Question

Let $f\left(x\right)=\{\begin{array}{ccc}2x+3,& & -3<x<-2\\ x+1,& & -2\le x<0\\ x+2,& & 0\le x<1\end{array}$.
Then the number of point(s) at which $f\left(x\right)$ is discontinuous in $(-3,1)$, is

Answer 1

We know that every polynomial function is continuous .
$\therefore f\left(x\right)$ is continuous in $(-3,-2),$ $(-2,0),$$(0,1)$.

As $f\left(x\right)$ changes it branches at $x=-2$ and $0.$ We need to check continuity at $-2$ and $0$
$f(-2)=f(-2^{+})=-1,$
$f(-2^{-})=\underset{h\to 0}{lim}f(-2-h)=\underset{h\to 0}{lim}\{2(-2-h)+3\}=-1$
$\therefore f(-2)=f(-2^{+})=f(-2^{-})$
So, $f\left(x\right)$ is continuous at $x=-2$
$f\left(0\right)=(0+2)=2,f\left(0^{+}\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}(h+2)=2$
$f\left(0^{-}\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}(-h+1)=1$
$\because f\left(0^{+}\right)\ne f\left(0^{-}\right)$
So, $f\left(x\right)$ is discontinuous at $x=0$
$\therefore f\left(x\right)$ is discontinuous at only $1$ point.