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skelet666 [1.2K]
3 years ago
13

PLEASE HELP ASAP!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

Twelve tickets cost $30 --> True

Thirty tickets cost $12 --> False

Each additional costs $2.50 --> True

The table is a partial rep --> True

ordered pairs --> False

Step-by-step explanation:

Twelve tickets cost $30 --> True, you can literally see that in the table

Thirty tickets cost $12 --> False, 30 is not in the table so you don't have that information. Besides, $12 is an unlikely low value for so many tickets.

Each additional costs $2.50 --> True, you can see the difference in the TotalCost column to be consistently 2.50.

The table is a partial rep --> True, values below 11 are not shown for example.

ordered pairs --> False --> Then the x value should be first, e.g., (11, 27.50), since the cost y is a function of the number x.

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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 f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}  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:

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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

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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