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

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

We have

$f'(x)=-2\sin2x-e^{-x}$

so

$-2\sin2c-e^{-c}=\dfrac{\cos6+e^{-3}-2}3\implies c\approx1.5418$