Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = e−5x, [0, 3] Yes, f is continuous and differentiable on double-struck R, so it is continuous on [0, 3] and differentiable on (0, 3) . No, f is continuous on [0, 3] but not differentiable on (0, 3). There is not enough information to verify if this function satisfies the Mean Value Theorem. No, f is not continuous on [0, 3]. Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.

Answer 1

Yes, and the first choice's reasoning is the only correct one.

The MVT guarantees the existence of at least one $c\in(0,3)$ such that

$f'(c)=\dfrac{f(3)-f(0)}{3-0}$

We have $f'(x)=-5e^{-5x}$, so that

$-5e^{-5c}=\dfrac{e^{-15}-1}3\implies c=-\dfrac15\ln\dfrac{1-e^{-15}}{15}$