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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2x^2 - 8x - 24
First, we can factor a 2 out of this expression to simplify it.
2(x^2 - 4x - 12)
Now, we can try factoring this two ways: by using the quadratic formula, or by using the AC method.
We're gonna try using the AC method first.
List factors of -12.
1 * -12
-1 * 12
2 * -6
-2 * 6 (these digits satisfy the criteria.)
Split the middle term.
2(x^2 - 2x + 6x - 12)
Factor by grouping.
2(x(x - 2) + 6(x - 2)
Rearrange terms.
<h3><u>(2)(x + 6)(x - 2) is the fully factored form of the given polynomial.</u></h3>
Answer:
D one solution.
Step-by-step explanation:
It's linear. It is going to have at most 1 solution
8 + x/6 = x/6 - 4 + 5x/12 Subtract x/6 from both sides
8 = -4 + 5x/12 Add 4 to both sides
8+4 = 5x / 12 Combine
12 = 5x / 12 Multiply both sides by 12
12 * 12 = 5x * 12 / 12 Cancel
144 = 5x Divide by 5
144/5 = 5x/5
28.8 = x
Answer:
ΔPTS≅ΔRTA by AAS axiom of congruency
Step-by-step explanation:
Consider ΔPQA and ΔRQS
∠PQA=∠RQS (Vertically Opposite Angles)
∠QAP=∠QSR (Complementary of two equal angles, ∠RAT and∠PST)
Due to angle sum property of a triangle, we come to the conclusion that
∠APQ=∠SRQ
Consider ΔPTS and ΔRTA
TA=TS (Given)
∠RAT=∠PST(Given)
∠APQ=∠SRQ (Proved above)
Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.
Answer:
196
Step-by-step explanation:
