link solutions manual for additional calculus topics 11th edition by raymond barnett karl byleen michael ziegler

Karl Byleen

x f’(x) = -x-2

(Using Power Rule)

6

f(3)

(x) = -6x-4

= - x 4

2 f(x) = ln(1 + x)

1

= (1 + x)-1

f'(x) = 1  x

f"(x) = (-1)(1 + x) (Using Power Rule)

f(3)

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

(Using Power Rule)

2

=

(1  x)3

f’(x) = -e

f”(x) = e

f (x) = -e

4 f(x) = ln(1 + 3x)

3 = 3(1 + 3x)

f’(x) =

1  3x

(Using Power

Rule) f(3)

(x) = (-9)(-2)(3)(1 + 3x)-3

= 54(1 + 3x)-3

f(4)

(x) = (54)(-3)(3)(1 + 3x)-4

= -486(1 + 3x)-4 486

=

(1  3x)4

5 f(x) = e5x

f(3)

= 53

e5x

f(4)

(x) = 53

(5)e5x

= 54

e5x

= 625e5x

96 TAYLOR POLYNOMIALS AND INFINITE SERIES

1

= (2 + x)-1

6 f(x) =

2  x

f’(x) = (-1)(2 + x)

f”(x) = (-1)(-2)(2 + x) = 2(2 + x)

f (x) = (2)(-3)(2 + x) = -6(2 + x)

f(4) (x) = (-6)(-4)(2 + x)-5 = 24(2 + x)-5

7 f(x) = e-x f(0) = e-0 = 1

-x -0 = -1

f'(x) = -e f'(0) = -e

f"(x) = e -x f"(0) = e-0 = 1

f(3) (x) = -e -x f(3) (0) = -e-0 = -1

f(4) (x) = e-x f (4) (0) = e-0 = 1

Using 2,

p (x) = f(0) + f'(0)x + f"(0) (3)

x + f (0) x + 4 2! 3!

Thus, 1 1 1

p (x) = 1 - x + x - x + x = 1 - x + 4 2! 3! 4!

8 f(x) = e4x f(0) = 1

f'(x) = 4e f'(0) = 4

f"(x) = 16e f"(0) = 16

f(3) (x) = 64e4x f (3) (0) = 64

f"(0) 2 (3) 3

f (0) Thus, p 3 (x) = f(0) + f'(0)x + x +

x 2! 3!

= 1 + 4x + 16 x2 + 64 x3 = 1 + 4x + 8x2 + 2! f(x) = (x + 1)3 , 3!

9 f(0) = 1

f'(x) = 3(x + 1)2 , f'(0) = 3

f"(x) = 6(x + 1), f"(0) = 6

f (x) = 6, f (0) = 6

f (x) = 0 f (0) = 0

p (x) = 1 + 3x + 6 x2 + 6 x3 = 1 + 3x + 3x2 + x3 4 2! 3!

(4)

f (0) x

4!

1 x - 1 x + 1 x

32

EXERCISE 2-1 97

10 f(x) = (1 - x)

4

, f(0) = 1

f'(x) = -4(1 - x) , f'(0) = -4

f"(x) = 12(1 - x) , f"(0) = 12

f(3)

(x) = -24(1 - x), f(3)

(0) = -24

12 24

4x + 6x2

- 4x3

x3 = 1 -

1 2 2

f'(x) = (2) =

1  2x 1  2x

-2

f"(x) = -2(1 + 2x) (2) =

= 2 f"(0) = -4

(1  2x)

(3) -3 16 (3)

f (x) = 8(1 + 2x) (2) = (1  2x) f (0) = 16

(3)

2 + f (0) 3

3

2! x

3! x

Thus, p (x) = 0 + 2x - 4 x2

+ 16 x3

+ 8 x3 3!

f(0) = 3

= 1

f'(x) = 3(x  1) f'(0) = 3

 2 2

f"(x) = 9(x  1)

f (x) = 27(x  1) 8/3

f"(0) = -9

f (0) =

27

p 3(x) = f(0) + f'(0)x +

Thus, p (x) = 1 + 1 x - 2

9

f"(0) (3)

f (0)

x2 + x3

x + 10 x = 1 + 1

x + 5 x

13 f(x) = 4

= (x + 16)1/4 f(0) = 2 x  16 1 1

f'(x) = -3/4 f'(0) = 4(x + 16) 32 f"(x) = - 3 (x + 16) f"(0) = - 3 16 1 3 2048

p (x) = 2 + x - x

2 32 4096

14 (A) f(x) = x - 1 f(0) = -1

f'(x) = 4x f'(0) = 0

f"(x) = 12x f"(0) = 0

f(3) (x) = 24x f(3) (0) = 0

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + 2 + f (0) 3 3

2! x 3! x

Thus, p3(x) = -1

Now |p 3(x) - f(x)| = |-1 - (x4 - 1)| = |x4 | = |x|4 and |x|4 < 0.1 implies |x| < (0.1)1/4 ≈ 0.562 Therefore, -0.562 < x < 0.562

(B) From part (A),

f(4) (x) = 24 f(4) (0) = 24 (3)

f"(0) Using 2, p (x) = f(0) + f'(0)x + x + f (0) x 4 24 2! 3!

Thus, p 4(x) = -1 + x 4 = -1 + x 4 = x 4 - 1 = f(x) 4!

and |p4(x) - f(x)| = 0 < 0.1 for all x 15 (A) f(x) = x f(0) = 0

f'(x) = 5x f'(0) = 0

f"(x) = 20x f"(0) = 0

f (x) = 60x f (0) = 0

f(4) (x) = 120x f(4) (0) = 0

p4(x) = 0

|p4(x) - f(x)| = |0 - x5

| = |x|5

=

< 0.01 or |x| < (0.01)1/5

= 0.398

Therefore, -0.398 < x < 0.398

(B) f(0) = f'(0) = f"(0) = f(3)

(0) = f(4)

(0)

= 0 f(5)

(x) = 120 and hence f(5)

(0) =

120 p5(x) = 5! x5

= x5

| = 0 < 0.01

(4)

+ f(0)x4

4!

EXERCISE 2-1 99

16 f(x) = x

3

f(1) = 1 f'(x) = 3x2

f'(1) = 3

(1) = 6

p (x) = 1 + 3(x - 1) + 6 (x - 1)2

+ 6 (x - 1)3

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

+ (x - 1)3

f'(x) = 2x - 6 f'(3) = 0

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

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

18 f(x) = ln(2 - x)

f'(x) = -(2 - x)

f"(x) = -(2 - x)

f (x) = -2(2 - x)

f(4)

(x) = -6(2 - x)-4

1

(x - 1)2

= -(x - 1) - 2

f(1) = 0 f'(1) = -1 f"(1) = -1

(1) = -6

-

3

-

4

3 (x - 1) 4(x - 1)

f"(x) = 4e -2x

(3) - 2x

f (x) = -8e

f(0) = 1 f'(0) = -2 f"(0) = 4

(0) = -8

f"(0) (3)

3

4 x 2

-

- 4 x 3

Thus, p (x) = 1 - 2x + 8 x3

= 1 - 2x + 2x2

3

Now, e- 0.5

= f(0.25) ≈ p 3 (0.25)

4

= 1 - 2(0.25) + 2(0.25)2

(0.25)3

= 0.60416667

- 3

20 f(x) = x = x1/2 f(1) = 1

1 x-1/2 1

f'(x) =

f'(1) =

2 2

f"(x) = - 1 x-3/2 f"(1) = - 1

4 4

3 3

f(3) (x) = x-5/2 f(3) (1) =

8 8

15 15

f(4) (x) = - f(4) (1) = -

16 x-7/2 16

Thus, (3) (4)

f"(1)

f (1) f (1) p4(x) = f(1) + f'(1)(x - 1) + (x - 1)2 + (x - 1)3 + (x - 1)4 2! 3! 4!

= 1 + 1 (x - 1) - 1 (x - 1)2 + 3 (x - 1)3 - 15 (x - 1)4 2 4 2! 8  3! 16  4!

1 1 1 5

= 1 +

(x - 1) - (x - 1)2 + 3

- (x - 1)4 2 8 16 (x - 1) 128 Now,

1.2 = f(1.2) ≈ p4(1.2) (1.2 - 1)2 + (1.2 - 1)3

(1.2 - 1)4 = 1 + 1 (1.2 - 1) - 1 1 - 5

2 8 16 128

= 1 + 1 (0.2) - 1 (0.2)2 + 1 (0.2)3 - 5 (0.2)4

2 8 16 128

= 1.0954375

21 f(x) = 1 = (4 - x)

-1

f'(x) = -1(4 - x)-3(-1) = (4 - x) -3

f"(x) = -2(4 - x) (-1) = 1·2(4 - x)

f(4)

(x) = 2·3(-4)(4 - x)-5

(-1) = 1·2·3·4(1 - x)-5

f(n)

(x) = n!(4 - x)-(n+1)

22 f(x) = 4 = 4(1 +x) -x

1  x f'(x) = -4(1 + x)

f"(x) = (-1) (4)(2)(1 + x) = 4(2!)(-1) (1 + x)

M

f(n)

(x) = 4(n!)(-1)n

(1 + x)-(n+1)

23 f(x) = e3x

f'(x) = e 3x

(3) = 3e 3x

f"(x) = 3e3x

e3x

f(3)

(x) = 9e3x

(3) = 27e3x

= 33

e3x

f(4)

(x) = 27e3x

(3) = 81e3x

= 34

e3x

f(n)

(x) = 3n

e3x

24 f(x) = ln(2x + 1)

-1

f'(x) = 2(2x + 1) = (-1) (2x + 1)

(2x +

1)-2

f(3)

(x) = (-1)4

23

(2!)(2x + 1)-3

f(n)

(x) = (-1)n+1

2n

25 f(x) = ln(6 - x)

f'(x) = 1 (-1) = - 1 = -(6 - x)

6 x -2 -2 6 f"(x) = (6 - x) (-1) x

= -(6 - x)

f(4)

(x) = 1·2·3(6 - x)-4

(-1) = -1·2·3(6 -

x)-4

f(n)

(x) = -(n - 1)!(6 - x)-n

26 f (x) = e x/2

1 x/2



f"(x) = ex/2 = 2 ex/2

 2  2

f(3)

(x) = 1 ex/2

3

f (x) = n e x/2

27 From Problem 31,

1 1 2! (3) 3! (n) n! f(0) = , f'(0) = , f"(0) = , f (0) = , … , f (0) =

Thus,

x

+ x + x + 1

x + … +

n 4 42

43

44

4n  1

From problem 32, f (x) = 4(n!)(-1) (1 + x) and

(n) n

The coefficient of xn f (0)



4(n!)( 1)

= 4(-1)n

is n! n!

29.From Problem 33,

=

Thus,

p (x) = 1 + 3x +

n

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

1) From problem 34,

3 2

(n ) n+1 n -n f (n) (0) n+1 n 2

f (x) = (-1) 2 ((n - 1)!)(2x + 1) and n! = (-1) n

Thus, p (x) = 2x - 22 x2 + 23 x3

31 From Problem 35,

1 1 (3) 2! (n) (n  1)!

f(0) = ln 6, f'(0) = - , f"(0) = - , f (0) = - , …, f (0) = -

6 6 6 Thus,

1

1 1 1

p (x) = ln 6 - x - x - x3 - … - x

n

6 2 3 

n  n

EXERCISE 2-1 105

From problem 36, f (n)

(x) = 1 (n) 1

e x/2 and f (0) =

2 n!2

x2

+ 1 x3 1

x2

+xn

2!2 2 3!2 3 n!2 n

f(1) = 2

f(3)

(x) = (-1)3

2(3!)x-4

2(n!)x -(n+1)

and

= 2(-1)

pn(x) = 2 - 2(x - 1) + 2(x - 1)2

- 2(x - 1)3

+ … + 2(-1)n

(x - 1)n

a

f(x) = ln x f(1) = 0 = f(1) = 0

1

f'(x) = x

1 f'(1) = 1 f"(1) 1 1

f"(x) = - f"(1) = -1 2 = 2! = - 2! = - 2

f(3)

(x) =

(1) = 2

(3)

(4) 2  3 (4) (4) 3! 1

4

a f (1)

= -3!

x

f (n)

(x) = f (n) (1) = a

n =

(  1)n  1

(n  1)!

(-1)n+1

(n - 1)! (n) n 1  1)! n  1

= (  1) (n = (  1)

n! n!

Step 4 The nth degree Taylor polynomial is:

1 (x - 1) 2 + 1 (x - 1) 3 - 1 (x - 1) n

p (x) = (x - 1) - (x - 1)4 + … +

35 f(x) = e

x

(n)

f (n) (x) = e x 1

f (  2)

n! n!e

Thus, p (x) = 1 + 1 (x + 2) + 1 (x + 2)2

+ … + 1 (x + 2)n

e2 e2

36.f(x) = x5

+ 1

(A) Fourth-degree Taylor polynomial p4(x) for f at 0 is:

p (x) = f(0) + F’(0) x + F’’(0) x2

(0) x3

(0) x4

f’(x) = 5x + 6x + 16x ; f’(0) = 0

f”(x) = 20x + 12x + 16 ; f”(0) = 16

f (x) = 60x + 12 ; f (0) = 12

f(4)

(x) = 120x ; f

(4)

(0) = 0

p 4(x) = 1 + 8x2

+ 2x3

= 2x3

+ 8x2

+ 1 (B) The degree of the polynomial is 3

+ 1

f”(x) = 30x + 12x ; f”(0) = 0

f (x) = 120x + 12 ; f (0) = 12

f (x) = 360x ; f (0) = 0

f (x) = 720x ; f (0) = 0

f (x) = 720 ; f (0) = 720

f(n)

(x) = 0 for n ≥ 7

Thus, for n = 0, 3 and 6, the nth-degree Taylor polynomial for f at 0 has degree n

f”(x) = 12x ; f”(0) = 0

f (x) = 24x ; f (0) = 0

f (x) = 24 ; f (0) = 24

f(n)

(x) = 0 for n ≥ 5

Thus, for n = 0 and n = 4; the nth degree Taylor polynomial for f at

0 has degree n

EXERCISE 2-1 10

9

39 f(x) = ln(1 + x) f(0) = 0

f"(x) = (1  x) f"(0) = -1

(3) 2 (3)

f (x) = f (0) = 2

(1  x)3

Thus,

x2

x2

x3

p1(x) = x, p 2(x) = x - 2 , p3(x) = x - 2 + 3

x p1(x) p2(x) p3(x) f(x) -0.2 -0.2 -0.22 -0.222667 -0.223144 -0.1 -0.1 -0.105 -0.105333 -0.105361

0.1 0.1 0.095 0.095333 0.09531 0.2 0.2 0.18 0.182667 0.182322

x p1(x) - f(x) p2(x) - f(x) p3(x) - f(x)

-0.2 0.023144 0.003144 0.000477 -0.1 0.005361 0.000361 0.000028

0.1 0.00469 0.00031 0.000023 0.2 0.017678 0.002322 0.000345

40 f(x) = ln(1 + x) f(0) = 0

f'(x) = (1 + x) f'(0) = 1

f"(x) = -(1 + x) f"(0) = -1

f(3)

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

f(3)

(0) = 2 Thus,

1 2 1 3

p (x) = x - x + x

Using a graphing utility, we find that

|p3(x) - ln(1 + x)| < 0.1 for -0.654 < x < 0.910

41 f(x) = e

x

f(a) = ea

f'(x) = e f'(a) = e

f"(x) = e f"(a) = e

f(3) (x) = ex f(3) (a) = ea

M M

f(n) (x) = ex f(n) (a) = ea

Thus, (3)

(a) p (x) = f(a) + f'(a)(x - a) + (x - a)2 + f 2! 3! (x - a) 3 n

(n) (a)

+ + f (x - a) n n!

= e a + e a (x - a) + ea (x - a) 2 + ea (x - a) 3 + + ea (x - a) n 2! 3! n! = e a  1  (x  a)  1 (x  a) 1 (x a)  1 (x  a) n   1 2! 3! n!  = e a n k

 (x - a) k 0

k ! 

42 f(x) = ln x -1 f'(x) = x f(3) (x) = (-1)4 (2!)x-3 f(4) (x) = (-1)5 (3!)x-4 f( n) (x) = (-1)n+1 ((n - 1)!)x-n Thus,

p (x) = ln a + 1 (x - a) - 1 n a 2a2 n (  1)k  1 f(a) = ln a

1

f'(a) = a

f"(a) = - 1 2

a

f(3) (a) = (-1)4 (2!) 1

3 a f(4) (a) = (-1)5 (3!) 1 4

a

f (n) (a) = (-1) n+1 ((n - 1)!) 1 n a (x - a)2 + 1 (x - a)3

3a n+1

n

- … + (-1) 1 n (x - a)

na

= ln a +  k (x - a)

43 f(x) = (x + c)n

f'(x) = n(x + c)

f"(x) = n(n - 1)(x + c)n- 2

f(0) = cn

f'(0) = nc

M

f (x) = n!(x + c) f (0) = n!c

f(n)

(x) = n! f(n)

(0) = n!

Thus,

x + n(n  1) c n! cx n!

= c + x + c n-2 x 2 + … + x

(n 1)! 2!(n  2)!

=

cn-kxk





k  0 k!(n k)!

44.Let f(x) be a polynomial of degree k, k ≥ 0 Then

f(x) = a 0 + a 1x + a 2 x2 + a3x3

+ … + a k xk f(0) = a 0

f'(x) = a 1 + 2a 2x + 3a 3x2 + … + ka kxk-1 f'(0) = a 1

f(3)

(x) = 6a 3 + … + k(k - 1)(k - 2)a k xk-3 f

(3)

(0) = 6a 3 = 3!a 3

In general, f (m)  m!am for m  0,1,2,K k

(0) = 

for m  k

 0

(3)

(n)

f"(0) Since p (x) = f(0) + f'(0)x + 2+ f (0) 3 + … + f (0) n

it follows that pn(x) = f(x) for all n ≥ k

45.f(x) = ex

f(0) = 1

f”(x) = ex

f(k)

(x) = ex

; f

(k)

p (x) = 1 + 1 x + 1 x + … + 1 x + 1x11

For x > 0, ex

> p 11 (x) and hence

Take x = 2(11!)1/11

, then

e x – p (x) > 1 (2(11!)1/11 )11 = 211 = 2048 10 11! So, there exist values of x for which |p 10(x) – ex | = |ex – p 10 (x)| ≥ 100 46.f(x) = 1 = x-1 f(1) = 1 f’(x) = -x-2 ; f’(1) = -1 = -1! or f (1) = -1

1!

f”(x) = (-1)(-2)x-3 ; f”(x) = 2 = 2! or f (1) = 1

2!

M

f(k) (1)

f(k) (1) = (-1)k k! or = (-1)k

k!

Therefore,

p 12 (x) = 1 – (x – 1) + (x – 1)2 - - (x – 1)11 + (x – 1)12 , and

1

|p12(x) – f(x)| = 1  (x  1)  (x  1)2  L  (x  1)12 x 1 |x – x(x – 1) + x(x – 1)2 - … + x(x – 1)12 – 1| = x If we take x = 0.001, then 1 = 1000 and every term involving x on x

the right-hand side of the above equation is positive

and smaller than x

Thus,

|p12(x) – f(x)| ≥ 1000(1 – 13x) = 1000(1 – 0.013) =

987 So there exist values of x ≠ 0 for

which |p12(x) – f(x)| ≥ 100

EXERCISE 2-1 115

47.ln 1.1

Let f(x) = ln(1 + x)

= (1 + x)-1

f’(x) =

f”(x) = -(1 + x)

f(3)

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

f(n)

(x) = (n – 1)!(-1)n-1

(1 + x)-n

Rn(x) = f(n  1)(t)xn  1

for some t between 0 and x

(t)| = |n!(-1)n

(1 + t)-(n+1)

| = n!(1 + t)-(n+1)

< n! for t >

0 Therefore,

|R (x)| = f(n  1)(t)xn 1

<

n! x n  1

=

x n  1

n (n  1)! (n  1)! n  1

and

(0.1)n 1

R (0.1) <

n (n  1)

For n = 4, |R (0.1)| <

4 = 0.000 002 < 0.000 005, and hence the

5

x2

+ x3

- x4

polynomial with the lowest degree is p (x) = x - 1 1 1

4 2 3 4 which has degree 4

ln(1.1) ≈ p (0.1) = 0.1 - 1 (0.1) + 1 (0.1) - 1 (0.1)

4 2 3 4

≈ 0.095308 CHAPTER 2 REVIEW 209