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Determine the location and value of the absolute extreme values of f on the given​ interval, if they exist. f(x) = (x-2)^(4/3) on [-7,7]

Question

Determine the location and value of the absolute extreme values of f on the given interval, if they exist.

Let
f(x)=(x2)43 f(x) = (x-2)^{\frac{4}{3}}
on the interval [7,7][-7, 7]

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Solution

To find the absolute extreme values of a function on a given interval, we need to find the critical points of the function, evaluate the function at these points and at the endpoints of the interval, and compare these values.

Step 1: Find the derivative of the function. The derivative of f(x) = (x-2)^(4/3) is f'(x) = (4/3)(x-2)^(1/3).

Step 2: Find the critical points. Set the derivative equal to zero and solve for x: (4/3)(x-2)^(1/3) = 0 This gives us x = 2 as a critical point.

Step 3: Evaluate the function at the critical points and at the endpoints of the interval. f(2) = (2-2)^(4/3) = 0 f(-7) = ((-7)-2)^(4/3) = (-9)^(4/3) = 81 f(7) = (7-2)^(4/3) = 5^(4/3) = 125^(1/3) = 5

Step 4: Compare these values. The absolute maximum value is 81 at x = -7, and the absolute minimum value is 0 at x = 2.

This problem has been solved

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