Find all points of discontinuity of f, where f is defined by

Solution:

The given function $f$ is $f(x)=\left\{\begin{array}{l}x^{3}-3, \text { if } x \leq 2 \\ x^{2}+1, \text { if } x>2\end{array}\right.$

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 $f(c)=c^{3}-3$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{3}-3\right)=c^{3}-3$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, f is continuous at all points x, such that x < 2

Case II:

If $c=2$, then $f(c)=f(2)=2^{3}-3=5$

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^{3}-3\right)=2^{3}-3=5$

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x^{2}+1\right)=2^{2}+1=5$

$\therefore \lim _{x \rightarrow 2} f(x)=f(2)$

Therefore, f is continuous at x = 2

Case III:

If $c>2$, then $f(c)=c^{2}+1$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2}+1\right)=c^{2}+1$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

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.