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• Karl Byleen

ziegler/

1.f(x) = 1 = x

-1 x

f’(x) = -x -2

(Using Power Rule)

f”(x) = 2x -3

6

f (3)

(x) = -6x -4

= -

x 4

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

f'(x) =

1 = (1 + x)

-1

1  x

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

(Using Power Rule)

f (3)

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

(Using Power Rule)

=

2

(1  x) 3

3. f(x) = e -x

f’(x) = -e -x

f”(x) = e -x

f (3)

(x) = -e -x

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

f’(x) = 1  3x

f”(x) = 3(-1)(3)(1 + 3x) -2

= -9(1 + 3x) -2

(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) = e 5x

f’(x) = 5e 5x

f”(x) = 5(5)e

5x = 5

2 e 5x

f (3)

(x) = 5 2 (5)e

5x = 5

3 e 5x

f (4)

(x) = 5 3 (5)e

5x = 5

4 e 5x

= 625e 5x

96 TAYLOR POLYNOMIALS AND INFINITE SERIES

= 3(1 + 3x) -1

• 6. f(x) =

1 = (2 + x)

-1

2  x

f’(x) = (-1)(2 + x) -2

f”(x) = (-1)(-2)(2 + x) -3

= 2(2 + x) -3

f (3)

(x) = (2)(-3)(2 + x) -4

= -6(2 + x) -4

f (4)

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

= 24(2 + x) -5

7. f(x) = e -x

f'(x) = -e

-x

f"(x) = e -x

f (3)

(x) = -e -x

f (4)

(x) = e -x

Using 2,

f(0) = e -0

= 1

f'(0) = -e -0

= -1

f"(0) = e -0

= 1

f (3)

(0) = -e -0

= -1

f (4)

(0) = e -0

= 1

f"(0) (3) (4)

p (x) = f(0) + f'(0)x + x 2 + f (0) x

3 + f (0) x

4

4

3!

Thus,

2! 4!

1

1

1

1

1

1

p (x) = 1 - x + x 2

- x 3 + x

4 = 1 - x + x 2

- x 3

+ x 4

4

24 2! 3! 4! 2 6

8. f(x) = e 4x

f(0) = 1

f'(x) = 4e 4x

f'(0) = 4

f"(x) = 16e 4x

f"(0) = 16

f (3)

(x) = 64e 4x

f (3)

(0) = 64

f"(0) (3)

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

x 2 +

f (0) x 3

3 2! 3!

= 1 + 4x + 16 x2 + 64 x3 = 1 + 4x + 8x

2 + 32 x3

2!

3

3!

9. f(x) = (x + 1) 3 , f(0) = 1

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

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

f (3)

(x) = 6, f (3)

(0) = 6

f (4)

(x) = 0 f (4)

(0) = 0

p (x) = 1 + 3x + 6 x 2 + 6 x3 = 1 + 3x + 3x

2 + x

3

4

2! 3!

EXERCISE 2-1 97

• 10. f(x) = (1 - x) 4 , f(0) = 1

f'(x) = -4(1 - x) 3 , f'(0) = -4

f"(x) = 12(1 - x) 2 , f"(0) = 12

f (3)

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

(0) = -24

Thus, 12

24

p

(x) = 1 - 4x +

x2 -

x3 = 1 - 11.

3 2! 3!

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

f'(x) =

(2) =

1  2x 1  2x

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

(2) = 4 (1  2x)

2

f (3)

(x) = 8(1 + 2x) -3

(2) = 16

(1  2x) 3

4x + 6x 2 - 4x

3

f(0) = ln(1) = 0

2

f'(0) = 1  2  0 = 2 f"(0) = -4

f (3)

(0) = 16

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + x2 + f (0) x3

3 2!

4

16

3!

8

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

3 = 2x - 2x

2 + x3

3

3!

2! 3

12. f(x) = 3

= (x + 1) 1/3

f(0) = 3

= 1

x  1 1

f'(x) =

1

f'(0) =

1

3(x  1) 2/3

3 2 2

f"(x) =

f"(0) = -

9(x  1) 5/3

9

f (3)

(x) = 10

f (3)

(0) = 10

27(x  1) 8/3

27

f"(0) (3)

p

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

x2 + f (0)

x3

3 2! 3!

1

2

10

1

1

5

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

+ x 3

= 1 + x - x 2 + x

3

3

3 9  2! 27 3! 3 9 81

• 13. f(x) = 4

= (x + 16) 1/4

f(0) = 2

x  16

f'(x) =

1 (x + 16)

-3/4 f'(0) =

1

4 32

3

3

f"(x) = - (x + 16) -7/4

f"(0) = -

16 1

3

2048 p (x) = 2 + x - x

2

2

4096

32

14. (A) f(x) = x 4 - 1 f(0) = -1

f'(x) = 4x 3

f'(0) = 0

f"(x) = 12x 2

f"(0) = 0

f (3)

(x) = 24x f (3)

(0) = 0

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + x2 + f (0) x3 3 2!

3!

Thus, p3(x) = -1

Now |p 3 (x) - f(x)| = |-1 - (x 4

- 1)| = |x 4 | = |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 2

+ f (0) x 3

4

24

2!

3!

Thus, p

(x) = -1 + x 4

= -1 + x 4 = x

4 - 1 = f(x) 4 4!

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

15. (A) f(x) = x 5

f(0) = 0

f'(x) = 5x 4

f'(0) = 0

f"(x) = 20x 3

f"(0) = 0

f (3)

(x) = 60x 2

f (3)

(0) = 0

f (4)

(x) = 120x f (4)

(0) = 0

p4(x) = 0

|p4(x) - f(x)| = |0 - x 5 | = |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! x 5 = x

5

|p5(x) - f(x)| = |x 5 - x

5 | = 0 < 0.01

Thus for all x, |p5(x) - f(x)| < 0.01.

(4)

+ f(0)

x4 4!

EXERCISE 2-1 99

• 16. f(x) = x 3

f(1) = 1

f'(x) = 3x 2

f'(1) = 3 f"(x) = 6x f"(1) = 6

f (3)

(x) = 6 f (3)

(1) = 6

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

3

3

2! 3!

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

3

17. f(x) = x 2 - 6x + 10 f(3) = 1

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

f"(x) = 2 2

f"(4) = 2

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

2

2

2!

• 18. f(x) = ln(2 - x) f(1) = 0

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

f'(1) = -1

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

f"(1) =