Answer:
Month 4 and 5
Step-by-step explanation:
This graph shows a very big increase in the change of the amt of money from 50 to 90 which is a 40 difference
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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All you would need to do for this equation is take the cubed root of 5,832
the answer ends up being b=18
Hello :
<span>1.5x - 3y = 6
3y = 1.5x - 6
y = (1.5x - 6)/3</span>