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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Think I know why don't you do the math cadence man all you said is "oh how many more pennies are produced in half a year really not even giving us any numbers it doesn't matter what device were on we can always look up the answer online your lucky
Try this option:
1] P(S)=0.84;
2] P(S or C, but not both)=0.4;
3] P(C)=0.76;
4] P(S∪C)=0.6;
5] P(C, but not S)=0.16;
6] P(S∩C)=1.00.
Answer:
E) we will use t- distribution because is un-known,n<30
the confidence interval is (0.0338,0.0392)
Step-by-step explanation:
<u>Step:-1</u>
Given sample size is n = 23<30 mortgage institutions
The mean interest rate 'x' = 0.0365
The standard deviation 'S' = 0.0046
the degree of freedom = n-1 = 23-1=22
99% of confidence intervals 
(from tabulated value).
using calculator
Confidence interval is
the mean value is lies between in this confidence interval
(0.0338,0.0392).
<u>Answer:-</u>
<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>
