# How do you find the relative extrema for #f(x) =2x- 3x^(2/3) +2# on the interval [-1,3]?

(This is a wonderful example of why you cannot ignore the second kind of critical numbers.)

For this function, we have:

decreasing on #(0,1) so

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To find the relative extrema for (f(x) = 2x - 3x^{\frac{2}{3}} + 2) on the interval ([-1, 3]), follow these steps:

- Find the critical points of the function by setting the derivative equal to zero and solving for (x).
- Evaluate the function at the critical points and endpoints of the interval.
- Identify the relative extrema by comparing the function values at these points.

First, find the derivative of (f(x)):

[f'(x) = 2 - 2x^{-\frac{1}{3}}]

Set (f'(x) = 0) and solve for (x):

[2 - 2x^{-\frac{1}{3}} = 0]

Solve for (x) to find the critical points:

[x^{-\frac{1}{3}} = 1]

[x = 1]

Evaluate the function (f(x)) at the critical point and endpoints of the interval:

[f(-1) = 2(-1) - 3(-1)^{\frac{2}{3}} + 2] [f(1) = 2(1) - 3(1)^{\frac{2}{3}} + 2] [f(3) = 2(3) - 3(3)^{\frac{2}{3}} + 2]

[f(-1) = -3 - 3 + 2 = -4] [f(1) = 2 - 3 + 2 = 1] [f(3) = 6 - 9 + 2 = -1]

Since (f(-1) = -4), (f(1) = 1), and (f(3) = -1), the relative extrema occur at (x = -1) and (x = 1).

Therefore, the relative maximum is (f(1) = 1) and the relative minimum is (f(-1) = -4).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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