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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?

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  • $\begingroup$ You can't apply Rolle's Theorem at infinities. $\endgroup$ – Jerry Jun 30 '18 at 15:48
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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.

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  • $\begingroup$ So what is the reason we use Rolle's theorem in interval between two finite values even when the concept is working here? Is there any counter example $\endgroup$ – Umesh shankar Jun 30 '18 at 15:51
  • $\begingroup$ No, there is no counterexample. If there was, I would not have written that the statement is correct. I'm just saying that Rolles's theorem is about functions defined on intervals $[a,b]$, with $a<b$. That's not the case here. $\endgroup$ – José Carlos Santos Jun 30 '18 at 15:55
  • $\begingroup$ You might note that one can prove it works by essentially repeating the proof of Rolle's theorem... $\endgroup$ – David C. Ullrich Jun 30 '18 at 16:08
  • $\begingroup$ @DavidC.Ullrich Indeed. And we can also use Rolle's theorem to prove it. $\endgroup$ – José Carlos Santos Jun 30 '18 at 16:11
  • $\begingroup$ In a cleaner way than I realized at first: Assume there exists $x$ with $f(x)\ne0$ (if not we're done). IVT shows that there exist $a<x$ and $b>x$ with $f(a)=f(x)/2=f(b)$. $\endgroup$ – David C. Ullrich Jun 30 '18 at 16:44
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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 $x_2(+\infty)=+0$.

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