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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Options
The circle at the new location has _____________ the original circle.
- the same center as
- twice the circumference of
- half the radius of
- the same area as
Answer:
the same area as
Step-by-step explanation:
When a circle is translated and reflected, the center of the circle will change; however, its area, circumference, radius and diameter remain the same.
This is so because, translation and reflection only affect the positioning of the circle not the size.
Considering the above analysis, we can conclude that option d answers the question correctly.
Answer:
You can use 5 to simplify 5 and 15, giving you 1 and 3
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I can only assume that you meant, "Solve for x:"
Apply the exponent 3/2 to both sides of this equation. The result will be
3/2
343 = x/6.
Multiplying both sides by 6 isolates x:
3/2
6*343 = x Since 7^3 = 343, the expression for x
can be rewritten as
3/2
6*(7^3) = x which can be further simplified, as follows:
x = 6^(3/2)*7^(9/2), or:
x = 6^(3/2)*7^(8/2)*√7, or
x = 6^(3/2)*7^4*√7
Answer:
x = -7
Step-by-step explanation:
2x+3 = x-4
2x-x = -4-3
x = -7
hope it helps ☺️
