Question

Show that the function $f\left(x\right)=\{\begin{array}{cc}|2x-3|\left[x\right],& x\ge 1\\ \sin \left(\frac{\pi x}{2}\right),& x<1\end{array}$ is continuous but not differentiable at $x=1$

Answer 1

$f\left(1\right)=|2-3|\left[1\right]=1-\left(i\right){lim}_{h\to 0}f(1+h)=|2+2h-3|[1+h]=1\times 1=1-\left(i\right){lim}_{h\to 0}f(1-h)=\sin \left(\frac{\pi (1-h)}{2}\right)=\sin \frac{\pi }{2}=1-\left(i\right)Hencef\left(x\right)iscontinuousat1.Fordifferentiability{lim}_{h\to 0}\frac{f(1+h)-f\left(1\right)}{h}mustbeequalto{lim}_{h\to 0}\frac{f\left(1\right)-f(1-h)}{h}\Rightarrow {lim}_{h\to 0}\frac{|2+2h-3|[1+h]-|2-3|\left[1\right]}{h}\frac{|2-3|\left[1\right]-\sin \left(\frac{\pi -h}{2}\right)}{h}\Rightarrow \frac{|2h-1|-1}{h}\frac{1-\sin \left(\frac{\pi -h}{2}\right)}{h}\left[\frac{0}{0}form\right]Sincehisverysmall,(2h-1)is-ve.\Rightarrow \frac{1-2h-1}{h}\Rightarrow -2\Rightarrow -\cos \left(\frac{\pi -h}{2}\right)\times -\frac{1}{2}\Rightarrow 0LHS\ne RHS$

Hence function is non-differentiable at 1.but continuous at 1.

Hence proved.