Nonnegative derivative bounded by function, show that $f$ is zero

$(e^{-x}f(x))'=e^{-x}(f'(x)-f(x)) \leq 0$ so $e^{-x}f(x)$ is decreasing. If $f(x_o)=0$ then $e^{-x}f(x)$ is non-negative and $\leq e^{-x_0}f(x_0) =0$ for $x>x_0$ so $f(x)=0$ for $x \geq x_0$. For $x \leq x_0$ you already have a proof.