Calculus for the Life Sciences Marvin L. Bittinger Solution Manuals.pdf (29)

57

Exercise Set 2.5 −2 ) 1) 50. dy = 2x+2(1−x ( Theslopeat(2,25 4)isdx |x=2=254 The equation of the line is

40. f′ (x)=2cos x The slops at (0, 0) is f′(0) = cos(0) = 1.

So the equation of the tangent lilne is y −y1 = m(x−x1) y −0 = 1(x−0) y = x

y−

41. f ′(x) = −3sin x

25 = 4 = y

The slope at (1,0) is dy dx|x=1 = 0 The equation of the line is

y −y1 = m(x−x1) y −4 = 0(x−0) −4 = 0 y y = 4

y −y1 = m(x−x1) y −0 = 0(x−1) y = 0

1 −1/2

52. Let D(x) = f(x)

42. f′(x)=3 1/2 2x +2x The slope at (4,10) is f′(4)=3 2(4) 1/2 +2(4) 1 −1/2 =3.25 So the equation of the line is

− g(x). Then

D(x + h) −D(x) h

= =

[f(x+h)−g(x+h)]−[f(x)−g(x)] h f(x+h) − f(x) −g(x+h)−g(x) h

y −10 −10 = = 3.25(x−4) =3.25x−13 y = 3.25x

′

D(x)

− 3

y −y1 − = m(x−x1) 17 20 = = 17(x−8) = y y y 6 68 17x − 3 +20 x 6 − 8 3

x2/31/3

lim

h→ 0

h

D(x+h)−D(x) h f(x+h)−()

g(x) (+)− h fx gx ] + h h g(x+h)−g(x) f(x+h) −f(x) h lim +lim h→0 h→ 0 h = f′(x) − g′(x) lim [

h→ 0

53. Let f(x) = cos x then ′

f(x)

44. y = x1/12 1 −11/12 dy dx = 12x ( x 3 / 2)

= = =

43. f′(x)=4 1/3 3 x + 3 x 2 −2/3 The slope at (8,20) is 2 −2/3 =617 f′(8)=4 3(8)1/3 +3(8) So the equation of the line is

45. Rewritey=

(x−2) 4 25 25 4x− 4

dy √−√1 51. dx=2 ( √ 1 +√1 ) ( x x ) 2x 2x3

The slope at (0,4) is f′(0) = 0.

So the equation of the tangent line is

y

25

= x −5/18

dy

− 5 −23/8 dx=18x

46 4sin x dy dx = 4cos x .

Use the trigonometric identity 1+tan2x = sec2x to rewrite 47 ydy= 3sec2x cos3x = 3cos x dx= −3sin x .48. y=√1sinx+√1cosx 2 2 √ 1 dy (cosx − sinx) dx= 2 √

49. Use the sum identity for cosine to rewrite y = 23 cos x− 1 2 sin −x√ 3 1 dy dx= 2 sinx − 2cosx

f(x+h)−f(x) limh →0 h cos(x + h) −cos(x) lim h−sinxsinh−cosx = h→ 0 cos xcos h lim = h→ 0 h cosh − ) sinh ) ( limcosx − sinx ( h →0 h h cosh−1 − sinh sinx·lim cosx · hlim h→0 →0 h h ·0−sin x·1 = cosx = = =

=

−sin x

54. The value of the slope of the tangent lines of the sine function posi-correspond to whether the cosine function is has

tive or negative values. For a positive slope of the 55. tangent

line means the cosine has a positive values at the 56. aprticular x value, and vice versa. Left to the student Left to the 57. ± 0.6922 student The tangent line is horizontal at x = 58. The tangent line is horizontal at x = 0

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Chapter 2: Differentiation

58 59.

The tangent line is horizontal at x = −0.3456 60 and at x = 1.9289

6.

The tangent line is horizontal at x = 1.9384, x = 3.6613 and at x = 5.6498

.

v( π ) = 3+sin( ) 3

Exercise Set 2.6 1.

π 3

2 = 3+sint −1 = sint t = sin−1( −1) 3π = +2nπ 2 7.

a) v(t) = ds(t) = −20t + 2 dt = −20 b) a(t) = dv(t)

= = 2996

c) v(1) = −20(1) + 2 = −20 + 2 = 18 feet per second −20 feet per squared seconds a(1) = = 2t −12 a) v(t) = dsdt dt=2

3 c) v(1) = 2(1) − 12=2 −12=2=1.5feetpersecond a(1) = 2 feet per squared seconds

a) v(t) = ds(t) dt =5+2cost b) a(t) = dv(t) = −2sin t dt c) Find v(π)4 and (π a 4) v( π ) = 5+2cos( ) 4

=

≈ 6.414 m/sec a( π) 4

= = =

π ) − 2sin( 42 √ 2· 2 √ 2

≈ 1.414 m/sec2

−(10)2 300(10) −100 ++3000 +6 +6

There will be 2996 units sold after spending $10000 on advertsing c) At a = 10, DNda= −2(10) + 300 = −20 + 300 = 280 units per thousand dollars spent on advertising d) The rate of change of the number of units sold depends on the amount spent on advertising according to equation da= −2athe + 300 whichdN means that for every a thousand dollars spent on advertising, the change in the units solds is−2a+300 units. If $10000 is spent on advertising, then there will be a 280 unit increase in the number of units sold.

b) a(t) = dv

=

da= −2a + 300 a) dN b) Since a is counted in thousands we need to find N(10) =

π √ 4 2 5+2 · 2 √ 5+ 2

1 2 =0.5m/sec2

d) Find t when v(t) = 2

dt

5.

a( π ) = cos( ) 3 =

a) v(t) = dsdt = 3 Note that the velocity is constant and does not depent on time b) a(t) = dvdt = 0 Note that the velocity is constant and does not depent on time c) v(2) = 3 miles per hour a(2) = 0 miles per squared hours d) Uniform motion means that the velocity and acceleration are constants namely, the velocity is the slope of the linear distance function and the acceleration is zero.

4.

3+

π 33

2 ≈ 3.866 m/sec

c) v(4) = 3(4)2 + 1 = 48 + 1 = 49 feet per second a(4) = 6(4) = 24 feet per squared seconds

3.

√

=

a) v(t) = ds(t)dt= 3 2 t+1 b) a(t) = dv(t) dt =6t

2.

a) v(t) = 3 + sin t b) a(t) = cos t and a(π3 ) c) Find v(π) 3

8.

t2 a) dw = 1.82 −0.11 92+0002274 t . dt b) w(0) = 8.15 + 1.82(0) − 0.0596(0)2 + 0.000758(0)3 = pounds c) 8.15 dtt0 dw 2 .

01192(0)+0002274(0) =1.82 . − 0.0596(12)2 | − =1.82− + 0.000758(12)3 pounds per month d) w(12) = 8.15+1.82(12) e) = 22.72 pounds dw 2 dtt=12 . =1.82−. 01192(12)+0002274(12)=0.717 | f) pounds per month Average rate of change = w(12)−w(0) = 22.72−8.15 = 12 12 g) 1.21 pounds per month

d) Find t when v(t) = 3 1.82 −0.1192t+0.002274t2 = 1.21 0.002274t2−0.1192t+0.61 = 0

3 = 5+2cost −2 = 2cost −1 = cost t = cos−1( = π+2nπ

−1)

Using the quadratic formula we get that t = 5.747 months.

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59

Exercise Set 2.6 9. a) dwdt =161 . − 0.0968t + 0.0018t2 b)w (0) = 7 60+1.61(0) −0.0484(0)2+0.0006(0)3 = 7.60 . p ounds c) dw|t=0= 1.61 − 0.0968(0) + 0.0018(0)2 = 1.61 pounds − 0.0484(12)2 + 0.0006(12)3 = dt per month d)w (12) = 7.60+1.61(12) 20. 987 pounds e) dw|12 = 1.61 − 0.0968(12) + 0.0018(12)2 = 0.708 dt pounds per month f) Average rate of change = w(12)−w(0) = 20.987−7.6= g

1. 1156 pounds per month

12

12

16.

17. First rewrite R(Q) as follows R(Q)= k 2 Q2 − 31 Q3 2 k dQ = 2(2Q) a) dR − 13(3Q)=kQ−Q

b) For every increase Q in the dosage there will be a change in the reaction of the body to that dosage change equal to the amount kQ− Q 2 18.

d) The population of the city will reach 300000 people after 10 years.

The population growth rate at 10 years will be 40000 people per year, that is, the population will increase by 40000 for the next year

dF = 2 feet per degrees a) dD

19.

dr = 2π a) dC b) For every increase of one centimeter of the radius the healing wound circumference increases by 2π centimeters

20.

a) dr = 2πr

dt= −0.2t + 1.2 a) dT b) At t = 1.5 T(1.5) =

c)

14.

15.

−0.1(1.5)2 + 1.2(1.5) + 98.6 = 100.2degrees

dT |t=1.5 = −0.2(1.5) + 1.2 = 0.9 degrees per day dt

d) The sign of T′(t) is significant because it indicates whether the rate of change in temperature is an increase (if positive) or a decrease (if negative). That is, whether the fever is increasing or decreasing during the illness dx = 0.1x a) dB − 0.9x2 b) The sensitivity is affected by the following rule: for every increase in the dosage of x cubic centimeters the blood pressure is changed by an amount equal to 0.1x − 0.9x2 a)

dT = 1.31W0.31 dW

b) For every increase of W in body weight the territorial area of an animal increases by an amount equal to 1.31W 0.31

1 −1/2 . dh=122(2h )= 0√.61 a)dV h √ b)V(40000)=1.22 40000=244miles √0.61 dhh=40000= c) dV =0.00305milesperfeet | 40000

dA

d) At a height of 40000 feet the view is 244 miles.

At a height of 40000, for every one foot increase in height corresponds to an increase in the view of 0.00305 miles

b) For every increase of one centimeter of the radius the circular area of a healing wound increases by 2πr squared centimeters 13.

dt = 0.08 a) dA b) The rate of change for the median age of women at first marriage is constant at 0.08. That is, each year the median age of women at first marriage in increasing by 0.08 years

b) For every increase of one degree there will be an increase of 2 feet in the stopping distance of an object on glare ice

12.

dt = 4000t a) dP c) dP dt |t=10 = 4000(10) = 40000 people per year

1.61 −0.0968t+0.0018t2 = 1.1156 0.0018t2−0.0968t+0.4944 =0 0.0968+ √ (0.0968)2−4(0.0018)(0.4944) 5.7147 = t 2(0.0018)

11.

2

b) P(10) = 100000 + 2000(10)2 = 300000 people

)

10.

a) dH =1.41W0.41 dW

b) For every increase of W in body weight the home range of an animal increases by an amount equal to 1.41W 0.41

21. The average rate of change of a function is the value of the difference quotient evaluated over a period of time, while the instantaneous rate of change is the value of the slope of the tangent line at that particular instant.

22. The derivative of a function at a point x can be thought of as the slope of the tangent line at that point. It could also be thought of as the instantaneous rate of change of teh function at that point.

23.

80 60 40 20 –4

–2

0 –20

2 t

4

Chapter 2: Differentiation

60 24.

28.

300

–4

a) 3

200

2

100

1

–2

–6

4

2 t

–4

–2

0

2

t

4

6

–1

–100

–2 –200 –3 –300

b) The acceleration function is the same as the product −1. That is it is a of the position function times reflection of the position function about the t axis.

25.

400 y 200

–200

Exercise Set 2.7 1. Method One: x3 ·x8=x3+8=x11.dy dx=11 x10

0

–100

200

100 t

–200 –400

2. Method One: x4 ·x4=x4+9=x13.dy dx=13 x12 Method Two (product rule):

26.

0.8

dy dx

0.6 0.4

√

0.2 –10

–5

0

5

–0.2 –0.4

15

10 t

3 2

–2

0

= 9x12+4x12 = 13x12

4. MethodOne:y=x8/3,dy dx=38x5/3 Method Two (product rule)

1 –4

x4 ·9x8+4x3·x9

1 dy = x( x−1/2)+(1)x1/2 2 dx 1 x1/2+x1/2 = = 2 x1/2 3 2

a)

–6

=

dx=23x1/2 3. MethodOne:xx=x3/2,dy Method Two (product rule):

20

–0.6

27.

Method Two (product rule): dy = x3 ·8x7+3x2·x8 dx = 8x10+3x10 = 11x10

2

t

4

6

–1 –2 –3

b) The acceleration function is the same as the product of the position function times −1. That is it is a reflection of the position function about the t axis.

dy = x2 2( x−1/3+2x(x2/3) 3 dx 2 x5/3+2x5/3 = = 3 x5/3 8 3 x5 dy,dx=3 x2 5. MethodOne:x8=x3 Method Two (quotient rule):

dy = = dx = =

x5(8x7) −5x4(x8) (x5)2 −5x12 8x12 x10 3x12 x10 3x2

61

Exercise Set 2.7 x5

,dx= − 3 x − 4 6. MethodOne: x8=x −3dy Method Two (quotient rule): dy dx

= = = =

12.

x8(5x4) −x5(8x7) (x8)2 −8x12 5x12 x16 −3x12 x16 −3x−4

7. Method One: y = x2 −25,dydx=2x Method Two (product rule): dy

′ f(x)

= = = =

=( x − 3), f′ (x)=1

(x+3)(2x)−(x2−9)(1) (x + 3)2 2x2 + 6x −x2+9 x2+6x+9 x2+6x+9 x2+6x+9 1

y = (8x5

− 3x2 + 20)(8x4 − 3x1/2) 3 x−1/2)+ dy = (8x5 −3x2+20)(32x3− 2 dx (40x4 − 6x)(8x4 − 3x1/2) 3 = (8x5 −3x2+20)(32x3− √ )+x 2 x4−√ (40x4 − x6)(8 3 x)

f′(x)

g′(x)

f(x) = (7x6+4x3 − 50)(9x10 − 7x1/2) f′ 7 x−1/2)+ (x) = (7x6+4x3 −50)(90x9− 2 (42x5 + 12x2)(9x10 − 7x1/2) 7 = (7x6+4x3 −50)(90x9− √ )+x −√2 (42x5 +12 x2)(9 x10 7 x) f(x) = (x1/2 − x1/3)(2x + 3) ′

= =

1 √ x − √3x)(2) (x−1/2 − 1 x−2/3)(2+3)+( x 2 3 x−√ 1 1 ( ) 3x)+(2x+3) 2( √ 2 x − 33x2

1 x−2/3+2

)

1 +2 ) 33x2

=

x1/2sec2x+tanx ( 1 x−1/2)

=

√ xsec2x+ (

)1 2√ x

√ xsecxtanx+secx

= =

tanx 1 3 (

x−2/3)

3

√3xsecxtanx+

secx √33x2

15. (2t + 3)2 = (2t + 3)(2t + 3) = 4t2 + 12t + 9 f′(t) = 8t + 12 16. r(t) = (5t − 4)2 = (5t − 4)(5t − 4) r′(t) = (5t = = 17. g(x) = (0.02x2 + 1.3x

− 4)(5) + (5)(5t − 4) 25t −20+25t−20 50t −40 − 11.7)(4.1x + 11.3)

g′(x) = (0.02x2+1.3x

− 11.7)(4.1) + (0.04x + 1.3)(4.1x + 11.3)

18. g(x) = (3.12x2 + 10.2x − 5.01)(2.9x2 + 4.3x − 2.1) g′(x) = (3.12x2+10.2x − 5.01)(5.8x + 4.3) + (6.24x + 10.2)(2.9x2 + 4.3x − 2.1)

19. g(x) = sec x csc x g′(x) = secx(

−csc x cot x) + csc x(sec x tan x) 1 − 1 = + sin2x cos2x = sec2x − csc2x

10.

f(x)

(3√ x+2x)sec2x+tanx

( (

13. f(x) = x1/2tan x

−5)

9.

11.

√ (3x+2x)sec2x+tanx

= =

14.

= (x+5)(1)+(1)(x dx = x + 5 + x − 5 = 2x

8. Method One: f(x) = (x+3)(x−3) x +3 Methods Two (quotient rule):

′ f(x)

20. p′(x) = cotx( =

−csc x cot x) + csc x(−csc2x)

−csc cot2x − −csc x x(cot2x + csc3x csc2x)

= 21. q(x)= sinx 1+cos x ′ q(x)

= = = =

(1+cosx)(cosx)−sinx(−sinx) (1 + cos x)2 cos x + cos2x + sin2x (1 + cos x)2 cosx+1 (1 + cos x)2 1 1+cosx

Chapter 2: Differentiation

62 22.

−1 31. w = t23t2+6 − t

′ f(x)

= = = =

sinx(sinx)−(1−cosx)(cosx) sin2 x

= dt = =

sin2 x−cosx+cos2x sin2 x 1 −cosx 1 − cos2x

32. q =t22t2+7t−1 −3t−7 dq dt

(2t2

= = 1

11t2

x

25. f(x)=(x+2

(2t2

2sectsecttant ( ) sec2t tan t 2

x−1)(2)+(1 )( x x 2 − x−2 x2− x−2 x2 = 2 +4+ − 3 −2 + 6 x2 = 3x2 − 1 + 6 x−2 6 3x2 − 1+ x2 =

q′ (x) = 5x + 313 + 32x2 − 3 q′ (x)

=

6 6 5 − 5 4 3

4x2+

x−

= dx = =

′ f(x)

34. f(x)=

1

x

′ f(x)

= =

2 (x+1)(0)−1(2x) (x2 + 1)2

−2x (x2 + 1)2

5 3

tant cost 35. y= 1+sec t Which can be rewritten as =y 1+ 1 mulcost tiplying every term by cos t to clear the fractions gives sint 1 t+1 which has a derivative of dy dx = 1+cos t (see y= cos problem 21 for details)

36. Rewrite y = cos sin tt+1 dy

(2x 2x2−1)(2x+3)−(x2+3x−4)(2) −1)2 (2x + 6x −2x−3−2x2−6x+8 (2x −1)2 −2x+5

30. y=(x−1)(x−3) (x −1)(x+1) = x+1 x−3

= dx =

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

1 x 1 = x2+1 x+

= dx = = =

(2x −1)2

dy

= = =

sint

29. y = x2+3x−4 2x − 1 dy

−38t−52 − 3t − 7)2

+1 = 1+x

3)

26. y=4x 4 − x −3 2 0 x2 +5 x dy = 16x 3 dx − 3x2 −40 x+5 27.

You could use the quotient rule, but a better technique to use would be to rewrite the function as follows:q( )x= 3 x2 − 65 x +45then 5

−3t−7)(2t+7)−(t2+7t−1)(4t−3) − 3t − 7)2 (2t2

x2

x

33. fx()=

x−1 )(x2 −3)

′ = ( +2 f(x)

28. q(x)=4x3 3

3t2

−3t2+2t+16 (t2 − 2t + 6)2

23. s( t) = tan2 t s′t( ) = 2 tan t sec2t 24. = =

−2t+6)(3)−(3t−1)(2t−2) − 2t + 6)2 −6t+18−[6t2−6t−2t+2] (t2 − 2t + 6)2 (t2

1 1+cosx

s′t( )

(t2

dw

(sint+1)( −sint)−cost(cost) (sint+1)2 −sin2t−sint−cos2t (sin t + 1)2 −(sint+1) (sin t + 1)2 −1 sint+1

sin x 37. w = tan x+√x x (x+1)(1) −(x−3)(1) (x+1)2 x + 1 −x+3 (x + 1)2

dw dx = = =

√x(sec2x+xcosx+sinx) ( √ x)2

−(tanx+√xsinx) 2x

2x(sec2x + xcos x + sin x)−tanx−xsinx 2x3/2 2xsec2x + 2x2cos x + x sin −tanx x 2x3/2

63 c) Since the graphs of both functions are similar then the average rate of change for the functions will be the same (that is why the answers in part (a) and part (b) are equal).

Exercise Set 2.7 38. √ ( sinx+x2cscxcotx 2xcscx) cosx−x2cscx √ x − − − 2 x √ ( x)2

dw = dx =

− sinx x3/2cscxcotx+2x5/2cscx √ − x − √ cosx xcscx − 2x3/2 2

x2

86.

a)f()= x

x2− 1

f′(x) = = =

(x2 2x3

−2x−2x3 −1)2 (x2

39. y = 1+t1/2 −1 t1/2

−2x (x2 −1)2

1/2 (1 − t1/2)( 2t11 / 2 ) (1+ t )( 2−t 11/ 2) − (1 − t1/22 ) 1 − t1/2 + 1 + t1/2 ) 2t1/2(1 − √ t2

dy

= = dx = =

b) g(x)=

dy dx = = =

= = = =

87. f(x) = sin2x + cos2x a) f′(x) = 2sinxcosx+2cosx( = 2sinxcosx

( tsint )( sec2t )+( tant )( tcost + sint) t sin t sec2t+ t tan t cos t+ sin t tan t sinttant 1 sint tsint cos2t+ t cost cost + ttantsect + tsint + sinttant

42.

=

=

( t csc t)( −sint )+(1+ cost)( − tcsctcott + csc t) tcot2t+ cott −t −tcsctcott + csct −

= 43. - 84. Left to the student 85. a)

f(x)= x x+1 ′

f(x)

= = =

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

b) g(x) = −1x+1 ′ g(x)

= = =

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

0

−sin x)

− 2sin x cos x

b) Since sin2x + cos2x = 1 (fundamental trigonometric identity) we would expect the derivative to be zero since we are taking a derivative of a constant, which is always zero.

88.

g′(t)

2 ( x−1)(0)−(1)(2x) (x2 −1)2 0−2x (x2 −1)2

c) Since the graphs of both functions are similar then the average rate of change for the functions will be the same (that is why the answers in part (a) and part (b) are equal).

15x2 +9

41. f (t) = t sin t tan t f()′t

= = =

−2x (x2 −1)2

2x − 3 x 2

(15x2 +9)(2 − 6)x − (2 x − 3x2)(30)x (15x2 +9) 2 30x2 − 90x3 +18 54− x − 60x2 − 90x3 (15x2 +9) 2 x3 x2 − 150 − 60 − 54x +18 (15x2 +9) 2

1 x2 − 1 ′ g(x)

2 √ 2 t(1 − √ t )2 1 ) √ − √t 2 t(1 40. Rewrite y =

−1)(2x)−x2(2x) −1)2 (x2

a) f′(x) = 2tan x sec2x b) g′(x) = 2sec x (sec x tan x) = 2tan xsec2x c) The answers in parts a) and b) are the same. The reason is that tan2x + 1 = sec2x and therefore when we take the derivative we get the same result since the derivative of 1 is zero.

89. y =

8 x2+4

dy

= dx = =

(x2+4)(0) −8(2x) (x2+4)2 0 −16x (x2 + 4)2

−16x (x2 + 4)2 (02+4)2=0. = m =−16(0) For the point (0,2), dy dxx=0 | tangent line is y −y1 = m(x−x1) −2 = 0(x−0) −2 yy y

= =

0 2

The

64

Chapter 2: Differentiation −16(−2) 2=m=((−2)2+4)2=8= For the point ( −2,−1),dy dx|x=− 4. The tangent line is

32

y −y1 m(x−x1) −1) == 4(x−(−2))

Whenx=1y 4, 16

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

= dx = =

92. y=

x2 +3 x− 1

dy

= = dx =

For the point (0,0), dy|x=0 = m = 4. The tangent line is dx

y −y1 = m(x−x1) y −0 = 4(x−0) y = 4 For the point ( −1, −2), dydx|x=−1 = 1. The tangent line is

91. y =

√x

x1/2

dy

= = dx = =

(x+1)(1−1/2 ) − x 1/2 (1) 2x (x+1)2 1 1/2

y −y1 = m(x−x1) y −6 = 0(x−3) −6 = 0 y y = 6

1

1 −1/2

Whenx=1,y=

√1

1

2x2 − 2 x − x 2 − 3 (x −1)2 (x −3)(x+1) −1)2 (x 22 +3

+3 32 dy 3−1 = 6 and dx|3 = 0. The tangent

Whenx=3,y= line is

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

−1)(2x)−(x2+3)(1) −1)2 (x

y −y1 = m(x−x1) y −7 −7== −1(x−2) −x+2 −x+9 = y y

x− 1

x+1 = x+1

(x

dy 2−1 = 7, and dx|x=2 = −1. The

Whenx=2,y= tangent line is

y −y1 = m(x−x1) y −(−2) = 1(x−(−1)) y+2 = x+1 y =

93. y = x sin x dy = dx

dy

1 1+1 = 2, and dxx=1 = m = 2√1(1+1)2 =

x cos x + sinx

When x = π

1

8. The tangent line is

4

y −y1 = m(x−x1) −1 = 1 1 y 8 (x−1) = = 1 = = 1 −1 8 x− 1 y − 2 1 y y 8 − 8 1 y 1+ 1x x 8+ 2 8 4 3 8 8 x+

2

y − y1 = m(x −x1) 2 16 1 y ) = (x y− 5 = 25 − 4 = = 2 = 16 4 − 5 25x x − 25 y 16 − 4 y 2 y 25 − 25 + 5 16 10 x 4 + 25 25 25 16 6 x + 25 25

4x 1+x2

dy

√ 11(+1

2

25. The tangent line is

y y −1 = 4x+8 y = 4x++8+1 y = 4x+9 90. y=

√ 14 dy 2 =1+1=5,anddxx=1=4 | 4

dy = 4 dx| x = π = = =

m π cos(π π )+ sin( ) 4 4√ 4 π√2 2 + 2 √42 2(π+4) 8

The tangent line is y −y1 =

m( x − x 1 )

=

Exercise Set 2.7 √ y−

2π 8

= =

y

=

y

=

y

65 √ ( π + 4 ) (x − π) 8 4 √2 √ 2 ( π + 4) 2π π +4 ) π ( + x − 8 8 8 4 √ ( π + 4 ) − π2√ √ 2π √ 2π x− − 8 + 8 8 16 2 2 ( ) π+4 π2 x √ √ 8 16 2

96.

a) P(t) = 10 +

50t+9 2t2

dP

(2t2+9)(50) −(50t)(4t) (2t2+9)2

= =

dt

−100t2+450 (2t2 + 9)2

b) When t = 8 years

2

94. y = x tan x

When x = π

P = 10+ 10+

dy √ = dx 2

x sec2x+ tanx

= ≈ 12.920

4

dy = 4 dx| x = π = = =

The population will reach approximately 12920 people after 8 years c) When t = 12

m πsec2π π 4 ( )+tan( 4 ) 4 π 4 ( )1+ 1 π +28 8

dP dt

−100(12)2+450 (2(12)2+9)2

= =

−14400 ≈−0.163 88209 −163peopleperyear

≈ The tangent line is y − y 1 = m ((x−x)1) π+8 (x − π ) y−π = 4( 4 8 = π+8 ( π+8 ) x − π ) + π y 4 = 8 4 ( 8 ) y x− 95.

a) T(t)=

97.

100= tant then st ( ) = 100 tan t a) Since s(t) b)

d s(t)

dt =100 sec2 t

c) 100sec2 t= sec2 t = 1 = cos2 t =

π+8 4t t2+1 + 98.6 8 dT (t2+1)(4) −4t(2t) = +0 = (t2+1)2 dt = 4t2 + 4 − 8 t 2 (t2 + 1)2 −4t2+4 (t2 + 1)2

cos2 t = cost = t =

98.

v( t) 4(2) +98.6 2 2+1 8 5 +98.6 = 100.2degrees

= =

c) When t = 2 hours dT dt

4(2)2+4 == − 2 (2+1)2 = − 12 5 − 24degreesperhour .

20 02 2 1 2 ±√12 ±π 2nπ 4 π 2 4 + nπ

a)

b) When t = 2 hours T

50(8) 2(8)2 + 9 400 137

= =

() s′√tt ( +1)(

−3 sin t) − (3 cos t)( 2t√1 ) √ ( t+1)2

b) v(1)

= =

√ ( 1+1)(

−3sin1)−(3cos1)(√1) √ ( 1+1)2 − 1465in/sec .

21

NOTE: the negative sign indicates a downward movement.

66

Chapter 2: Differentiation ( −xsinx−cosx)(xcosx+sinx (xsinx+cosx)2 x sin2x + x cos2x 2 (x sin x + cos x)2

c) v(1) = =

√(

1

3)(√2 ) π 3 − (3 cosπ

π3+1)(3sinπ3) − π

− 1.487in/sec

√ ( 3+1)

103.

(2

x 99. g( x) = x +1)tan (x2 −1) g′x ( )

=

=

= = =

=

2

2(2+1)+2 secxx x( 2 1)2 2x( x2+1) t a n x− x( 2−1)2 (2x 1)−

4

xtanx

g ′ (x) = − =

2

( x − 1) secx 2xtanx + − (2 x 1)2 x(2 1)2 − 3 2xtanx 2xtan x− − (2 (2 x−1)2 x 1)2 − 2 (4x 1)− secx 23 tanx − (2 (2 x−1)2 x 1)2 − x 4 3 2 sec x( x − x2 sinxcosx−1) x( 2−1)2 2 3 (4 sec x x −x sin x(2)− 1) x( 2−1)2

=

(t)

104. f (x) =

+1)[( cscxcotx )( t3 − 1)+ cscx (3 t2)] − (t3 + 1)2 − 3t2(t3−1)cscx (t3 + 1)2 [( −cscxcotx)(t3 − 1)+cscx(3t2)] − (t3 + 1) 3t2(t3−1)cscx (t3 + 1)2

=

t 101. s(t) = ttan cos t s′

(t)

= = = = = = =

102. Rewrite fx()= f′ (x)

=

x 1/2 sin x − x 3 / 2 cos x x 2+2x+3

105. Let y=(x

xsinx−cosx x sin x+cos x

,

( xsinx + cosx)(xcosx + sinx + sinx) ( xsinx + cosx2 )

(x2 + 2x + 3)[x1/2cos x + si√n x 2 ]x (x2 + 2x + 3)2 ( x3/2sinx + 3x/2 cosx) −− 12 2 x2 ( +2+3) x √ ( xsinx − x3/2cos x)(2x + 2) (x2 + 2x + 3)2

− 1)(x − 2)(x − 3)

a) dy

tcostsec2t− tant ( − tsint + cost) ( tcost2) + tsect tsinttant − sint t2 cos2t sec2ttsect ( + tsinttant − sint t2 tsec3t+ tsinttantsec2t − sintsec2t t2 sec t( t sec2t + t sin t tan t sec t− sin t sec t) t2 sec t( t sec2t + t tan2 t− tan t) t2 sec t( t sec2t + tan t( t tan t− 1)) t2

−x2sin2x − sec x − sin x tan x+ tan x ( x + c o s x) 2

f′(x) =

(t3

=

( + sinx−sec2x] ) (x + cosx)[ −xsinx + cosxxcosx + c o s x 2 (x ) )[ ] −(1−sinxxsinxcosx−tanx ( x + cosx2 ) −x2sin2x + x2cos2x+ xcosxsinx−xsec2x ( x + cosx2 ) xcos3x cos2xsinx−secx + + −xcosxsin2x cosx2 (x + ) xsin2 + + xcosx−sinxtanx xsinxcosx tanx − ( x + cosx2 ) xcos2x + ( x2 + sinxcos2x ) − x sec2x ( x + c o s x) 2 +

100. g′

− sin x)

= dx = =

(x −1)[(x−2)(1)+(x−3)(1)]+[(x−2)(x (x − 1)(x − 2) + (x − 1)(x − 3) + (x − 2)(x 3x2

− 12x + 12

b) y = (2x + 1)(3x − 5)(−x + 3) dy dx

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

−5)(−1)+(2x+3)(−x+3)(3)

+(3x − 5)(−x + 3)(2) −(2x + 1)(3x − 5) + 3(2x + 1)(−x + 3) +2(3x − 5)(−x + 3) −18x2 + 50x − 16

c) The derivative of a product of three functions is the sum of all possible combinations consisting of the product of two functions and the derivative of the third function.

− 3)](1) − 3)

67

Exercise Set 2.8 d) The derivative of more than three function is the sum of all possible combinations consisting of the product of three of the functions and the derivative of the fourth function.

Let y = x(x+1)(2x+3)( −x+1) = −2x4−3x3+2x2+ 3x which has a derivative y′ = −8x3 − 9x2 + 4x + 3.

109. (0.2, 0.75) and ( −0.2, −0.75)

6

4

Let us use the rule to find the derivative of y: −x+1)+ − x(x+1)(2x+3)+x(2x+3)( 2x(x + 1)( −x + 1) + (x + 1)(2x + 3)(−x + 1) − 2x3 −5x2−3x−2x3−x2+3x

y′ = =

2

− 2x3 + 2x −2x3−3x2+2x+3 − 8x3 −9x2+4x+3

=

–2

0

–1

1 x

2

1 x

2

1 x

2

110. (0, 0) and ( ± 0.213, 0.0164)

106. ( ± 1.414,−4)

0.6

20

0.4 y10

–4

–2

0.2

0

2 x

–2

4

0

–1

–0.2 –10

–0.4 –0.6

–20

111.

107. No horizontal tangent lines

(1, 2) and ( − 1 , −2) 4

200

3

150

2

y100

1

50

–2

1 x

–1

2

–2

–1 –1

–50

–2

–100

4x 4− 4 x 112. y = (x2 y = x 2 +1 +1) 2 seems to be the correct derivative of 2

108. (0, −1) 10

Exercise Set 2.8

8

1. y = (2x + 1)2 Method One (chain rule):

6 y 4 2 –10 –8

–6

–4

–2 0 –2 –4 –6 –8 –10

dy 2

4 x 6

8

10

= 2(2x+1)(2) dx = 4(2x+1) = 8x+4 Method Two (product rule): y = (2x + 1)(2x + 1) dy = (2x+1)(2)+(2x+1)(2) dx = 4x+2+4x+2 = 8x+4

68

Chapter 2: Differentiation 9. y = 3x22 +1 =2(3 x 2 + 1)−1

Method Three (expand first): y = 4x2+4x+1

dy = dx

dy dx = 4(2x)+4 = 8x+4

10. y = 3(2x + 4)− 1/ 2 d y d x

= 2(3 2x)( − 2) = =

− 4(3 2x) − − 12+8x

1 11.

y = dy dx = = =

(3 −2x)(3 2x) − −6+4x 6+4x − −12 + 8x

−2)

′ s(t)

y = 9 12x+4x2 − =

2 (2t+3)−1/2(2) ) +(2t+3)1/2(1)

=

t(

=

√t √ + 2t+3 2t+3

12.

Method Three: d y d x

s(t) = t(2t + 3)1/2 ′ s(t)

2x)(− − (3 −2x)( 2)+(3

3( − 1 (2x+4)−3/2(2)) 2 √−3 (2x+4)3

dy = dx =

−

Method Two:

− x )55

−12x (3x2 + 1)2

=

2. y=(3 − 2x)2 Method One:

3. y=(1

2( −1)(3x2 + 1)− 2 (6x)

t2 ( 1 (3t+4)−2/3(3) ) +2t(3t+4)1/3

= =

− 12+4x

√ t2 √ +2t33t+4 33t+4

13. s(t)=sinπt+π (6

dy = 55(1 − x)54(−1) dx = −55(1 − x)54 2,dydx=2 tanxsec2x 4. y = tanx

3) ′

s(t)

= =

π) π cos( π( t + 6 3 · π t+ π)6 π cos

6 14. s′(t) = −sin(3t − 4)(3) = −3sin(3t − 4)

2

5. secx

15. g(x) = (1 + x3)3 − (1 + x3)4 dy

g′(x) = 3(1+x3)2(3x2)

= 2secxsecxtanx dx = 2tanxsec2x 6. y = (2x + 3)10 9 dy dx=10(2x+3)(2)=20(2 +3) x 7. √ = 1 − 3x = (1 − 3x)1/2 y dy

= dx =

= 9x2(1+x3)2 16.

9

1

dy dx

= dx =

= =

1

(1+secx)−1/2(secxtanx) 2 s√ecxtanx 2 1+secx

(1 −3x)−1/2(−3) 2 √−3 2 1−3x

17. y= √ 1 − csc x = (1 − csc x)1/2

1 (x2+1)−2/3(2x) 3√ 2 x

18.

dy dx

√ 8. y=3x2+1=(x2+1)1/3 dy

− 4(1 + x3)3(3x2) − 12x2(1 + x3)3

33(x2+1)2

= =

1 (1 −cscx)−1/2(−(−cscxcotx)) 2 c√scxcotx 2 1 − csc x

g′(x) = =

1 x−1/2+3(x 3)2 − 2 √ 1 +3(x 2 − 3)2 x

69

Exercise Set 2.8 ( )=(2x 1)1/3+(4 19. g x − g(x) =

1

′

−x)2 −1)−2/3(2)+2(4−x)(−1)

f ′ ( x )=

(2x

3 √2− 33(2x

=

28.

1)2

=

−2(4−x)

1 (5 x− 2) ( 2 (3x +4) −1/2 (3) )+5(3x+4)1/2 3(5 √ x − 2) + 5 √3 + 5 x 2 3x + 4

29.

20. dy

= dx =

dy

x2(−sinx)+2xcosx−2xcosx +2sin x − 2(−csc2x) −x2sin x + 2sin x + 2csc2x

dx

dy = x3cosx+3x2sinx+5x( −sinx) dx +5cos x + 4sec x tan x = x3cosx+3x2sinx − 5x sin x +5cos x + 4sec x tan x

dy sec2(sin x)( cos x) 30. dx=

dy = dx =

22. = =

1 (2x+3)−1/2(2)(x2+3x+1) 2 x2 (2x+3) 3/2 + √+3x+1 2x+3 (2x+3)1/2(2x+3)+

dy = (x2+x3)1/2(4x+3) dx 1 + (x2+x3)−1/2(2x+3x2)(2x2+3x+5) 2 x2 x2 +3x+5) (2+3√)(2 x 2 = (4+3)x+ x √ 2 x3 + 2x2+x3

f′t()

= =

√

26. ′ f(t)

=

cost (

=

√ cos √t 2 t

1 t−1/2 ) 2

=

=

1

(2x+5)−1/2(2) ) +3(2x+5)1/2

=

(3x+2) (

=

√3x+2 +3 √ 2x+5 2x+5

5 3 x2 ( x3 − 1)[4( +3)(2)] x x2 − 10 x3 ( − 1) 4 ( x2 +3)[5( 4 x3 − 1)(3 )] 10 ( x3 − 1) 5x2 3 ( +3)4(x3 8(xx3 − − 15x2x2 1)( +3) ( x3 − 1) 10

−1)4

(2x4 + 1)2[3(x3 + 2x + 3x 1)2(3x2 + 4x + 3)] − (2x4 + 1)4 (x3 + 2x2 + 3x 1)3[2(2x4 + 1)(8x3)] − (2x4 + 1)4 3(x3+2x2+3x 1)2(2x4+1)2(3x2+4x+3) − (2x4 + 1)4 16x3(x3 + 2x2 + 3x 1)3(2x4 + 1) − (2x4 + 1)4 /

34. y= ( 2x+3 3x − 5 )13 dy = dx = =

−2/3 1 2 x+3 (3x 5)(2) − − (2x+3)(3) ( ) [ 3 3x − 5 (3x − 6x − 10 6x−9 − 3(3x − 5)4/33(3x √ 5)2− 19 − 3(3x − 5)4/33(3x √ 5)2−

]

/

− 4)12 35. y= ( 3x 5x+3

dy = dx

27. f (x)+(3x+2)(2x+5)1/2 () f′ x

f ′ x( )

√ 1 t−1/2 √ ) − sint ( 2 √ − si√nt 2 t

( cos (4)) t −1/2 ( −sin (4)(4)) t 2 2 (4) −√sint (4) cost

(x3 +2x2+3 x 1) 3 − (2 x4 +1)2

csc(1/x)

dy 24. dx= − csc(1/x−)(1/x)=

25. f ()=cost t

= 33. f (x) =

23. y = ( 2x + x3)1/2(2x2 + 3x + 5)

1

32. f ′ ( x )=

′

=

cos( cos x)( −sinx ) ( −sin x cos cos x)

31. y = ( cos(4t))1/2

21. y = x3sin x + 5x cos x + 4sec x

f(x)

=

= =

−1/2 13 x4 (5x+3)(3) (3x− − 4)(5) ] − ) ( [ 2 5x+3 (5x+3)2 1√5x+9 − 15x+20 2 3x − 4(5x+3)3/2 29 2√ 3x − 4(5x+3)3/2

−

−

Chapter 2: Differentiation

70 36.

44. f′(x)

dy dx

1 √ sec2x ( x −1) ( x x −1)−1/2 ]+( x −1) 1/2 (1) ) 2 1 x sec2 ( x√x −1) √ √ ( x −1) + x − sec2x 2√ x − 1

= =

37. rx( )= (001 x +2391 . x2

x − 8.51)

.

5

sin3

4

(001 +2391. . x2 x2 = (001. +2391 .

(001 +2391. . x2 x2 = (001. +2391 .

r′x( ) = (321

4

x −851)(01 . +11955 + .

x

3

x −5 87) . (2 36 . x −5 .45)

(x2 + 5)2

46. dx

3

4

. x −5 87) . [5(2 36 . x −5 45) . (2 36)]+ .

5 2 (236 . x − 5 .45) [3(3 21 . x − 5 .87) (3 21)] .

4

47. f(x) = cot3(x sin(2x + 4)) f′(x) = 3cot2(xsin(2x+4))(

=

dy

dy dx = =

− cscx −

=

( ) − cscx + 2 sinxcosxcoscos2x

8x sec4( x2) tan( x2) cos( sec4 ( x2 )) 1/2

((2 x +3) 1/2 +1) −1/2 (

2√

√

1 2x + 3 + 1

√

1

(2 x +3) −1/2 (2) )

2x + 3

/ x2 43. y =(( +2)14+1)1/3

dy dx

= =

1

49. y=

((x2+2)1/4 +1)−2/3 (

√x √ 64(x2+2)33(4x2+2+1)2 √

1

(x2+2)−3/4(2x) )

1 ·2cost(−sint) √ 2 2+cos2t −√costsint 2+cos2t

√ sec4x+x 1 ·(4sec3x(secxtanx)+1) √ 2 sec4x+x 2 sec4 √xtanx+1 dy sec4x+x dx

dy dx = =

50.

cossec4x2 ( )( secx2tanx2 )·2 x ( ( )) · 4sec3x2

1

= dx =

sin(5)(5)) x

coscos2x )(2 cosx (− sinx )) (

=

1/2 42. y =((2 x +3) +1)

= 48.

41. y = sin (sec4( x2)) dy dx =

−csc2(x sin(2x + 4)) × [x cos(2x + 4)(2) + sin(2x + 4)] −3cot2(x sin(2x + 4))csc2(x sin(2x + 4)) × [2x cos(2x + 4) + sin(2x + 4)]

2

40. dx

sec2√ 1 √ ( t+2) · √ 2 tan( t +2) 2 t +2 sec2√ √ + ( t+2) tan( t √2) t+2

= =

5

1 x−4/5 (− csc2(5)(5)+ x 5 ( cot5x − cos5) 5 ( sin 5 x − csc x ) cot5x − cos5x4/5

dy

(x2+5)3sin2xcosx −sin3x(2x) (x2+5)2 3x2 sin2x cos x + 15sin2x cos −2xsin3x x (x2 + 5)2 sin2x(3x2 cos x−2xsinx+15cosx

= dx = =

dy

39. y = ( cot5x −cos x 5 )1/5 dx

sinx)

dy

x − 5 .45) (3 21 . x − 5 .87) (11 8) .

(321 . x − 5 .87) + 9 63(2 . . x − 5 .45)) 36 4 2 = (236. x −545) . (321. x −587) . (606 . x −12175) .

= =

(x + sinx)2 + ( −2x2−2x2sinx)(1+xcosx+ (x + sin x)4

5

= (236.

dy

sinx)]

x − 8.51) 4 x −8 51) . (5 (0x 02. +2x 391)+ . .x2 x − 8. 51))

001 +2391. . x2 x − 8 .51) 4 x2 = (001. +2391 . x −851)(011 . .+ x2 14. 346x − 8.51)

38. r ( x)=(321 .

−x[2(x+xsinx)(1+xcosx+ (x+sinx)4

45. y = x2+5x

+2391. x [5(001 . x2 x − 8 .51) (0 02 . +x 2 391)] . + 5 (001 +2391. . x2 x − 8 .51) (1) 4 = 5 (002 +2391 x .+2391)(001 x . . x2 . x −851)+ .

r′x() =

(x+sinx)2(1)

= =

= =

1 ·(1−cscxcotx) 2√ x+cscx 1 −√cscxcotx 2 x+cscx

u=x2 − 1 51. y √ = u=u1/2, 1 1 dy = u1/2−1= 1 u−1/2= 2 2 √u du 2 du =2x dx dy dy ·du = 1 2x dx = du dx = 2√ u · = 2 2x √√ x2 1 x − x2 − 1

71

Exercise Set 2.8 15 =15 u−3, =2 u x +1 u3 45 dy = 45u−4, or − u4 du −

52. y =

du =2 dx dy 45 dy ·du − 90 = = − 2= · u4 (2 +1)4 dx du dx x 53. y = u50, u = 4 x3 − 2x2

d u d x d y d x

dy du =50 u49 du dx =12 x2 − 4x

dy dx = dy · du du dx =50 u49 (12 x2 − 4)x =50(4 x3 − 2x249 ) (12x2 − 4)x

12

+1 u 54. y = , u= 1 + √ x = 1 + x / 1u − ( u − 1)(1) 1−( +1) dy · u = = (u − 1) 2 du = = = u− 1− u − 1 ( u − 1)2 −2 ( u − 1) 2 1 x−1/2 du dx 2 dy 1 12 dy du −2 dx = · · x−/ du · dx ( u − 1)2 2 1 −1/2 −2 = x = = (1+ x 1/2 − 1) 2 2 = −1 x−/1 2 ( x1 /2)2 · 1 1 2 − · x−/ x 3 2 − x−/ 55. y = u( u+1), u = x 3 − 2x d y = u · 1+1( ·u +1) u d u = + u+ 1 = 2 u+ 1 du dx = 3 x2 − 2 dy dx = dy · du =(2 u +1)(3x2 − 2) du dx x2 − 2) =[2( x3 − 2)+1](3 x =(2 x3 − 4x +1)(3 x2 − 2)

56. y = (u + 1)(u −1),u=x3+1 dy = (u + 1)(1) + 1 ·(u−1) du = u + 1 + u − 1 =2u

57. y= dy dx

=3x2 =

dy ·du =2u·3x2 du = 2(x3 + 1)(3x2) = 6x2(x3 + 1)

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

=2 √2x+3 x2+3x dy 21+3 Whenx=1, = √· 2 12+3 · 1 dx 2 =2 √+3 4 5 = = 2· 2 5 4 5 Thus, at (1,2), m = . We use point-slope equation.

4 y − y1 = m(x − x1) 5 (x−1) y y −2= −2= 4 y= 5 5 x− 4x + 4 3 5 4 4 58. y = (x3 − 4x)10 dy = 10(x3 dx

Whenx=2,

−4x)9(3x2−4) dy =10(23 −4·2)9(3·22−4) dx = 10(8 − 8)9(12 − 4) =10 · 0 · 8

=0 Use the point-slope equation: −0 = 0(x−2) y y=0 √ 59. y=x2x+3=x(2x+3)1/2 1 (2x+3)−1/2(2)+1·(2x+3)1/2 dy dx = x · 2 =

√x √ + 2x+3 2x+3

72

Chapter 2: Differentiation 3 d √ + 2·3+3 y = √ d 23+3 · x = √3 √ = 9+ 9 3+3 3

Whenx=3,

a) ′ f(x)

2 5 ( 1+ x )(2x) − x(5(1+x)) (1 + x)10 (1 + x)4[2x + 2x2 −5x2] (1 + x)10 2x − 3 x 2 (1 + x)6

= = =

=1+3=4 Thus, at (3,9), m = 4. We use point-slope equation. y − y1 = m(x − x1)

b)

y −9 −9 = = 4(x−3) 4x−12 y 60. y − (

y=4x

f′(x) = x2[ =

− 3

2x+3 ) 3 x( − 1

= =

2x +3 ) 2 [ ( x− 1)(2) − 1(2x+3) · ] (x − 2 x− 1)2( 3 2x +3 2 x− 2 − x2− 1) ) =3 ( 2 x − )1) x− 1)2( (5 2x +3 − =3 ( (x− 1) 2 x −1 dy 22+3 ) 2 ( − 5 ) Whenx=2, =3 ( · (2−1)2 −1 dx = 3(72)( −5) = −735 Use the point-slope equation: dy

=3

y − 343 = −735(x − 2) − 343 = −735x + 1470 y −735x + 1813

dy = − π6,dx=2

π sin− (6)π cos− (6)=

−√3

Use the point-slope equation: −√3x 2( − ( − π6)) y − 14= √ √ y −14 = − 3x − 3 √2 √12 −3x 3 x − √14 y = 2− √ − + 12 = 3π+3) 1 (6 3 y 12 62. f(x) = x sin 2x dy dx = 2x cos 2x + sin 2x When x = π, dy dx = 2π cos 2π + sin 2π = 2π Use the point-slope equation: y −0=2π(x−π) y = 2πx − 2 π 2 63. f(x)=

x2 (1+x)5

−5(1 + x)−6(1)] + 2x(1 + x)− 2x −5x2 + (1+x)6 (1+x)5 −5x2+2x(1+x) (1 + x)6 2x − 3 x 2 (1 + x)6

c) The results in the previous parts are the same.

64. a) g(x) = (x3 + 5x)2 g′(x) = 2(x3 + 5x)(3x2 + 5) = 2(3x5 + 20x3 + 25x) = 6x5 + 40x3 + 50x b) g(x) = x6 + 10x4 + 25x2 g′(x) = 6x5 + 40x3 + 50x c) The answers are the same 65. Using the Chain Rule: Let y = f(u). Then dy dy du = · dx du dx = 3u2(8x3)

y= 61. f(x) = sin2x dy dx =2sinxcosx When x

4

2

= 3(2x4 + 1)2(8x3) Substituting 2x4 +1 for u When x = − 1,

dy =3[2(−1)4+1]2[8(−1)3] = 3(2 + 1)2(

= 3 · 32(−8) = −216 Finding f(g(x)): f g(x)=f(g(x))=f(2x4+1)=(2x4+1)3 ◦

Then(f g)′(x)=3(2x4+1)2(8x3)and ◦ (f g)′( ◦ 1)= −

−216 as above.

−8)

5

73

Exercise Set 2.8 68. Using the Chain Rule: Let y = f(u). dy dy ·du dx = du dx ·(4+x)(−1)−(1)(3−x) 4 =10u =10u (4+x)2 (4 ·−4−x−3+x

66. Using the Chain Rule: Let y = f(u). dy dy du dx = = du · dx = = (1)(u+1) 1 = (u 1)(1) − −

·2 (u − 1)2 u − 1 − u − 1 1 x−1/2 ·2 (u 1)2 − 1 −2 · (u 1)2 − 2x √ √ −1 √ x( x 1)2 −

Whenx=4,

dy √ −1 = 1)2 dx √ 4( 4 −

Finding f(g(x)):

f ◦g(x)=f(g(x))=f(

√ x)=

x−1/2

=10 Whenx=− =

√ x+1 √ x)1 −

=

+1 x =x1/2

Finding f(g(x)):

(

′

3−x)5 (+x 3−x

4

·4+x(−1)−(1)(3−x) (4+x) 2 4+x 3−x)4·−4−x−3+x =10 ( (4+x) 2 (4+x 3−x)4· −7 =10 (4+x) 2 (4+x) 3+10− 4 −7 and(f ◦g′)(10)=10 = − 10 · Then( f ◦ g)(x)=25

·

)

−70 · 134 ≈−42.8513asabove.

66 69. A = 1000(1 + i)3 a)

dy du = · dx du dx 1 = u−2/3 ( 6x) · − 3 = 2x u−2/3 − ·

dA = 1000(3(1+i)2) di = 3000(1+i)2 di represents the rate at which the amount of investb)dA ment is changing with respect to an annual interest rate i.

= 2x(1 Substituting1 3x2foru − 3x2)−2/3 − − =22(1 − 322)−2/3 · · = 4( 11)−2/3 − − ≈ − 0.8087

Finding f(g(x)): f g(x)=f(g(x))=f(1 3x2)=31 3x2,or ◦ −

√ −

(1 3x2)1/3 −

g)′ Then(f ◦ (x)=

1 (1 3x2)−2/3(6x)= − 3 − − 2x(1 3x2)−2/3and − (f g)′(2)= 4( 11)−2/3 ◦ − −

−70·134 6 ≈−42.8513 6

f ◦g(x)=f(g(x))=24

dy

dx

(4+ x ) 2 4 −7 2 · (4+ (x) dy 3+10 −7 ) 4· =10 2 (4 dx − 10) (4− 10 13)4(−7) =10−6 (−6)2

1/2

67. Using the Chain Rule: Let y = f(u) = 3 u√= u1/3. Then

dy

10,

− 12 = − .1 21· 2

−1 1 x−1/2 1 x−1/2(x1/2+1) (x1/21) − (2 −2 Then(f ◦g)′(x)= (x1/2 − 1)2 1 1 x−1/2 1 1x−1/2 − 2− 2 − 2 = (x1/2 − 1)2 2 x−1/2 == − (x1/2 − 1)2 √ −1 √ x( x 1)2 − g)′ 1 and(f ◦ (4)= − asabove.

2

Whenx=2,

3−x)

≈ − 0.8087asabove.

70.

a) dA

1 i 1000( 20(1+ )19() ) 4 = 5000(1+ )19i 4 =

di represents the rate at which the amount of investb)dA ment is changing with respect to an quarterly componded interest rate i.

71. D = 0.85A(c + 25), c = (140

w − y)72x

a) To find D as a function of c, we substitute 5 for A in the formula for D. D = 0.85A(c+25)

74

Chapter 2: Differentiation = 0.85(5)(c+25) = 4.25(c+25) = 4.25(c+25) = 4.25c+106.25 To find c as a function of w, we substitute 45 for y and 0.6 for x in the formula for c. w c = (140 − 45)72(0.6) =

95 ·

e)

73.

a) January 2009 corresponds to t = 52 C′(t) = 0.74+0.02376t C′(52) = 0.74+0.02376(52) =

w

dD b) c) dc =4.25 dc =2.199 dw d) First we find D ◦ c(w).

D ◦ c(w) = D(c(w)) = 4.25(2.199w) + 106.25 = 9.34575w + 106.25 Then we have = 9.34575 ≈9.346. dw e) dD representstherateofchangeofthedosagewith dw respect to the patient’s weight. For each additional kilogram of weight, the dosage is increased by about 9.35 mg. w − y)72x

a) To find D as a function of c, we substitute 5 for A in the formula for D. D = A(c+25)

To find c as a function of w, we substitute 45 for y and 0.6 for x in the formula for c. w (140 − 45)72(0.6) 95 ·

w

≈ 2.199w b) c)

dD

=5

dc dc =2.199 dw d) First we find D ◦ c(w).

D ◦ c(w) = D(c(w)) = 5(2.199w) + 125 = 10.995w + 125 Then we have dD dw

= 10.995.

− 1.0814πcos(105π)

= 5.3847ppmv/yr π 74. T′(t) = 0 .82πcos 12 ( 12[t+2] ) ′

T(8)

75.

y=((x2+4)8 dy dx

0.82π cos( π [8+2] ) 12 12 −0.1859 degrees/hr

= =

√+3 x)4

= 4((x2+48 )+3 = = 4((x2+48 )+3

√ x)3[8(x2+4)7(2x)+ √ x)3 ( 16x(x2+4)7+

3 √] 2 x) 3 √ 2x

76.

1

dy

2)−1/2 × ((5+ x +1)3+7 x secx 2 [3(5+ x x +1)2(54+1)+14 x secxsecxtanx ] 2 3(5+ x x√+1)2(54+1)+14 x secxtanx 2 2(5+x x +1)3+7 secx

dx = 77. Lety=

= 5(c+25) = 5c+125

=

− 1.0814πcos(104π)

−1.4218 ppmv/yr

C′(52.5) = 0.74+0.02376(52.5)

dD

c =

− 1.0814π cos(2πt)

b) July 2009 corresponds to t = 52.5

≈ 2.199w

72. D = A(c + 25), = (140 c

dD representstherateofchangeofthedosagewith dw respect to the patient’s weight. For each additional kilogram of weight, the dosage is increased by about 11 mg.

( sin x ))then sin( sin dy dx = = cos( sin( sin x)) ·cos( sin x) · cos x ( sin x )cos( sin( sin x)) cos x cos

78. Lety=

cos( (sec sin2))then x

dy = dx =

( ( x2))· − sinsecsin

79. Lety=

( tan( cot sec3))then x

dy = dx = 80. Lety= dy =− dx

−2

( ( x 2)·2 secsin x2)tansin

cos2x

2 ( 2) ( 2) ( ( 2)) cos xsecsin x tansin x sinsecsin x

( 3)) · − csc2(sec3)·3 sec2(cotsec x sec3xtan 3x x 2( x3) 2( ( x 3)) sec cotsex − 3 sec3xtan xcsc sec 3 ( √ ( sec ( x 2+1)))then csc cos 2 ( ( √ ( √ x 2 +1)))√( cotcossex ( ( √ x√+1))) csccossec

( ( − sinsec √2x 2 x 2+1

x 2+1))·

sec(

x 2+1) tan(

×

x 2+1) ×

75

Exercise Set 2.9 3π2 81.y = ( sin ( +3 ) )1/5isaconstant,whichmeansdy dx = 0.

–4

2 x

–2

4

0

82. The derivative of a constant is 0 83.

–50

sin(a+x) = sinacosx+cosasinx d d (sin(a+x)) x =

y–100

d d (sinacosx+cosasinx) x = −sin a sin x + cos a cos x = cosacosx − sin a sin x

–150

–200

= cos(a+x) 89.

84. cos(a+x) = cos a cos x d

(cos(a+x)) = dx

d

=

f ′ ( x )= = =

− sin a sin x

(cosacosx −sinasinx) dx −cos a sin x + sin a cos x

= sinacosx

− cos a sin x

= sin(a+x)

90.

D(x). Then we can write 85. Let Q(x) = N(x) Q(x) = N(x)

f ′ ( x )= = =

· [D(x)]−1

using the property of negative exponents. Now we use the product differentiation rule Q′(x) = N(x) = = =

· −1[D(x)]−2 · D′(x) + [D(x)]−1 · N′(x) N′(x) + D(x) −N(x)·D′(x) + N′(x)·D(x) [D(x)]2 [D(x)]2 N ′ ( ·D(x)−N(x)·D′(x) x) [D (x)]2

−N(x)·D′(x) [D(x)]2

86. One might use the following: Composition of functions could be expressed as a process through which one function is evaluated using another function that falls in the original function’s domain.

1 0 –10 –20

–30

88. ( − 1.475, 9.488)

x

2

x − 10 2 x x − 10) − ( x − 10)3/2 − 2x 40 ( x+10)3/2 4(

1. y=3x+5 dy dx = = d2y dx2

3 0

2. y = −4x + 7 dy = dx = d2y dx2

–3 –1

4 · √ x − 10 − x· 4 √12 x−10

Exercise Set 2.9

87. ( −2.145, −7.728) and (2.145, 7.728) –2

x · √ − −2x + √ 4−x2 2 4 x2 2 4−x2 √−x + √ 2 4− x 4−x2 2 √4 − x2 4 − x2

−4 0

3

3. y =− 3(2x+2)−1 dy = 3(2x+2)−2(2)=6(2x+2)−2= dx d2y 24 = −12(2x + 2)−3(2) = (2x +−2)3 dx2

6 (2x + 2)2

4. y = − (3 x − 4)−1 3 dy = (3x 4)−2(3)= − (3x 4)2 dx − 18 d2y = 3( 2(3x −4)−3(3))=(3x 4)3− − dx2 −

Chapter 2: Differentiation

76 12. g(x) = tan(2x)

5. y = (2 + x 1)1/3 dy dx

1 2 (2x+1)−2/3(2)= 3 3(2x+1)2/3 2 −2 −8 ( (2x+1) −5/3 (2))= 3 3 9(2x+1)/ 5 2

= =

d2y dx2 6. f (x)=(3 +2)−3 x f ′ x( )

g′(x) = sec2(2x)(2)=2sec2(2x) g′′(x) = 2(2sec(2x)sec(2x)tan(2x)(2)) = 8sec2(2x)tan(2x) 13. g(x) = sec(3x + 1) g′(x) = sec(3x+1)tan(3x+1)(3)

=

−3(3x + 2)−4(3) = −9(3x + 2)−4 108 f ′ (′ )x= 36(3x+2)−5(3)= (3x + 2)5 7. f (x)=(4

= 3sec(3x+1)tan(3x+1) g′′(x) = 3sec(3x+1)sec(3x+1)(3)+ tan(3x + 1)(3 sec(3x + 1) tan(3x + 1)(3) = 9sec3(3x+1)+9sec(3x+1)tan2(3x+1)

− 3x)−4

f′(x) = f′′ (x) =

−4(4 − 3x)−5(−3) = 12(4 − 3x)− 180 −60(4 − 3x)−6(−3) = (4 6 −3x)

= 9sec(3x+1)[sec2(3x+1)+tan2(3x+1)]

5

14. f(x) = 13x2 + 2x + 7 − csc x f′(x) = 26x+2

− (−csc x cot x) = 26x+2+cscxcotx f′′(x) = 26+cscx( −csc2x) + cot x(−csc x cot x)

8. y= √ x − 1 = (x − 1)1/2 dy

= = =

dx d2y = dx2 = =

1

=

(x −1)1/2−1·1 2 (x −1)−1/2 1 2 1 1) (x−1)−1/2−1·1 2 ·( −

15.

26 − csc3x − csc x cot2x

f(x) = sec(2x + 3) + 4x2 + 3x − 7 f′(x) = sec(2x+3)tan(2x+3)(2)+8x+3 = 2sec(2x+3)tan(2x+3)+8x+3 f′′(x) = 2sec(2x+3)sec2(2x+3)(2)+

−1(x−1)−3/2 4 −4(x 1 −

tan(2x + 3) sec(2x + 3) tan(2x + 3)(2) + 8 = 4sec3(2x+3)+2sec(2x+3)tan2(2x+3)+8

− 1)3/2 1

16. g(x) = mx + b

√4(x−1)3

g′(x) = m

9. y= √ x+1=(x+1)1/2

g′′(x) = 0

1 dy = = (x+1)−1/2 dx = 2 (x+1)−1/2 · 1 1 d2y = 2 2 == − 1(x+1) −3/2 ·1 dx −1(x+1)−3/2 4 1 − 4(x + 1)3/2

17. y=ax2+bx+c dy = a ·2x+b+0 dx d2y = 2ax+b 2 = 2a+0 dx = 2a √ 18. y=32x+4=(2x+4)1/3

1 − √4(x+1)3

1 dy = = (2x+4)−2/3(2) dx = 3 (2x+4)−2/3 = =2 3

10. f (x)=(3x+2)10

d2y = dx2

f′(x) = 10(3x+2)9(3)=30(3x+2)9 f′′(x) = 270(3x+2)8(3)=810(3x+2)8 11.

f (x)=(2x+9)16 f′(x) = 16(2x+9)15(2)=32(2x+9) f′′(x) = 480(2x+9)14(2)=960(2x+9)

− 2 · 2 (2 +4) −5/3 (2) x 3 3 −8(2x+4)−5/3 9 8 −

15 14

−

√8 93(2x+4)5

77

Exercise Set 2.9 4 26. y = 1π2 tanπt + π2 sec πt

19. y= ( +1)3=(x2+1)3/4 √4 x 2 3 dy = = (x2+1)−1/4(2x) dx = 4 = 3 x2 +1) −1/4 2[ x ( 2 1 3 ( x − ) (x2+1)−5/4(2x)+1·(x2+1)−/ dy dx 2 2 3x2 3 + 2(x2+1)1/4 − 4(x2+1)5/4 √3x2 44(x2+1)5

=−

dy = dx = =

14 ]

3 +√ 24x2+1

d2y dx2 =

1 4 2 2sec πt·π+ 2 sec πt tan πt·π π π 4 1 sec2πt+ sec πt tan πt π π (2 πt 1 sec πt sec πt tan ·π)+ π( sec πt sec2πt ·π+secπttanπttanπt·π) 4 2πsec2πt tan πt + 4 sec3πt + 4 sec πt tan2πt

27. y = x 4 dy

= 4x3 dx = 2 d2y 43x · dx2 d3y = 12x2

20. y = 13x3.2 + 12x1.2 − 5x−0.25 dy

= 41.6x2.2+14.4x0.2 −1.25x−1.25 dx 1.2 −0.8 +1.5625x 2.25 dx2 d2y= 91.52x +2.88x

3 = 12·2x dx d4y = 24x 4

21. f(x) = (4x + 3) cos x f′(x) = (4x+3)( =

−sin x) + 4 cos x

−(4x + 3) sin x + 4 cos x

dy

= 5x4 dx 3 d2y 2 = 20x dx 2 d3y 3 = 60x dx d4y 4 = 120x dy

s′(t) = cos(at) a = a cos(at) s′′(t) = a( =

−sin(at) a) −a2 sin(at)

23. s(t) = cos(at + b) s′(t) = −sin(at + b) a = −a sin(at + b) −a(cos(at −a2 cos(at++b) b)a) s′′(t) =

29. y = x 6 −x3+2x dy dx = 6x5 −3x2+2 d2y 2 = 30x4−6x dx d3y 3 = 120x3−6 dx d4y 4 = 360x2 dx d5y 5 = 720x dx

= 24. y = x5/2 + x3/2− x1/2 dx d2y dx2 t2

25. y=(+3) 7= = dy dx d2y 2= dx

=

1/2

24

28. y = x 5

22. s(t) = sin(at)

dy

=

dy

= =

2 +(3t+1)

5 x3/2+ x1/2 3 1 x−1/2 − 2 2 2 3 −1/2 1−3/2 /2 151 + 4x + x x 4 4 1/3

1 1 1 (3t2+1)−2/3(6t) · (t2+3)−1/2(2t)+ 7 2 3 t (t2+3)−1/2+2t(3t2+1)−2/3 7 t −1 2 2 −3/2 −1/2 (2t))+(2)(t+3) + 7 2 ( (t+3) (3t2+1)−5/3(6t))+(2)(3t2+1)−2/3 2t( − 2 3 2 −t2 + 7(t2+3)3/2 (t2+3)1/2 8t2 2 − (3t2+1)5/3 +

30. y = x 7 −8x2+2 dy dx = 7x6 −16x d2y 2 = 42x5−16 dx d3y 3 = 210x4 dx d4y 4 = 840x3 dx d5y 5 = 2520x2 dx d6y 6 = 5040x dx

Chapter 2: Differentiation

78 31.

y=(x2 5)10 − dy dx 2

d y dx 2

= 10( x2 − 5) 92 · x = 2 0 ( xx2 − 5 )9 9 x 2− 5)8·2 x +20( x 2− 5) 8 x2− 5)9 = 360 (x2x2 − 5)+20( 8 x2 x2− 5)] = 20(x2 − − 5)[18 +( 8 = 20(x2 5)(19 x 2 − 5)

= 20 9( x·

32. y=xk dy dx = = k xk−1 d2y k−2 dx2 = = kk ( − 1) x d3y − k−3 dx3 = kk ( − 1)( k − 2) x d4y − − k−4 dx4 ( 1)( k − 2)( k − 3) x kk d5y − k−5 dx5 ( 1)( k 2)( k 3)( k − 4) x kk 33. y=sec (2 x+3)

37. s(t) = t3 + t2 + 2t v(t) = s′(t)=3t2+2t+2 a(t) = s′′(t)=6t+2 38. s(t) = t4 + t2 + 3t v(t) = s′(t)=4t3+2t+3 a(t) = s′′(t)=12t2+2 39. w(t) = 0.000758t3 − 0.0596t2 − 1.82t + 8.15 The acceleration of a function that depends on time is the second derivative of the function with respect to time. w′(t) = 0.002274t2 − 0.1192t + 1.82 w′′(t) = 0.004548t

w′(t) = 0.0018t2 w′′(t) = 0.0036t

tan(2 x+3)(2))+ sec2 4 sec(2 x +3)[2 tan(2 x +3) (2 x +3)(2)]+ tan2 (2 x +3) sec(2x +3) tan(2 x +3)(2) 40 sec3(2 x +3) tan(2x +3)+

P′(t) = 100000(0.6+2t) = 200000

sec(2 x+3)

34. y = cot(3 x − 1) dy (3 x− 1) = −csc2 3 dx d2y = 3[2 (3 csc x − 1)( − csc(3x − 1) cot(3x − 1) (3))] − dx2 csc2 (3 x − 1) cot(3x− 1) d3y = 18 =

42. P (t)100000(1 + 0.4t + t2) P′(t) = 100000(0.4+2t) P′′(t) = 100000(2)

(3+2) t 4[ (3+2)(3) v()t= 10[ − cos t −sin (3+2)(3)] = − 30 sin(3+2) 12 t cos(3+2) t −

a t

= 200000 x 43. y = (x −1)1/2

y′

2 18 csc2(3x − 1)( −csc (3x − 1)(3))+ (3 x− 1)(3))] (3 18[2 csc2(3 x − 1) cot cot x −1) csc4 x− csc2 x− (3 1) 108 (3 1) cot2(3x −1) − 54 −

35. st()10 cos(3+2) t 4 − sin t

() = − 30[cos (3+2)(3)] 12[ − t −sin (3+2)(3)] t = − 90 cos(3+2) t 36− sin(3+2) t = 9[10 cos(3+2) t 4 − sin t (3+2)] = 9 s( t)

− 0.0968t + 1.61 − 0.068

P′′(t) = 100000(2)

8 sec(2x +3) tan3(2 x +3)

dx3 =

− 0.1192

41. P (t)100000(1 + 0.6t + t2)

[2sec(2 x +3) tan(2 x +3)(2)] tan(2x +3) 4 sec3(2 x +3)+4 sec(2 x +3) tan2(2 x +3)

d3y 2 dx3 = 4(3 sec (2 x+3)

=

− 1)(2.6) = 17.68 sec2(2.6t − 1) a(t) = 17.68[2sec(2.6t − 1) sec(2.6t − 1) tan(2.6t − 1)(2.6)] = 91.936sec2(2.6t − 1) tan(2.6t − 1)

40. w(t) = 0.0006t3 − 0.0484t2 − 1.61t + 7.60

dy = (2 x +3) tan(2x +3)(2) sec dx (2 = 2 sec x +3) tan(2x +3) dy 2 dx2 = 2 sec(2 x+3) sec (2 x+3)(2)+ =

36. s(t) = 6.8 tan(2.6t −1) v(t) = 6.8sec2(2.6t

y′′

= = = =

√x = = =

−1(1)−

1

√ x · 2x−1

x− 1 2(x − 1√ ) −x 2(x − 1) x−1 x −2 2(x − 1)3/2

32 2(x − 1)/(1)

1/2 − ( x − 2)2[ · 32 (x − 1) ] −1)3 4(x 2(x 1)3/2 −3(x−2)(x−1)1/2 − 4(x −1)3 (x 1)1/2[2(x −1)−3(x−2)] − (x −1)3 4− x (x 1)5/2 −

Exercise Set 2.9 4(x− 152− )/(1)

y′ ′ ′ = = = =

44. =

79 − (4 − x )4 [ · 52( x− 1) 16(x −1)5

3/2

(x −1)3/2[4(x−1)−10(4−x)] 16(x −1)5 4x −4−40+10x 16(x − 1)7/2 3x −18 16(x − 1)7/2

]

47. y = sinx a) b) c) d) e) f)

√

g)

√x−1x+1 ′

= = =

y

( √ x +1)(

√√x

1

√1 ) −√ ( 2

√x ( x+1)2

x−

1)( 2x√ 1 )

2

x √(√+1) x+1)2]−1 [x(

dy = cosx dx d2y = −sin x dx2 −cos sinx x d3y = sinx dx3 d4y = dx4 d8y = dx8 d 1 0 y = −sin x dx10 = d 8 3 7 y cosx dx837

48. y = cosx a) b) c)

dy dx

= −sin x

d2y dx2 = −cos x sinx d3y cosx cosx = dx3 d4y dx4 = d8y = dx8 d11y = sinx dx10 d523 =

d) 1 − 1[ √ √(x +1)] 2−2√[ x[2(x+1)(√)]+ √ y x e) 2 x x − 1[ √ √(x +1)] 2−2√1 ( √ x+1)2] f) ( 2x g) sinx 1 2−2 ) x+2+ = [ − √ √( x x +1)] y 2 x that have the form f(x) = Asin x + Bcos x 49. Functions d 3x52 √ where and areconstants,willsatisfytheconditionof A B −√ √ their second derivative being the negative of the original 3 1 x−3/2]+ x( x+1)2]−2[x−1/2+ y′ ′ ′ = function.

2 [ 1 √ 50. f(x) = x−1 √ √(x x +1)] 2−3√ 2[ [ x[2(x+1)( x+2 2√)]+ x (x+2)(1)−(x−1)(1) − √ x( √ x+1)2]−2√] 1 ′ f(x) = 2 x (x + 2)2 = 3 1 3 − √ √ x−3/2]+ x( x+1)2]−2[x−1/2+ =3(x+2)−2 = 2 (x + 2)2 √ 2−3√xx − √[ x( √ x+1)2] − 2 √12 ] −6 2[ √(x x +1)] [ (+1)] f ′ ()′ x = −6(x+2)−3 =(x+2)3 x 18 f′′′(x) = 18(x+2)−4= 45. f (x) = x x (x + 2)4 −1 72 − ′ (x−1)(1)−x(1) −72(x+2)−5 =(x+2)5 f(4)(x) = f(x) = (x −1)2 360 f(5)(x) = 360(x+2)−6= = −1 (x + 2)6 (x −1)2 −(x − 1)−2 = f(x) = x+3 51. x ′′

=

f′′(x) = −(−2(x − 1)−3) 2 = (x −1)3 x2 )−1

46. f (x)=(1+

f′(x) = −1(1 + x2)−2(2x) −2x(1 + x2)−2 = f ′′(x) = =

−2x[−2(1 + x2)−3(2x)] + (−2)(1 + x2)−2 4 −8x2 + (1+x2)3 (1+x2)2

−2

(x − 2)(1) − (x+3)(1) (x 2)2 − − 5 = 5(x 2)−2 − − (x 2)2 − f′′ 10 (x) 10(x − 2)−3= (x−2)3 − 30 2)−4= f′′′(x) = −30(x − (x 2)4 − 120 f(4)(x) = 120(x 2)−5= − (x 2)5 − − 600 f(5)(x) = − 600(x −2)−6= (x 2)6 − ′ f(x)

= = =

Chapter 2: Differentiation

80 52.

56.

40

20 15

20 y10

–4

–2

2 x

5

4 –1.5

–20

–1

–0.5

0.5 x

1

1.5

0.5 x

1

1.5

–5

–40

–10

57.

53.

20 15

10 –3

–2

–1

x

1

2

y10

3

5

–10

–1.5

–1

–0.5 –5

–20 –10

54.

120

Chapter Review Exercises

100 80 60 40 20 –3

–2

–1

0 –20 –40

1

x

2

3

1.

–60

55.

40

x f(x) -8 -11 -7.5 -10.5 -7.1 -10.1 -7.01 -10.01 -7.001 -10.001 a) -7.0001 -10.0001 -6 -9 -9.5 -6.5 -9.9 -6.9 -9.99 -6.99 -9.999 -6.999 -6.9999 -9.9999 b) lim f(x)=

20

x→ − 7 −

lim f(x)=

–3

–2

–1

0

–20

1

2 x 3

4

5

x→ − 7 +

lim f(x)=

x→ − 7

−10 −10 −10

2. From the graph below, we can see that lim f (x) = −10 7 x →−

–40 5 –10 –8

–6

–4

–2

2 0

–5

–10

4 x 6

8

10

81

Chapter Review Exercises (x+7)(x−3) (x+7)

3. f x()=

=( x −

3),solimf(x)= →−7 x

−10

8

=4 4. lim x→ − 2 x − 5. lim(2 x4 − x32+ +4)=2(1) x

20. dy = −2x + 8 d x

−2x+8 = 0 −2x = x=4

−3(1)+(1)+4=4

x→ 1

6. ( 6)( x +11) 7. lim x − =17 x→ 6 ( x− 6) 8.

2 lim √ x2 +9= √ 4+9= 9. x→4

= −(4)2 8(4) − 11 −16 ++32 − 11

y √

=

25=5

The function is not continuous at x = − 2

= line is horizontal at (4, 5) The slope of the tangent

10x −49 = 1

x→ 1

10x = 50 x=5

11.g(1)= − 4 ) is continuoous at x = 1 12. Yes, (gx lim ( gx ) does not exist

y = 5(5)2

x→ − 2

14.g(2)= −

5

dy = 10x −49 21. dx

The function is continuous 10. limgx( )= − 4

13.

−8

=

−2

)isnotcontinuousatx= 15. No, (gx

−2

16.f (2) = 8, ( f −1) = 2), average rate of change is 8 −2

2 −(−1)

−3(x+h)+5−(−3x+5) h

= =

h

−3x−3h+5+3x−5 h −3h h −3

= =

−108 = The slope of the tangent line is 1 at (5, −108) dy 22. dx=4 · 5x4 = 20x4

6= =2 3

17. f(x+h) −f(x)

− 49(5) + 12 125 − 245 + 12

23. y = 3x1/3 dy = x 21/ 3 3 dx=3 ·1x−2/3 24. y = −8x−8 dy −64 − x−9 = x9 25. dx = 64 y = 21x4/3 26. dy =28x 1/3 3 dx=21 ·4x1/3 27. y = sec 5x dy dx = 5 sec 5x tan 5x y = x cot(x2)

18. f(x+h) −f(x) h

= = =

2(x+h)2−3−(2x2−3) h

dy = dx =

2x2 + 4xh + 2h2 −3−2x2+3

4xh + 2h2 h = 4x+2h

h

dy 19. dx=2x+3 The slope at ( −1, −2) is

m = =

dy |x= − 1 dx = 2( 1)+3

28. y=2.3 √ 0.4x+5.3+0.01sin(0.17x dy

= dx =

f′

Now we use the slope-point equation 30. y=x2+1 dy dx = 2x

− 0.31)

1 (0.4)+0.01cos(0.17x−0.31)(0.17) 2.3 √ 2 0.4x+5.3 0.46 √ 0.4x+5.3 +0.0017cos(0.17x−0.31)

+ x4 29. f(x) = 1x6 8 6

1−

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

x − (2−csc2(x2)(2x))+cot(x2) x2 csc2(x2) + cot(x2)

(x)

−5x 1 ·6x5+84x3−5 · 6 = x5+32x3 5 − =

Chapter 2: Differentiation

82 +8 31. y=x28

39. s(t) = t + t4

−x

(8

dy

= = dx =

16x

a) v(t) = s′(t) = 1 + 4t3

−x)(2x)−(x2+8)(−1) −x)2 (8

−2x2+x2+8 (8 − x ) 2 −x2+16x+8 (8 − x ) 2

b) a(t)=v′(t)=12t2 c) v(2) = 1 + 4(2)3 = 1 + 4(8) = 33 a(2) = 12(2)2 = 12(4) = 48 40.

= =

dx

x ·sec2x−tanx·(1) x2 x sec2x −tanx x2

c) v(2.5) = −1.62521 a(2.5) = 0.17992 41.

33. y = (sin2x + x)1/2 dy dx = =

a) Amplitude = 135−1 = 67, period = 1 = 2π 2 2 b → b = 4π, mid-line= 67 + 1 = 68 and the capsule begins at the top means we are using a cosine model. Therefore h(t) = 67 cos(4πt) + 68 b) Five minutes after reaching the bottom corrspond to t=20min=1 3 hours since it is five minutes after half a period.

1 sin2x x−1/2 + ) (2 sinxcosx +1) 2( 2 si√nxcosx + 1 2 sin2x + x

h(

34. fx( ) = tan2xcosx ( ) f′x()

= =

36. fx( )= (4 x2 +3) x f′x()

dx d2

=

)

b) P(20) = 10000 + 50(400) = 30000 people c) P′(20) = 100(20) = 2000 people/year 43. (f ◦g)(x) = (1−2x)2+5

3x2 + 2 x−2

= 6x 4x − 3 − y dx2 = 6+12 x − 4 d3 y = − 48x − 5 dx3 d4 = 240 − 6 = 240 x y x6 5+ dx x2 38. y = 142x7− 10x3 +13 +28 x− cosx +28 4 dy 1 = d5 y 6 x6 − 30x2 +26 x −sinx dx = dx5 5 d2y x− 60x +26 −cosx dx2 4 d3y = 5 x− 60+ sinx dx3 3 d4y = 20x + cosx dx4

4π

a) P′(t) = 100t

37. y=x3 − x2− 1 dy

− 268πsin(

=

= 729.1473ft/hr 42.

3 3/4 = = x2[(4 x +3) −1/4 (4)]+2(4x x +3) 4 3 x2 +2(4x x +3) 3/4 (4 x +3) 1/4

1 67cos( 4π ) +68

=

1 h′ ( ) 3

( )(1 − −sinx ( )) −csc2x−cosx ( )(1+ sinx ) −csc2x−cosx 3/4

)

c) h′(t) = −67 sin(4πt)(4π) = −268π sin(4πt)

35. fx( ) = cotx ( − cos x) = =

1

= 34.5ft

2 tanxcosx ( ) sec2xcosx ( )((x −sinx )+ cosx (1)) 2 sec2( x cos x ) tan( x cos x)( cos xx sin x)

f ′ x()

cos(π15t) −0.002π2 225

b) a(t) = 0.072t

32. y=tanx x dy

sin(π15t) a) v(t) = 0.036t2 −1.85−0.002π15

=

1 −4x+4x2+5

=

−4x+6 4 x 2

(g ◦f)(x) = 1−2(x2+5) =

1 − 2x2 − 10 −2x2 − 9

= 44. dy

= = dx = =

(1+x3

)+ √ 1+3x(1)] − √x 1+3(3x x2) (1+x3)2 3x(1 + x3) + 2(1 + 3x)(1 + x3) 6x3(1 √ − +3) x x4 2 1+3x(1+x3)2 √3 )[(x)( 21+3

x

3x+3x4+2+2x3+6x+6x4 6 √ 2 1+3x(1+x3)2 9x4 4x3 + 9x +2 −√ − 2 1+3x(1+x3)2

45. Limitapproaches − 1 4

− x 3 − 18

83

Chapter 2 Test 1

46. Limit approaches6

f(x)=4 8. lim →−1 x

100

47.

9. lim f(x) = 1

80

x→ 1

60

10. lim f(x) = 1 x→ 2

y 40

11. lim f(x) = 1 x→ 3

20 –3

0 –1 –20 –40

–2

1

x

2

12. Function is continuous

3

13. Function is not continuous at x = 3 14. lim f(x) does not exist

–60

x→ 3

–80

15. f(3) = 1

–100

16. No, the function is not continuous at x = 3

Chapter 2 Test

1.

x 5 5.7 5.9 5.99 5.999 5.9999

a)

7 6.5 6.1 6.01 6.001 6.0001

17. lim f(x) = 3 x→ 4

18. f(4) = 3

f(x) 11 11.7 11.9 11.99 11.999 11.9999 13 12.5 12.1 12.01 12.001 12.0001

19. Yes, the function is continupus at x = 4 20. lim (3x4 21.

x→ 1

−2x2+5=3(1)−2(1)+5=6

x −2 lim →2+x(x− 2)(x+2) 1 1 = 8 lim x →2+x(x+2)

7 22. lim doesnotexist x →0x 23. f(x) = 2x2 + 3x −9,f(x+h)−f(x) h

6.00001 12.00001 x→ 6 −

lim f(x) = 12

x→ 6 +

lim f(x) = 12

x→ 6

2. From the graph below, we can see that lim f(x) = 12 x→ 6

15

2(x + h)2 + 3(x + h)

−9−(2x2+3x − 9) h 2x2 + 4xh + 2h2 + 3x + 3h − 9 − 2 x 2 − 3x+9 h 4xh + 2h2 + 3h h = 4x+3+2h = = =

b) lim f(x)=12

24. First find dy dx

10

–10 –8

–6

–4

dy

= 1+4(−1x−2) 4

dx

= 1− x 2 Next find the slope at (4, 5)

5

–2 0

=

x

2

4 x 6

8

m = = =

10

3. f (x) = (x−6)(x+6) =(x+6),solimf(x)=12 (x −6) x→6 4. lim f(x) does not exist x→ − 5

5.

lim f(x)=0

x→ − 4

Finally use the point-slope equation y

− 5=

6. lim f(x) does not exist

y =

7.

y =

x→ − 3

lim f(x)=2

x→ − 2

4 1 1 − (4)2 1 − 4 3 4 3 3 (x−4) 4 3 x −3+5 4x + 2

Chapter 2: Differentiation

84 25.

dy dx = 3x 2 −6x. The tangent line is horizontal when dydx = 0 so

3x2

36. ′ f(x)

−6x = 0 0 −2) =

3x(x either 3x = x = or −2 = = x x

0 0 0 2

2 2 (c osx−x)(2tanxsecx) − 2 2 (cosx − x) (tan2 x)(−2cos x sin x − 1) (cos2 x− x ) 2

=

2

= =

tan −2xtanxsec2x+2sin2xtanx+tan2x (cos2 x− x ) 2 tan −2xsec2x+2sin2x+tanx)

x

(cos2 x− x ) 2

x(2 37. f(x) = sin(cos(x2))

Whenx=0,y=03 Whenx=2,y=23 −4) point (2, dy 26. dx = 10 · 2√=x1 √ 5x

− 3(0)2 = 0, we have the point (0, 0) − 3(2)2 = 8 − 12 = −4, we have the

f′(x) = cos(cos(x2)) = 38.

27. y = −10x−1 dy dx= −10 · −1x−2 = 10x2 28. dy dx=45 x1/4 29. y = −0.5x2+0.61x+90 30. dx= dy −0.5(2x) + 0.61 = −x + 0.61 31. dy dx = (sec 22x)(2) = 2 sec2 2x

′ f(x)

= =

2x+ ( √

39. y=x4 − 3 x 2

= = =

dy 32. dx= −5(x5 − 4x3 + x)−6(5x4 − 12x2 + 1) +1) = −5(5x4−12x2 (x5

40.

√

x2

x2+5 2x2 + 5 √x2 + 5

+√

x2+5

x2+5

√ sin√(x)

1 dy √ − 34. dx= − sin( x )· 2x√ = 2x

35. f(x) = tan 2x sec 3x f′(x) = tan2x[(sec3xtan3x)(3)]+[(sec22x)(2)]sec3x = sec3x(3tan2xtan3x+2sec22x)

4x3 − 6 x

b) a(t)

=

= =

+ √ x2+5

√ 2 2 x ·√x+5 √ x2+5 + √ x2 +5 x2+5

√ cos(2xx+2)

√ x+2

a) v(t) = 2t cost2 2t−sin 2t

√ 33. f(x)=xx2+5

= = =

)

2 = 12x −62 dx d3y = 24x dx3

−4x3 +x)6

′ f(x)

+2

dx d2y =

(5−x)(1)−x(−1) (5 − x ) 2 5 −x+x (5 − x ) 2 5 (5 − x ) 2

2x =x·√ 2x2+5

x x+2

√ +(2)x+2

dy

−x

(x)

1 ( · 2x · √ x+2 )2

√ 2x+cos(2xx+2)

f(x)= x 5

f′

· −sin(x2) · 2x

−2x sin(x2) cos(cos(x2))

t2 [2t(−sin2t)(2)+2cos2t−(cos2t)(2)]

t4 [(2t cos 2t− sin 2t)(2t)] t4 −4t3sin2t−4t2cos2t−2tsin2t t4 2(1 −2t2)sin2t−4tcos2t t4 (7π

c ) s (7 π ) = s i n7 π 3 ) = 6

6

√ 37π3 (7π

(7π) 49π2 v(7π)=3cos3−sin3) 6 36 7π

7

a(π)= 6

√

=6(33−7π)

(7π)14π sin3−3cos3) 291−2(7π)2 (7π)4 6 6

√ √ 3π2) −12(− 18 3+42π+49 343π 3

49π2 (7π

=

41. M = −0.001t3 + 0.1t2 a) M′(t) = −0.003t2 + 0.2t b) M(10) = 0.001(10)3 + 0.1(10)2 =

−1 + 10 = 9 words c) M′(10) = −0.003(10)2 + 0.2(10) = −0.3 + 2 = 1.7 words/minutes

−

85

Extended Life Science Connection 42.

• Page 109: −3 ′ 1. f′ x( ) = − 3x2, f′(− 2)=4,f(−1/2)=−12 2.y = −34x−3, y=−12x−12

( f ◦g)(x) = (2x3)2 − (2x3) = 4x6 − 2 x 3 (g ◦f)(x) = 2(x2−x)3 = 2(x6 − 3x5 + 3x4 − x3)

2.f′( −5)=−96, f′(0)=−11, f′(7)=24, f ′ (12) = −11, f′(15)=−56

− 3x)2/3(1 + 3x)1/3 dy

44. lim x→3

1.f′(20)=60, f′(37)=26, f′(50)=0, f ′ (90) = −80

2x6 − 6x5 + 6x4 − 2x3

= 43. y=(1

• 112:

(x

limt

(1−3x)2/3 (

1

(1+3x)−2/3(3) ) + 3 ( 2(1−3x)−1/3(−3) ) (1+3x)1/3 (3− 1/3 3 x)2/3 1+3x ) = − 2( 1 −3x −3)(x2+3x+9) =

3.f′( −2)=−36, f′(0)=0, f′(2)=12, f ′ (4) = 0, f′(6.3) = −43.47 f′(−1.3) = 0.4079, 4.f′( −2) does not exist, −05) = 1.8074, f′(0) = 2, f′(1) = 1.1547, f′( . f ′ (2) does not exist √ 5.x =3isnotinthedomainofthefunctionx4 − x • Page 118: 1.f′(24) = 152, f′(138) = −76, f′(150) = −100, f′ −180 (190) = 2.f′(5)= −96, f′(0)=−11, f′(7)=24, − (12) = −11, f′(15)=−56 f′ −36, f′(0)=0, f′(2)=12, − 3.f′(2)= (4) = 0, f′(6.3) = −43.47 f′ (0)=1.61, f′(12)=0.7076, f′(24)=0.3236, 4.f′ (6 3) = 0.458 f′ .

=lim(x2+3x+9)=27

x− 3

x→3

45. → 0 tant+sint sint ) ( tant t) =(limt→0 = ) +t t ( tant sint lim +lim =1+1=2 t→0 t t t →0 46. (1.0836,25.1029) and (2.9503,8.6247)

• Page 137:

100

1. Part “c” is the correct derivative 2. Left to the student 3. Left to the student

80 60 y

Extended Life Science Connection

40

1. f ()= x

20

√ 4+x f′

0

1

2

x

3

4

5

f′

–20

47. Limit approaches 0

2.

1. 3 2. -0.2 • Page 87: 1. 53 2. 2.8284 3. Limit does not exist • Page 95: 1.fx ( +h)=64, f(x)+h=38 2.fx ( +h)=218.78, f(x)+h=208.1

=

(0)

= =

1 √ 4+x 2 √ 2 1 4+0 4 1 √2

= 0.25

Technology Connection • Page 73:

(x)

a) b)

1 2

1

1

1 16

1

1

1

1 16

h 4 8 F+(h) 0.242641 0.24621 0.24808 0.24903 h 2 4 8 F−(h) 0.258342 0.25403 0.25198 0.25098

3.

a) Forward difference:: 350−264 (2/7) =301Backwarddifference: 264−167 (2/7) =339.5 b) Forward difference:: 306−264 (1/7) = 294 Backward dif264−219 ference: (1/7) =315Itseemsthattheanswers from part a) are more accurate since they use points closer to the point under consideration (March 19) and therefore represent the slope of the tangent line more closely than points away from the point corresponding to March 19.

Chapter 2: Differentiation

86 4.

Fc = = = = =

1

[ 2 (F+(h)+F−(h)] 1 f(h) − f(0) f(0) − f(h) − ) + 2 h 1 ((f(h)−f(0)+f)(0)−f(−h)) 21 2 f(h)−f(−h) f(h) −f(−h) 2h

5. Since Fc(h) cover more of the function domain containing x = 0 than does F+(h) or F −(h), then it makes sense that Fc(h) be closer to f′(0) than either of them.

6. Using the results from question 2 parts a) and b) and the definition of Fc we get 1

1

1

h 2 4 8 Fc(h) 0.25049 0.25012 0.25003 0.250005

1 16

7. The answers in Exercise 6 are closer to f′(0) = 0.25 than 1 16 using the those from Exercise 2. The value of h = central difference quotient gives the closest value to 0.25. This is expected since the central difference quotient is the most accurate out of the three difference 16 quotients and 1value of h to be considered. is the smallest 8.

a)F(17)=f c /

(17)−(−17) / 2(1/7) f

/

b)F c(2/7)=

f (27)−(−27) / 2(2/7) f

/

= =

306−219 =304.5 =320.25 (2/7) 350−167 (4/7)

c) The answer from part a) gives the more accurate rate of change.

9. Fc(4/7)= Fc(4/7)= 10.

f (47)−(−47) / 2(4/7) f

/

f (27)−(−27) / 2(2/7) f

/

7739−7053 (8/7) =600.25 = = 7628−7296 (4/7) =581

a) f(x) = x3 + x f′(x) = 3x2 + 1, f′(0) = 1 Close to answer given by nDeriv, which is 1.000001 b) f(x) = 1000x3 + x f′(x) = 3000x2, f′(0) = 1 Close to answer given by nDeriv, which is 1.001 c) f(x)=1000000x3+x f′(x) = 3000000x2 + 1, f′(0) = 1 Does not match the answer from nDeriv, which is 2. d) f(x)=1000000000x3+x f′(x) = 3000000000x2 + 1, f′(0) = 1 Does not match the answer from nDeriv, which is 1001. e) f(x) = | x | f′(x) does not exist at x = 0 Does not match the answer given by nDeriv, which is 0.

Chapter 3

Application of Differentiation Exercise Set 3.1

1. f(x) = x 2 − 4x + 5. First, find the critical points (values of x at which the derivative is zero or undefined). f′(x) = 2x − 4 f′(x) exists for all real numbers. We solve f′(x) = 0:

2x − 4 = 2x = x =

2x − 6 = x =

The only critical point is at x = 3. Use it to divide the real number line into two intervals, A: (−∞,3) and B: (3,∞).

A: Test 0, f′(0) = 2 B: ·0−6=−6<0 ·4−6=2>0 Test 4, f′(4) = 2 Since f(x) is decreasing on ( −∞,3) and increasing on ∞), there is a relative minimum at x = 3.

(3, f(3) = 32 −6·3−3=−12

0 4 2

Thus, there is a relative minimum at (3, −12). We sketch

The only critical point is at x = 2. We use 2 to divide the real number line into two intervals, A: (−∞, 2) and B: (2, ∞): ✲

the graph.

5 –6

✛

︸A︸

–4

–2

2

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test 0, f′(0) = 2 ·0−4=−4<0 ·3−4=2>0 B: Test 3, f′(3) = 2 We see that f(x) is decreasing on ( −∞, 2) and increasing ∞), and the change from decreasing to increasing on (2, indicates that a relative minimum occurs at x = 2. We f(2) = 22 substitute into− the 4 · 2original + 5 = 1 equation to find f(2): Thus, there is a relative minimum at (2,1). We use the information obtained to sketch the graph.

x4

6

0

︸B︸

︷ 2︷

0 3

–5 y –10

–15

3. f(x)=5+x

−x2

First, find the critical points, f′(x) = 1 − 2 x f′(x) exists for all real numbers. We solve f′(x) = 0:

10

1 −2x = 0

8

1 = 1 2 =

6 y 4

2x x

2 –6

–4

–2

0 –2 –4

2

x4

6

–6 –8

1 The only critical point is at x = . We u 2 the (realn)umberlineintotwointervals,A: 1 ,∞: B:

1

tod)ivide 2 1 and −∞,

(se

2

–10

✛ 2. f (x)=x2 f ′ x( )=2x

−6x−3

− 6 f ′ x( ) exists for all real numbers. Solve f ′ x( )=0:

︸A︸

︷︷

︸B︸

✲

2 We use a test value in each interval to determine the sign of the derivative in each interval.

88

Chapter 3: Application of Differentiation A: Test 0, f′(0) = 1 B: Test 1, f′(1) = 1

−2·0=1>0

5

−2·1=−1

(<01 anddecreasing We 2) (seetha)tf(x)isincreasingon−∞, 1 1 on x =.We 2(,∞,sothereisarelativemaximumat 2 1) findf ():21 1 21 =5+ f − ( 12 ) 2 = 2 Thus,thereisarelativemaximumat (121 , 24 informationobtainedtosketchthegraph.

) .Weusethe

–4

–4

–2

2

–2

2

x4

6

0

6

–5 y –10

–15

5. f(x) = 1 + 6x + 3x2 f′(x) = 6 + 6x f′(x) exists for all real numbers. We solve f′(x) = 0:

–5

6+6x = 0 6x = − 6 x = − 1

y –10

The only critical point is at x = −1. We use −1 to divide −∞,−1) and the real number line into two intervals, A: ( B:( − 1 , ∞ ) :

–15

4. f(x) = 2

x4

0

First, find the critical points.

5 –6

–6

− 3x − 2x2

f ′(x) = − 3 − 4 x

✛

︸A︸

f′(x) exists for all real numbers. Solve f′(x) = 0: −3−4x = 0 −3 = 4

x

3 −.Useittodividethereal 4 3) n ( − ∞ , − 4 andB: (umberl)ineintotwointervals,A: −3,∞.

4 A: Test − 1, f′(−1) = −3 − 4(−1) = 1 > 0 f′(0)= − 3 − B:Test0, (4·0=−3)<0 3 anddecreasingon − ∞ , − (Sincef(x))isincreasingon 3 −3 4 ( , ∞ ,)thereisa 4 (relati)vem(aximu)matx=−.4 3 =2−3−3− 3 2 25 2− = f− 4 (8325) Thus,thereisarelativemaximumatx= .We − , 48 The only critical point is

︸B︸

︷︷

✲

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test − 2, f′(−2) = 6 + 6(−2) = −6 < 0 f′(0)=6+6 B:Test0, ·0=6>0 We see that f(x) is decreasing on ( −∞, −1) and increasing −1, ∞), so there is a relative minimum at x = −1. We on( −1): findff(( −1) = 1 + 6(−1) + 3(−1)2 = −2

Thus, there is a relative minimum at (−1, −2). We use the information obtained to sketch the graph.

10 8 6 y 4 2 –6

–4

–2

2

x4

6

–2 –4

6. f (x)=0.5x2 f ′()=x x

−2x−11

− 2

f ′( x) exists for all real numbers. Solve f ′( x)=0.

Exercise Set 3.1

89

x − 2 = x =

A: Test −1, f′(−1)=3(−1)2−2(−1)−1= −1=4>0 3+2 B: Test 0, f′(0) = 3(0)2 C: −2(0)−1=−1<0 −2(2)−1=12−4−1=

0 2

Test 2, f′(2) = 3(2)2 7>0 The only critical point is at x = 2. Use it to divide the real number line into two intervals, A: (−∞,2) and B: (2,∞).

A: Test 0, f′(0) = 0 B: −2=−2<0 −2=1>0 Test 3, f′(3) = 3 Since f(x) is decreasing on ( −∞,2) and increasing on ∞), there is a relative minimum at x = 2.

(2, f(2) = 0.5(2)2 −2·2−11=−13 Thus, there is a relative minimum at (2, −13). We sketch the graph.

20

1) We ( − ∞ , − 3 ,decreasing (seethat)f(x)isincreasingon on −1,1,andincreasingagainon(1,∞),sothereisa 3 1 relative maximum andarelativeminimumat (atx)=− x=1.

3 (Wefin)df(−3):3 f−1 1 −1+2 ( −1 = − ) 2− ( − 3) = 1 = 27 − 1 1+ +2 − 9 59 27 Then we find f(1):

y10

f(1) = 13 –6

–4

–2 0

2

4 x 6

8

=

10

−12−1+2 1− 1 − 1 + 2

= 1 Thereisarelativemaximumat

–10

159) ( − , 327 ,andthereis arelativeminimumat(1,1).Weusetheinformationobtained to sketch the graph.

–20

7. f(x) = x3 − x 2 − x + 2 First, find the critical points.

10 8

f′(x) = 3x2 − 2 x − 1 f′(x) exists for all real numbers. We solve f′(x) = 0:

6 y 4 2 –4

3x2

–2

−2x−1 =0 0 −1) = 3x+1 = 0

(3x + 1)(x

0 –2 –4

2 x

4

–6 –8

x3x 0r − 1 x

= = = =

−1 1 −3 0 1

x

The critical points are at x =

–10

1 8. f(x) = x3 + x2 2 −2x+5 f′(x) = 3x2 + x − 2 f′(x) exists for all real numbers. Solve f′(x) = 0.

−1 and x = 1. We use 3 berlineintothreeintervals,

the (mtodivid)ethe(realnum) 1 −∞,−1 ,B: −,1,andC:(1,∞).

A: 3 3 ✛

︸A︸

︷︷ −1

︸B︸ ︷︷ 1

(3x

3x2 + x − 2 = 0 −2)(x+1) = 0 x = Or = x

︸ C ︸✲

We use a test value in each interval to determine the sign of the derivative in each interval.

Thecriticalpointsareatx= them to divide th A:(

− ∞ , − 1),B:

2 3 −1

2 − 1and x = 3 . Use

(erealn)umberlin(einto)threeintervals, 2 2 ,andC: ,∞ . −1,

90

Chapter 3: Application of Differentiation −∞, −1),so decreasing −1,1), and increasing (1,∞), there isat a −1again and aon relative minimum A: Test − 2, f′(−2) = 3(−2)2 + (−2) − 2 = 8 > 0 B:Test0, f′(0)=3 ·02+0−2=−2<0 ·12+1−2=2>0 C:Test1,

f′(1)=3

(Sincef)(x)isincreasingon(−∞,−1),decreasingon 2 2 ) − 1, ,andincreasingagainon ( ,∞,thereisarelativemaximumatx= −1andarelativeminimumat 2 x = 3.

1

f ( −1)=(−1)3+

2

13

(− 1)−2(−1)+5=2 2 113 2) 1 22 2 = (2) + f( (3 ) − 2( ) +5= 2713 3 3 3

We see that f(x) is increasing on ( on( relative maximum at x = −1): x = 1. find f( f ( We −1)=(−1)3−3(−1)+6=−1+3+6=8 Then we find f(1): f(1) = 13 − 3 · 1 + 6 = 1 − 3 + 6 = 4 There is a relative maximum at ( −1, 8), and there is a relative minimum at (1, 4). We use the information obtained to sketch the graph.

15

,andthereisa There is a relative maximum at ( − 1 , 2) 2 113 relativeminimumat , ( .Wesketchthegraph.

3 27 )

10 y 5

20 –4

15

–2

2 x –5

y10

–10

5

–4

–2

4

2 x

4

–5

10. f(x) = x3 − 3 x 2 f′(x) = 3x2 − 6 x f′(x) exists for all real numbers. Solve f′(x) = 0.

–10

= x3 9. f(x) −3x+6

3x2

First, find the critical points. f′(x) = 3x2 − 3

x2

f′(x) exists for all real numbers. We solve f′(x) = 0: 3x2 − 3 = 0 0 −−11 ) == 0 x2 x + 1 = 0 (x + 1)(x x = Or

−1

x−1 = =0 x = 1 The critical points are at x = −1 and x = 1. We use them to divide the real number line into three intervals, A:( −∞,−1), B: ( 1,1), − and C: (1,∞). ✛ ︸A︸

︸B︸ ︸ C ︸✲ ︷︷ ︷︷ −1 We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test −2, f′( 2)=3( − − − 2)2 3=12−3=9>0 B:Test0,f′(0)=302 · − 3=0 − 3= −3<0 C:Test2,f′(2)=322 · − 3=12 − 3=9>0

x(x

−6x = 0 −2x = 0 0 −2) = 0 x = 2 Or = x

The critical points are at x = 0 and x = 2. Use them to divide the real number line into three intervals, A:−∞, ( 0), B:(0,2),andC:(2, ). ∞ A: Test −1,f′( 1)=3( − 1)2 6(−1)=9>0 − − f′(1)=312 61= 3<0 · − · − B:Test1,

f′(3)=332 63=9>0 C:Test3, · · − Since f(x) is increasing on ( − ∞ ,0), decreasing on (0,2), and increasing again on (2, ),∞there is a relative maximum at x = 0 and a relative minimum at x = 2. f(0) = 03 −3·02=0 f(2) = 23 −3·22= − 4

There is a relative maximum at (0, 0), and there is a relative minimum at (2, 4). − We sketch the graph.

Exercise Set 3.1

91 10

−3x2 = 0 x2 = x =

5 –4

2 x

–2

4

0

The only critical point is at x = 0. We use 0 to divide the real number line into two intervals, A: (−∞, 0) and B: (0, ∞): ✲

–5 y –10

✛

︸A︸

–15

11.

f(x) = 2x3 First, find the critical points. f′(x) = 6x2 f′(x) exists for all real numbers. We solve f′(x) = 0:

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test -1, f′( −1) = −3(−1)2 = −3 < 0 B: Test 1, f′(1) = −3(1)2 = −3 < 0

0 0

increasing or from increasing to decreasing. Therefore the function does not has a relative extrema.

40

The only critical point is at x = 0. We use 0 to divide the real number line into two intervals, A: (−∞, 0) and B: (0, ∞): ✲

y 20

–4

︸A︸

︷ 0︷

We see that f(x) is increasing on ( −∞,0) and increasing on (0, ∞), therefore there is no change from decreasing to increasing or from increasing to decreasing. Therefore the function does not has a relative extrema.

0

–2

︸B︸

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test -1, f′( −1) = 6(−1)2 = 6 > 0 B: Test 1, f′(1) = 6(1)2 = 6 > 0

0

4

–40

13. f(x) = 0.02x2 + 1.3x + 2.31 First, find the critical points. f′(x) = 0.04x + 1.3 f′(x) exists for all real numbers. We solve f′(x) = 0: 0.04x+1.3 = 0 0.04x = −1.3 = −1.3 x 0.04 = −32.5 x

y 20

–2

2 x

–20

40

–4

︸B︸

︷ 0︷

We see that f(x) is decreasing on ( −∞, 0) and decreasing ∞), therefore there is no change from decreasing to on (0,

6x2 = 0 x2 = x =

✛

0 0

2 x

4

–20 –40

12. f(x) = 1 − x 3 First, find the critical points. f ′(x) = − 3 x 2 f′(x) exists for all real numbers. We solve f ′ x( )=0:

The only critical point is at x =

−32.5. We use −32.5

to divide the realB:number line into two intervals, A: ( −∞,−32.5) and ( 32.5,∞): − ︸B︸ ✲ ︷ 32.5 ︷ − We use a test value in each interval to determine the sign of the derivative in each interval. f′(0)=0.04 ·0+1.3=1.3>0 A:Test0, ✛

︸A︸

B: Test -100, f′( 100) − = 0.04

· (−100) + 1.3 = −2.7 < 0 We see that f(x) is decreasing on ( −∞,−32.5) and in− ∞ creasing on ( 32.5, ), and the change from decreasing

92

Chapter 3: Application of Differentiation to increasing indicates that a relative minimum occurs at x = −32.5. We substitute into the original equation to find f( −32.5): −32.5) = 0.02(−32.5)2+1.3(−32.5)+2.31

30 20 y 10

f( = 21.125 − 42.25 + 2.31 = −18.815

20

Thus, there is a relative minimum at −32.5, ( −18.815). We

–10

use the information obtained to sketch the graph.

–20

40

x

60

80

100

20

15. f(x) = x4 − 2 x 3 f′(x) = 4x3

10y

−6x2

f′(x) exists for all real numbers. Solve f′(x) = 0.

–100

–80

–60 x –40

–20

0

4x3 −6x2 = −3x2 = 0

–10

2x3 x2 (2x

–20

14. f (x) = −0.03x2 + 1.8x − 3.45 First, find the critical points. f ′(x) = −0.06x + 1.8 f′(x) exists for all real numbers. We solve f′(x) = 0: −0.06x+1.8 = 0

−0.06x = −1.8 −1.8 x = −0.06 x = 30

0

−3) = 0 x = 0 = xOr 3 2

3 The critical points are at x = 0 and x = . Use them to 2 div − ∞ , 0), (ideth)erealnumberlineintothreeintervals,A:( 3 ,∞.

B:0, ,andC: ( 3 ) A: Test − 1, f′(−1) = 4(−1)3 − 6(−1)2 = −10 < 0 B:Test1, f′(1)=4 ·13−6·12=−2<0 ·23−6·22=8>0

The only critical point is at x = 30. We use 30 to divide the real number line into two intervals, A: ( −∞,30) and B: (30, ∞):

C:Test2, f′(2)=4 3) Sincef(x)isdec ( 2 and (reasin)gonboth(−∞,0)and0, 3 ,∞,thereisnorelativeextremumat isincreasingon

✛

x=0 ︸A︸

︷ 30︷

︸B︸

✲

We use a test value in each interval to determine the sign of the derivative in each interval.

A:Test0, f′(0)= −0.06 · 0 + 1.8 = 1.8 > 0

3 3)isarelativeminimumatx= (bu)tthe(re 4 27.2 3 81 27 = = ( 3 ) 3= f − 2 4)−16 2 2 1( − 6 3 27 Thereisarealtiveminimumat .Wesketchthe ,−

B: Test 100, f′( −100) = −0.06 · (100) + 1.3 = −4.7 < 0 We see that f(x) is increasing on ( −∞, 30) and decreasing ∞), and the change from increasing to decreasing on (30,

10

indicates that a relative maximum occurs at x = 30. We substitute into the original equation to find f(30):

2

f(30) = = = 23.55 −0.03(30)2 + 1.8(30) − 3.45 −27 + 54 − 3.45

8 6 y 4

–4

–2

0 –2 –4 –6 –8

Thus, there is a relative maximum at (30,23.55). We use the information obtained to sketch the graph.

–10

2 x

4

93

Exercise Set 3.1 16. f(x) = x4 −8x2+3 First, find the critical points.

10 8 6 y 4

f′(x) = 4x3 −16x f′(x) exists for all real numbers. We solve f′(x) = 0:

2 –4

4x3 −16x = 0 x3 −4x = 0 −4) −2) = = 0 0 x(x2 0 x(x + 2)(x x = Or x+2 = 0

x

x = = Or = − 2 x

The critical points are at x =

−2 0 2

–8

√ 17. f(x)=x 8 x2 − The domain of this function is between − [ √ 8, √ 8] First, find the critical points. f′

12>0 ·13−16·1=4−16=−12<0 ·33−16·3=108−48=60>0

We see that f(x) is decreasing on ( −∞, −2), increasing on −2, 0), decreasing again is ona (0, 2), and increasing ∞). Thus, there relative minimum at again x = −2,on a ( (2, relative maximum at x = 0, and another relative minimum −2): atx=2. f ( −2)f( = (−2)4 − 8(−2)2 + 3 = 16 − 32 + 3 = −13 We find Then we find f(0): f(0) = 04 −8·02+3=0−0+3=3 Finally, we find f(2): f(2) = 24 −8·22+3=16−32+3=−13 There are relative minima at (

=

x(

−2,−13) and (2,−13), and

there is a relative maximum at (0, 3). We use the information obtained to sketch the graph.

1

(8−x2)−1/2(2x) −

−x2

) +(8 − x2)1/2(1)

+(8 − x2)1/2

f′(x) does not exist for x = ± √ 8. We solve f′(x) = 0: −x2 +(8 x2)1/2 =0 − (8 − x2)1/2 8 − x2 = − x2 8 = 2x2 4 = x2

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test − 3, f′(−3) = 4(−3)3 − 16(−3) = −108 + 48 =

D: Test 3, f′(3) = 4

(x)

(8 − x2)1/2

︸A︸ ︷︷︸B︸︷︷︸C ︸︸ D✲ ︷ ︸︷ 0 2 −2

C: Test 1, f′(1) = 4

4

–10

−2,x=0,andx=2.We

−60 < 0 B: Test −1, f′(−1) = 4(−1)3 − 16(−1) = −4 + 16 =

2 x

–6

=

use them toB: divide theC:real number into four intervals, A:( −∞,−2), (−2,0), (0,2), and D:line (2,∞). ✛

0 –2 –4

–2

±2 =

x

±√ The critical points are at x = 8, x = 2−and x = 2. We use them to divide the real number line into three intervals, C: (2, 8). √ A:( −√8,−2), B: ( 2,2), − ✛

︸A︸

︸B︸ ︸ C ︸✲ ︷︷ ︷︷ 2 −2 We use a test value in each interval to determine the sign of the derivative in each interval.

2 2 1/ 2 − (25) − .

A: Test −2.5,f′ (2.5)= +(8 − (25)) < 0 − . − / B:Test0,f′ (8 − (2.5)2)12 − 02)1/2 +(8 2 − (0) (0)= +(8 > 0 (8 − 02)1 2 − C:Test2.5,f′ . − (25) (2.5)= − (2.5)2)1/2 < 0 (8 − (2.5)2)1/2 8, 2), and on −√ We see that f(x) is decreasing on ( − √ (2, 8),andincreasingon( 2,2),sothereisarelative − minimum at x = 2−and a relative maximum at x = 2.

Wefindf( 2):− f( −2)=( 2)−8 ( 2)2= 4 − √ − − Then we find f(2): f(2)=(2) 8 (2)2=4Thereisarelativeminimum √ − at( 2, We − 4),andthereisarelativemaximumat(2,4). −

94

Chapter 3: Application of Differentiation use the information obtained to sketch the graph.

10 8 6 y 4

√ Then we find f( √8): =(8)16−√ √ f(2) (8)2=8Thereisarelativemin−√ 8,8),andthereisarelativemaximumat im √umat( − (8,8).Weusetheinformationobtainedtosketchthe graph.

10 8

2 –2

0

–1

1

2

x

3

4

6 y 4

5

–2

2

√ 18. f(x)=x 16 x2 −

The domain of this function is between−[ 4,4]

–2

–1

First, find the critical points.

f′(x)

1 (16−x2)−1/2(2x) − 2

) +(16 − x2)1/2(1)

=

x(

=

−x2 +(16 − x2)1/2 (16 − x2)1/2

−x2 +(16 x2)1/2 =0 − (16 − x2)1/2 x2 16 −x2 = − 16 = 2x2 8 = x2 √8 = x ± − √ 8andx= √ 8.

Thecriticalpointsareatx= ± 4,x= We use them to divide the real number line into three intervals,A:( 4,− − √ 8),B:( − √ 8, √ 8),C:( 8,4).√ ︸A︸

−︷√ ︷ 8

︸B︸ ︷√ ︷ 8

︸ C ︸✲

We use a test value in each interval to determine the sign of the derivative in each interval. − ( − 3)2 <0 A:Test-3,f′(3)= − +(16 − (3)2)1/2 − (16 − ( −3) ) 2/21 − (0)2 02)1/ > B:Test0,f′(0)= +(16 0 (16 − 02)1/2 +(16 − 32)1/ − 2 < 2 0 (3)2 − C:Test3,f′(3)= (16 − 32)1/2 √ We see that f(x) is decreasing on ( 4, − − 8), and on √ −√−√ √ ( 8,4), and increasing on ( 8, 8) ,so there is a relative minimum at x = 8 and a relative maximum at x= √ 8. −√ Wefindf( 8): f( − √ 8)=( − √

√

1

2

x

3

5

4

–2

f′(x) does not exist for x = 4.±We solve f′(x) = 0:

✛

0

−−√ ( 8)2= − 8

19. f(x) = 1 − x2/3 First, find the critical points. −2 x−1/3=−√ 2 3 33x f′(x) does not exist for x = 0. The equation f′(x) = 0 has no solution, so the only critical point is at x = 0. We use it to divide the real number line into two intervals: A: ( −∞,0) and B: (0,∞). ′ f(x)=

︸A︸

✛

︸B︸

︷ 0︷

✲

We use a test value in each interval to determine the sign of the derivative in each interval.

2 2 2 = = A: Test −1,f′(−1)=−√ > 0 B: Test 1, f′(1) = 3(− 1) 3 33 − 1 − = 2 2 2 =− · < 0 −√ 3 1 −3 331

We see that f(x) is increasing on ( −∞, 0) and decreasing ∞), so there is a relative maximum at x = 0. on (0, We find f(0): f(0) = 1 −02/3=1−0=1 There is a relative maximum at (0,1). We use the information obtained to sketch the graph.

2 –2

–1

1 0 –2 y –4 –6

2

x

3

4

5

95 −∞, 0) and increasing

Exercise Set 3.1 20. f(x) = (x + 3)2/3 − 5 2 2 f′(x)= (x+3)−1/3= √ 3 33x+3 f′(x) does not exist for x = −3. The equation f′(x) −3. =0 has no solution, so the only critical point is at x =

Use it toand divide real number line into two intervals, A: ( −∞,−3) B: the (−3,∞).

2 2 <0 A: Test −4,f′(−4)= √ 4+3 =− B: Test −2, f′(−2)= 33 3 − 2 2 √ 2+3 = 3 >0 33 − Since f(x) is decreasing on ( −∞,−3) and increasing on ( −3, ∞), there is a relative minimum at x = −3. f ( −3) = (−3 + 3)2/3 − 5 = −5

We see that f(x) is decreasing on ( on (0, ∞), so there is a relative minimum at x = 0.

We find f(0): 2− 8 = − 8 =−8 0+1 1 There is a relative minimum at (0, −8). We use the information obtained to sketch the graph.

f(0)=

10 8 6 y 4 2 –10 –8

–6

There is a relative minimum at ( −3,−5). We sketch the

–4

–2 0 –2 –4

graph.

–6

10

–8

8

–10

6 y 4

22. f(x)=

x2

5 =5(x2+ +1

2

4 x 6

8

10

1) − 1

2 0 –2 –4

–5

5

10

x 15

20

25

f′(x) = 5( =

–6

=

–8 –10

−1)(x2 + 1)−2(2x) −10x(x2 + 1)−2 −10x (x2 + 1)2

f′(x) exists for all real numbers. Solve f′(x) = 0.

− 8 = 8( x2 +1) − 1 − x2 +1 First, find the critical points. f′(x) = −8(−1)(x2 + 1)−2(2x)

21. f(x)=

−10x =0 (x2 + 1)2 −10x = 0 x = 0

= 16x(x2 + 1)−2 16x = (x2 + 1)2 f′(x) exists for all real numbers. We solve f′(x) = 0: 16x (x2 + 1)2 x

=

0

16x = 0 = 0

The only critical point is at x = 0. We use it to divide the real number line into two intervals, A: (−∞, 0) and B: (0, ∞). ✲ ✛

︸A︸

︷ 0︷

︸B︸

We use a test value in each interval to determine the sign of the derivative in each interval.

16(−1) −16 A:Test −,f′− 1 (1)= 2= =−4<0 2 [(1)+1] 4 − 16·1 16 2= =4>0 B:Test1, f′ (1)= 2 (1+1) 4

The only critical point is at x = 0. Use it to divide the real number line into two intervals, A: (−∞,0) and B: (0,∞). − 10( 1) 2= 5 >0 A: Test − 1 , f ′ ( − 1)= 2 B:Test1, [(1)+1] 2 − − 1 f′ 5 − · 10 2 = − < 0 (1)= 2 2 (1+1) Since f(x) is increasing on ( −∞,0) and decreasing on ∞), there is a relative minimum at x = 0.

(0, 5 f(0)= =5 02+1 There is a relative maximum at (0,5). We sketch the graph.

96

Chapter 3: Application of Differentiation −1 and a relative maximum at relative minimum at x = x=1.

10 8

We find f( −1):

6 y 4

–10 –8

–6

–4

–2 0 –2 –4

2

4 x 6

8

10

–6

There is a relative minimum at (

–8

−1,−2), and there is a

relative maximum at (1,2). We use the information obtained to sketch the graph.

–10

23. f (x) =

4(−1) (1)+1 = −24 =−2 2 − Then we find f(1): 4 4 f(1)= 12 ·1 +1 = 2 =2 − f (1)=

2

4x x2+1

4

First, find the critical points.

′ f(x)

= = =

y 2

2

(x+1)(4)−2x(4x) (x2 + 1)2

4x2 + 4 − 8 x 2 (x2 + 1)2 −4x2 4 (x2 + 1)2

f′(x) exists for all real numbers. We solve f′(x) = 0: 4 −4x2 (x2 + 1)2 4 1 (1

=0

−4x2 −x2 = =0 0

−x)(1+x) = 0 0 1 − x = x 1 = Or 1+x = 0 = −1

x The critical points are at x =

−1 and x = 1. We use them to divide the real number line into three intervals, A:( −∞,−1), B: (−1,1), and C: (1,∞). ✛ ︸A︸

︸B︸ ︸ C ︸✲ ︷︷ ︷︷ 1 −1 We use a test value in each interval to determine the sign of the derivative in each interval.

4 4(2) 2 12 −2,f′(− <0 2)=[( − 2 − 2= − A: Test 2)+1] 25 − ′ f(0)= B:Test0, C:Test2,f′

(2)=

4 − 4 · 02 (02+12= )

4 1 =4>0

12 4 − 4 · 22 <0 2= − (22+ 1) 25

, 1), increasing We see that f(x) is decreasing on ( − ∞ − on( −1,1), and decreasing again on (1, ),∞so there is a

–10 –8

–6

–4

–2 0

2

4 x 6

8

10

–2 –4

24. f (x) =

x2 x2+1 f ′ ( x )= = =

(x2 + 1)(2x) −2x(x2) (x2 + 1)2 2x3 + 2x − 2 x 3 (x2 + 1)2 2x (x2 + 1)2

f′(x) exists for all real numbers. Solve f′(x) = 0.

(x2

2x = +1)2 = 2x = x

0 0 0

The only critical point is at x = 0. Use it to divide the real number line into two intervals, A: (−∞,0) and B: (0,∞).

1 2(−1) ′(− 2= −2 =− 2 <0 1)= A: Test − 1 ,f 2 [(1)+1] 4 − ′ 2·1 2 1 B:Test1, >0 f(1)= 2= = 2 (1+1) 4 2 Since f(x) is decreasing on ( −∞,0) and increasing on ∞), there is a relative minimum at x = 0.

(0, 02 f (0) = 2 =0 0+1 There is a relative minimum at (0, 0). We sketch the graph.

Exercise Set 3.1

97 Since f(x) is increasing on both intervals, there are no relative extrema. We sketch the graph.

4 y 2

–10 –8

–6

–4

–2 0

4

2

4 x 6

8

y 2

10

–2 –10 –8

–6

–4

–4

27. f(x)=

f′(x) does not exist for x = 0. The equation f′(x) = 0 has no solution, so the only critical point is at x = 0. We use it to divide the real number line into two intervals, A: ( −∞,0), and B: (0,∞). ︸A︸

︸B︸

︷ 0︷

8

10

4 y 2

–4

–2 0

2

4 x 6

8

10

–2

First, find the critical points. x+1 1 2 ′ (x+2x+5)−1/2(2x+2)= f(x)= √ x2+2x+5 2 f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 x+1 = 0 √ x2+2x+5 x+1 = 0 x

=

−1

So the only critical point is at x = −1. We use it to divide −∞, −1), and the real number line into two intervals, A: ( B:( − 1 , ∞ ) . ✛ ︸A︸

︸B︸ ✲ ︷︷ −1 We use a test value in each interval to determine the sign of the derivative in each interval. (−2)+1 −1 A:Test − 2,f(2)= =√<0 ′ − (2)2+2(2)+5 − √ − 0+1 B:Test0,f′(0)= 1=√>0 5 √ (0)2+2(0)+5 Weseethatf(x)isdecreasingon(−∞,−1)andincreasing on( −1,∞), −1. so the function has a relative minimum at x= We find f( −1):

–4

26. f(x) = (x + 1)1/3 f′ 1 (x+1)−2/3= (x)= 3

√ x2+2x+5=(x2+2x+5)1/2

✲

We use a test value in each interval to determine the sign of the derivative in each interval.

1 1 1 >0 A: Test − 1, f′(−1) = = = 3·1 3 33(−1)2 √ B:Test1,f′ 1 1 1 = = (1)= > 0 3 33√ 12 3 · 1 We see that f(x) is increasing on both intervals, so the function has no relative extrema. We use the information obtained to sketch the graph.

–6

4 x 6

–4

First, find the critical points. f′ 1 1 (x)= 3 x−2/3= √ 33x2

–10 –8

2

–2

√ 25. f(x)=3x=x1/3

✛

–2 0

1

33(x+1)2 √ f′(x) does not exist for x = −1. The equation f′(x) −1. =0 has no solution, so the only critical point is at x = We use it toand divide the real number line into two intervals, A:( −∞,−1) B: (−1,∞).

1 1 >0 A: Test −2,f′(−2)= = B:Test0, 3 33( −2+1)2 √ 1 1 f′(0)= >0 = 33(0+1)2 3 √

f(−1)=√(−1)2+2(−1)+5 √ = √1−2+5 = 4=2 There is a relative minimum at ( −1,2). We use the information obtained to sketch the graph.

98

Chapter 3: Application of Differentiation

–10 –8

28. f(x)=

–6

–4

15

15

10 y 5

10 y 5

–2 0

2

4 x 6

8

10

–4

–2 0 –5

–10

–10

–15

–15

√ x2+6x+10=(x2+6x+10)1/2

29. f(x)=√1

2x +6 √ x2+6x+10 f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 2x+6 = √ x2+6x+10 2x+6 = 0 2x =

︸A︸

x2 +1

2

(2 =x+1)

4 x 6

2 − 1(x+1)

10

−1/2

First, find the critical points. ′ f(x)=

8

−3/2

2x

(2x)=

√ (x2+1)3 2 f′(x)existsforallxvalues.Wesolvef′(x)=0, f′(x) = 0

0

√2x

− 6 − 3

︸B︸

=

= 0 0

x

x = So the only critical point is at x = −3. We use it to divide −∞, −3), and the real number line into two intervals, A: ( B:( − 3 , ∞ ) . ✲

︷︷ −3 We use a test value in each interval to determine the sign of the derivative in each interval.

2(−4)+6 √ − 2<0 = A: Test −4,f′− (4)= 2 (2)2+6(2)+10 − √ − 6 B:Test0,f′ >0 (0)= =√ 10 √ (0)2+6(0)+10 Weseethatf(x)isdecreasingon(−∞,−3)andincreasing on( −3,∞), so the function has a relative minimum at x = −3.

We find f( −3): f(−3)

–6

–5

First, find the critical points.

1 (x2 +6x+10)−1/2 f′(x)= (2x+6)= 2

✛

–10 –8

=√(−3)2+6(−3)+10 √ √9−18+10 1=1

= =

There is a relative minimum at ( −3,1). We use the information obtained to sketch the graph.

So the only critical point is at x = 0. We use it to divide the real number line into two intervals, A: ( −∞,0), and B: (0, ∞). ✛

︸A︸

︷︷

︸B︸

✲

We use a test value in each interval to determine the sign of the derivative in each interval. √ − 2<0 A:Test − 1,f(1)= = ′ − √2(−1) 8 ((1)2+1)3 − 2=√>0 B:Test1,f′(1)= √2(1) 8 ((1)2+1)3 Weseethatf(x)isdecreasingon(−∞,0)andincreasing on (0, ∞), so the function has a relative minimum at x = 0.

We find f(0): f(0)= =

2

√ √0+1 1=1

There is a relative minimum at (0,1). We use the information obtained to sketch the graph.

Exercise Set 3.1

–10 –8

–6

99

–4

3

3

2

2

y

y

1

1

–2 0

2

4 x 6

8

10

–10 –8

–6

–4

–1

30. f(x)=√1

x2 +9

–2 0

2

4 x 6

8

10

–1

= (x2+9)−1/2

31. f(x) = sin x First, find the critical points.

First, find the critical points. ′ (x2+9)−3/2 f (x)= − 1 (2x)=

2x

f′(x) = cos x

√ (x2+9)3 2 f′(x)existsforallxvalues.Wesolvef′(x)=0,

f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 cosx = 0

f′(x) = 0

x = and = x

= 0 = 0

√2x

π 2 3π 2

x So the only critical point is at x = 0. We use it to divide the real number line into two intervals, A: ( −∞,0), and B: (0, ∞). ✛

︸A︸

︷ 0︷

︸B︸

✲

We use a test value in each interval to determine the sign of the derivative in each interval. = √ − 2 <0 A:T est − 1, f′(−1) = √2(−1) 10 ((1)2+9)3 − (1) 2 >0 B:Test1,f′(1)= =√ √2 10 ((1)2+9)3 Weseethatf(x)isdecreasingon(−∞,0)andincreasing on (0, ∞), so the function has a relative minimum at x = 0.

We find f(0): f(0)

= =

2 Sotheonlycriticalpointsareatx=πand =3π x 2 and there might be extrema points at the end points x = 0 and x = 2π. We use them to divide the real number line into three intervals, A:2 [0, π), B: (π 3π), and C: (3π 2,2π].

2,2

2

√(0)+9 √ √0+9 9=3

︷

︸A︸ ︷︷ π

︸B︸ ︷3π︷ 2

2

︸C︸ ︷ 2π

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test 0, f′(0) = cos(0) = 1 > 0 B: Test π, f′(π) = cos(π) =

−1<0

C: Test 2π, f′(2π) = cos(2π) = 1 > 0 We see that f(x) is decreasing on ( −∞, π2 ) and increasing 3π on(π2, 2 ), we also see that f(x) is increasing on ( 23π , 2 and∞) a so the function has a relative maximum at x = π relative minimum at x = 3π 2.

We find f(π2): f( )π= sin( ) 2 = 1

There is a relative minimum at (0,3). We use the information obtained to sketch the graph.

π 2

We find f ( 3π 2): f(

3π 3π ) = sin( ) 2 2 = −1

and there is a relThere is a relative maximum at (π2,1) −1). We use the information obative minimum at (3π 2, tained to sketch the graph.

100

Chapter 3: Application of Differentiation There is a relative maximum at (0, 1) and (2π, 1) and there is a relative minimum at (π,

2

obtained to sketch the graph. −1). We use the information

y1 2 0

1

2

3 x

5

4

6

y1

–1 0

1

2

3 x

5

4

6

–2

32.

–1

f(x) = cos x First, find the critical points. f ′(x) =

−sin x f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 −sinx = 0 x = 0

–2

33.

f(x) = sin x − cos x First, find the critical points. f′(x) = cos x + sin x f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 cosx+sinx = 0

and = π x = 2π and x So the only critical points are at x = 0, x = π and x = 2π.

There might be extrema points at the end points x = 0 and x = 2π. We use them to divide the real number line into two intervals, A: [0, π), B: (π, 2π]. ︷

︸A︸ ︸B︸ ︷︷ ︷ 2π

We use a test value in each interval to determine the sign of the derivative in each interval. π ( ) = −sin( 2 )=−1<0 A :Testπ,2 f′π 2 3π ′3π 2,f( ) = − sin( )=1>0 B:Test 3π 2 2 We see that f(x) is decreasing on [0,π) and increasing on (π,2π] so the function has a relative minimum at x = π and the function has a relative maximum at the endpoints x=0andx=2π.

We findf(0): f(0) = cos(0) =

1

We find f(π):

cos x = −sin x 3π = x = 4 and 7π x 4

So the only critical points are at x = 3π4and x = 7π 4.We use them to divide the real number line into three intervals, 7π A:[0,3π),B:(3π,7π 4),C:(4,2π].

4 4 ︷

︸A︸ ︸B︸ ︸C︸ ︷3π︷ ︷7π︷ ︷ 2π 4

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test 0, f′(0) = cos(0) + sin(0) = 1 > 0 B:Test5π,4 f( ′

C: Test 2π, f′(2π) = cos(2π) + sin(2π) = 1 > 0 We see that f(x) is increasing on [0, 3π) and on (7π 4,2π] 4 and decreasing on (3π, 7π 4)sothefunctionhasarelative 4 maximum at x = 3π and a relative minimum at x = 7π4 .

4 We findf ( 3π 4): f(

−1

We find f(2π): f(2π) = cos(2π) =

3π ) = sin( 4 = =

f(π) = cos(π) =

5π 5π 5π )+sin( )= −1.414<0 )=cos( 2 2 2

1

3π 3π ) cos( ) 4 − 4 1 1 ( √ ) √2−− 2 2 √ ≈ 1.414 2

We find f ( 7π 4): f(

7π ) = sin( 4

7π 7π ) cos( ) 4 − 4

Exercise Set 3.1

101 1 −√1 −(√) 2 2 2 − √ ≈−1.414 2

= =

3 x

4

5

π · 3 )=−3<0

2π ′2π )=3cos(3 · 3 )=3>0 3

D: Test π, f′(π) = 3cos(3 E:Test F:Test

y1

2

π ′π B:Test C:Test 3,f( )=3cos(3 3

3,f(

2

1

·0)=3>0

2π

There is a relative maximum at (3π,4 √2 2 )andthereisa relative minimum at 7π, √2 .Weusetheinformationob4 2 tained to sketch the graph.

0

A: Test 0, f′(0) = 3cos(3

·π)=−3<0

4π ′4π )=3cos(3 · 3 )=3>0 3 ′ 5π 5π 3 , f (π) = 3cos(3 · 3 )=−3<0 4π

3,f(

G: Test 2π, f′(2π) = 3cos(3 ·2π)=3>0 π 5π 6),(2, 6),(6, 7π 9π6), Weseethatf(x)isincreasingon(0,π 5π 7π 9π 11π π 6,2π)anddecreasingon(π 6,2),(6,6),(6, 6), 5π and ( 11π so the function has a relative maximum at x = π 6,x= 6, and x = 9π6 and the function has a relative minimum at x=π 2,x= 67πand 11π 6 . π We find f(π6):

6

–1

–2

π f( ) = sin(3 · ) 6 6 π = sin( ) 2 = 1

34. f(x) = sin 3x First, find the critical points. f′(x) = 3cos 3x f′(x) exists for all x values. We solve f′(x) = 0, We find f(π2):

f′(x) = 0 3cos3x = 0 π 2 x3x = = = π and = 6 x π = 2 and = 5π 6 = 7π x 6 9π 6 and x and x and x

π f( )π= sin(3 · ) 2 π2 = sin(3 ) 2 = −1 We find f ( 5π 6): f(

5π ) = sin (3 · 5 π) 6 6 π = sin(5 ) 2 = 1

We find f ( 7π 6): f( 11π 6

Sothecriticalpointsareatx=π 11π x=9π6,andx= 6

6,x=π2,x=

5

π6,x=

7

π6,

We use them to divide the real number line into seven π5 5π 7 ππ 6),D:(6,π6),E: intervals,A:(0,π 6),B:(6,2),C:(2,π 11π (7π,9π),F:(9π,11π 6),andG:(6,2π).

6 6 6

7π ) = sin (3 · 7 π) 6 6 π = sin(7 ) 2 = −1

We find f ( 9π 6): f(

9π ) = sin (3 · 9 π) 6 6 π = sin(3 ) 2 = 1

We find f ( 11π 6 ): ︷︸A︸︷︷︸B︸︷︷︸C︸︷︷︸D︸︷︷︸E︸︷︷︸F︸︷︷︸G︸︷ 11π π 5π 7π 9π 11π f( π 6 ) = sin (3 ·11π ) 0 2π 6 6 π We use a test value in each interval to determine the sign = sin(11 ) 2 of the derivative in each interval. = −1

102

Chapter 3: Application of Differentiation 6 , 1), and ( 9π ,1) There is a relative maximum at 6(π,1), (5π 61),and and there is a relative minimum at (π, 2 −1),(7π6,− 1 − ( 16π , 1).Weusetheinformationobtainedtosketchthe graph.

(3π2 , 2π) so the function has a relative maximum at x = 0 (end point), x = π and x = 2π (end point) and a relative 3π x minimumatx=π 2and =2.

We findf(0):

2

f(0) = cos(2 = cos(0) = 1

y1

·0)

We find f(π): 0

1

2

3 x

4

5

6

f(π) = cos(2 =

–1

·π)

1

We find f(2π): f(2π) = cos(2

–2

=

35. f(x) = cos 2x

We find f(π2):

First, find the critical points. f ′(x) = −2sin 2x

π · )2 = cos(π)

f(2π) = cos(2

f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 −2sin2x = 0 2x = 0 x and x and x and x and x

= 0 = = π = 2 = π 3π 2 2π

2,x=π, So the only critical points are at x = 0, x = π x=3π 2 and x = 2π. We use them to divide the real number line into four intervals, A: (0, π),2B: (π 2,π),C: (π, 3π) and D: (3π 2,2π).

2 ︷

·2π)

1

︸A︸︸B︸︸C︸ ︸D︸ ︷π ︷ ︷ ︷ ︷3π︷ ︷ 2 2 2π π

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test πover4, f′(πover4) = −2sin(πover2) = −2 < 0 3 3 B: Test 3π, f(′ π )= − 2sin( π)=2>0 4 4 )= 2 )=−2<0 )= )=2>0 3 5 π ′π C: Test 5π, f( 4 − 2sin( 2 4 7 5π D:Test 7π, f(′ π 4 − 2sin( 2 4 We see that f(x) is decreasing on (0, π)2and (π, 3π 2)and the function is increasing on (π 2,π) and

=

−1

We find f ( 3π 2): f(2π) = cos

π (2 · 3)2

= cos(3π) = −1 There is a relative maximum at (0,1), (π,1) and (2π,1) and there is a relative minimum at (π, 2 −1)and(3π 2,−1).

We use the information obtained to sketch the graph.

2

y1

0

1

2

3 x

4

5

6

–1

–2

36. f (x) = x + 2 sin x First, find the critical points. f ′( x)=1+2cosx f ′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 1+2cosx = 0 1 cos x= = −2 = 2π 3 x 4π and 3 x

Exercise Set 3.1

103

4π So the only critical points are at x= 2π 3 and =x3 and there might be extrema points at the end points x = 0 and x = 2π. We use them to divide the real number line 4π threeintervals,A:[0,2π 3),B:(3 2π,4π into 3),andC:(3,2π].

︷

f′(x) exists for all x values. We solve f′(x) = 0, f′(x) = 0 −2sin2x = 0 1 sin2x= = 2 = = π x = 12 5π 12 and 13π 12 17π x 12

1

︸A︸ ︸B︸ ︸C︸ ︷︷ ︷4π︷ ︷ 2π 2π

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test 0, f′(0) = 1 + 2cos(0) = 3 > 0 B: Test π, f′(π) = 1 + 2cos(π) =

and x

−1<0

and

C: Test 2π, f′(2π) = 1 + 2cos(2π) = 3 > 0 Weseethatf(x)isincreasingon[0,2π 3 ) and ( 4π 3 ,2π] and 4π decreasing on (2π 3,3 ) so the function has a relative maximum at x = 2π3 and 2π and a relative minimum at x = 0 and x = 4π 3.

We findf(0):

x So the only critical points are at x = π12, =x5π, = 13π 12, 12 x andand x = there 17π might be extrema points at the end 12 points x = 0 and x = 2π. We use them to divide the real number line into five intervals, A: [0, π ), B: ( π 12,12),C: 5π 12

13π 17π),andE:(17π12,2π].

(5π :(13π 12,12),D 12,12

f(0) = 0+2sin(0) =

0

︷︸A︸ ︸︷ B︷ ︸ ︸ C ︸ ︸ D︷︸︷ ︸ E ︸ ︷ ︷5π︷ ︷13π ︷ 17π π 0 2π

We find f ( 2π 3): f(

2π )= 3

2π 2π +2sin( ) 3 3 ≈ 3.826

We find f ( 4π 3):

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test 0, f′(0) = 1 −2sin(2·0)=1>0 π ( B:Test4 , f ′ π4)=1

f(

4π )= 3

4π 4π +2sin( ) 3 3 ≈ 2.457

We find f(2π): f(2π) = 2π+2sin(2π) =

2π

There is a relative maximum at (2π3 ,3.826) and (2π,2π) and there is a relative minimum at (4π3 ,2.457). We use the information obtained to sketch the graph.

C: Test π, f′(π) = 1 −2sin(2·π)=1>0 ′15π 15π D:Test12,f( )=1 − 2sin(2 ·15π )=−1<0 12 12 E: Test 2πf′(2π) = 1 − 2sin(2 · 2π) = 1 > 0 We see that f(x) is increasing on [0, π ),12(5π, 12 13π12),and 17π 5π),(13π (17π 12,2π]an ddecreasingon(π 12,12 12and12)sothe function has a relative maximum at x = π12 and x = 13π 12 and a relative minimum at x = 5π, 12 = x 17π12,andx=2π.

We find f( π12):

f(

10

π π ) = + cos(2 · π ) 12 12 12 ≈ 1.128

We find f ( 5π 12):

8

f(

6 y 4

5π ) 12

5π +cos(2 · 5 π) 12 12 ≈ 0.443 =

We find f ( 13π 12):

2 0

(2·π )=−1<0 − 2 sin 4

f( 1

2

3 x

4

5

6

–2

37. f (x ) = x + cos2xFirst,findthecriticalpoints. f ′()=1 x − 2sin 2x

13π 13π = +cos(2 ·13π ) 12 ) 12 12 ≈ 4.269

We find f ( 17π 12): f(

17π 17π = +cos(2 ·17π ) 12 ) 12 12 ≈ 3.585

104

Chapter 3: Application of Differentiation We find f (2 π):

5

f (2pi) = 2π+cos(2

·2π)

≈ 7.2832 There is a relative maximum at 12, ( π 1. 128) and (13π 12,4.269) and a relative minimum at (5π (17π 12,3.585),and .

12,0 443), (2π,7.283). We use the information obtained to sketch the graph.

10

4 3 y 2 1 0

1

3 x

2

4

5

6

–1 –2

8

x 39. f (x) = x 3 + cos2 3 . Find the critical values

6 y

f ′()=1 x

4 2 0

1

3 x

2

5

4

6

–2

38. f (′ x) = x 4+ sin x2 . First, find the critical points.

4 + 12cosx2 . We solve f′(x) = 0 f ()=1 x 1 4

+

1

x co 2 sco s 2 1 co s x 2 2 x 2

= = = = =

x and x

0 − −1 4 1 2

3 −

2x 3 We solve f(x)=0 ′ 3sin 1 −2 2x = 0 sin 3 3 2x 3 = 2 1 3 sin sin 2x = 3 1 3 2 2x 3 2x 1 = sin−1( ) 3 2 pi = = 6 = = π x 4 5π and 6 5π 2x 3 4 2

2π 3 4π 3

x A

︷︸︸︷︷ π 0 A

B

︸︸ ︷ ︷︸ ︸︷ ︷ ︷ 2π 4π 3 0 3

C

︸︸

︷ 2π

We use a test value in each interval to determine the sign of the derivative in each interval.

1 1 π =0.683>0 A:Testπ, 3 f( )=′π+ cos 3 4 2 6 = 0 . 2 5 > 0 ′ 1 1 π B:Test + cos2 C:Test , 4 2 1 1 5π ′5π 5π 3,f( )= + cos = −0.183>0 3 4 2 6 We see that f(x) is increasing on [0, 4π 3 ), and decreasing on (4π,2π] so the function has a relative maximum at x = 4π 3 3 4π )= 3

2π 1 4π ( )+sin( ) 3 3 4 = 0.1812

There is a relative maximum at (4π3 ,0.1812). We sketch the graph.

C

︸︸

︷︷ 5π 4

︸︸

︷ 2π

We use a test value in each interval to determine the sign of the derivative in each interval.

1 2 π A:Testπ, 6 f ′π ( ) = 3 − 3 sin 9 =0.105>0 6 1 2 2π B:Test , πfπ′( )= − sin 3 =−0.323<0 3 3 1 2 ′5π 2,f( C:Test 3π ) = 3 − 3 sinπ=0.333>0 3 We see that f(x) is increasing on [0, π) and (5π 4,2π])and 4 decreasing on (π, 5π 4)sothefunctionhasarelativemax4 imum at x = π 4and a relative minimum at = 5pi 4.We x find f(π4 )

We find f ( 4π 3)

f(

B

f( )π= 4

π +cos( ) 6 12 = 1.128

5π )= 4

5π 5π +cos( ) 12 12 = 0.443

π

We find f ( 5π 4) f(

Exercise Set 3.1

105

There is a relative maximum at (π4 , 1.128) and a relative minimum at (5π 4 ,0.443). We sktch the graph

We find f ( 43π ) f(

5

4π ) 3

sin4π 3

= =

2+cos4π3 −0.5774

4

There is a relative maximum at (2π 3 , 0.5774) and a relative minimum at (4π 3 ,−0.5774). We sketch the graph

3 y 2

2

1 0

1

3 x

2

5

4

6

y1

–1 –2 0

40. f (x) =

1

sinx 2+ cosx

3 x

2

5

4

6

–1

f ′ ( x )= = =

(2+ cosx )(cosx ) − sinx( − sin x) (2+ cosx )2 2 2 2 cosx + + cosx sinx (2+ cosx )2 2 cosx + 1 (2+ cosx )2

–2

41. f(x)=

cosx 2 −sin x

Find the critical values ′

f(x) =

We solve f ′()=0 x

= = 2cosx+1

=

(2 + cos x)2 2cosx+1 = 0

0

cos x = = = x and x

A

︷

︸︸

B

︷︷ 2π

1 −2 2π 3 4π 3

︸︸ ︷︷ 4π

−1<0

3π )=0.25>0 C: Test 2,f′( 2 We see that f(x) is increasing on [0, 2π) and (4π 3,2π])and 3 decreasing on (2π , 4π 3)sothefunctionhasarelativemax3 imum at x = 2π3 and a relative minimum at x = 4π3 .

We find f ( 2π 3) 2π ) 3

1 −2sinx = 0 (2 − sin x)2 1 −2sinx = 0 sin x= = = x and x

︸︸ ︷ 2π

3π

f(

We solve f′(x) = 0

C

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test π4, f′ π( )=0.329>0 4 B: Test π, f′(π) =

(2− sinx )(− sinx )− cosx( − cos x) (2 − sinx )2 2 2 + + cosx − 2 sinx sinx (2 − sinx )2 1 − 2 sinx (2 − sinx )2

=

sin2π 3

2+cos2π3

= 0.5774

A

︸︷ ︷ ︸︷ π 6 0

B

︸︸

1 2 π 6 5π 6

︸C︸ ︷︷ ︷ 5π 2π 6

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test π18,f′π ( )=0.196>0 18 )= 4,f′π( B: Test π −0.248<0 )=0.159>0 4 11π 12,f′ C: Test 11π ( 12 Weseethatf(x)isincreasingon[0,π 6) and ( 5π 6 ,2π]) and decreasing on 6(π, 6)sothefunctionhasarelativemaxi5π 3 mumatx=πand arelativeminimumatx=6. 5π

106

Chapter 3: Application of Differentiation We use a test value in each interval to determine the sign of the derivative in each interval.

We find f ( π6 ) f(

π ) 6

cos(π) 6 2 − sin( π 6 ) 2√ 3

= = =

′π A: Test π,6 f( )=0.5>0 6 ′ π B: Test π,2 f ( ) = −1<0 2

√ −2 213

C: Test π, f′(π) =1 > 0

3 We find f ( 5π 6) f(

5π ) 6

cos(5π) 6

= = =

2 − sin( 5π6 ) −√3 2

2− 1

3 D:Test3π,2 f (′ π ) = −1>0 2 ′ 15π E:Test15π, 8 f( )=0.707>0 8 We see that f(x) is increasing on [0, π), (3π, 5π 4), and 4 4 ( 7π , 2π] and decreasing on ( π , 3π ) and ( 5π , 7π 4)sothefunc4 4 4 4 tion has a relative maximum at x = π4and x = 5π 4 anda relative minimum at x = 3π and x = 7π.

We find f(π 4 ) 4 4 f( )π= sin( )cos( )π 4 4 1 = 2

−√23 3 There is a relative maximum at (π, 6

√3

3)andarelative minimum at ( 5π , 3).Wesketchthegraph − 3 6 √

π 4

We find f ( 3π 4)

2

f(

3π ) = sin( 4 =

y1

−1 2

3π 3π )cos( ) 4 4

We find f ( 5π 4) 0

1

3 x

2

5

4

6

f(

5π ) = sin( 4 =

–1

5π 5π )cos( ) 4 4

1 2

We find f ( 7π 4)

–2

f(

42. f (x) = sin x cos x f ′(x) = − sin2x + cos2x = 1− 2 sin2x We solve f′(x) = 0

2

= = 2 sinx = = sinx = = = x

1 −2sinx

0 1 2 ±√1 2 π 4 3 π

and x

4

and

5π 4

x and

7π 4

x D

︷ ︸ A ︸︸ ︷B ︷︸ ︸C︸ ︷︷︸E︸︷ ︷︷ ︷︷︸︸ 3π 5π 7π4 π4 0 2π 4 4

7π ) = sin( 4 =

−1 2

7π 7π )cos( ) 4 4

There is a relative maximum at (π, 1) and (5π, 1) and a 42 42 1 relative minimum at (3π, )and(7 4π ,− 1 4 − the graph 2

y1

0

1

2

3 x

4

5

6

–1

–2

43. f(x) = sin x − sin2 x Find the critical values f′(x) = cosx − 2 sin x cos x = cosx (1 − 2 sin x)

Exercise Set 3.1

107 We find f ( 3π 2)

We solve f ′()=0 x cosx (1

−2sinx) = 0 cosx = 0 π x = 2 = 3π 2 and x

1 −2sinx = 0 sin x= = = x and

1 2 π 6 5π 6

x

f(

1 There is a relative maximum at ( π6 , 14) and (5π, 6 4)anda 3π relative minimum at (π2,0)and( 2 ,− 2). We sketch the graph 2

y1

0

︷ ︸ A ︸ ︷ ︸ B ︸ ︷︸︷D︸ ︸C ︸ ︸ ︷ E︷︸ ︷3π︷ ︷ π 5π π 2 0 6 2π 2 6

–1

We use a test value in each interval to determine the sign of the derivative in each interval.

A:Test π 18, f′ π ( )=0.643>0 18 B:Testπ f′4 π , ( )= −0.293<0 )=0.366>0 4 C: Test 2π f′ 2π 3, ( 3

–2

D: Test π, f′(π) =

−1<0

1

f ′()x

We solve f x

)=1.707<0 4 We see that f(x) is increasing on [0,π 6), (2,π 5π 6 ), and 5π 3π π π 2 (3π,2π]an ddecreasingon(6,2)and(6,2)sothefunc5π tionhasarelativemaximumatx=π 6andx=6anda 3π relativeminimumatx=π 2 andx= 2.

′()=0

sin( π) − sin2(π) 6 1 −1 2 1 4

x 1 + 2 sinx cosx x and

5π ) = sin 6

(

5π

5π ) − sin2( )

= =

6 1 −1 2 4 1 4

= =

π π sin( ) − sin2( ) 2 1− 1

We find f(π2 ) f( π) 2

=

0

6

= − sinx +2 cosx (− sinx) − sinx (1+2 cosx ) =

We find f ( 5π 6) f(

5

4

− sinx (1+2 cosx ) = − sinx = x = and = x and =

We find f(π6 ) = = =

3 x

2

2 44. f (x) = cosx + cos x

E: Test 7π 4,f′( 7π

π f( ) 6

23π 3π 3π ( ) ) = sin( ) 2 2 −sin 2 = −1−1 −2 =

C

= =

π 2π 0 −1 22π 3 4π 3

x A

= =

0 0 0

D

︸︸ ︷︷ ︸ ︸ ︷ ︷ ︸ B ︸ ︷ ︷ ︸︸ ︷ ︷ 2π 4π 2π π 3 3 We use a test value in each interval to determine the sign of the derivative in each interval.

108

Chapter 3: Application of Differentiation A:Tes t π3 , f′(π ) = −1.732<0 3 B:Test5π6,f′( 5π)=0.366>0 6 C:Test7π6 , f ′ ( 7π) = −0.366<0 6 D:Test5π3,f′( 5π)=1.732>0 3 We see that f(x) is decreasing on [0, 2π) and (π, 4π 3),and 3 increasing on (2π,π) and (4π 3,2π])sothefunctionhasa 3 relative maximum at x = 0 (end point that is also a critical value), x = π and x = 2π (end point that is also a critical value) and a relative minimum at x = 2π and x = 4π3 .

3

45. f(x) = 9 sinx − 4 sin3 x Find the critical values =

f ′()x

=

9 cosx − 12 sin2x cos x 3 cosx (3 − 4 sin2 x)

Wesolve ′(f )=0 x 2 ) 3 cosx (3 − 4 sinx

= cosx = x = = and = = = = x = =

We find f(0) f(0) = cos(0)+cos2(0) =

3 −4sinx2

2

2

sinx

We find π

sinx

f(π) = cos(π)+cos2(π) =

x

0

and

We find f(2π)

x

f(2π) = cos(2π)+cos2(2π) =

0 0 π 2 3π 2

0 3 4

√3

± 2 π 3 2π 3 5π 3

and

2

x

We find f ( 2π 3) f(

2π ) = cos( 3 =

1 −4

2π 2π +cos2( ) 3 3

4π ) = cos( 3 =

1 −4

4π 4π +cos2( ) 3 3

There is a relative maximum at (0, 2),(π, 0) and (2π, 2) and 4).Wesketch a relative minimum at (2π, 3 −1)and(4π,−1 4 3 the graph 3 2 y 1

0 –1 –2

1

2

3 x

4

0

3

We find f ( 4π 3) f(

A

5

6

D

︷︸︸︷︷︸B︸ ︸︸ ︷E ︸ ︷ ︸ F ︸ ︷ ︸ G ︸ ︷ ︸︷ ︸ ︸︷C︷ π 3

π 2

2π 3

4π 3π 5π 2 3

2π

We use a test value in each interval to determine the sign of the derivative in each interval.

A: Test π,4 f′ π ( )=2.121>0 4 B: Test 5π, 1 2 f′ π 5 ( )= −0.568<0 12 7 C: Test 7π, 12 f′ π ( )=0.568>0 12 D: Test π, f′(π) = −9<0 17π )=0.568>0 12 19π F: Test 19π, 1 2 f′ ( )= −0.568<0 12 11π G:Test11π,f′ 6 ( )=5.760>0 6 We see that f(x) is increasing on [0, π), (π 2 2π), , 3 (4π 3π 3 , a2n)d 3 4π and (5π 3,2 ]πand decreasing on (π 3,2π), (2π3,3) (3π,5π 3)sothefunctionhasarelativemaximumatx=π3, 2 x = 2π3and x = 3π2 and a relative minimum at x = π2 , x=4πand =5π 3.

3 3x ) We find f(π E: Test 17π, 12 f′

(

f( )π= 9sin( ) 3

π π −4sin3( ) 3 3

Exercise Set 3.1

109

= =

√ √ 9 3 12 3 2 − 8 √ 3 3

46. f (x) = 3 cosx − 2 cos3 x

We find f ( 2π 3) f(

2π 2π ) = 9sin( ) 4sin( 3 3 − √ 12√ 3 9 3 = − 2 8 3 = √ 3

32π ) 3

f′(x) = −3 sin x + 6 cos2x sin x −3 sin x (1 − 2cos2 x) = We solve f′(x) = 0 − 3 sin x(1 − 2cos2x ) = sinx = x = and = x = = = and = = = x = = 2 1 − 2 cosx

We find f ( 3π 2) f(

3π 3π ) = 9sin( ) 4sin( 2 2 − = − 9 −4(1) − = −5

33π ) 2

2

cosx

3 π 2

f( )π= 9sin( ) 4sin( )π − 2 2 = 9− 4 = 5

x and x

We find f ( 4π 3) 4π 4π ) 4sin3( ) = 9sin( − 3 √3 9 3 12√ 3 = − + 8 √2 = 3 −3

5π 5π 5π ) ) = 9sin( −4sin3( 3 √3 ) 3 √ 9 3 12 3 = − + 8 √2 = −3 3

y4 2

–4 –6

1

and

1 ±√2 π 4 3π 4 5π 4 7π 4

︷ ︸ A ︸︸ ︷B ︷︸ ︷︷︸C︸︷︷ ︸︸ E︸ D︷︸︷︷︸︷F ︸ ︷

6

0

x

x

√ √ 2π 3,3 3), ( 3 ,3 −√3)and Thereisarelativemaximumat(π 4π π (3π 3 2, − 5)andarelativeminimumat(2,5),(3, (5π 3, 3−3).Wesketchthegraph √

–2

and

4π ) 3

We find f ( 5π 3) f(

2

cosx

We find f(π2 )

f(

0 0 0 π 2π 0 1

2

3 x

4

5

6

5π 7π 2π π We use a test value in each interval to determine the sign of the derivative in each interval. π A:Testπ 6, f′( )=0.75>0 6 B:Testπ 2 , f ′ ( π ) = −3<0 2 5π 6, f′( )=0.75>0 C: Test 5π D: Test 6 E: Test F: Test 7π 7π 6,f′( ) = −0.75<0 6 9π 9π 6, f′( )=3>0 6 11π 11π 6,f′( )= −0.75<0 6 We see that f(x) is increasing on [0, π), (3π 4,π), and 4 7π π,3π (5π,7π)and 4),(π,5π 4)and(4,2π] decreasingon(4 4 4,x=π 4 so the function has a relative maximum at x = π 3π andx=7πa4 ndarelativeminimumat = 4xandx=5π4.

0

3π

π

We find f(π4 ) f( )π= 3cos( ) 4

π π −2cos3( ) 4 4