Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x) = -x^2 +9 on [-2,3]
Question
Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x) = -x^2 +9 on [-2,3]
Solution
Sure, here are the steps to find the absolute extreme values of the function f(x) = -x^2 + 9 on the interval [-2,3]:
Step 1: Find the derivative of the function. The derivative of f(x) = -x^2 + 9 is f'(x) = -2x.
Step 2: Set the derivative equal to zero and solve for x. Setting -2x = 0 gives x = 0.
Step 3: Evaluate the function at the critical points and the endpoints of the interval. The critical point is x = 0. The endpoints of the interval are x = -2 and x = 3.
So, we have: f(0) = -(0)^2 + 9 = 9, f(-2) = -(-2)^2 + 9 = 5, f(3) = -(3)^2 + 9 = 0.
Step 4: The largest of these values is the absolute maximum, and the smallest is the absolute minimum. So, the absolute maximum value is 9 at x = 0, and the absolute minimum value is 0 at x = 3.
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