Explanation Calculation: Assume f ( x ) = x n + 1 3 + 1 to be the required function and a = 0 . Here, n is a positive integer. The first derivative of function is, f ′ ( x ) = d d x ( x n + 1 3 + 1 ) = d d x x n + 1 3 + d d x 1 = ( n + 1 3 ) x n + 1 3 − 1 = ( n + 1 3 ) x n − 2 3 Substitute 0 for x in f ′ ( x ) = ( n + 1 3 ) x n − 2 3 to find the value of f ′ ( 0 ) . f ′ ( 0 ) = ( n + 1 3 ) ( 0 ) n − 2 3 = 0 The second derivative of the function is, f ″ ( x ) = d d x f ′ ( x ) = d d x ( n + 1 3 ) x n − 2 3 = ( n + 1 3 ) d d x x n − 2 3 = ( n + 1 3 ) ( n − 2 3 ) x n − 2 3 − 1 Simplify it further. f ″ ( x ) = ( n + 1 3 ) ( n − 2 3 ) x n − 5 3 Substitute 0 for x in f ″ ( x ) = ( n + 1 3 ) ( n − 2 3 ) x n − 5 3 to find the value of f ″ ( 0 ) . f ″ ( 0 ) = ( n + 1 3 ) ( n − 2 3 ) ( 0 ) n − 5 3 = 0 The third derivative of the function is, f ‴ ( x ) = d d x f ″ ( x ) = d d x ( n + 1 3 ) ( n − 2 3 ) x n − 5 3 = ( n + 1 3 ) ( n − 2 3 ) d d x x n − 5 3 = ( n + 1 3 ) ( n − 2 3 ) ( n − 5 3 ) x n − 5 3 − 1 Simplify it further. f ‴ ( x ) = ( n + 1 3 ) ( n − 2 3 ) ( n − 5 3 ) x n − 8 3 Substitute 0 for x in f ‴ ( x ) = ( n + 1 3 ) ( n − 2 3 ) ( n − 5 3 ) x n − 8 3 to find the value of f ‴ ( 0 )

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

10th Edition

ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Chapter 3.5, Problem 49E

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A function f that has derivatives of order up through and including n at point a but doesn’t have (n+1)st derivative at point a.

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Chapter 3.5, Problem 48E

Chapter 3.5, Problem 50E

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Chapter 3 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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A function f that has derivatives of order up through and including n at point a but doesn’t have ( n + 1 ) st derivative at point a .

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A function f that has derivatives of order up through and including n at point a but doesn’t have (n+1)st derivative at point a.

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