The critical values of the given function are x = 6, -6. If the second derivative is positive at one of these points, the function will have a minimum value there; if the second derivative is negative, the function will have a maximum value. Take the second derivative of the function. 36 – x2 f'(x) (x2 + 36)2 (x? + 36)(( 2 f"(x) x) - (36 – x)(2(x2 x3(2(x² + 36)(2x) |-2 (x² + 36)4 2x(x² + 36) – 2(36 – x?)| + 36 (x² + 36)* 2x(x? - 108 (x2 + 36)3 Now, evaluate the second derivative at the critical values x = 6, -6. f"(6) f"(-6) Enter a fraction, integer, or exact decimal. Do not approximate.

The critical values of the given function are x = 6, -6. If the second derivative is positive at one of these points, the function will have a minimum value there; if the second derivative is negative, the function will have a maximum value. Take the second derivative of the function. 36 – x2 f'(x) (x2 + 36)2 (x? + 36)(( 2 f"(x) x) - (36 – x)(2(x2 x3(2(x² + 36)(2x) |-2 (x² + 36)4 2x(x² + 36) – 2(36 – x?)| + 36 (x² + 36)* 2x(x? - 108 (x2 + 36)3 Now, evaluate the second derivative at the critical values x = 6, -6. f"(6) f"(-6) Enter a fraction, integer, or exact decimal. Do not approximate.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter1: Functions

Section1.2: Functions Given By Tables

Problem 32SBE: Does a Limiting Value Occur? A rocket ship is flying away from Earth at a constant velocity, and it...

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