Question

$f\left(x\right)=e^x(\sin x-\cos x)$ in $(\frac{\pi }{4},\frac{5\pi }{4}).$ Is Rolle's theorem applicable?

Answer 1

We have $f\left(x\right)=e^x(\sin x-\cos x)in(\frac{\pi }{4},\frac{5\pi }{4}).$
$f^{\prime }\left(x\right)=2e^x\sin x$ ...(1)
$R{f}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{2e^{x+h}\sin (x+h)-2e^x\sin x}{h}$
$=2e^x(\cos x+\sin x)$
$L{f}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x-h)-f\left(x\right)}{-h}=2e^x(\cos x+\sin x)$
$\Rightarrow L{f}^{\prime }\left(x\right)=Rf\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)$ exists for all values of x in $[\pi /4,5\pi /4]$
Now, $f\left(\frac{\pi }{4}\right)=e^{\frac{\pi }{4}}[\sin \left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{4}\right)]=0$
$f\left(\frac{5\pi }{4}\right)=e^{\frac{5\pi }{4}}[\sin \left(\frac{5\pi }{4}\right)-\cos \left(\frac{5\pi }{4}\right)]=0$
$\therefore f\left(\frac{\pi }{4}\right)=f\left(\frac{5\pi }{4}\right).$
Since $f\left(x\right)$ is differentiable for all values of $x$ in $[\frac{\pi }{4},\frac{5\pi }{4}].$
Hence, all three condition of Rolle's theorem are satisfied. Therefore, there exists a point $cϵ(\frac{\pi }{4},\frac{5\pi }{4})$ such that $f^{\prime }\left(c\right)=0$.