Prove that there exists some $c>0$ such that $f'(c)=-e^{-c}$

Question

Let $f$ be a differentiable function on $[0,\infty)$ and $f(0)=1$.

Further given that $|f(x)|\leq e^{-x}$ for all $x\geq 0$.

Prove that there exists some $c>0$ such that $f'(c)=-e^{-c}$

My Attempt:

Let $g(x)=f(x)-e^{-x}$

$g(0)=0$

I am trying to invoke Rolle's theorem but I need at least one $x$ other than $0$ for which $g(x)=0$

Answer 1

If the $c$ doesn’t exist then $g’$ always $<0$ or $g’$ always $>0$, for otherwise we can use the intermediate property of $g’$ to find such a $c$.

Since $g(0)=0$ and $g(x)\leq0$, we must have $g’(x)<0$ for all $x>0$, so $g(x)<g(1)<g(0)=0$ for all $x>1$. Letting $x\to\infty$ gives that $\limsup_{x\to\infty}g(x)\leq g(1)<0$, contradicting to the fact that $\lim_{x\to\infty}g(x)=0$.