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
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](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%28x%29%26%3D%5Csqrt%7B%282-x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%5E3%7D%5C%5C%20f%28x%29%26%3D%282-x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5C%5C%20f%27%28x%29%26%3D%5Cfrac%7B3%7D%7B2%7D%282-x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Ccdot%20%5Cfrac%7B2%7D%7B3%7D%5Ccdot%20%28-x%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%20f%27%28x%29%26%3D%5Cfrac%7B-%5Csqrt%7B2-x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cend)
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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Answer:
Step-by-step explanation:
-2+ (-7) = -2 - 7
Answer:
30/7
Step-by-step explanation:
2 x 10 -13
20 - 13
7
4 x 10 -10
40 - 10
30
Answer:
i think the answer is -y^2/20 + 2y/3 - 1/4
Answer:
The index of a number says how many times to use the number in a multiplication. It is written as a small number to the right and above the base number. In this example: 82 = 8 × 8 = 64. The plural of index is indices. (Other names for index are exponent or power.)
Step-by-step explanation: