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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 4x2 − 3x + 2, [0, 2] Yes, it does not matter if f is continuous or differentiable, every function satifies the Mean Value Theorem. Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on double-struck R. No, f is not continuous on [0, 2]. No, f is continuous on [0, 2] but not differentiable on (0, 2). There is not enough information to verify if this function satifies the Mean Value Theorem. Correct: Your answer is correct. If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisify the hypotheses, enter DNE). c =

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Answer:

c = 1

Step-by-step explanation:

The formula for the MVT is

[tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]

where f'(c) is the derivative of the function:  8x - 3;

f(b) is the function evaluated at an x value of 2 (the second number in the interval):  f(b) = 12;

f(a) is the function evaluated at an x value of 1 (the first number in the interval:  f(a) = 2

Setting up:

[tex]8c-3=\frac{12-2}{2-0}=\frac{10}{2}=5[/tex]

So basically what we end up with is

8c - 3 = 5 and

8c = 8 so

c = 1

This function does in fact satisfy the requirements for the MVT:  it is continuous on the interval and it is differentiable on the interval as polynomials by nature are continuous and differentiable on all values of x.

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