Question

$f\left(x\right)=\tan x$ in $0\le x\le \pi$

Is Rolle's theorem applicable?

Answer 1

$f\left(x\right)=\tan x$ in $0\le x\le \pi$
At $x=\frac{\pi }{2}$
$f(\frac{\pi }{2}+)\ne f(\frac{\pi }{2}-)\ne f\left(\frac{\pi }{2}\right)$

$\Rightarrow f\left(x\right)$ is not continuous at $x=\frac{\pi }{2}$ and not differentiable,
therefore Rolle's first two conditions are not satisfied.
Hence, Rolle's theorem is not applicable.