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1 and f"(-6) 432 1 In the previous step, we determined that f"(6) = 432 Thus, at x = 6, the second derivative is negative and the function will have a maximum value. Similarly, at x = -6, the second derivative is positive and the function will have a minimum value. Now, evaluate the given function f(x) at the critical values x = 6, -6 to determine the values of x + 36 the relative extrema. f(6) = 6/72 f(-6) -6/72 %3D To summarize, state the relative extrema of the function. relative maximum (х, у) = ) relative minimum (х, у)

1 and f"(-6) 432 1 In the previous step, we determined that f"(6) = 432 Thus, at x = 6, the second derivative is negative and the function will have a maximum value. Similarly, at x = -6, the second derivative is positive and the function will have a minimum value. Now, evaluate the given function f(x) at the critical values x = 6, -6 to determine the values of x + 36 the relative extrema. f(6) = 6/72 f(-6) -6/72 %3D To summarize, state the relative extrema of the function. relative maximum (х, у) = ) relative minimum (х, у)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In the previous step, we determined that f"(6) and f"(-6) = 432 432 Thus, at x = 6, the second derivative is negative and the function will have a maximum v value. Similarly, at x = -6, the second derivative is positive and the function will have a minimum value. Now, evaluate the given function f(x) at the critical values x = 6, -6 to determine the values of x2 + 36 the relative extrema. f(6) 6/72 f(-6) -6/72 = To summarize, state the relative extrema of the function. relative maximum (х, у) relative minimum (х, у) %D
Transcribed Image Text:In the previous step, we determined that f"(6) and f"(-6) = 432 432 Thus, at x = 6, the second derivative is negative and the function will have a maximum v value. Similarly, at x = -6, the second derivative is positive and the function will have a minimum value. Now, evaluate the given function f(x) at the critical values x = 6, -6 to determine the values of x2 + 36 the relative extrema. f(6) 6/72 f(-6) -6/72 = To summarize, state the relative extrema of the function. relative maximum (х, у) relative minimum (х, у) %D
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