Question

Discuss the continuity of f(x) on its domain where,
$f\left(x\right)=x^2-4$, for $0\le x\le 2$
$=2x+5$, for $2<x\le 4$
$=x^2-5$, for $4<x\le 6$

Answer 1

$f\left(x\right)$ is a poly. ftn so continuous in each particular domain so we have to check continuity at only $2$ pts i.e at $x=2,4$

(i) $\underset{x\to 2^{-}}{lim}f\left(x\right)=(x^2-4{)}_{x=2}=0$

$\underset{x\to 2^{+}}{lim}f\left(x\right)=(2x+5{)}_{x=2}=9$

As $LHL\ne RHL\therefore notcontinuousat2$

(ii) $\underset{x\to 4^{-}}{lim}f\left(x\right)=(2x+5{)}_{x=4^{-}}=13$

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

is $LHL\ne RHL\therefore notcontinuousat4$