Question

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

Answer 1

Answer: 5

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

and mark $y=f\left(x\right)$
$f(x)=-x-2,x<-3$
$=1,-3\le x<-1$
$=x+2,-1\le x<0$
$=2-x,0\le x<1$
$=1,1\le x<3$
$=x-2,3\le x$
The points, at which function $f$ may not be differentiable, are $x=-3,-1,0,1,3$
Differentiability at $x=-3:$
$L.hlim.=\underset{x\to -3^{-}}{\lim }\frac{f(x)-f(-3)}{x+3}$
$=\underset{x\to -3^{-}}{\lim }\frac{(-x-2)-1}{x+3}=-1$
$R.h.lim.=\underset{x\to -3^{+}}{\lim }\frac{f(x)-f(-3)}{x+3}$
$=\underset{x\to -3^{+}}{\lim }\frac{1-1}{x+3}$
$=0$
$\Rightarrow f$ 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 $f=5$