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Consider the following function.

f(x) = 4 ? x2/3
Find
f(?8) and f(8).
f(?8) = f(8) = Find all values c in (?8, 8) such that
f?'(c) = 0.
(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
c =
Based off of this information, what conclusions can be made about Rolle's Theorem?
This contradicts Rolle's Theorem, since f is differentiable, f(?8) = f(8), and f?'(c) = 0 exists, but c is not in (?8, 8).This does not contradict Rolle's Theorem, since f?'(0) = 0, and 0 is in the interval (?8, 8). This contradicts Rolle's Theorem, since f(?8) = f(8), there should exist a number c in (?8, 8) such that f?'(c) = 0.This does not contradict Rolle's Theorem, since f?'(0) does not exist, and so f is not differentiable on (?8, 8).Nothing can be concluded.

User Bato
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2 Answers

4 votes

Final answer:

The function values f(-8) and f(8) are evaluated using the given function. The value of c in the interval (-8, 8) such that f'(c) = 0 is found by solving the derivative equation. The conclusion made about Rolle's Theorem is that it does not contradict, as f'(0) = 0 and 0 is in the interval (-8, 8).

Step-by-step explanation:

First, we need to evaluate f(-8) and f(8) using the given function f(x). For f(-8), substitute -8 into the function:

f(-8) = 4 - (-8)^(2/3) = 4 - (-2) = 6.

Similarly, for f(8), substitute 8 into the function:

f(8) = 4 - 8^(2/3) = 4 - 4 = 0.

Now, let's find the value of c in the interval (-8, 8) such that f'(c) = 0. To do this, we need to find the derivative of f(x).

f'(x) = (d/dx)(4 - x^(2/3)) = -2/3 * x^(-1/3).

To find c, we need to solve the equation -2/3 * c^(-1/3) = 0.

Since the derivative is equal to 0 when c = 0, we have a value of c in the interval (-8, 8) such that f'(c) = 0.

Therefore, the correct conclusion about Rolle's Theorem is: This does not contradict Rolle's Theorem, since f'(0) = 0, and 0 is in the interval (-8, 8).

User Sergei Vasilenko
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3 votes

Final answer:

After calculating, we found that f(-8) = f(8) = 0, and no real value of c exists in the interval (-8, 8) for which f'(c) = 0, because f'(x) is not defined at x = 0. This does not contradict Rolle's Theorem, as the function is not differentiable over the entire interval, particularly at x = 0.

Step-by-step explanation:

First, let's find the values of f(-8) and f(8) for the function f(x) = 4 - x2/3.

f(-8) = 4 - (-8)2/3 = 4 - (64)1/3 = 4 - 4 = 0

f(8) = 4 - (8)2/3 = 4 - (64)1/3 = 4 - 4 = 0

Since f(-8) = f(8), we see that the function has the same value at both ends of the interval [-8, 8]. Next, we will find the values of c in the interval (-8, 8) where the derivative f'(c) equals 0.

The derivative is f'(x) = -½ x-1/3. Setting f'(x) to 0 and solving for x gives us no real solution since x-1/3 cannot equal 0. Therefore, the list of values c where f'(c) = 0 is DNE (Does Not Exist).

Considering Rolle's Theorem, we see that it does not apply here because although f(-8) = f(8), the function is not differentiable over the entire interval [-8, 8], specifically at x = 0. Rolle's Theorem states that if a function is continuous on [a, b] and differentiable on (a, b), and f(a) = f(b), then there exists at least one c in (a, b) such that f'(c) = 0. However, because f'(0) does not exist, the function f(x) is not differentiable at x = 0, which is within the interval. Thus, the theorem's conditions are not fully met.

User Martin Reiner
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7.7k points

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