Rolle's Theorem

**VonNemo19** · June 28th 2009, 09:49 AM

Determine whether Rolle's theorem can be applied to f on the interval and, if so, find all values of c in the open interval (a,b) such that $f'(c)=0$ .

$f(x)=4x-\tan{\pi{x}}, [-\frac{1}{4},\frac{1}{4}]$

Here's what I've done:

$4x$ is everywhere cont./diff. and $\tan{\pi{x}}$ is everywhere cont. on $[-\frac{1}{4},\frac{1}{4}]$ and everywhere diff. on $(-\frac{1}{4},\frac{1}{4})$ , therefore $f(x)$ is everywhere cont./diff. on $(-\frac{1}{4},\frac{1}{4})$ . Furthermore, $f(-\frac{1}{4})=0=f(\frac{1}{4})$ , so Rolle's theorem applies.

$f'(x)=4-\pi\sec^2\pi{x}=0$
$\pi\sec^2\pi{x}=4$

$\sec{\pi{x}}=\pm\frac{2}{\sqrt{\pi}}$

${x}=\frac{1}{\pi}cos^{-1}\left(\pm\frac{\sqrt{\pi}}{2}\right)\Rightarrow{ c}=\frac{1}{\pi}cos^{-1}\left(\pm\frac{\sqrt{\pi}}{2}\right)$

Could someone please verify my result?

**Amer** · June 28th 2009, 10:36 AM

I think it is correct I did not see any mistakes but you should take the solution which is in the interval $[\frac{-1}{4},\frac{1}{4}]$

$c=\frac{1}{\pi} cos^{-1}\left(\frac{\sqrt{\pi}}{2}\right)\approx 0.1533172924205\in [\frac{-1}{4},\frac{1}{4}]$

$c=\frac{1}{\pi}cos^{-1}\left(-\frac{\sqrt{\pi}}{2}\right)\approx 0.8466827075795 \ne [\frac{-1}{4},\frac{1}{4}]$

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