[tex]f(x)[/tex] is continuous on [0, 3] and differentiable on (0, 3), so the mean value theorem applies here. It says that there is some [tex]c[/tex] in the open interval (0, 3) such that
[tex]f'(c)=\dfrac{f(3)-f(0)}{3-0}[/tex]
We have
[tex]f'(x)=-2\sin2x-e^{-x}[/tex]
so
[tex]-2\sin2c-e^{-c}=\dfrac{\cos6+e^{-3}-2}3\implies c\approx1.5418[/tex]
