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Does the following function satisfy the hypotheses of the Mean Value Theorem on the given...

Question:

Does the following function satisfy the hypotheses of the Mean Value Theorem on the given interval? Provide a reason for the answer.

$$f(x) = x^{\frac{2}{3}}, \, [-1, 8] $$

Mean Value Theorem:

According to the Mean Value Theorem, if f(x) is a continuous and differentiable function over an interval (a,b),

then there exists a point x0 in (a,b) such that

{eq}\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x) dx = f(x_0) {/eq}

Answer and Explanation:

The function

$$\displaystyle f(x) = x^{\frac{2}{3}}, \, [-1, 8] $$

does satisfy the hypotheses of the Mean Value Theorem on the given interval because it is continuous

but not differentiable at x=0 since left and right derivatives goes to infinity

$$\displaystyle f'(x) = \frac{2}{3}x^{-\frac{1}{3}} \\ \displaystyle f'(0^-) = -\infty \\ \displaystyle f'(0^+) = \infty $$


Learn more about this topic:

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What is the Mean Value Theorem?

from Math 104: Calculus

Chapter 8 / Lesson 3
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