Explanation Formula used: Quotient rule : If f and g are differentiable functions, then ( f g ) ′ = f ′ ⋅ g − g ′ ⋅ f g 2 . Calculation: The given function is f ( x ) = x 2 + 2 x − 5 x 3 + 1 and the value of point x = 2.1. Calculate the first derivative of the above function using the quotient rule. f ′ ( x ) = d f d x = d d x ( x 2 + 2 x − 5 x 3 + 1 ) = ( x 3 + 1 ) d d x ( x 2 + 2 x − 5 ) − ( x 2 + 2 x − 5 ) d d x ( x 3 + 1 ) ( x 3 + 1 ) 2 = ( x 3 + 1 ) ( 2 x + 2 ) − ( x 2 + 2 x − 5 ) ( 3 x 2 ) ( x 3 + 1 ) 2 On further simplification, f ′ ( x ) = ( 2 x 4 + 2 x 3 + 2 x + 2 ) − ( 3 x 4 + 6 x 3 − 15 x 2 ) ( x 3 + 1 ) 2 = 2 x 4 + 2 x 3 + 2 x + 2 − 3 x 4 − 6 x 3 + 15 x 2 ( x 3 + 1 ) 2 = − x 4 − 4 x 3 + 15 x 2 + 2 x + 2 ( x 3 + 1 ) 2 = − ( x 4 + 4 x 3 − 15 x 2 − 2 x − 2 ) ( x 3 + 1 ) 2 Calculate the second derivative of the above function f ′ ( x ) using the quotient rule. f ″ ( x ) = d f ′ d x = d d x ( − ( x 4 + 4 x 3 − 15 x 2 − 2 x − 2 ) ( x 3 + 1 ) 2 ) = { − ( x 3 + 1 ) 2 d d x ( x 4 + 4 x 3 − 15 x 2 − 2 x − 2 ) + (

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 3.3, Problem 5TE

To determine

The second derivative of the function f(x)=x2+2x−5x3+1 at the given value of x = 2.1.

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Chapter 3.3, Problem 4TE

Chapter 3.3, Problem 6TE

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Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The second derivative of the function f ( x ) = x 2 + 2 x − 5 x 3 + 1 at the given value of x = 2.1.

Solution Summary: The author calculates the first derivative of the above function using the quotient rule.

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The second derivative of the function f(x)=x2+2x−5x3+1 at the given value of x = 2.1.

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Chapter 3 Solutions