find the extrema of f(x)=x^(2/5), -3<=x<1. I find zero critical points, but the answer says there is a maxima and minima at 0 and -3.

To find the extrema of the function f(x) = x^(2/5) on the closed interval -3 ≤ x < 1, you need to follow these steps:

1. Identify the critical points: A critical point occurs when the derivative of the function is either zero or undefined. To find the derivative of f(x), you can use the power rule for derivatives:

f'(x) = (2/5) * x^(-3/5)

Set f'(x) equal to zero and solve for x:

(2/5) * x^(-3/5) = 0

Since x^(-3/5) cannot be zero (as it would make the function undefined), there are no critical points in this case.

2. Determine the endpoints of the interval: In this case, the interval is given as -3 ≤ x < 1. The endpoints of the interval are x = -3 and x = 1.

3. Evaluate the function at the critical points and endpoints: Calculate the function values of f(x) at the critical points and the endpoints.

f(0) = (0)^(2/5) = 0
f(-3) = (-3)^(2/5) ≈ 1.2457

4. Analyze the function values: Compare the function values calculated in the previous step to determine the maximum and minimum within the given interval. In this case, since f(0) = 0 and f(-3) ≈ 1.2457, we can conclude that f(x) has a minimum at x = 0 and a maximum at x = -3.

Therefore, the extrema of f(x) = x^(2/5) on the closed interval -3 ≤ x < 1 are a minimum at x = 0 and a maximum at x = -3.