Question

Find all points of discontinuity of $f$, where $f$ is defined by
$f\left(x\right)=$ $\{\begin{array}{c}{x}^3-3,ifx\le 2\\ x^2+1,ifx>2\end{array}$

Answer 1

The given function $f$ is $f\left(x\right)=$ $\{\begin{array}{c}{x}^3-3,\text{if}x\le 2\\ x^2+1,\text{if}x>2\end{array}$
The given function $f$ is defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I :
If $c<2$, then $f\left(c\right)=c^3-3$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x^3-3)=c^3-3$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$, such that $x<2$
Case II
If $c=2$, then $f\left(c\right)=f\left(2\right)=2^3-3=5$
$\underset{x\to 2^{-}}{lim}=\underset{x\to 2^{-}}{lim}(x^3-3)=2^3-3=5$
$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(x^2+1)=2^2+1=5$
$\therefore \underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$
Therefore, $f$ is continuous at $x=2$
Case III :
if $c>2$, then $f\left(c\right)=c^2+1$
$\underset{x\to c}{lim}=\underset{x\to c}{lim}(x^2+1)=c^2+1$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$, such that $x>2$
Thus, the given function $f$ is continuous at every point on the real line.
Hence, $f$ has no point of discontinuity.