Question

Find the absolute extrema of the function f(x) = (x^2 + 15)^(2/3) on the interval [-1,4]. The absolute maximum occurs at x = (Simplify your answer. Use a comma to separate.)

Answer 1

To find the absolute extrema of the function f(x) = (x^2 + 15)^(2/3) on the interval [-1,4], you can follow these steps: find the critical points by taking the derivative and setting it equal to zero, evaluate the function at the critical points and endpoints of the interval, and compare the values to determine the absolute extrema.

Step-by-step explanation:

To find the absolute extrema of the function f(x) = (x^2 + 15)^(2/3) on the interval [-1,4], we need to examine the critical points and endpoints within the given interval.

1. First, we find the critical point(s) by taking the derivative of f(x) and setting it equal to zero:
2. f'(x) = (2/3)(x^2 + 15)^(-1/3)(2x) = 0
3. Simplifying, we get 2x = 0, which gives us x = 0 as the only critical point.
4. Next, we evaluate f(x) at the critical point and the endpoints of the interval:
5. f(-1) = (-1^2 + 15)^(2/3) = 14^(2/3) ≈ 6.937
6. f(0) = (0^2 + 15)^(2/3) = 15^(2/3) ≈ 6.708
7. f(4) = (4^2 + 15)^(2/3) = 31^(2/3) ≈ 8.084
8. Comparing these values, we can see that the absolute minimum occurs at x = -1, where f(x) ≈ 6.937, and the absolute maximum occurs at x = 4, where f(x) ≈ 8.084.