Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 2x³6x² - 18x + 8, [-2, 4] Step 1 The absolute maximum and minimum values of f occur either at a critical point inside the interval or at an endpoint of the interval. Recall that a critical point is a point where f '(x) = 0 or is undefined. We begin by finding the derivative of f. f'(x) = 6x² - 12x - 18 Step 2 We now solve f '(x) = 0 for x, which gives the following critical numbers. (Enter your answers as a comma-separated list.) x= -1,3 f(-2) = 2 Step 3 We must now find the function values at the critical numbers we just found and at the endpoints of the interval [-2, 4]. f(-1) 16 f(3) -48 f(4) = -34 X X X X X 6x² 12x - 18 -1,3 X

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 2x³6x² - 18x + 8, [-2, 4] Step 1 The absolute maximum and minimum values of f occur either at a critical point inside the interval or at an endpoint of the interval. Recall that a critical point is a point where f '(x) = 0 or is undefined. We begin by finding the derivative of f. f'(x) = 6x² - 12x - 18 Step 2 We now solve f '(x) = 0 for x, which gives the following critical numbers. (Enter your answers as a comma-separated list.) x= -1,3 f(-2) = 2 Step 3 We must now find the function values at the critical numbers we just found and at the endpoints of the interval [-2, 4]. f(-1) 16 f(3) -48 f(4) = -34 X X X X X 6x² 12x - 18 -1,3 X

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter4: Calculating The Derivative

Section4.2: Derivatives Of Products And Quotients

Problem 35E

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