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To verify that the function f(x)=eˣ satisfies the hypotheses of the Mean Value Theorem on the interval [0,ln4] and find all numbers c that satisfy the conclusion of the Mean Value Theorem, please select the correct option:

A. The function does not satisfy the Mean Value Theorem on this interval.
B. The function satisfies the Mean Value Theorem, and c=ln2 is a number that satisfies the conclusion.
C. The function satisfies the Mean Value Theorem, and c=2 is a number that satisfies the conclusion.
D. The function satisfies the Mean Value Theorem, and there is no c that satisfies the conclusion.

1 Answer

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Final answer:

The function f(x)=e^x satisfies the Mean Value Theorem on the interval [0,ln4] because it is continuous on the closed interval and differentiable on the open interval. However, none of the given options provide the correct value of c that satisfies the theorem; the correct c is ln(3/ln4), not given in the choices.

Step-by-step explanation:

To verify that the function f(x)=ex satisfies the hypotheses of the Mean Value Theorem on the interval [0,ln4], we first need to ensure that the function is continuous on the closed interval [0,ln4] and differentiable on the open interval (0,ln4). Since the exponential function ex is continuous and differentiable everywhere, it satisfies these conditions.

According to the Mean Value Theorem, there exists at least one number c in the open interval (0,ln4) such that f'(c) = [f(ln4) - f(0)] / (ln4 - 0). Calculating the right side gives us:

  • f(ln4) = eln4 = 4
  • f(0) = e0 = 1
  • Thus, [f(ln4) - f(0)] / (ln4 - 0) = (4 - 1) / ln4 = 3 / ln4

The derivative of f(x)=ex is f'(x)=ex. We set f'(c) = ec equal to the previous result to find c:

ec = 3 / ln4

By taking the natural logarithm of both sides, we get:

c = ln(3/ln4)

Therefore, the function satisfies the Mean Value Theorem, but c=ln(3/ln4) is not among the given options. Thus, option D is correct; the function satisfies the Mean Value Theorem, but there is no c that satisfies the conclusion among the provided options.

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