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Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 - 9x² – 216x + 4, [-4, 5)

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 - 9x² – 216x + 4, [-4, 5)

Calculus For The Life Sciences
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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Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 – 9x² – 216x + 4, [-4, 5] 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) 18x – 18x – 216 18:2 – 18x 216 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 = -3,4 -3, 4 Step 3 We must now find the function values at the critical numbers we just found and at the endpoints of the interval [-4, 5]. f(-3) = 409 409 f(4) = -620 -620 f(-4) = 340 340 f(5) = -551 -551 Step 4 Therefore, on the interval [-4, 5], the absolute minimum of f(x) is and the absolute maximum is
Transcribed Image Text:Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 – 9x² – 216x + 4, [-4, 5] 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) 18x – 18x – 216 18:2 – 18x 216 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 = -3,4 -3, 4 Step 3 We must now find the function values at the critical numbers we just found and at the endpoints of the interval [-4, 5]. f(-3) = 409 409 f(4) = -620 -620 f(-4) = 340 340 f(5) = -551 -551 Step 4 Therefore, on the interval [-4, 5], the absolute minimum of f(x) is and the absolute maximum is
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