Finding another root to apply Rolle's theorem.

Question

Let $f(x)$ be continuous on $[0,+\infty]$, continuously differentiable on $(0,+\infty)$ and $f(0)=1$; $\underset{x\to +\infty}{\lim} f(x)=0.$ Prove that there exists $c> 0$ such that $f'(c)+e^{-c}=0$.
My idea is to apply Rolle's theorem for $g(x)=f(x)-e^{-x}$. I found that $g(0)=0$ and stuck in finding another root (which is not equal to $0$) of $g(x)=0$ to complete my proof using Rolle's theorem. Can someone help me?

Answer 1

For this problem, I would forget Rolle's theorem, since $g(x)$ as you have defined it need not have any other root. For example, if you start with $f(x)=e^{-x}+xe^{-x}$, then $g(x)=xe^{-x}$ which has no root on $(0,\infty)$.

However, since $g(0)=0$, and $\lim_{x\to\infty} g(x)=0$, you can argue that $|g(x)|$ achieves a maximum at some $c\in(0,\infty)$, giving a point where $g'(c)=0$.

To see this, note that either $g$ is identically equal to $0$ (in which case every point on $(0,\infty)$ achieves the maximum), or $\sup_{x\in(0,\infty)}|g(x)|\neq 0$.

In the latter case, by the limiting condition, since $g(x)\to 0$ as $x\to\infty$, we have $|g(x)|< \frac{1}{2} \sup_{t\in(0,\infty)}|g(t)|$ for all $x\in [M,\infty)$ for some $M$. Therefore $|g|$ approaches its supremum on $[0,M]$, and thus achieves its supremum at some $c\in (0,M)$, since $[0,M]$ is a compact interval and $g$ is continuous (we know it doesn't realize its supremum on the endpoints since $g(0)=0$ and $|g(M)|<\frac{1}{2}\sup_{t\in(0,\infty)}|g(t)|$).

Since $|g(c)|$ is a maximum, $g'(c)=0$.

Answer 2

For $g(x)=f(x)-e^{-x}$ we have $g(0)=0$ and $\lim_{x\to\infty}g(x)=0.$ Therefore the function $$ h(x)=\begin{cases} g((1-x)^{-1}-1) & 0\le x<1\\ 0 & x=1\end{cases}$$ is continuous on $[0,1]$ and differentiable in $(0,1)$ and $h(0)=h(1)=0.$ By the Rolle theorem $$0=h'(b)=-(1-b)^{-2}g'((1-b)^{-1}-1)$$ for some $0<b<1.$ Thus for $c=(1-b)^{-1}-1$ we get $$0=g'(c)=f'(c)+e^{-c}$$