Question

Show that the function $f\left(x\right)=|x-3|,xϵR$, is continuous but not differentiable at $x=3$.

Answer 1

$f\left(x\right)=|x-3|;x=3$

$LHD=f^{\prime }\left(3^{-}\right)={lim}_{h\to 0}\frac{f(3-h)-f\left(3\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{|3-h-3|-|3-3|}{-h}$

$=\underset{h\to 0}{lim}\frac{|-h|}{-h}$

$=\underset{h\to 0}{lim}\frac{h}{-h}$ $=-1$

$RHD=f^{\prime }\left(3^{+}\right)=\underset{h\to 0}{lim}\frac{f(3+h)-f\left(3\right)}{h}$

$=\underset{h\to 0}{lim}\frac{|3+h-3|-|3-3|}{h}$

$=\underset{h\to 0}{lim}=\frac{\left|h\right|}{h}$ $=\underset{h\to 0}{lim}\frac{h}{h}$ $=1$

As $LHD\ne RHD$

Therefore, f is not differentiable.

Again, $LHL=\underset{x\to 3^{-}}{lim}|x-3|$
$=\underset{h\to 0}{lim}|3-h-3|$

$=\underset{h\to 0}{lim}|-h|=0$

$RHL=\underset{x\to 3^{+}}{lim}|x-3|$

$=\underset{h\to 0}{lim}|3+h-3|$

$=\underset{h\to 0}{lim}\left|h\right|=0$

Since $LHL=RHL$

Therefore, f is continuous at $x=3.$