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  • Karl Byleen

    Link full download: https://getbooksolutions.com/download/solutions-manual-for-

    additional-calculus-topics-11th-edition-by-raymond-barnett-karl-byleen-michael-

    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

    https://getbooksolutions.com/download/solutions-manual-for-additional-calculus-topics-11th-edition-by-raymond-barnett-karl-byleen-michael-ziegler/ https://getbooksolutions.com/download/solutions-manual-for-additional-calculus-topics-11th-edition-by-raymond-barnett-karl-byleen-michael-ziegler/ https://getbooksolutions.com/download/solutions-manual-for-additional-calculus-topics-11th-edition-by-raymond-barnett-karl-byleen-michael-ziegler/

  • 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) =