ap calculus ab chapter 4 solutions

Chapter 4 Solutions Calculus

Calculus Chapter 4 Solutions 1. a. The derivative or rate of the distance function is the velocity function. The area under

the velocity function is the reverse of the derivative: it gives the position function. b. By taking the limit of slopes of smaller and smaller secants. c. As we use more rectangles with smaller widths, we can try to take a limit of the area

approximations, which they give us. 2. a. We could take the limit of the approximations of area as the number of rectangles

goes to infinity.

b. As Δx gets smaller, potentially approaching 0.

3. limn!"

b#an

f a + b#ani( )

i=0

n#1

$ (using left rectangles)

a. No, we cannot have rectangles with a width of 0. b. Each individual rectangle’s area would approach 0. c. This is because the number of rectangles is increasing toward infinity, so the total

area need not approach 0. 4. a. The numbers b and a represent the right and left boundaries for x, respectively.

b. Instead of Δx, we now have dx in the integral.

c. f (x) represents the heights of all rectangles and dx represents the bases. 5. a. The region covers two cycles of the sine curve

(compressed by a factor of two), and it has equal

area above and below the x-axis: sin(2t)dt0

2!

" = 0

b. See graph below left. The region is a trapezoid with area 2+4.5

2!5 = 16.25

c. See graph below center. The region is a 4 X 6 rectangle with area 24.

Solution continues on next page. →


5. Solution continued from previous page. d. See graph below right. The region has width 0 and has area 0. b. c. d. 6. These are areas of trapezoids: a. 10+25

2!1 = 17.5 miles b. 10+70

2! 4 = 160 miles

c. 10+17.5

2!

1

2= 6.875 miles d. 10+10+15t

2! t = (10 + 7.5t)t = 10t + 7.5t2 miles

7. a. Even

b. f (x)dx!2

2

" will be twice f (x)dx0

2

! , i.e., it is 20, because the area from x = !2 to 0

must equal the area from 0 to 2 due to symmetry. c. This area must be the difference of the other two: !2 !10 = !12 .

d. We can halve f (x)dx!2

2

" to get f (x)dx0

2

! , and then we have:

f (x)dx2

3

! = f (x)dx0

3

! " 12

f (x)dx"2

2

!

8. a. !

4

5x!3

! 4 b. !x!1/2

c. !6 sin x d. f (x) =4(x2

+1)

x2+1

= 4,! !f (x) = 0

9. a. 1 ! f ("8 + i) =

i=0

15

# 2("8 + i)1/3 +1

i=0

15

# = 12

b. 1

4! f "8 + i

4( ) = 1

42 "8 + i

4( )1/3

+1#$%

&'(= 15

i=0

63

)i=0

63

)

10. f (x) = x2

+ x !12,! "f (x) = 2x +1 , f (3) = 0, !f (3) = 7 , tangent line: y = 7(x ! 3)


11. 3

nf 2 + 3

ni( )

i=0

n!1

" = 3

n3 2 + 3

ni( ) + 5( )

i=0

n!1

"

Larger n-values give better estimates, e.g., n = 30 yields 46.05. Using midpoint rectangles would give the exact area of 46.5 for any n.

12. He is wrong because the triangle farthest from the pole sweeps out more volume when it is

rotated.

13. a. 3 and 3!40sin 2 "! + 3!

40i( )#

$%& ' 2.94

i=0

19

( , fairly close

b. 531

3 and

1

2!5 + i

2( )2

! 3"#

$% & 53.75

i=0

19

' , fairly close

14. See graph at right. Note: y-axis is scaled by 10’s.

They are next to each other and form (9x ! 2)dx = 250!2

8

"

15. See graph at right. Note: y-axis is scaled by 5’s. a. Geometrically it looks like it would have area 50,

but the calculator gives the (correct) answer of –50. b. The bounds have been switched. 16. They will not be equal (unless they are both zero): they are opposites, as seen in the

limiting Riemann sum, in which b!an

becomes ! b!a

n=a!b

n if we were to switch a and b.

So f (x)dx = ! f (x)dxb

a

"a

b

" .

17. a. f (x)dx1

3

! b. f (x)dx9

10

!

c. f (x)dxe

d

! d. f (t)dtx

x+h

!

Sample conjecture: f (x)dx + f (x)dxb

c

! = f (x)dxa

c

!a

b

! (which is basically part (c)).

18. All equal –4.5.


19. a. F(x) = !2x + C b. F(x) = 3x1/2 + C

c. F(x) = !x3+ 3x2 + C d. F(x) = (x + 3)2 + C

20. a. !f (x) = 18(x " 2)3

b. !f (x) = 2 cos x

c. f (x) = 2x2 + 9x ! 5!"! #f (x) = 4x + 9

d. f (x) = x2 ! 6x + 2!"! #f (x) = 2x ! 6 for x $ 0 21. A slope function returns the slope of the tangent to f(x) for any value of x, i.e., the exact

slope of f(x).

22. a. x sin x dx0

!

" = ! , 2 x sin x dx = 2!0

!

" , x sin x = 2!"!

!

#

(because x sin x is an even function) 23. a. b.

24. a. limx!2

x2"4

x"2= limx!2

(x+2)(x"2)

x"2= limx!2

(x + 2) = 4

b. limh!0

(1+h)2 "1

h= limh!0

1+2h+h2 "1h

= limh!0

(2 + h) = 2

25. Similar triangles help us get the radius: r = 10!h

10" 4 = 2

5(10 ! h)

A = ! r2 = ! 25(10 " h)#$ %&

2=4!25(10 " h)2

26. The rectangles still provide a good estimate of area.


27. a. A = 0 because the width is zero.

b. The regions may be combined: f (x)dxa

b

! + f (x)dxb

c

! = f (x)dxa

c

!

c. We are subtracting this area from itself: f (x)dxb

a

! + f (x)dxa

b

! = 0

d. We are amplifying the function by k, which multiplies the area

by k: kf (x)dxa

b

! = k " f (x)dxa

b

!

28. a. xdx0

2

! + 3xdx0

2

! =12x2 2

0+32x2 2

0(2 " 0) + (6 " 0) = 8

b. (4x)dx0

2

! = 2x22

0= 8 " 0 = 0

c. ( f (x) + g(x))dxa

b

!

29. a. No, adding k to f(x) shifts the whole region up, so the total area will be

f (x)dxa

b

! + (b " a)k .

b. No, these are different parts of the curve.

c. f (x)dxa

b

! = f (x " c)dxa+c

b+c

! is true because it shifts the curve and shifts the integral

so as to take the same area. d. No, the second integral shifts the curve, so it takes a different area.

30. • f (x)dxa

a

! = 0

• f (x)dxa

b

! + f (x)dxb

c

! = f (x)dxa

c

!

• f (x)dxa

b

! + g(x)dxa

b

! = ( f (x) + g(x))dxa

b

!

• f (x)a

b

! dx = " f (x)b

a

! dx

• k ! f (x)b

a

" dx = k ! f (x)b

a

" dx

• f (x)dxa

b

! = f (x " h)dxa+h

b+h

!

31. a. y = 1+ x!1 , !y = "x"2 b. !y = " sin x + cos x

c. y = x5/3, ! !y =53x2/3 d. y = 6 !17x +10x2, "y = !17 + 20x


32. a. A trapezoid, A =16+72

2! 7 = 308

b. The curve is translated up by 5, A = 308 + 5 ! 7 = 343

c. A rectangle, A = 5 ! 7 = 35 The area in (b) is the sum of those in (a) and (c), because its function is the sum of the function in (a) and (c).

33. a. 3+7

2!2 = 10 b. 10

c. They are equivalent expressions; they just use different dummy variables. 34. a. !f (x) = 3x2 " 2x +1 = 2, 3x2 " 2x "1 = 0 , (3x +1)(x !1) = 0 , x = !

1

3, 1;

f !1

3( ) = !1

27!1

9!1

3+1 = 14

27, f (1) = 1!1+1+1 = 2 ,

The lines are y ! 14

27= 2 x + 1

3( ) or y = 2x + 32

29, and y ! 2 = 2(x !1) or y = 2x

b. y = !1

2x + 1

3( ) + 14

27 and y = !

12(x !1) + 2 = !

12x + 5

2

35. a. f (g(x)) b. g( f (x))

c. h( f (x)) d. g(h( f (h(x)))) 36. a. Negative b. Zero c. Positive d. Zero e. Negative 37. The secant line goes through (0, 0) and (2, 0) and has slope 0, so we

set !f (x) = 0!" 3x2 ! 4x = 0

x(3x ! 4) = 0!"!x = 0,! 43

f (0) = 0 , so at (0, 0) and

f 4

3( ) = 4

3( )3

! 24

3( )2

= 64

27!32

9"3

3= !

32

27, so at 4

3, ! 32

27( ) . 38. See graph at right. y-axis is scaled by 30’s, x-axis is scaled by 2’s.

a. !f (x) = 3x2 + 6x " 45

!f ("2) = 3("2)2 + 6("2) " 45 = 12 "12 " 45 = "45

b. This is where !!f (x) = 6x + 6 = 0 , x = !1 , f (x) = 55 ; The inflection point is (–1, 55).


39. a. !2(3) ! 5 = !11 b. 2(3)2 ! 4 = 14

c. f(x) is discontinuous and non-differentiable at x = 3 and limx!3

f (x) does not exist.

40. a. 0 b. 5 c. 10 d. 5x e. cx 41. a. (trapezoid) 5+13

2!2 = 18 b. 5+41

2!2 = 207

c. 5+(4x+5)

2! x = (2x + 5)x = 2x2 + 5x

d. The area under the curve for 0 ! t ! x . e. A(x) = m

2x2 + bx ; mt + b is the equation of a line of slope m through (0, b) .

42. b. B(x) = 13+4x+5

2! (x " 2) = (2x + 9)(x " 2) = 2x2 + 9x " 4x "18 = 2x2 + 5x "18 , and it

differs from A(x) by a constant. c. B(x) = (2x2 + 5x) !18 = A(x) ! A(2) .

These two coincide to the areas (4t + 5)dt =2

x

! (4t + 5)dt " (4t + 5)dt0

2

!0

x

! .

d. Yes, they grow at the same rate because any incremental change in x (Δx) adds and equal amount of area to each.

e. G(x) = A(x) ! A(c) is the area under f(x) from c to x. 43. We can get distance from velocity by taking the area function (for area under the velocity

curve), and we can get velocity from distance by taking the slope function of the distance.

44. a. sin x + C b. !2 "x!1

+C

!1= 2x

!1+ C

c. !9 "x4/5

4/3+ C = !

27

4x4/3

+ C

45. a. 1

52 ! 4 i

5( )3/2"

#$% & !40.

i=0

19

' 04 , integral is –43.2

b. 7i

204 1+ 7i

20( ) + 3i=0

19

! " 30.85 , integral is 31.42

46. Any constant has a derivative of zero, so any constant can be added to one antiderivative

without changing its derivative, yielding infinite possibilities for the antiderivative.


47. f (x) = x3 ! 4x2 + x ! 4 , !f (x) = 3x2 " 8x +1, f (!3) = !27 ! 36 ! 3! 4 = !70 , !f (3) = 27 + 24 +1 = 52 , tangent line is y + 70 = 52(x + 3) or y = 52x + 86 .

48. No, for example for problem 4-47, d

dx(x2 +1) = 2x and d

dx(x ! 4) = 1 and 2x !1 = 2x is

not the derivative. 49. a. f (x) = x and g(x) = sin(x2 ) +1

b. f (x) = x2 + 2 and g(x) = 3x3 !12

50. Sample answer: 1

xdx

0

1

!

51. 5dx0

6

! = 5x6

0= 30 15 = 5dx

0

x

!15 = 5x

3 = x

No, this is not the same line that will divide 5xdx0

6

! =52(6)2 = 90 in half.

52. f (t)dt2

3

! = f (t)dt0

3

! " f (t)dt0

2

! = 9(3)2 " 2 " [9(2)2 " 2] = 79 " 34 = 45

53. a. Definite b. Indefinite c. Definite d. Indefinite e. Indefinite 54. a. A(x) = 2(x ! c) = 2x ! 2c = 2x + C

b. A(x) =(5c+2)+(5x+2)

2! (x " c)

=12(5x + 5c + 4)(x " c)

=52x2+ 2x " 5

2c2" 2c

=52x2+ 2x + C

c. A(x) =3 2!c( )+ 3 2!x( )

2" (x ! c) = 1

2(3! c ! x)(x ! c)

= !12x2 + 3

2x !

32c + 1

2c2 = !

12x2 + 3

2x + C


55. a. 2 !8 " 2 ! 3 = 10 b. 5 !8 " ("2) !8 = 56

c. 3 52!62 + 2 !6 " 5

2!22 + 2 !2( )( ) = 3 ! (102 "14) = 264

d. (5x + 2)dx = 52! 32 + 2 ! 3"

1

3

# 52!12 + 2 !1( ) = 28.5 " 4.5 = 24

e. 32! k( ) dk

4

10

" = 32k ! 1

2k2 10

4= 30

2! 100

2( ) ! 122! 162( ) = !35 ! (!2) = !33

f. 32! k( ) dk

0

!2

" = 32k ! 1

2k2 !2

0= !6

2! 42( ) ! (0) = !5

56. a. g(m)dm0

3

! =4(3)+1

3+2=135

b. g(m)dm!1

0

" = ! g(m)dm0

!1

" =4(!1)+1!1+2

= ! !31= 3

c. g(m)dm!1

5

" =4(5)+1

5+2!4(!1)+1!1+2

=217! !31= 6

57. Ignacio is correct. 58. a. !d (t) = sin t , !d (1) = sin1 " 0.84 , !d (" ) = 0 , !d (50) = sin 5 " #0.96 This uses the derivative as the velocity, particularly the derivative of

cos t , and also the derivative of a multiple (in this case of –1) of a function and the derivative of a sum of two functions.

b. See graph at right. Area =1

2!5 !15 =

75

2= 37.5 ft

c. Velocity is the derivative of distance and distance is the area under the velocity curve. 59. a. f (1) = 50 miles

b. 50 miles1 hour

= 50 mph

c. 75 mph, at 0.5 hours (30 minutes)


60. (x1/3 ! 4)dx " !13.14!2

1

#

61. Yes, f must be continuous at x = c to be differentiable there, so f (c)must exist. 62. !f (x) = cos x , f !

3( ) = sin 3

2, !f "

3( ) = cos "3 = 1

2,

tangent line is y ! 3

2= 1

2x ! "

3( ) or y = 1

2x +

3

2!

"

6

63. a. 7

3x2/3 b. !21m

!8! 21m

2

c. 0 d. d

dt(6t2 +15t) = 12t +15

64. a. 3x5 + 2x2 ! 3x + C b. !2 sin x + C

c. !x4+10x + C

65. i. F(x) = 2x + C , A(x) = 2x

ii. F(x) = x2 + C , A(x) = x2

A(x) + C = F(x)

66. a. cos x dx! /6

! /3

"

b. Answers vary, you may suggest Tommy use the antiderivative F(x) = ! sin x + C . c. F(x) = ! sin x + C

d. (! sin "

3( ) + C) ! (! sin "

6( ) + C) , ! 3

2+1

2=1! 3

2un2

e. The +C is added and subtracted when evaluating, therefore it is eliminated. 67. a. (2x + 5)dx! = x

2+ 5x

b. !A (x) = f (x) ; Geometrically, f (x) is the slope function of area, A(x) . c. A(x) = 1

2mx2 + bx + C and !A (x) = mx + b


68. a. As x increases, there is more area under the curve. f (x)dx! = A(x)!!"!! d

dxf (x)dx! =

ddxA(x)!!"!! f (x) = #A (x)

b. Since F(c) is a constant, this expression becomes f (x) + 0 = f (x) .

c. (1) Since f (x) is continuous, the anti derivative exists, or F(x) = f (t)dta

x

! .

(2) F(x) means geometrically, the area under the curve from a to x.

(3) Using (1), take the derivative of both sides: ddxF(x) = d

dxf (t)dt

a

x

! .

(4) Rewrite the left side using the definition of the derivative: limh!0

F(x+h)"F(x)

h

(5) The definition of the derivative for F(x): F(x + h) ! F(x) " hf (x) . (6) Divide by h and apply the limit as h! 0 . The derivative becomes f(x).

(7) Simplify your results: f (x) = ddx

f (t)dta

x

! .

d. !f (x)dx" =ddxf (x)dx" = f (x) + C

69. A(x) = F(x) + C

70. F(x) = 2x3 + 7x + C , (6x2 + 7)dx = F(2) ! F(1) = (16 +14 + C) ! (2 + 7 + C) = 211

2

"

71. a. F(x) = x2 + 5 + C , (2x + 5)dx = x2 ! 5x + C ! (02 + 5 "0 + C) = x2 + 5x0

x

#

b. (2x + 5)dx = F(9) ! F(4) = (92 + 5 "9) ! (42 + 5 " 4) = 81+ 45 ! 36 = 904

9

#

72. F(x) = x2 + 5x + C ,

(2x + 5)dx4

9

! = (92 + 5(9) + C) " (42 + 5(4) + C) = (81+ 45) " (16 + 20) = 90

73. a. (6x3 + 9x)dx!5

2

"

b. h(x)5

9

! dx " h(x)3

9

! dx = " h(x)dx3

5

! !!or !! h(x)dx5

3

!

c. ![(2x)2 + ( x )2 ]dx = ! " (4x2 + x)dx2

7

#2

7

#

d. (x + 5)2dx + xdx = x2+ xdx = (x2 + x )dx

6

9

!6

9

!6

9

!6

9

!1

4

!

(by translating the first integral)


74. f (t)dta

b

! represents total weight loss, in pounds, between day a and day b.

75. a. ∞ or does not exist b. does not exist

c. 2 d. –2 e. 2 f. 0 g. At x = –2, –1, and 2 (and below –2) 76. a. 17

1

3

b. The area under v(t) from t = 0 to 4, and this is the distance at t = 4 , i.e. s(4) .

c. acceleration d. feet

77. a. ! f (x)dx4

2

" = f (x)dx2

4

" = 10

b. ( f (x ! 8) + 4)10

12

" dx = f (x)dx + 4"2

4

" dx = 10 + 4x12

10= 10 + (48 ! 40) = 18

c. f (x + 8)dx10

12

! = f (x)18

20

! ; not enough info

78. 2xdx0

10

! = x2 10

0= 100 Solve: x2 x

0= 50!!!!!x2 = 50!!!!!x = 50 = 5 2

79. a. !s (t) = 4 " 1

2t#1/2

= 2t#1/2 , !s (1) = 2 , !s (4) = 2

4= 1, !s (16) = 2

16=12

This uses derivative to obtain velocity, the Power Rule for this particular derivative, and the rule d

dx(c ! f (x)) = c ! d

dx( f (x)) .

b. s(1) = v(t)dt = (9t + 32)dt0

1

!0

1

! , v(t)dt = 92t2 + 32t + C! ,

(9t + 32)dt = 92

(1)2+ 32(1) ! 9

2(0)2

+ 32(0)"# $% = 36.5 miles0

1

&

This uses a definite integral and the fundamental theorem to use the antiderivative to calculate this integral.

c. Distance is an antiderivative or area function of velocity, and velocity is the derivative or slope function of area.


4-80. a. See graph at right.

i. g(!3) = f (t)dt0

!3

" = ! f (t)dt = ! 12# !3 # 3( ) = 4.5

!3

0

"

ii. 2g(!7) = 2 f (t)dt0

!7

" = !2 f (t)dt!7

0

" = !2 12(!4)(4) + (!4)(3)( )

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = !2(!20) = !40

iii. g(9) + g(!9) = f (t)dt0

9

" ! f (t)dt!9

0

"= 1

2(4)(4) + 4(5)( ) ! 1

2(4)(!4) + (!4)(5)( ) = 28 ! (!28) = 56

b. See graph at right.

i. g(!3) = f (t)dt0

!3

" = ! f (t)dt = ! 12# 3 # 3( ) = !4.5

!3

0

"

ii. 2g(!7) = 2 f (t)dt0

!7

" = !2 f (t)dt!7

0

" = !2 12(4)(4) + (4)(3)( )

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = !2(20) = !40

iii. g(9) + g(!9) = f (t)dt0

9

" ! f (t)dt!9

0

"= 1

2(4)(4) + 4(5)( ) ! 1

2(4)(4) + (4)(5)( ) = 28 ! 28 = 0

82. a. The areas under the curve from 0 to a and from 0 to b, respectively.

b. f (x)dxa

b

! = f (x)dx "0

b

! f (x)dx = F(b) " F(a)0

a

!

c. !F (t) = d

dt(2t + 3)

0

t

" dt =d

dt(t2 + 3t # 0) = 2t + 3

83. a. 2 hours: s(2) = (2t + 5)dt0

2

! = t2 + 5t2

0= 22 + 5(2) " (0 + 0) = 14miles

5 hours: s(5) = (2t + 5)dt0

5

! = t2 + 5t5

0= 52 + 5(5) " (0 + 0) = 50miles

x hours: s(x) = (2t + 5)dt0

x

! = t2 + 5tx

0= x2 + 5x

b. s(4) ! s(2) = (2t + 5)dt2

4

" = t2 + 5t4

2= 42 + 5(4) ! (22 + 5(2)) = 22miles

c. The derivative is the slope of the tangent line and slope is the rate of change. The rate of change involving distance is velocity.

x

y

x

y


84. a. 3x ! 5

b. d

dx(3x ! 5)dx

3

x

" = 3dx3

x

" = 3xx

3= 3x ! 9

c. 0; The integration will yield a constant. The derivative of a constant is 0.

d. d

dx(3x ! 5)dx" = 3" dx = 3x + C

e. cos(x2 )

f. d

dx(cos x2 )dx

1

4

! = cos x24

1= cos(16) " cos(1)

85. Note: The FTC gives exact results, whereas calculators give very close estimates.

a. 2 !1

8x"2

"2= "x

"2

1

8

= "164

" ("1) = 6364

b. ! cos x" /4

" /2= ! cos "

2! (! cos "

4) = 0 +

2

2=

2

2

c. 3x3/2

3/24

9

= 2x3/2

4

9

= 2 !27 " 2 !8 = 54 "16 = 38

d. x3! 3x

2+ 2x

0

2

= 8 ! 3 " 4 + 2 "2 ! 0 = 8 !12 + 4 = 0

e. 2 ! 3 ! x"1"1

1

3#$%

&'(= "6x"1

1

3= " 6

3" ("6) = 4

f. x4/3

4/3!1

8

=34x4/3

!1

8=34"16 ! 3

4(1) = 45

4

g. Combine: (2x ! 20)dx = x2 ! 20x!2

9

" !2

9

= (81!180) ! (4 + 40) = !99 ! 44 = !143

h. 9 m1/3

1/3! 2 m

!2

!21

27

= 27m1/3 + m!2

1

27= 81+ 1

272( ) ! (27 +1) " 53.001

86. He did not distribute the negative. So it should have ended in line 2, with –2 – C. Then the

C’s would subtract out giving 68.


87. a. 3

2x4! x

2+ 5x + C

b. 3

2x4! x

2 + 5x2

4

= 3

2"256 !16 + 20( ) ! 3

2"16 ! 4 +10( ) = 384 + 4 ! 30 = 358

c. 3t3! t + C

d. 3t 3 ! t!2

2= (3 "8 ! 2) ! (3(!8) ! (!2)) = 22 ! (!22) = 44

e. ! cosm +1

9m3+ C

f. ! cosm + 19m3

!"

"

= ! cos" + 19" 3( ) ! ! cos(!" ) + 1

9(!" )3( ) = 1+ " 3

9+ (!1) + " 3

9= 2" 3

9

88. a. ! f (x)dx!5

!3

" + g(x)dx!5

!3

" = (g(x) ! f (x))dx!5

!3

"

b. (3 f (x) + 5g(x))dx1

6

!

c. f (x)dx11

11

! or f (x)dx6

6

! or just zero.

d. f (t)dt9

10

!

89. !f (x) = limh"0

f (x+h)# f (x#h)

2h= limh"0

4(x+h)2 #9(x+h)#1#[4(x#h)2 #9(x#h)#1]

2h=

limh!0

4x2 +8xh+4h2 "9x"9h"1"4x2 +8xh"4h2 +9x"9h+1

2h= limh!0

16xh"18h

2h= limh!0

8x " 9 = 8x " 9

90. a. cos(x ! 5) b. 82x

81

c. !23

2( )m1/2 = !3m1/2 d. 3(x + 6)2

91. !f (x) = 3x2 + 24x + 36 = 3(x + 2)(x + 6) , and !!f (x) = 6x + 24 . From !f (x) , we know that

f (x) is increasing in (–∞, –6] and [–2, ∞) (where !f (x) " 0 ), and f (x) is decreasing in [–6, –2] ( !f (x) " 0 ). From !!f (x) , the point of inflection is at x = !4 and is (!4, !22) , and f is concave down in (–∞, –4] and concave up in [–4, ∞).

92. Its inflection point (from #78) was (–4, –22), where the slope is

!f ("4) = 3("4)2 + 24("4) + 36 = "12 . The tangent line is y + 22 = !12(x + 4) or y = !12x ! 70 .


93. a. The tangent is vertical; the slope goes to infinity and no actual numerical derivative exists.

b. No, the tangent is still vertical (and in this case there is a cusp). c. This is just f (x) = x and it is differentiable everywhere.

d. Because the graph changes from concave up to concave down.

94. (g(x) + 3)dx2

2

! = 0 (because of the bounds)

(3! 5g(x))dx4

2

" = !3x4

2+ 5 g(x)

2

4

" = !3(4) + 3(2) + 5(6) = 24

96. The defendant traveled 12 +15 + 21+ 26 +10 +16 = 100 miles in 2 ! 18

60= 2.3 hours, at an

average velocity of 43.48 mph, so the officer is incorrect, the teacher may be innocent. But this does not prove innocence.

97. This is at the inflection point !s (t) = "93.75 # 3t2 + 93.75 #2t + 45 ,

!!s (t) = "93.75 #6t + 93.75 #2 = 0 , t = 93.75!2

93.75!6=

1

3 hour , and this velocity is

!s1

3( ) = "93.75 # 31

3( )2

+ 93.75 #2 1

3( ) + 45 = 76.25mph, which confirms the prosecution’s claim.

98. a. cos x is even, so cos xdx = 2 ! cos xdx = 2 ! sin x0

"= 2 ! (0 # 0) = 0

0

"

$#"

"

$

b. 5 !y3/2

3/2+ cos y + C =

10

3y3/2 + cos y + C

c. The antiderivative of 1x

has not yet been taught, but with a calculator this is ! 1.

d. 1dx = x + C! 99. a. The graph represents distance and !s (t) is velocity.

b. The graph represents sales rate (in sales/hr) and r(t)dt! represents and indefinite area of total sales.

c. The graph represents a weight rate in lbs/inch, and w(i)di! represents an indefinite area or lbs.


100. a. v(t) =! 203t + 20, 0 " t " 4.5

!!20t !100, 4.5 < t " 5

#$%

&%

b. For 3 hours, the triangle has area 12!20 ! 3 = 30miles. For the 4th hour we should

count the area as positive, 12!20

3!1 =

10

3, for a total of 33 1

3miles.

c. The car was traveling south, towards its starting point.

d. v(t)dt0

5

!

e. v(t) dt0

5

!

101. a. The backwards/forwards nature of finding velocity from position and vice versa is

accomplished using the derivative and integral, as proscribed in the Fundamental Theorem.

b. We must consider integrating both parts of the function:

s(t) =! 103t2+ 20t, 0 " t " 4.5

22.5 +10t2 !100t + 247.5 = 10t2 !100t + 270, 4.5 < t " 5

#$%

&%

The 22.5 is ! 103t2+ 20t evaluated at t = 4.5 , and 247.5 is the opposite of

10t2!100t evaluated at 4.5, in order to make s(t) continuous at 4.5.

102. !y = 3x2 "14x +15 = (3x " 5)(x " 3) is non-negative in

!", 5

3( )! [3,") , which is where y

is increasing. It is non-positive in 53, 3!" #$ , where y is decreasing.

103. a. !y = cos(x " 3) , differentiable everywhere

b. !f (x) ="2x, x < 1

3(x "1)2, x > 1

#$%&

, it is not differentiable at x = 1 because the limits (of

secants) are different from the left and right sides.

104. 6x dx ! 2

n"621/n

i=0

n#1

$0

2

%


105. a. f (x) = 2x3 + 6x2 ! 7x + C

b. 25 = f (!2)

25 = 2(!2)3 + 6(!2)2 ! 7(!2) + C

25 = !16 + 24 +14 + C

C = 3

Therefore f (x) = 2x3 + 6x2 ! 7x + 3

c. The integral of the derivative gives the original function plus a constant. Using the point provided, we can solve for the constant.

106. a. The velocity is negative in the second interval, and s(1.6) = 84 and s(1.8) ! 74 .

b. Look at the inflection point in each section: 1st section: !s (t) = "291.67 # 3t2 +1125 #2t "1360 , !!s (t) = "291.67 #6t +1125 #2 = 0 , t = 2!425

291.67!6" 1.286 , !s (1.286) " 86.4mph

2nd section: !s (t) = 2500 " 3t2 #12750 "2t + 21600 , !!s (t) = 2500 "6t #12750 "2 = 0 , t =

12750!2

12750!6= 1.7 , !s (1.7) = "75mph so max speed is 75 mph

3rd section: !s (t) = "176.3t2 +1142.4 #2t " 2402 , !!s (t) = "176 #6t +1142.4 #2 = 0 , t =

1142.4!2

176!6" 2.164 , !s (2.164) " 69.7 mph

Basically the inspector is right, there are two sections where 70 mph is exceeded. 107. Exhibit B does not meet the starting point of Exhibit C, portraying a discontinuity. At time

t = 1hour, Exhibit C shows the teacher at a position of 48 miles, whereas the function of Exhibit B shows a position 45 miles. Exhibit B shows that the teacher arrived in D’exdete before an hour. (But if the teacher stopped in D’exdete, she should have left at an initial velocity of 0 for her distance function to be differentiable, but Exhibit C’s function starts with a derivative of 15mph).

108. a. (2y ! 3)2dy" = (4y2 !12y + 9)dy" =

43y3 ! 6y2 + 9y + C ; Expanded the function

and then the reverse Power Rule was used for each term. b. ! 26.93 , using a calculator

c. 9 !m"1

"1+ 7 !

m12

12+ C = "9m"1

+7

12m12

+ C ; The reverse Power Rule was used for each term.

d. ! 7.069 , using a calculator


109. y = x(x2 + 2x ! 3) = x(x !1)(x + 3) crosses the x-axis at x = !3 , so this is

(x3 + 2x2 ! 3x)dx = x4

4+2x3

3! 3x2

2!3

0

= !(!3)2

4!2(!3)3

3+3(!3)2

2= 11.25

!3

0

" un2

110. a. [ f (x)2 ! g(x)2 ]dx2

9

"

b. f (x)dx3

9

!

c. f (x)dx!2

1

" + f (x)dx1

4

" + f (x)dx4

9

" = f (x)dx!2

9

"

d. (2k(x) + j(x))dx2

8

!

111. a. !f (x) = limh"0

#(x+h)2 +3(x+h)+1#(#x2 +3x+1)

h= limh"0

#2xh#h2 +3hh

= limh"0

# 2x # h + 3 = #2x + 3

b. !f (x) = "2x + 3 , the same

c. !f (0) = 3 , !f (1) = 1 112. a. When !f (x) is non-negative:

(!", !5]! [1, 6]

b. When !f (x) is increasing, roughly [–2.5, 3.7].

c. Negative, because at x = 8 the slope of f '(x) is negative and !!f is the slope function of !f .

113. a. It is a cylinder with cones carved out of its top and bottom. b. ! r2h " 2 ! r2 # 4 # 1

3( ) = ! #25 #8 " 83! #25 = 2

3(! #200) = 400!

3un3

114. 15(!4 + i

5) "2!2+i /5

i=0

19

# $ !6.448

115. (h(x) + j(x))dx5

3

! = " (h(x) + j(x))dx3

5

! = " h(x)dx3

5

! " j(x)dx =3

5

! " (4 + 2) = "6

(h(x ! 2) + 2)dx5

7

" = h(x)dx + 2dx5

7

"3

5

" = 4 + 2x7

5= 2 + (14 !10) = 8


116. All of the velocities basically connect between the pieces. The total area v(t)dt0

2.3

! can be obtained by adding each area separately:

(!187.5t2 +150t + 45)dt0

0.7

" + (43.75t + 27.5)dt0.7

1.2

" + (!500t2 +1200t ! 640)dt1.2

1.6

" +

(7500t2 ! 25500t + 21600)dt1.6

1.8

" + (!528t2 + 2284.8t ! 2402)dt =1.8

23

"

!187.5 " t3

3+150

t2

2+ 45t

0

0.7#$%

&'(+ 43.75 ! t

2

2+ 27.5t

0.7

1.2"#$

%&'+ !500 t

3

3+1200

t2

2! 640t

1.2

1.6"#$

%&'+

7500 ! t3

3" 25500 ! t

2

2+ 21600t

1.6

1.8#$%

&'(+ !528 " t

3

3+ 2284.8

t2

2! 2402t

1.8

2.3#$%

&'()

46.81+ 34.53+ 21.33+ (!10) + 25.96 " 118.63 miles. This represents the total displacement and should be 100 miles according to Exhibit A, and represents a contradiction in the data.

117. a. (!x2 + 6x ! 3)dx1

4

" = ! x3

3+ 6 # x

2

2! 3x

1

4

= ! 643+ 3 #16 !12( ) ! 1

3+ 3! 3( ) = 15

b. (x +1)dx = x2

2+ x

1

4

=162+ 4 ! (1

2+1) = 10.5

1

4

"

c. ( f (x) ! g(x))dx =1

4

" (!x2 + 6x ! 3! x !1)dx =1

4

" (!x2 + 5x ! 4)dx1

4

"

= ! x3

3+ 5 # x

2

2! 4x

1

4

= 2 23! (!1 5

6) = 4.5

This can also be seen as part (a) minus part (b). 118. a. Length! f (x) " g(x) , width= !x , area! "x( f (x) # g(x))

b. ( f (x) ! g(x))dxa

b

"

c. The x-values of the points where the curves intersect. 119. a. Length! g(x) " f (x) = "x + 8 " ((x " 3)2 "1) = "x2 + 5x , width= !x ,

area! "x(#x2 + 5x)

b. (!x2 + 5x)dx0

5

" = ! x3

3+5x2

22

5

= ! 1253+1252

=1256

= 20 56

120. a. !x ! (x2 ! 6)dx0

2

" =223un2

b. Because we are finding the area between two curves. When g(x) is subtracted, the negative area becomes positive.


121. a. width = !x , length= !(x ! 3)2 + 9 ! (x + 6) = !x2+ 5x ! 6 = !(x ! 2)(x ! 3) ,

Area= (!x2 + 5x ! 6)dx = ! x3

3+ 5 " x

2

2! 6x

2

3

=2

3

# !27

3+ 45

2!18 ! !

8

3+ 20

2!12( )

= !9

2! !

14

3( ) = 1

6un2

b. width = !x , length= sin x ! (x2 !1) ,

Area= (sin x ! x2 +1)dx!0.637

1.410

" = ! cos x ! 13x3+ x

1.410

!0.637

= 0.315 ! (!1.355) # 1.671un2

123. a. (3m2 + 2m ! 9)dm!5

!2

" = m3+ m

2 ! 9m!5

!2

!!!!!= !8 + 4 +18 ! (!125 + 25 + 45) = 14 ! (!55) = 69

b. (2t2 + 3t)dt!1

2

" = 2 t3

3+ 3 t

2

2!1

2

= 163+ 6 ! ! 2

3+ 32( ) = 10.5

c. 1+ 2

x+ 1

x2( ) dx!4

!1

" , of which the middle term requires a calculator: ! 0.977

d. ax2

2+ bx

2

3

= a !9

2+ 3b( ) " a !

4

2+ 2b( ) = 5a

2+ b

124. From the fundamental Theorem, F(x) is an antiderivative of f(x), so f(x) must be the

derivative of F(x): f (x) = ddx(3(x ! 4)3 + 6) = 9(x ! 4)2 .

125. a. A hollow sphere. b. !

43" r1

3, ! 43" r2

3=43" #64 ! 4

3" #27 = 4

3" # 37 = 148"

3un3

126. Yes, because: (1) It is continuous at x = 2 (both functions approach 1).

(2) ddx((x !1)2 ) = 2(x !1) approaches 2 and d

dx2 sin(x ! 2) = 2 cos(x ! 2) also approaches 2.

127. a. Distance is an antiderivative: (!32 !18)dt = !16t2 !18t + C = s(t)" and

500 = s(0) = !16(0)2 !18(0) + C = C ,so s(t) = !16t2 !18t + 500 . So s(1) = !16(1)2 !18(1) + 500 = 466 ft, s(2) = !16(2)2 !18(2) + 500 = 400 ft, and s(3) = !16(3)2 !18(3) + 500 = !144 ! 54 + 500 = 302 ft.

b. s(t) = !16t2 !18t + 500 is an antiderivative of v(t).


128. a. f (x) =(x!3)(x+1)

x!2= x ! 3

x!2: As x! "#, f (x)! "# . As x! 2" , f (x)!# .

As x! 2+ , f (x)! 0 ; as x!", f (x)!" . End behavior: y = x .

b. As x! "#, f (x)! 0 ; as x! 0" , f (x)! "# ; as x! 0+ , f (x)!" ; as x!", f (x)! 0 .

129. No, d

dx((x ! 3)(2x ! 9)) = d

dx(2x2 + 3x ! 27) = 4x + 3 , but d

dx(x ! 3) " d

dx(2x + 9) = 1 !2 = 2 .

130. a. lim

x!9

x "3

x"9#

x +3

x +3= limx!9

x"9

(x"9)( x+3)= limx!9

1

x +3=16

b. limh!0

2+h" 2

h#2+h+ 2

2+h+ 2= limh!0

2+h"2

h( 2+h+ 2 )= limh!0

1

2+h+ 2=

1

2 2=

2

4

c. limx!"

2 x +1

5# x= limx!"

2 x

x

+1

x

5

x

#x

x

= limx!"

2

#1= #2

d. Cos(x) oscillates between 1 and –1, therefore the limit does not exist. 131. a. Width= !x , length = !2(x2 !1) ! (!x2 +1) = 1! x2 ,

Area = (1! x2 )dx!1

1

" = x ! x3

3!1

1

= 23! ! 2

3( ) = 43un2

b. Width = !x, length = sin x " 34x + 7 ,

Area = sin x ! 3

4x + 7( ) dx

"

2"

# = ! cos x ! 3

4$ x2

2+ 7x

"

2"

= !1!3"2

2+14" ! 1!

3

8" 2 + 7"( ) # 8.89un2

132. This would give a negative area; it uses rectangles with negative height. 133. a. The upper bound has two parts, one of which is x2 and the other is !x + 6 .

b. (x2 ! x )dx1

2

" = (!x + 6 ! x )dx2

4

" =x3

3! x3/2

3/21

2#$%

&'(+ ! x

2

2+ 6x ! x

3/2

3/22

4"#$

%&'=

83!2 2

3/2! (1

3!23) + (! 16

2+ 24 ! 8

3/2) ! (! 4

2+12 !

2 x

3/2) ! 1.114 + 2.552 ! 3 2

3un2


134. a. The functions cross paths, so the first part requires an integral of g(x) ! f (x) and the second requires an integral of f (x) ! g(x) .

b. (x3 ! 2x2 +1! (x !1))dx!1

1

" + (x !1! (x3 ! 2x2 +1))dx1

2

"

= (x3 ! 2x2 ! x + 2)dx!1

1

" + (!x3 + 2x2 + x ! 2)dx1

2

"

=x4

4! 2 x

3

3! x

2

2+ 2x

!1

1"#$

%&'+ ! x

4

4+ 2

x3

3+ +

x2

2! 2x

1

2"#$

%&'( 3.083 or 3

1

12

135. a. (2 ! x + 3)dx0

2

" + (x ! 2 + 3)dx2

5

" = ! x2

2+ 5x

0

2"#$

%&'+

x2

2+ x

2

5"#$

%&'=

!2 +10 ! (0) + (252+ 5) ! (2 + 2) = 8 +13.5 = 21.5 , by splitting the integral into two

parts so as to remove the absolute value.

b. (4m!3 ! 3 cosm)dm" = 4 m!2

!2! 3sin m + C = !2m

!2! 3sin m + C by

antidifferentiating each term.

c. ! 2.050 by calculator integration

d. ! 2x + C , because ! 2 is just a constant. 136. The most evident solutions are symmetric across the line y = x , but other solutions do exist

(e.g. the rectangle with 0 ! x ! 3 , 1 ! y ! 2 ). 137. a. 0

b. limx!"

x2#2x+1

x3

= limx!"

1#4x+(1 x2 )

x= 0

c. limx!"

cos x+1x#"

= limx!"

# cos(x#" )+1

x#"= limx!"

#(cos(x#" )#1)

x#"= #0 = 0

138. a. lim

x!1

2x "2

3x "3

# 0.42 b. limx!0

1"cos x

x2

# 0.50

139. a. sin( x ! 2

2) = sin(x ! 2) for x ! 2

b. sin1

x! 2( ) for 1

x! 2 , i.e., for 0.5 ! x > 0

c. 1sin(x!2)

for x ! 2 (and x cannot be 2 + n! for integers n).

d. x for x ! 0


140. a. y =3(x2 !4)

x+2=3(x!2)(x+2)

x+2= 3(x ! 2) (for x = !2 ) and has derivative of 3, with

y(1) = !3 ; tangent line is y + 3 = 3(x !1) or y = 3x ! 6 .

b. y + 3 = !13(x !1)!!" y = !

13x + 1

3! 3!!"!!y = !

13x ! 8

3

141. a. It is hottest after ~7 months on July 1st and coldest after ~1 month on February 1st, by

looking at the maximum and minimum peaks on the graph. b. The inflection points are about 4 months in, around May 1st, and about 10 months in,

around October 1st. 142. The derivative of each piece must be equal at x = 1; d

dx((x + 2)2 ! 3) = 2(x + 2) is 6 at

x = 1 , so ddx(a sin(x !1) + 6) = a cos(x !1) must be 6 at x = 1 ; a cos(0) = 6, a = 6 .

143. f(x) is increasing in

[!3, 2]! [4,"] (where !f is non-negative), f(x) is decreasing in

(!", !3]! [2, 4] (where !f is non-positive), f(x) is concave up in about (!", !1.2]! [3.1,") (where !f (x) increases), and f(x) is concave down in about [!1.2, 3.1] (where !f (x) decreases).

144. The 2

n indicates that the width of the interval is 2 units. The function is y = x2 . Possible

answers include x2dx

1

3

! and (x +1)2dx0

2

! . 145. a. A: first rectangle is 9 ! (!8x + 9) = 8x by !x , and the second is 9 ! x

2 by "x .

B: !x by y "y"9

8= y +

y"9

8

C: x +x!9

8 by "x

b. Adam broke the region into two integrals:

8xdx0

1

! + (9 " x2 )dx1

3

! = 4x2

0

1( ) + 9x ! x3

31

3"#$

%&'= 4 + (27 ! 9) ! (9 ! 1

3) = 13 1

3

Becky integrated over y-values:

y +y!98( ) dy

1

9

" =y3/2

3/2+y2

16!9

8y1

9

= 27

312+ 81

16!81

8!

2

3+ 1

16!9

8( ) =

18 ! 51

16!

!19

48= 13+

16

48= 13

1

3

Cathy reflected the region across the line y = x and integrated over x-values:

x + x!98( ) dx

1

9

" , essentially the same integral as Becky’s, = 13 13

Each method is valid and gives the same answer.


146. The three possibilities are (4 ! x )dx0

16

" , y2dy,0

4

" and x2dx

0

4

! , and each equals 64

3= 21

1

3.

147. The curves intersect at 2 x = x!!!!!2 = x !!!!!x = 4 .

(2 x ! x)dx0

4

" =43x3/2 ! 1

2x2 4

0=4343 ! 1

2(42 ) = 32

3! 8 = 8

3= 2 2

3un2

148. The functions cross at x = ! . So students must take

(sin x + sin x)dx0

!

" + (! sin x ! sin x)dx"

2"

# or double the first integral:

2 (sin x + sin x)dx0

!

" . (Or 4 (sin x + sin x)dx0

! /2

" works too). All methods yield 8 un2.

149. a. Using a calculator, the minimum of f is at (4, –13.5), so this is 13.5 in. This can also

be obtained by factoring the derivative:

!f (x) = x3

128+3x2

32" 2 = 1

128(x + 8)(x + 8)(x !14) is 0 at x = 4

b. It is ! 2

225x2 + 20 ! x

4

512+ x

3

32! 2x ! 8( ) = !

x4

512!x3

32!2x2

225+ 2x + 28 by !x .

c. The bounds are obtained using a calculator: ! "15.326 and ! 9.753

Area! " x4

512" x3

32" 2x2

225+ 2x + 28( ) dx"15.326

9.753

# = !x5

5"512!

x4

4"32!2x3

3"225+ x

2+ 28x

!15.326

9.753

! 544.60 in2

150. a. x2dx

!1

1

" = 2 xdx0

1

" = 2x2

20

1#$%

&'(= 1 , using symmetry to take double the second part

of the integral (or two integrals may be used), followed by the Fundamental Theorem.

b. 8 !x4

4"1

2!x2

2+ C = 2x

2"x2

4+ C , by antidifferentiating each term.

c. (3x+1)(x!2)3x+1

dx1

5

" = (x ! 2)dx1

5

" =x2

2# !2

1

5

=25

2!10 !

1

2! 2( ) = 4

d. y + C , by the second part of the Fundamental Theorem. 151. Possible answers: a. f (x) = x5 , g(x) = cos x b. f (x) = x3, g(x) = 3x cos(x2 )

c. f (x) = x +1, g(x) = 0 d. f (x) = x +1, g(x) = x !1


152. limx!3"

f (x) = 4 " 3(3) = 5 , so a(3)2 + 2 = 5, 9a = 3, a =13

153. !y = "4x , so !y (1) = "4 , so the linearization is y = !4(x !1) + 2 = !4x + 6 . Therefore y(1.15) ! "4(1.15) + 6 = 1.4 154. a. The left bound is 0 and the right bound is ! 2.163 (using a calculator)

(sin x ! (x !1)4 +1)dx0

2.163

" = ! cos x !(x!1)5

5+ x

0

2.163

! 2.296 " "4

5( ) ! 3.096

b. The left bound is 0 and the right bound is ! 3.532 (using a calculator)

( x ! x(x ! 3))dx0

3.532

" =x3/2

3/2! x3

3+3x2

20

3.532

# 8.451! 0 = 8.451.

155. y = x2 !1 156. a. y = 2x2

b. y = 0

c. y = x2 !3x!10

x2 !1= 1+

!3x!11

x2 +1, so it is y = 1. The idea is that as x! 0 or x!" , the

terms with a higher power of x in the denominator will approach zero. 157. a. lim

x!"

y = "

b. limx!"

y = 0

c. limx!"

y = limx!"

1+#3x#11

x2 +1= 1

These limits are the same as the limits of the respective end-behavior functions. 158. Answers may vary but should be below 2. One method would be to estimate it as a

quadratic. 159. a. x1 = 1

b. x2 ! 0.7 ; x2 is a better approximation

c. x3 ! 0.6

d. Each successive estimate is better than the previous one.


160. a. !f (x) = 3x2 + 2x ; f (1) = 1 , !f (1) = 5 , tangent is y = 5(x !1) +1 = 5x ! 4 . x2 = 0.8 , f (0.8) = 0.152 , !f (0.8) = 3.52 , tangent is y = 3.52(x ! 0.8) + 0.152 = 3.52x ! 2.66 x3 =

2.6643.52

! 0.757 , f (x3) ! 0.006 , "f (x3) ! 3.232 , tangent is y ! 3.232(x " 0.757) + 0.006 ! 3.23x " 2.440 , x4 ! 0.755

b. xn

keeps getting smaller and closer to the actual root.

161. If f (x) = x2 ! N , then xn+1 = xn !(xn )

2 !N2xn

"#$

%&'= xn !

xn

2+ N

2xn=

xn

2+ N

2xn= 12xn +

N

xn( ) .

162. a. Since f (x) is a continuous function, f (!1) = !0.4597 , and f (0) = 1 , by the

Intermediate Value Theorem there must be a root between x = –1 and x = 0.

b. !f (x) = 1" sin x , x2 = "1"f ("1)

!f ("1)# "0.750 , x3 # "0.750 "

f ("0.750)

!f ("0.750)# "0.739

c. The root is ! "0.739 (using a calculator), and the error ! 0.00003 ! 0 . 163. !f (x) = 48x2 " 48x +12 ; e.g. for x1 = 0.2 , x2 ! 0.0685 , x3 ! 0.100 , x4 ! 0.103 " x5 164. a. 5 sin(6x)

b. sin(x ! 3x) = sin(!2x)

c. cos2(1! sin2 x) ! sin2 x(1! cos2 x) = cos2 x ! cos2 x sin2 x ! sin2 x + sin2 x cos2 x = cos2 x ! sin2 x = cos(2x)

Or, (cos2 x ! sin2 x)(cos2 x + sin2 x) = (cos2 x ! sin2 x) "1 = cos(2x)

d. sin xcos x

+cos xsin x

=sin2 x+cos2 xcos x!sin x

=1

1 2 sin 2x= 2 csc(2x)

165. a. Let y = sin(x +1) . Then 2y2 ! y !1 = 0 = (2y +1)(y !1) , y = 1, ! 1

2

So sin(x +1) = 1 or ! 12

; x +1 = !

2+ 2!n or 7!

6+ 2!n or 11!

6+ 2!n ;

x = !

2"1( ) + 2!n, 7!

6"1( ) + 2!n, 11!

6"1( ) + 2!n for integers n.

b. Let y = x2 ! 2x +1 . Then y2 ! 3y = !2 , y2 ! 3y + 2 = 0 = (y ! 2)(y !1) , y = 1, 2 . So x2 ! 2x +1 = 1 , x2 ! 2x = 0 = x(x ! 2) , x = 0, 2 .

Or, x2! 2x +1 = 2 = (x !1)2 , x !1 = ±2 , x = 3, !1 ; x = !1, 0, 2, 3


166. limh!0

2/(x+h)2 "2/x2

h= limh!0

2 #x2"(x+h)2

hx2 (x+h)2

= limh!0

2 # "2xh"h2

hx2 (x+h)2

=

limh!0

2 " #2x#h

x2 (x+h)2

= 2 !"2x

x2!x2= "4x

"3= "

4

x3

167. minimums: (!4.49, !0.217) and (4.49, !0.217) The maximum does not exist because y(0) is undefined (is discontinuous) at x = 0 .

Inflection points: (!5.940, !0.057) , (5.940, !0.057) , (!2.082, 0.42) , (2.082, 0.42) . 168. This is when !!y is non-negative: !y = 3x2 + 6x " 24 , !!y = 6x + 6 ; on [!1,") .

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