Can $a$ and $b$ in Rolle's Theorem use Infinities?

Question

I am actually analyzing the behavior of the function $$f(x)=\frac{x}{1+x^2}$$

we have $f$ is Continuous and Differentiable over $\mathbb{R}$

Also

$$\lim_{x \to -\infty} f(x)=\lim_ {x \to \infty}f(x)=0$$

So by Rolle's Theorem in $(-\infty \:,\: \infty)$ we have there exists atleast one $c$ in $(-\infty \:,\: \infty)$ such that $$f'(c)=0$$ $\implies$

$$\frac{1-c^2}{(1+c^2)^2}=0$$

$$c=\pm1$$

Is this application of Rolle's theorem correct?

Answer 1

The statement is correct. That is, if $f\colon\mathbb{R}\longrightarrow\mathbb{R}$ is differentiable and $\lim_{x\to\pm\infty}f(x)=0$, then there is a real number $c$ such that $f'(c)=0$. However, this is not Rolle's theorem, although it is similar to it.

Answer 2

No. This is not a correct application of Rolle's Theorem. You have $$ f(x)=\dfrac{x}{1+x^2}. $$ Then:

(1) $f(+0)=+0$ and $f(+\infty)=+0$.

(2) $f$ is continuous in $[0,+\infty)$.

(3) $f$ is differentiable in $[0,+\infty)$.

Now, let $\varepsilon<1/2$ be a sufficiently small positive number. Using (1): there exist $x_1,x_2\in [0,+\infty)$, such that $0<x_1<x_2$ and $$ f(x_1)=f(x_2)=\varepsilon. $$ The above, (2), (3) and Rolle's Theorem imply the result.

Remark. Of course $x_1=x_1(\varepsilon)$ and $x_2=x_2(\varepsilon)$. Moreover $x_1(+0)=+0$ and $x_2(+\infty)=+0$.