Question

Discuss the continuity of the following functions in its domain, where $f\left(x\right)=\{\begin{array}{c}{x}^2-4,for0\le x\le 2\\ 2x+3,for2<x\le 4\\ x^2-5,for4<x\le 6\end{array}$

Answer 1

$f\left(2\right)=2^2-4=0$

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

$\Rightarrow f\left(2^{+}\right)={lim}_{h\to 0}f(2+h)=2(2+h)+3=7$

$\Rightarrow f\left(4\right)=2\left(4\right)+3=11$

$\Rightarrow f\left(4^{-}\right)={lim}_{h\to 0}f(4-h)=11$

$\Rightarrow f\left(4^{+}\right)={lim}_{h\to 0}f(4+h)={(4+h)}^2-5=11$

$f\left(x\right)$ is continuous at $x=4$ and discontinuous at $x=2$

Hence, solved.