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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 4x2 − 2x + 3, [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. 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

Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on double-struck R.

[tex]c = 1[/tex]

EXPLANATION

We want to determine whether ,

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

satisfies the Mean Value Theorem, on the interval,

[0,2].

We check to see if the function satisfies all the hypotheses of the Mean Value Theorem.

1. The function is continuous on the closed interval, [0,2]

2. The function is differentiable on the open interval (0,2).

3. There is a c, such that,

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

Since the given function is a polynomial it is continuous on [0,2] and differentiable on (0,3) because polynomials are continuous and differentiable on the entire real numbers.

We now differentiate the function to obtain;

[tex]f'(x) = 8x - 2[/tex]

This implies that;

[tex]f'(c) = \frac{f(2) - f(0)}{2 - 0} [/tex]

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

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

[tex]f(2) = 15[/tex]

[tex]f(0) = 4( 0) - 2(0) + 3 = 3[/tex]

This implies that:

[tex]8c - 2 = \frac{15 - 3}{2 } [/tex]

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

[tex]8c - 2 =6[/tex]

[tex]8c = 6 + 2[/tex]

[tex]8c = 8[/tex]

[tex]c = 1[/tex]

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