Question

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)

Answer 1

Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is $f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}$  and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

• f(a)=f(b)
• f(x) must be continuous in [a,b].
• f(x) must be differentiable in (a,b).

To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

$\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end$

From this point of view we can see that the given function is not defined for x=0.

Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.

Learn more about Rolle's Theorem from here brainly.com/question/12279222

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