Question

If $f\left(x\right)=\{\begin{array}{c}-x^2,whenx\le 0\\ 5x-4,when0<x\le 1\\ 4x^2-3x,when1<x<2\\ 3x+4,whenx\ge 2\end{array}$ ,
The no. of values of $x$ where$f\left(x\right)$ is not continuous is

Answer 1

At $x=0$

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}(-x^2)=0$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}(5x-4)=0-4=-4$

LHL$\ne$RHL $\therefore$Discontinuous

At $x=1$

${lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1^{-}}(5x-4)=5-4=1$

${lim}_{x\to 1^{+}}f\left(x\right)={lim}_{x\to 1^{+}}(4x^2-3x)=4-3=1$

LHL$=$RHL $\therefore$ continuous

At $x=2$

${lim}_{x\to 2^{-}}f\left(x\right)={lim}_{x\to 2^{-}}\left((4x^2-3x)\right)=4\left(4\right)-3\left(2\right)=10$

${lim}_{x\to 2^{+}}f\left(x\right)={lim}_{x\to 2^{+}}(3x+4)=3\left(2\right)+4=10$

LHL$=$RHL $\therefore$ continuous

$\Rightarrow$ Only at $x=0$ this function is continuous.