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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 . 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. 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).

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Answer:

a) Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on R

b) c=1

Step-by-step explanation:

a)The given function is;

[tex]f(x)=4x^2-3x+2[/tex]

For this function to satisfy the Mean Value Theorem, it must;

i) be continuous on [0,2]

ii)be differentiable on (0,2)

Since [tex]f(x)=4x^2-3x+2[/tex] is a polynomial function, it is continuous on [0,2] and differentiable on (0,2) because polynomial functions are continuous and differentiable on the entire real numbers.

b)

Since the above two hypotheses are satisfied,it implies that, there is a certain [tex]c\in (0,2)[/tex] such that; [tex]f'(c)=\frac{f(2)-f(0)}{2-0}[/tex]

[tex]f'(x)=8x-3[/tex]

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

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

[tex]8c-3=5[/tex]

[tex]8c=5+3[/tex]

[tex]8c=8[/tex]

[tex]c=1[/tex]

Therefore the number that satisfies the conclusion of the Mean Value Theorem is c=1.

We can see from the graph that at x=1, the secant line is parallel to the tangent on the interval [0,2]

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