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Determine the absolute extrema of the function f(x)=x³-12x² on the closed interval [0,12]. What are the x-values of the maxima or minima (if any)?

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User Qendrim
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1 Answer

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Final answer:

To determine the absolute extrema of the function f(x)=x³-12x² on the closed interval [0,12], we need to find the highest and lowest points of the function within that interval. The maximum value of the function is f(8) = -128, which occurs at x = 8, and the minimum value is f(0) = 0, which occurs at x = 0.

Step-by-step explanation:

To determine the absolute extrema of the function f(x)=x³-12x² on the closed interval [0,12], we need to find the highest and lowest points of the function within that interval.

We start by finding the critical points of the function, which are the points where the derivative is equal to zero or undefined. Taking the derivative of f(x), we get f'(x) = 3x² - 24x. Setting this equal to zero, we find the critical points x = 0 and x = 8.

Next, we evaluate the function at the end points of the interval, which are f(0) = 0 and f(12) = -144.

Finally, we compare the values of the function at the critical points and end points to determine the absolute extrema. The maximum value of the function is f(8) = -128, which occurs at x = 8, and the minimum value is f(0) = 0, which occurs at x = 0.

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User Jhonson
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