Question

Verify Rolle's theorem\quad for the function $f\left(x\right)=e^{-x}\sin x,xϵ[0,\pi ]$

Answer 1

$f\left(x\right)=e^{-x}\sin x,x\in (0,\pi )$

For Rolle's Theorem, $f\left(0\right)=f\left(\pi \right)$ & $f\left(x\right)$ must be continuous & differentiable over $[0,\pi ]$

Let us check if $f\left(0\right)=f\left(\pi \right)$

$\Rightarrow f\left(0\right)=e^{-0}\sin \left(0\right)=0\Rightarrow f\left(\pi \right)=e^{-\pi }\sin \left(\pi \right)=0$

Therefore, $f\left(0\right)=f\left(\pi \right)$

The function $e^{-x}$ & $\sin x$ are both continuous & differentiable over $[0,\pi ]$

Therefore, Rolle's Theorem can be applied for the function given.

There exists ${}^{\prime }{c}^{\prime }$ such that $f^{\prime }\left(c\right)=0$

$f^{\prime }\left(x\right)=-e^{-x}\sin x+e^{-x}\cos x{f}^{\prime }\left(c\right)=-e^{-c}\sin \left(c\right)+e^{-c}\cos \left(c\right)=0\Rightarrow e^{-c}[\cos c-\sin c]=0\Rightarrow \cos c=\sin c\Rightarrow tanc=1\Rightarrow c=\frac{\pi }{4}$

Hence, Rolle's Theorem is verified.