Question

Let f(x)= cos(2x) +e^(-x). for what value of x on the interval (0,3) will f have the same instantaneous rate of change as the average rate of change of f over the interval?

Answer 1

`f(x)` is continuous on [0, 3] and differentiable on (0, 3), so the mean value theorem applies here. It says that there is some
`c` in the open interval (0, 3) such that

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

We have

`f'(x)=-2\sin2x-e^(-x)`

so

`-2\sin2c-e^(-c)=\frac{\cos6+e^(-3)-2}3\implies c\approx1.5418`