = ¡162¡2

27= ¡4376

27

23. f(x) =4x+ 2

3x¡ 6f 0(x) =

(3x¡ 6)(4)¡ (4x+ 2)(3)(3x¡ 6)2

=12x¡ 24¡ 12x¡ 6

(3x¡ 6)2

=¡30

(3x¡ 6)2= ¡30(3x¡ 6)¡2

f 00(x) = ¡30(¡2)(3x¡ 6)¡3(3)

= 180(3x¡ 6)¡3 or180

(3x¡ 6)3

f 00(1) = 180[3(1)¡ 6]¡3 = ¡203

f 00(¡3) = 180[3(¡3)¡ 6]¡3 = ¡ 4

75

24. f(x) =1¡ 2x4x+ 5

f 0(x) =(4x+ 5)(¡2)¡ (1¡ 2x)(4)

(4x+ 5)2

=¡8x¡ 10¡ 4 + 8x

(4x+ 5)2

=¡14

(4x+ 5)2= ¡14(4x+ 5)¡2

f 00(x) = ¡14(¡2)(4x+ 5)¡3(4)

= 112(4x+ 5)¡3 or112

(4x+ 5)3

f 00(1) =112

[4(1) + 5]3=112

729

f 00(¡3) = 112

[4(¡3) + 5]3 = ¡16

49

25. f(t) =pt2 + 1 = (t2 + 1)1=2

f 0(t) =1

2(t2 + 1)¡1=2(2t) = t(t2 + 1)¡1=2

f 00(t) = (t2 + 1)¡1=2(1)

+ t

·μ¡12

¶(t2 + 1)¡3=2(2t)

¸

= (t2 + 1)¡1=2 ¡ t2(t2 + 1)¡3=2

=1

(t2 + 1)1=2¡ t2

(t2 + 1)3=2=t2 + 1¡ t2(t2 + 1)3=2

= (t2 + 1)¡3=2 or1

(t2 + 1)3=2

f 00(1) =1

(1 + 1)3=2=

1

23=2¼ 0:354

f 00(¡3) = 1

(9 + 1)3=2=

1

103=2¼ 0:032

26. f(t) = ¡p5¡ t2 = ¡(5¡ t2)1=2

f 0(t) = ¡12(5¡ t2)¡1=2(¡2t) = t(5¡ t2)¡1=2

f 00(t) = (1)(5¡ t2)¡1=2 + t·¡12(5¡ t2)¡3=2(¡2t)

¸

= (5¡ t2)¡1=2 + t[t(5¡ t2)¡3=2]

= (5¡ t2)¡3=2(5¡ t2 + t2) = 5

(5¡ t2)3=2


f 00(1) =5

(5¡ 1)3=2 =5

8

f 00(¡3) = 5

(5¡ 9)3=2This value does not exist since (¡4)3=2 does notexist.

27. f(x) = ¡2x3 ¡ 12x2 + x¡ 3

Domain is (¡1;1)The graph has no symmetry.

f 0(x) = ¡6x2 ¡ x+ 1 = 0(3x¡ 1)(2x+ 1) = 0

Critical numbers: 13 and ¡12

Critical points:¡13 ;¡2:80

¢and

¡¡12 ;¡3:375

¢f 00(x) = ¡12x¡ 1

f 00μ1

3

¶= ¡5 < 0

f 00μ¡12

¶= 5 > 0

Relative maximum at 13

Relative minimum at ¡12

Increasing on¡¡1

2 ;13

¢Decreasing on

¡¡1;¡12

¢and

¡13 ;1

¢f 00(x) =¡12x¡ 1 = 0

x = ¡ 1

12

Point of in‡ection at¡¡ 1

12 ;¡3:09¢

Concave upward on¡¡1;¡ 1

12


¡¡ 112 ;1

¢y-intercept:

y = ¡2(0)3 ¡ 12(0)2 + (0)¡ 3 = ¡3

28. f(x) = ¡43x3 + x2 + 30x¡ 7


f 0(x) = ¡4x2 + 2x+ 30= ¡2(2x2 ¡ x¡ 15)= ¡2(2x+ 5)(x¡ 3) = 0



2 ;¡54:91¢and (3; 56)

f 00(x) = ¡8x+ 2

f 00μ¡52

¶= 22 > 0

f 00(3) = ¡22 < 0Relative maximum at 3


Increasing on¡¡5

2 ; 3¢

Decreasing on¡¡1;¡5

2

¢and (3;1)

f 00(x) =¡8x+ 2 = 0

x =1

4

Point of in‡ection at¡14 ; 0:54

¢Concave upward on


¡14 ;1

¢y-intercept:

y = ¡43(0)3 + (0)3 + 30(0)2 ¡ 7

= ¡7


29. f(x) = x4 ¡ 43x3 ¡ 4x2 + 1


f 0(x) =4x3 ¡ 4x2 ¡ 8x = 04x(x2 ¡ x¡ 2) = 0

4x(x¡ 2)(x+ 1) = 0Critical numbers: 0; 2; and ¡1Critical points: (0; 1);

¡2;¡29

3

¢and

¡¡1;¡23

¢f 00(x) = 12x2 ¡ 8x¡ 8

= 4(3x2 ¡ 2x¡ 2)f 00(¡1) = 12 > 0f 00(0) = ¡8 < 0f 00(2) = 24 > 0

Relative maximum at 0Relative minima at ¡1 and 2Increasing on (¡1; 0) and (2;1)Decreasing on (¡1;¡1) and (0; 2)

f 00(x) = 4(3x2 ¡ 2x¡ 2) = 0

x =2§p4¡ (¡24)

6

=1§p73

Points of in‡ection at³1+p7

3 ;¡5:12ánd³

1¡p73 ; 0:11

Ćoncave upward on

³¡1; 1¡

p7

3

ánd

³1+p7

3 ;1´

Concave downward on³1¡p73 ; 1+

p7

3

ý-intercept:

y = (0)4 ¡ 43(0)3 ¡ 4(0)2 + 1 = 1

30. f(x) = ¡23x3 +

9

2x2 + 5x+ 1


f 0(x) = ¡2x2 + 9x+ 5 = 0¡(2x+ 1)(x¡ 5) = 0



2 ;¡0:29¢and (5; 55:17)

f 00(x) = ¡4x+ 9

f 00μ¡12

¶= 11 > 0

f 00(5) = ¡11 < 0Relative maximum at 5


Increasing on¡¡1

2 ; 5¢

Decreasing on¡¡1;¡1

2

¢and (5;1)

f 00(x) =¡4x+ 9 = 0

x =9

4

Point of in‡ection at¡94 ; 27:44

¢Concave upward on


¡94 ;1

¢y-intercept:

y = ¡23(0)3 +

9

2(0)2 + 5(0) + 1 = 1


31. f(x) =x¡ 12x+ 1

Domain is¡¡1;¡1

2

¢ [ ¡¡12 ;1

¢The graph has no symmetry.

f 0(x) =(2x+ 1)(1)¡ (x¡ 1)(2)

(2x+ 1)2

=3

(2x+ 1)2

f 0 is never zero.f 0(¡1

2) does not exist, but ¡12 is not a critical

number because ¡12 is not in the domain of f:

Thus, there are no critical numbers, so f(x) hasno relative extrema.Increasing on

¡¡1; 12¢ and ¡12 ;1¢

f 00(x) =¡12

(2x+ 1)3

f 00(0) = ¡12 < 0f 00(¡1) = 12 > 0

No in‡ection points

Concave upward on¡¡1;¡1

2


¡¡12 ;1

¢x-intercept:

x¡ 12x+ 1

= 0

x = 1

y-intercept: y =0¡ 12(0) + 1

= ¡1



32. f(x) =2x¡ 5x+ 3

Domain is (¡1;¡3) [ (¡3;1)

f 0(x) =2(x+ 3)¡ (2x¡ 5)

(x+ 3)2

=11

(x+ 3)2

f 0 is never zero.f(x) has no extrema.

f 00(x) =¡22

(x+ 3)3

f 00(¡4) = 22 > 0f 00(¡2) = ¡22 < 0

Increasing on (¡1;¡3) [ (¡3;1)

Concave upward on (¡1;¡3)Concave downward on (¡3;1)

x-intercept:2x¡ 5x+ 3

= 0

x =5

2

y-intercept:2(0)¡ 50 + 3

= ¡53

Vertical asymptote at x = ¡3Horizontal asymptote at y = 2


33. f(x) = ¡4x3 ¡ x2 + 4x+ 5Domain is (¡1;1)The graph has no symmetry.

f 0(x) = ¡12x2 ¡ 2x+ 4=¡2(6x2 + x¡ 2) = 0(3x+2)(2x¡1) = 0

Critical numbers: ¡23 and

12


3 ; 3:07¢and

¡12 ; 6:25

¢f 00(x) = ¡24x¡ 2

= ¡2(12x+ 1)

f 00μ¡23

¶= 14 > 0

f 00μ1

2

¶= ¡14 < 0



Increasing on¡¡2

3 ;12

¢Decreasing on

¡¡1;¡23

¢and

¡12 ;1

¢f 00(x) =¡2(12x+ 1) = 0

x = ¡ 1

12

Point of in‡ection at¡¡ 1

12 ; 4:66¢

Concave upward on¡¡1;¡ 1

12


¡¡ 112 ;1

¢y-intercept:

y = ¡4(0)3 ¡ (0)2 + 4(0) + 5 = 5

34. f(x) = x3 +5

2x2 ¡ 2x¡ 3


f 0(x) = 3x2 + 5x¡ 2 = (3x¡ 1)(x+ 2) = 0

Critical numbers: 13 and ¡2

Critical points:¡13 ;¡3:35

¢and (¡2; 3)

f 00(x) = 6x+ 5

f 00μ1

3

¶= 7 > 0

f 00(¡2) = ¡7 < 0

Relative maximum at ¡2

Relative minimum at 13

Increasing on (¡1;¡2) and ¡13 ;1¢Decreasing on

¡¡2; 13¢f 00(x) = 6x+ 5 = 0

x = ¡56

Point of in‡ection at¡¡5

6 ;¡0:18¢

Concave upward on¡¡5

6 ;1¢

Concave downward on¡¡1;¡5

6

¢y-intercept: ¡3


35. f(x) = x4 + 2x2

Domain is (¡1;1)f(¡x) = (¡x)4 ¡ 2(¡x)2

= x4 + 2x2 = f(x)


f 0(x) = 4x3 + 4x= 4x(x2 + 1) = 0

Critical number: 0Critical point: (0; 0)

f 00(x) = 12x2 + 4 = 4(3x2 + 1)f 00(0) = 4 > 0

Relative minimum at 0Increasing on (0;1)Decreasing on (¡1; 0)

f 00(x) = 4(3x2 + 1) 6= 0 for any x

No points of in‡ection

f 00(¡1) = 16 > 0f 00(1) = 16 > 0

Concave upward on (¡1;1)x-intercept: 0; y-intercept: 0

36. f(x) = 6x3 ¡ x4


f 0(x) = 18x2 ¡ 4x3 = 2x2(9¡ 2x) = 0Critical numbers: 0 and 9

2

Critical points: (0; 0) and¡92 ; 136:7

¢f 00(x) = 36x¡ 12x2 = 12x(3¡ x)f 00(0) = 0

f 00μ9

2

¶= ¡81 < 0


No relative extrema at 0

Increasing on¡¡1; 92¢

Decreasing on¡92 ;1

¢f 00(x) = 12x(3¡ x) = 0

x = 0 or x = 3

Points of in‡ection at (0; 0) and (3; 81)

Concave upward on (0; 3)Concave downward on (¡1; 0) and (3;1)x-intercepts: 6x3 ¡ x4 = 0

x3(6¡ x) = 0x = 0; x = 6

The x-intercepts are 0 and 6.y-intercept: 0

37. f(x) =x2 + 4

x

Domain is (¡1; 0) [ (0;1)

f(¡x) = (¡x)2 + 4¡x

=x2 + 4

¡x = ¡f(x)The graph is symmetric about the origin.

f 0(x) =x(2x)¡ (x2 + 4)

x2

=x2 ¡ 4x2

= 0

Critical numbers: ¡2 and 2Critical points: (¡2;¡4) and (2; 4)

f 00(x) =8

x3

f 00(¡2) = ¡1 < 0f 00(2) = 1 > 0

Relative maximum at ¡2Relative minimum at 2


Increasing on (¡1;¡2) and (2;1)Decreasing on (¡2; 0) and (0; 2)f 00(x) = 8

x3 > 0 for all x:

No in‡ection pointsConcave upward on (0;1)Concave downward on (¡1; 0)No x- or y-interceptsVertical asymptote at x = 0Oblique asymptote at y = x

38. f(x) = x+8

x

Domain is (¡1; 0) [ (0;1)

f(¡x) = ¡x+ 8

¡x= ¡

μx+

8

x

¶= ¡f(x)


f 0(x) = 1¡ 8

x2

=x2 ¡ 8x2

= 0

Critical numbers: x = §2p2Critical points: (2

p2; 4p2); (¡2p2;¡4p2)

f 00(x) =16

x3

f 00(¡2p2) = ¡p2

2< 0

f 00(2p2) =

p2

2> 0

Relative maximum at ¡2p2Relative minimum at 2

p2

Increasing on (¡1;¡2p2) and (2p2;1)Decreasing on (¡2p2; 0) and (0; 2p2)f 00(x) = 16

x3 > 0 for all x:

No in‡ection points

Concave upward on (0;1)Concave downward on (¡1; 0)Vertical asymptote at x = 0Oblique asymptote at y = x

39. f(x) =2x

3¡ xDomain is (¡1; 3) [ (3;1)The graph has no symmetry.

f 0(x) =(3¡ x)(2)¡ (2x)(¡1)

(3¡ x)2

=6

(3¡ x)2f 0(x) is never zero. f 0(3) does not exist, but since3 is not in the domain of f; it is not a criticalnumber.No critical numbers, so no relative extrema

f 0(0) =2

3> 0

f 0(4) = 6 > 0

Increasing on (¡1; 3) and (3;1)

f 00(x) =12

(3¡ x)3

f 00(x) is never zero. f 00(3) does not exist, but since3 is not in the domain of f; there is no in‡ectionpoint at x = 3:

f 00(0) =12

27> 0

f 00(4) = ¡12 < 0

Concave upward on (¡1; 3)Concave downward on (3;1)x-intercept: 0; y-intercept: 0




40. f(x) =¡4x1 + 2x

Domain is¡¡1;¡1

2

¢ [ ¡¡12 ;1

¢The graph has no symmetry.

f 0(x) =¡4(1 + 2x)¡ 2(¡4x)

(1 + 2x)2

=¡4¡ 8x+ 8x(1 + 2x)2

=¡4

(1 + 2x)2

f 0(x) is never zero.No critical values; no relative extrema

f 0(0) = ¡4 < 0f 0(¡1) = ¡4 < 0Decreasing on

¡¡1;¡12

¢and

¡¡12 ;1

¢f 00(x) =

16

(1 + 2x)3

f 00(x) is never zero; no points of in‡ection.

f 00(0) = 16 > 0f 00(¡1) = ¡16 < 0Concave upward on

¡¡12 ;1


¡¡1;¡12

¢x-intercept: 0; y-intercept: 0



41. f(x) = xe2x

Domain is (¡1;1).f(¡x) = ¡xe¡2x

The graph has no symmety.

f 0(x) = (1)(e2x) + (x)(2e2x)= e2x(2x+ 1)

f 0(x) = 0 when x = ¡12 .


Critical point:¡¡1

2 ;¡ 12e

¢f 0(¡1) = e2(¡1)[2(¡1) + 1] = ¡e¡2 < 0f 0(0) = e2(0)[2(0) + 1] = 1 > 0

No relative maximum

Relative minimum at¡¡1

2 ;¡ 12e

¢Decreasing on

¡¡1;¡12

¢and increasing on

¡¡12 ;1

¢f 00(x) = 2e2x(2x+ 1) + e2x(2)

= 4e2x(x+ 1)

f 00(x) = 0 when x = ¡1:f 00(¡2) = 4e2(¡2)[(¡2) + 1] = ¡4e¡4 < 0f 00(0) = 4e2(0)[(0) + 1] = 4 > 0

In‡ection point at (¡1;¡e¡2)Concave upward on (¡1;1)Concave downward on (¡1;¡1)x-intercept: xe2x = 0

x = 0

y-intercept: y = (0)e2(0) = 0

Since limx!¡1 xe2x = 0, there is a horizontal as-

ymptote at y = 0.


42. f(x) = x2e2x

Domain is (¡1;1).f(¡x) = (¡x)2e2(¡x) = x2e¡2x

The graph is not symmetric about the y-axis ororigin.

f 0(x) = (2x)(e2x) + (x2)(2e2x)= 2(x2 + x)e2x

= 2x(x+ 1)e2x

f 0(x) = 0 when x = ¡1 and x = 0.Critical numbers: ¡1 and 0Critical points: (¡1; e¡2) and (0; 0)f 0(¡2) = 2[(¡2)2 + (¡2)]e2(¡2) = 4e¡4 > 0

f 0μ¡12

¶= 2

"μ¡12

¶2+

μ¡12

¶#e2(¡1=2)

= ¡12e¡1 < 0

f 0(1) = 2[(1)2 + (1)]e2(1) = 4e2 > 0

Relative maximum at (¡1; e¡2)Relative minimum at (0; 0)

Increasing on (¡1;¡1) and (0;1)Decreasing on (¡1; 0)f 00(x) = 2(2x+ 1)e2x + 2(x2 + x)e2x(2)

= 2(2x2 + 4x+ 1)e2x

f 00(x) = 0 when

2x2 + 4x+ 1 = 0

x =¡4§p16¡ 4(2)(1)

2(2)

=¡4§p8

4

= ¡1§p2

2

f 00(¡2) = 2[2(¡2)2 + 4(¡2) + 1]e2(¡2) = 2e¡4 > 0f 00(¡1) = 2[2(¡1)2 + 4(¡1) + 1]e2(¡1) = ¡2e¡2 < 0f 00(0) = 2[2(0)2 + 4(0) + 1]e2(0) = 2 > 0

In‡ection points at³¡1¡

p22 ;¡32 +

p2¢e¡2¡

p2´¼ (¡1:707, 0:09588)³

¡1 +p22 ;¡32 ¡

p2¢e¡2+

p2´¼ (¡0:293, 0:04775)

Concave upward on³¡1;¡1¡

p22

´and

³¡1 +

p22 ;1

´

Concave downward on³¡1¡

p22 ;¡1 +

p22

´

x-intercept: x2e2x = 0x = 0

y-intercept: y = (0)2e2(0) = 0

Since limx!¡1 x2e2x = 0, horizontal asymptote at

y = 0.

43. f(x) = ln(x2 + 4)

Domain is (¡1;1).f(¡x) = ln[(¡x)2 + 4] = ln(x2 + 4) = f(x)The graph is symmetric about the y-axis.

f 0(x) =2x

x2 + 4

f 0(x) = 0 when x = 0.

Critical number: 0

Critical point: (0; ln 4)

f 0(¡1) = 2(¡1)(¡1)2 + 4 = ¡

2

5< 0

f 0(1) =2(1)

(1)2 + 4=2

5> 0

No relative maximum

Relative minimum at (0; ln 4)

Increasing on (0;1)Decreasing on (¡1; 0)

f 00(x) =(x2 + 4)(2)¡ (2x)(2x)

(x2 + 4)2

=¡2(x2 ¡ 4)(x2 + 4)2


f 00(x) = 0 when

x2 ¡ 4 = 0x = §2

f 00(¡3) = ¡2[(¡3)2 ¡ 4][(¡3)2 + 4]2 = ¡ 10

169< 0

f 00(0) =¡2[(0)2 ¡ 4][(0)2 + 4]2

=1

2> 0

f 00(3) =¡2[(3)2 ¡ 4][(3)2 + 4]2

= ¡ 10

169< 0

In‡ection points at (¡2; ln 8) and (2; ln 8)Concave upward on (¡2; 2)Concave downward on (¡1;¡2) and (2;1)Since f(x) never equals zero, there are no x-intercepts.

y-intercept: y = ln[(0)2 + 4] = ln 4

No horizontal or vertical asymptotes.

44. f(x) = x2 lnx

Domain is (0;1).The graph is not symmetric about the y-axis ororigin since f(x) is not de…ned for x · 0:

f 0(x) = 2x(lnx) + x2 ¢ 1x

= 2x lnx+ x

= x(2 lnx+ 1)

f 0(x) = 0 when

2 lnx+ 1 = 0

lnx = ¡12

x = e¡1=2

Critical number: e¡1=2

Critical point:¡e¡1=2;¡ 1

2e

¢

f 0μ1

2

¶=1

2

μ2 ln

1

2+ 1

¶¼ ¡0:1931 < 0

f 0(1) = 1(2 ln 1 + 1) = 1 > 0

No relative maximum

Relative minimum at¡e¡1=2;¡ 1

2e

¢Increasing on

¡e¡1=2;1¢

Decreasing on¡0; e¡1=2

¢f 00(x) = (1)(2 lnx+ 1) + x

μ2

x

¶

= 2 lnx+ 3

f 00(x) = 0 when

2 lnx+ 3 = 0

lnx = ¡32

x = e¡3=2

f 00(e¡2) = 2 ln e¡2 + 3 = ¡1 < 0f 00(1) = 2 ln 1 + 3 = 3 > 0

In‡ection point at (e¡3=2;¡ 32e3 )

Concave upward on (e¡3=2;1)Concave downward on (0; e¡3=2)

x-intercept: x2 lnx = 0lnx = 0

x = 1

Since f(x) is not de…ned at x = 0, there is no y-intercept.No horizontal or vertical asymptotes

45. f(x) = 4x1=3 + x4=3

Domain is (¡1;1).f(¡x) = 4(¡x)1=3 + (¡x)4=3 = ¡4x1=3 + x4=3



f 0(x) =4

3x¡2=3 +

4

3x1=3

f 0(x) = 0 when

4

3x¡2=3 +

4

3x1=3 = 0

4

3x¡2=3(1 + x) = 0

x = ¡1

f 0(x) is not de…ned when x = 0

Critical numbers: ¡1 and 0Critical points: (¡1;¡3) and (0; 0)

f 0(¡8) = 4

3(¡8)¡2=3 + 4

3(¡8)1=3 = ¡7

3< 0

f 0μ¡18

¶=4

3

μ¡18

¶¡2=3+4

3

μ¡18

¶1=3=14

3> 0

f 0(1) =4

3(1)¡2=3 +

4

3(1)1=3 =

8

3> 0

No relative maximum

Relative minimum at (¡1;¡3)Increasing on (¡1;1)Decreasing on (¡1;¡1)

f 00(x) = ¡89x¡5=3 +

4

9x¡2=3

f 00(x) = 0 when

¡89x¡5=3 +

4

9x¡2=3 = 0

4

9x¡5=3(¡2 + x) = 0

x = 2


f 00(¡1) = ¡89(¡1)¡5=3 + 4

9(¡1)¡2=3 = 4

3> 0

f 00(1) = ¡89(1)¡5=3 +

4

9(1)¡2=3 = ¡4

9< 0

f 00(8) = ¡89(8)¡5=3 +

4

9(8)¡2=3 =

1

12> 0

In‡ection points at (0; 0) and (2; 6 ¢ 21=3)Concave upward on (¡1; 0) and (2;1)Concave downward on (0; 2)

x-intercept: 4x1=3 + x4=3 = 0x1=3(4 + x) = 0

x = 0 or x = ¡4y-intercept: y = 4(0)1=3 + (0)4=3 = 0

No horizontal or vertical asymptotes

46. f(x) = 5x2=3 + x5=3

Domain is (¡1;1).f(¡x) = 5(¡x)2=3 + (¡x)5=3 = 5x2=3 ¡ x5=3


f 0(x) =10

3x¡1=3 +

5

3x2=3

f 0(x) = 0 when

10

3x¡1=3 +

5

3x2=3 = 0

5

3x¡1=3(2 + x) = 0

x = ¡2


Critical numbers: ¡2 and 0Critical points: (0; 0) and (¡2; 3 ¢ 22=3)

f 0(¡8) = 10

3(¡8)¡1=3 + 5

3(¡8)2=3

= 5 > 0

f 0(¡1) = 10

3(¡1)¡1=3 + 5

3(¡1)2=3

= ¡53< 0

f 0(1) =10

3(1)¡1=3 +

5

3(1)2=3

= 5 > 0

Relative maximum at (¡2; 3 ¢ 22=3)Relative minimum at (0; 0)

Increasing on (¡1;¡2) and (0;1)Decreasing on (¡2; 0)

f 00(x) = ¡109x¡4=3 +

10

9x¡1=3


f 00(x) = 0 when

¡109x¡4=3 +

10

9x¡1=3 = 0

10

9x¡4=3(¡1 + x) = 0

x = 1


f 00(¡1) = ¡109(¡1)¡4=3 + 10

9(¡1)¡1=3

= ¡209< 0

f 00μ1

8

¶= ¡10

9

μ1

8

¶¡4=3+10

9

μ1

8

¶¡1=3

= ¡1409< 0

f 00(8) = ¡109(8)¡4=3 +

10

9(8)¡1=3

=35

72> 0

Concave upward on (1;1)Concave downward on (¡1; 0) and (0; 1)In‡ection point at (1; 6)

x-intercept: 5x2=3 + x5=3 = 0x2=3(5 + x) = 0

x = 0 or x = ¡5y-intercept: y = 5(0)2=3 + (0)5=3 = 0

No horizontal or vertical asymptotes.

47.

Other graphs are possible.

48.

Other graphs are possible.

49. (a)-(b) If the price of the stock is falling fasterand faster, P (t) would be decreasing, so P 0(t)would be negative. P (t) would be concave down-ward, so P 00(t) would also be negative.

50. (a)-(b) When a stock reaches its highest priceof the day, P (t) is at a maximum. Maxima andminima occur when P 0(t) = 0: Since this is a max-imum, the graph would be concave down. There-fore, P 00(t) < 0:

51. (a) Pro…t = Income¡CostP (q) = qp¡C(q)

= q(¡q2 ¡ 3q + 299)¡ (¡10q2 + 250q)= ¡q3 ¡ 3q2 + 299q + 10q2 ¡ 250q= ¡q3 + 7q2 + 49q

(b) P 0(q) = ¡3q2 + 14q + 49= (¡3q ¡ 7)(q ¡ 7)= ¡(3q + 7)(q ¡ 7)q = 7

3 (nonsensical) or q = 7

P 00(q) = ¡6q + 14P 00(7) = ¡6 ¢ 7 + 14 = ¡28 < 0

(indicates a maximum)

7 brushes would produce the maximum pro…t.

(c) p = ¡72 ¡ 3(7) + 299= ¡49¡ 21 + 299 = 229

$229 is the price that produces the maximum pro…t.

(d) P (7) = ¡73 + 7(72) + 49(7) = 343The maximum pro…t is $343.

(e) P 00(q) = 0 when ¡6q + 14 = 0

q =7

3

P 00(2) = ¡6(2) + 14 = 2 > 0P 00(3) = ¡6 ¢ 3 + 14 = ¡4 < 0The point of diminishing returns is q = 7

3 (between2 and 3 brushes).


52. (a) Since the second derivative has many signchanges, the graph continually changes from con-cave upward to concave downward. Since there isa nonlinear decline, the graph must be one thatdeclines, levels o¤, declines, levels o¤, etc. There-fore, the …rst derivative has many critical numberswhere the …rst derivative is zero.

(b) The curve is always decreasing except at fre-quent points of in‡ection.

53. (a) Y (M) = Y0Mb

Y 0(M) = bY0Mb¡1

Y 00(M) = b(b¡ 1)Y0Mb¡2

When b > 0, Y 0 > 0, so metabolic rate and lifespan are increasing function of mass.When b < 0, Y 0 < 0, so heartbeat is a decreasingfunction of mass.When 0 < b < 1, b(b ¡ 1) < 0, so metabolic rateand life span have graphs that are concave down-ward.When b < 0; b(b¡1) > 0, so heartbeat has a graphthat is concave upward.

(b)dY

dM= bY0M

b¡1 =b

MY0M

b

=b

MY

54. Sketch the curve for l1(v) = 0:08e0:33v

l01(v) = 0:0264e0:33v

e0:33v 6= 0

l01(v) has no critical points.

l001 (v) = 0:008712e0:33v

e0:33v 6= 0

l001 (v) has no in‡ection points.

Sketch the curve for l2 = ¡0:87v2+28:17v¡211:41

l02(v) = ¡1:74v + 28:17¡1:74v + 28:17 = 0

v ¼ 16:19

Critical point: (16:19; 16:62)

l002 (v) = ¡1:74

l0(v) has no in‡ection points.

l02(v) has a relative maximum at (16:19; 16:62):

55. Let a = 0:25:

f(v) = v(0:25¡ v)(v ¡ 1)f 0(v) = ¡v3 + 1:25v2 ¡ 0:25v

= ¡3v2 + 2:5v ¡ 0:25By the quadratic formula, f 0(v) = 0 whenv = 0:12 and v ¼ 0:72:Critical numbers v ¼ 0:12 and v ¼ 0:72:

f 0(0:1) = ¡0:03 < 0f 0(0:5) = 0:25 > 0f 0(1) = ¡0:75 < 0

The function is decreasing on (0; 0:12) and (0:72; 1)and increasing on (0:12; 0:72):

f(0:12) ¼ ¡0:01¼ 0:09

Relative minimum at (0:12;¡0:01); relative max-imum at (0:72; 0:09):

f 00(v) = ¡6v + 2:5f 00(v) = 0 when

¡6v + 2:5 = 0v ¼ 0:42

f 00(0:1) = 1:9 > 0f 00(0:5) = ¡0:5 < 0

Concave upward on (0; 0:42)Concave downward on (0:42; 1)


56. y = 34:7(1:0186)¡x(x¡0:658)

In chapter 4, the function was originally de…nedas

log y = 1:54¡ 0:008x¡ 0:658 logxso, 0 < x <1; and 0 < y <1:The function will have a vertical asymptote atx = 0 and a horizontal asymptote at y = 0:

dy

dx= ¡34:7(1:0186)¡xx¡0:658

·ln(1:0186) +

0:658

x

¸

For every value in the domain, dydx < 0; so y hasno critical points and is decreasing on (0;1):

57. (a) Set the two formulas equal to each other.

1486S2 ¡ 4106S + 4514 = 1486S ¡ 8251486S2 ¡ 5592S + 5339 = 0

Take the derivative.

2972S ¡ 5592 = 0

Solve for S:S ¼ 1:88

For males with 1.88 square meters of surface area,the red cell volume increases approximately 1486ml for each additional square meter of surface area.

(b) Set the formulas equal to each other.

995e0:6085S = 1578S995e0:6085S ¡ 1578S = 0

Take the derivative.

605:4575e0:6085S ¡ 1578 = 0e0:6085S ¼ 2:60630:6085S ¼ ln 2:6063

S ¼ 1:57 square meters

By plugging the exact value of S into the two for-mulas given for PV; we get about 2593 ml (Hurley)and 2484 for Pearson et al.

(c) For males with 1.57 square meters of surfacearea, the red cell volume increases approximately1578 ml for each additional square meter of sur-face area.

(d) When f and g are closest together, their ab-solute di¤erence is minimized.

d

dxjf(x0)¡ g(x0)j = 0

jf 0(x0)¡ g0(x0)j = 0f 0(x0) = g0(x0)

58. y(t) = Act

y0(t) = (lnA)Act ¢ ddtct

= (lnA)(ln c)ctAct

y00(t) = (lnA)(ln c)¢ [(ln c)ctAct + ct(lnA)(ln c)ctAct ]

= (lnA)(ln c)2ctAct

[1 + (lnA)ct]

y00(t) = 0 when 1 + ln(A)ct = 0

ct = ¡ 1

lnA

t ln c = ln

μ¡ 1

lnA

¶

t = ¡ ln(¡ lnA)ln c

= ¡ ln[¡ ln(0:3982 ¢ 10¡291)]

ln 0:4152

By properties of logarithms,

¡ ln(0:3982 ¢ 10¡291) = ¡[ln(0:3982) + ln(10¡291)]= ¡[ln(0:3982)¡ 291 ln(10)]= ¡ ln(0:3982) + 291 ln(10)

So,

t = ¡ ln[¡ ln(0:3982) + 291 ln(10)]ln(0:4152)

¼ 7:405

At about 7.405 years the rate of learning to passthe test begins to slow down.


59. (a) P (t) = 325 + 7:475(t+ 10)e¡(t+10)=20

P 0(t) = 7:475£1 ¢ e¡(t+10)=20

+ (t+ 10)e¡(t+10)=20 ¢ ¡120

¸

= 7:475

·1¡ 1

20(t+ 10)

¸e¡(t+10)=20

= 7:475

μ1

2¡ t

20

¶e¡(t+10)=20

P 0(t) is zero when

1

2¡ t

20= 0

t = 10

P 0(9) ¼ 0:39P 0(11) ¼ ¡0:36P (t) is increasing at 9 and decreasing at 11. So, arelative maximum occurs at t = 10. The popula-tion is largest in the year 2010.

(b) P 0(t) = 7:475μ1

2¡ t

20

¶e¡(t+10)=20

P 00(t) = 7:475·¡ 1

20e¡(t+10)=20

+

μ1

2¡ t

20

¶e¡(t+10)=20 ¢ ¡1

20

¸

= 7:475

μ¡ 1

20¡ 1

40+

t

400

¶e¡(t+10)=20

= 7:475

μt

400¡ 3

40

¶e¡(t+10)=20

P 0(t) is zero when

t

400¡ 3

40= 0

t = 30

P 0(29) = ¡0:0027P 0(31) = 0:0024

P 0 is decreasing at 29, and increasing at 31. So, arelative minimum occurs at t = 30. The popula-tion is declining most rapidly in the year 2010.

(c) As time t approaches in…nity, the populationP approaches

limt!1 P (t) = lim

t!1 (325 + 7:475(t+ 10)e¡(t+10)=20)

= 325 + 7:475 limt!1

t+ 10

e(t+10)=20

= 325 + 7:475(0)

= 325:

The population is approaching 325 million.

60. (a) The U.S. stockpile was at a relative maximumbetween 1965 and 1967, at 1974, 1980, 1984, andat 1987.

(b) The U.S. stockpile was at its largest relativemaximum from 1965 to 1967. During this period,the Soviet stockpile was concave upward. Thismeans that the stockpile was increasing at an in-creasingly rapid rate.

61. (a) s(t) = 512t¡ 16t2v(t) = s0(t) = 512¡ 32ta(t) = v0(t) = s00(t) = ¡32

(b) The maximum height is attained whenv(t) = 0.

512¡ 32t = 0t = 16

v(0) = 512 > 0

v(20) = 512¡ 640 = ¡128 < 0The height reaches a maximum when t = 16:

s(16) = 512 ¢ 16¡ 16(162) = 4096

The maximum height is 4096 ft.

(c) The projectile hits the ground when s(t) = 0:

512t¡ 16t2 = 016t(32¡ t) = 0t = 0 or t = 32

v(32) = 512¡ 32(32) = ¡512The projectile hits the ground after 32 secondswith a velocity of ¡512 ft/sec.

grosse pointe public schools · chapter 5 graphs and the derivative 5.1 increasing and decreasing...

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