Question

Find all numbers c that satisfy the conclusion of the Mean Value Theorem for the following function and interval.

f(x)=e^(-3x), [0,9]

Answer 1

`f(x)=e^(-3x)\implies f'(x)=-3e^(-3x)`


`f` is differentiable across its domain, so there is some
`c\in(0,9)` such that


`f'(c)=(f(9)-f(0))/(9-0)`

Solve for
`c`:


`-3e^(-3c)=\frac{e^(-27)-1}9`

`e^(-3c)=(1-e^(-27))/(27)`

`-3c=\ln(1-e^(-27))/(27)`

`c=-\frac13\ln(1-e^(-27))/(27)\approx1.0986`