Consider the function $$f(x) = \frac 1 {x^2+1}$$ $f(x) \to 0$ as $x \to \pm\infty$, thus we can say that $f(x)$ has roots at $\pm\infty$. Can we apply Rolle's theorem to prove $f'(c)=0, c \in (-\infty, \infty)$?
NOTE: $c=0$ does exist as specified above. This also works for other functions like $\frac{x}{x^2+1}$. However, is using Rolle's theorem a valid proof in this case?
The derivative $f'$ has IVP. If it is never $0$ then it has constant sign. This means $f$ is strictly monotonic. But $f(x)\to 0$ as $x \to \pm \infty$ so we must have $f \equiv 0$!. Hence the answer is YES.