[tex]f(x)=e^{-5x}[/tex] is continuous on [0, 1] and differentiable on (0, 1), so yes, the MVT is satisfied.
By the MVT, there is some [tex]c\in(0,1)[/tex] such that
[tex]f'(c)=\dfrac{f(1)-f(0)}{1-0}[/tex]
The derivative is
[tex]f'(x)=-5e^{-5x}[/tex]
so we get
[tex]-5e^{-5c}=e^{-5}-1\implies e^{-5c}=\dfrac{1-e^{-5}}5\implies-5c=\ln\dfrac{1-e^{-5}}5[/tex]
[tex]\implies\boxed{c=-\dfrac15\ln\dfrac{1-e^{-5}}5}[/tex]