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What are the relative extrema of the function f() = (8-)(+1)²?

User Carmela
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Final answer:

To find the relative extrema of the function f(x) = (8-x)(x+1)², we take the derivative of the function, set it equal to zero, and use the second derivative test to determine the nature of the critical point.

Step-by-step explanation:

The function is given by f(x) = (8-x)(x+1)². To find the relative extrema, we need to find the critical points of the function, where the derivative is equal to zero or undefined. Let's take the derivative of f(x):

f'(x) = -(x+1)² - 2(8-x)(x+1) = -2(x+1)(9-x)

Setting f'(x) equal to zero, we get -2(x+1)(9-x) = 0. This equation is zero when x = -1 or x = 9. The function is defined for 0 ≤ x ≤ 8, so we only consider the critical point x = -1. To determine whether it's a relative maximum or minimum, we can use the second derivative test. Taking the second derivative f''(x), we get f''(x) = -2(18-2x)

f''(-1) = -2(18-2(-1)) = -2(18+2) = -40. Since the second derivative is negative at x = -1, this critical point represents a relative maximum.

User Ron Srebro
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