[tex]f(x)=e^{-3x}\implies f'(x)=-3e^{-3x}[/tex]
[tex]f[/tex] is differentiable across its domain, so there is some [tex]c\in(0,9)[/tex] such that
[tex]f'(c)=\dfrac{f(9)-f(0)}{9-0}[/tex]
Solve for [tex]c[/tex]:
[tex]-3e^{-3c}=\dfrac{e^{-27}-1}9[/tex]
[tex]e^{-3c}=\dfrac{1-e^{-27}}{27}[/tex]
[tex]-3c=\ln\dfrac{1-e^{-27}}{27}[/tex]
[tex]c=-\dfrac13\ln\dfrac{1-e^{-27}}{27}\approx1.0986[/tex]
