1. ## Rolle's Theorem

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 $\displaystyle f'(c)=0$.

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

Here's what I've done:

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

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

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

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

Could someone please verify my result?

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

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

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