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)=(f(3)-f(0))/(3-0)`

We have
`f'(x)=-5e^(-5x)`, so that

`-5e^(-5c)=\frac{e^(-15)-1}3\implies c=-\frac15\ln(1-e^(-15))/(15)`