Suppose that $\,f:[a,b]\to \mathbb{R}$, where $\,b-a\ge 4,\,$ is differentiable in $(a,b)$ and continuous in $[a,b]$. Prove that there is $x_0\in (a,b)$, such that


But, I could not make the slightest approach towards the solution of this problem. Please help. Thank you.

  • $\begingroup$ Is it continuous at $a$ and $b$? $\endgroup$ – Hasan Saad Mar 9 '15 at 8:48
  • $\begingroup$ $f$ is continous at $a$ and $b$. Yes. $\endgroup$ – Swadhin Mar 9 '15 at 8:50

First proof. (Assuming that $\,f$ is continuously differentiable). Assume that for every $x\in[a,b]$: $$ f'(x)\ge 1+f^2(x), $$ then $$ \frac{d}{dx}\tan^{-1}\big(f(x)\big)=\frac{f'(x)}{1+f^2(x)}\ge 1, $$ and thus integrating in $[a,b]$ $$ \tan^{-1}\big(f(b)\big)-\tan^{-1}\big(f(a)\big)\ge b-a\ge 4. $$ But $\tan^{-1}(x)\in(-\pi/2,\pi/2)$, and hence $$ \tan^{-1}\big(f(b)\big)-\tan^{-1}\big(f(a)\big)<\pi. $$

Alternative proof. (Assuming that $f$ is ONLY differentiable). As $\tan^{-1}: \mathbb R\to (-\pi/2,\pi/2)$, then, using the mean value theorem for $g(x)=\tan^{-1}\big(f(x)\big)$, with $g'(x)=f'(x)/\big(1+f^2(x)\big)$, we obtain that there exists a $\xi\in (a,b)$, such that $$ \pi>\tan^{-1}\big(f(b)\big)-\tan^{-1}\big(\,f(a)\big)= (b-a)\frac{f'(\xi)}{1+f^2(\xi)}, $$ and hence $$ 1+f^2(\xi)>\frac{\pi}{b-a}\big(1+f^2(\xi)\big)= f'(\xi). $$

  • $\begingroup$ Wow... I didn't see that at all... $\endgroup$ – Hasan Saad Mar 9 '15 at 8:50
  • $\begingroup$ Well, that's very advanced and nice stuff...thank you. $\endgroup$ – Swadhin Mar 9 '15 at 8:56

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