Question

The number of non-differentiability points of function $f$ defined as $f\left(\right.x\left.\right)=max\cdot \left(\right.\left|\right.\left|\right.x\left|\right.-2\left|\right.,1\left.\right)$ is equal to

Answer 1

$f\left(\right.x\left.\right)=max\left(\right.\left|\right.\left|\right.x\left|\right.-2\left|\right.,1\left.\right)=max\cdot \left(\right.g\left(\right.x\left.\right),1\left.\right)$
where $g\left(\right.x\left.\right)=\left|\right|x\left|- 2\right|$
$\left|\right.\left|\right.x\left|\right.-2\left|\right.=\left|\right.-x-2\left|\right.,x < 0$
$=\left|\right.x-2\left|\right.,x\geq 0$
$\Rightarrow \left|\right.\left|\right.x\left|\right.-2\left|\right.=-x-2,x < -2$
$=x+2,-2\leq x < 0$
$=2-x,0\leq x < 2$
$=x-2,2\leq x$
We draw the graphs of $y=g\left(x\right)$ and $y=1$ .

and mark $y=f\left(x\right)$
$f\left(\right.x\left.\right)=-x-2,x < -3$
$=1,-3\leq x < -1$
$=x+2,-1\leq x < 0$
$=2-x,0\leq x < 1$
$=1,1\leq x < 3$
$=x-2,3\leq x$
The points, at which function $f$ may not be differentiable, are $x=-3,-1,0,1,3$
Differentiability at $x=-3:$
$L.hlim.=\underset{x \rightarrow - 3^{-}}{lim}\frac{f \left(\right. x \left.\right) - f \left(\right. - 3 \left.\right)}{x + 3}$
$=\underset{x \rightarrow - 3^{-}}{lim}\frac{\left(\right. - x - 2 \left.\right) - 1}{x + 3}=-1$
$R.h.lim.=\underset{x \rightarrow - 3^{+}}{lim}\frac{f \left(\right. x \left.\right) - f \left(\right. - 3 \left.\right)}{x + 3}$
$=\underset{x \rightarrow - 3^{+}}{lim}\frac{1 - 1}{x + 3}$
$=0$
$\Rightarrow f$ is not differentiable at $x=-3$

	$x=-1$	$x=0$	$x=1$	$x=3$
L.h.lim	$0$	$1$	$-1$	$0$
R.h. lim	$1$	$-1$	$0$	$1$

From the table, it is evident that $f$ is not differentiable at $x=-1,0,1,3$ also.
$\Rightarrow $ The number of non-differentiability points of function $f=5$