grosse pointe public schools · chapter 5 graphs and the derivative 5.1 increasing and decreasing...
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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):
2. By reading the graph, f is
(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):
8. By reading the graph, f is
(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 :
Test a point in each interval.
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:
Test a point in each interval.
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:
Test a point in each interval.
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
Test a point in each interval.
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:
Test a point in each interval.
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:
Test a point in each interval.
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):
300 Chapter 5 GRAPHS AND THE DERIVATIVE
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.
Section 5.1 Increasing and Decreasing Functions 301
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.
Test a point in each interval.
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 :
Test a point in each interval.
f 0(0) =1¡ 3(0)e3(0)
= 1 > 0
f 0(1) =1¡ 3(1)e3(1)
= ¡ 2e3< 0
302 Chapter 5 GRAPHS AND THE DERIVATIVE
(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.
Test a point in each interval.
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 :
Test a point in each interval.
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:
Test a point in each interval.
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 .
Section 5.1 Increasing and Decreasing Functions 303
Test a point in each interval.
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¢:
304 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.1 Increasing and Decreasing Functions 305
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.
306 Chapter 5 GRAPHS AND THE DERIVATIVE
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.
308 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.2 Relative Extrema 309
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
310 Chapter 5 GRAPHS AND THE DERIVATIVE
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:
Section 5.2 Relative Extrema 311
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
312 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Find the critical numbers:
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
Find the critical numbers:
(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
Section 5.2 Relative Extrema 313
30. f(x) = 3xex + 2f 0(x) = 3xex + 3ex
= 3ex(x+ 1)
Find the critical numbers:
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
Find the critical numbers:
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
Find the critical numbers:
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
314 Chapter 5 GRAPHS AND THE DERIVATIVE
34. f(x) = x+ 8¡x
f 0(x) = 1 + (ln 8)8¡x(¡1)
= 1¡ ln 88x
Find the critical numbers:
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.
Section 5.2 Relative Extrema 315
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
Find the critical numbers:
¡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
316 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.2 Relative Extrema 317
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.
318 Chapter 5 GRAPHS AND THE DERIVATIVE
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
320 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 321
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
322 Chapter 5 GRAPHS AND THE DERIVATIVE
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
¢
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 323
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
¢
324 Chapter 5 GRAPHS AND THE DERIVATIVE
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)
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 325
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)
326 Chapter 5 GRAPHS AND THE DERIVATIVE
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
´and
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
´and
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
´and³
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.
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 327
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:
328 Chapter 5 GRAPHS AND THE DERIVATIVE
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.
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 329
Use the …rst derivative test.
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.
Use the …rst derivative test.
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.
330 Chapter 5 GRAPHS AND THE DERIVATIVE
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.
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 331
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.
332 Chapter 5 GRAPHS AND THE DERIVATIVE
(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):
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 333
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:
334 Chapter 5 GRAPHS AND THE DERIVATIVE
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%:
Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test 335
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);
336 Chapter 5 GRAPHS AND THE DERIVATIVE
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:
338 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.4 Curve Sketching 339
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)
340 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.4 Curve Sketching 341
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
´Concave upward on
³¡ 3p
2; 0´and
³3p2;1´
Concave downward on³¡1;¡ 3p
2
´and
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)
342 Chapter 5 GRAPHS AND THE DERIVATIVE
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
The graph is not symmetric about the y-axis orthe origin.
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
Section 5.4 Curve Sketching 343
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
The graph is not symmetric about the y-axis orthe origin.
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
Vertical asymptote at x = 2
344 Chapter 5 GRAPHS AND THE DERIVATIVE
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)
Section 5.4 Curve Sketching 345
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
346 Chapter 5 GRAPHS AND THE DERIVATIVE
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).
Section 5.4 Curve Sketching 347
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)
The graph is symmetric about the origin.
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)
348 Chapter 5 GRAPHS AND THE DERIVATIVE
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
´and
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
Horizontal asymptote at y = 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
´and
³2p3; 316
´Concave upward on
³¡1;¡ 2p
3
´and
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
Horizontal asymptote at y = 0
Section 5.4 Curve Sketching 349
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)
The graph is symmetric about the y-axis.
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)
The graph is symmetric about the origin.
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)
350 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Section 5.4 Curve Sketching 351
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
Vertical asymptote at x = 0
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
352 Chapter 5 GRAPHS AND THE DERIVATIVE
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)
The graph is symmetric about the y-axis.
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:
Section 5.4 Curve Sketching 353
25. f(x) = xe¡x
Domain is (¡1;1):
f(¡x) = ¡xex
The graph has no symmetry.
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
Horizontal asymptote at y = 0
26. f(x) = x2e¡x
Domain is (¡1;1)f(¡x) = (¡x)2ex
= x2ex
The graph has no symmetry.
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:
354 Chapter 5 GRAPHS AND THE DERIVATIVE
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
The graph has no symmetry.
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
Horizontal asymptote at y = 0
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.
Section 5.4 Curve Sketching 355
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
The graph has no symmetry.
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
¢Concave downward on
¡¡15 ;1
¢In‡ection point at
¡¡15 ;
655=3
¢ ¼ (¡0:2; 0:410)
356 Chapter 5 GRAPHS AND THE DERIVATIVE
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):
The graph has no symmetry.
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
Section 5.4 Curve Sketching 357
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:
For Exercise 16, the default window does not allowthe graph to properly display the vertical asymp-totes.
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 Exercise 20, the default window does not allowthe graph to properly display the vertical asymp-totes.
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.
358 Chapter 5 GRAPHS AND THE DERIVATIVE
38. (a) indicates that the curve may not contain breaks.
(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):
39. (a) indicates that the curve may not contain breaks.
(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.
Critical number: ¡92
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.
Critical numbers: ¡2 and 23
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
360 Chapter 5 GRAPHS AND THE DERIVATIVE
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
Critical numbers: ¡3 and 2
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
Chapter 5 Review Exercises 361
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
362 Chapter 5 GRAPHS AND THE DERIVATIVE
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
¢Concave downward on
¡¡ 112 ;1
¢y-intercept:
y = ¡2(0)3 ¡ 12(0)2 + (0)¡ 3 = ¡3
28. f(x) = ¡43x3 + x2 + 30x¡ 7
Domain is (¡1;1)The graph has no symmetry.
f 0(x) = ¡4x2 + 2x+ 30= ¡2(2x2 ¡ x¡ 15)= ¡2(2x+ 5)(x¡ 3) = 0
Critical numbers: ¡52 and 3
Critical points:¡¡5
2 ;¡54:91¢and (3; 56)
f 00(x) = ¡8x+ 2
f 00μ¡52
¶= 22 > 0
f 00(3) = ¡22 < 0Relative maximum at 3
Relative minimum at ¡52
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
¡¡1; 14¢Concave downward on
¡14 ;1
¢y-intercept:
y = ¡43(0)3 + (0)3 + 30(0)2 ¡ 7
= ¡7
Chapter 5 Review Exercises 363
29. f(x) = x4 ¡ 43x3 ¡ 4x2 + 1
Domain is (¡1;1)The graph has no symmetry.
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 ;¡5:12´and³
1¡p73 ; 0:11
´Concave upward on
³¡1; 1¡
p7
3
´and
³1+p7
3 ;1´
Concave downward on³1¡p73 ; 1+
p7
3
´y-intercept:
y = (0)4 ¡ 43(0)3 ¡ 4(0)2 + 1 = 1
30. f(x) = ¡23x3 +
9
2x2 + 5x+ 1
Domain is (¡1;1)The graph has no symmetry.
f 0(x) = ¡2x2 + 9x+ 5 = 0¡(2x+ 1)(x¡ 5) = 0
Critical numbers: ¡12 and 5
Critical points:¡¡1
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
Relative minimum at ¡12
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
¡¡1; 94¢Concave downward on
¡94 ;1
¢y-intercept:
y = ¡23(0)3 +
9
2(0)2 + 5(0) + 1 = 1
364 Chapter 5 GRAPHS AND THE DERIVATIVE
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
¢Concave downward on
¡¡12 ;1
¢x-intercept:
x¡ 12x+ 1
= 0
x = 1
y-intercept: y =0¡ 12(0) + 1
= ¡1
Vertical asymptote at x = ¡12
Horizontal asymptote at y = 12
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
Chapter 5 Review Exercises 365
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
Critical points:¡¡2
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
Relative maximum at 12
Relative minimum at ¡23
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
¢Concave downward on
¡¡ 112 ;1
¢y-intercept:
y = ¡4(0)3 ¡ (0)2 + 4(0) + 5 = 5
34. f(x) = x3 +5
2x2 ¡ 2x¡ 3
Domain is (¡1;1)The graph has no symmetry.
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
366 Chapter 5 GRAPHS AND THE DERIVATIVE
35. f(x) = x4 + 2x2
Domain is (¡1;1)f(¡x) = (¡x)4 ¡ 2(¡x)2
= x4 + 2x2 = f(x)
The graph is symmetric about the y-axis.
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
Domain is (¡1;1)The graph has no symmetry.
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
Relative maximum at 92
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
Chapter 5 Review Exercises 367
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)
The graph is symmetric about the origin.
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
Vertical asymptote at x = 3
368 Chapter 5 GRAPHS AND THE DERIVATIVE
Horizontal asymptote at y = ¡2
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
¢Concave downward on
¡¡1;¡12
¢x-intercept: 0; y-intercept: 0
Vertical asymptote at x = ¡12
Horizontal asymptote at y = ¡2
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 number: ¡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.
Chapter 5 Review Exercises 369
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
370 Chapter 5 GRAPHS AND THE DERIVATIVE
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
The graph is not symmetric about the y-axis ororigin.
Chapter 5 Review Exercises 371
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(x) is not de…ned when x = 0
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
The graph is not symmetric about the y-axis ororigin.
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
f 0(x) is not de…ned when x = 0
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
372 Chapter 5 GRAPHS AND THE DERIVATIVE
f 00(x) = 0 when
¡109x¡4=3 +
10
9x¡1=3 = 0
10
9x¡4=3(¡1 + x) = 0
x = 1
f 00(x) is not de…ned when x = 0
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).
Chapter 5 Review Exercises 373
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)
374 Chapter 5 GRAPHS AND THE DERIVATIVE
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.
Chapter 5 Review Exercises 375
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.