Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = e−5x, [0, 1] Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem. Yes, f is continuous and differentiable on double-struck R, so it is continuous on [0, 1] and differentiable on (0, 1) . There is not enough information to verify if this function satisfies the Mean Value Theorem. No, f is not continuous on [0, 1]. No, f is continuous on [0, 1] but not differentiable on (0, 1). 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 satisfy the hypotheses, enter DNE). c =

Answer 1

$f(x)=e^{-5x}$ is continuous on [0, 1] and differentiable on (0, 1), so yes, the MVT is satisfied.

By the MVT, there is some $c\in(0,1)$ such that

$f'(c)=\dfrac{f(1)-f(0)}{1-0}$

The derivative is

$f'(x)=-5e^{-5x}$

so we get

$-5e^{-5c}=e^{-5}-1\implies e^{-5c}=\dfrac{1-e^{-5}}5\implies-5c=\ln\dfrac{1-e^{-5}}5$

$\implies\boxed{c=-\dfrac15\ln\dfrac{1-e^{-5}}5}$