$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^3} - 3,}&{ifx \le 2}\\{{x^2} + 1,}&{ifx > 2}\end{array}} \right.$ }

We observe that f(x) is continuous at all real numbers $x < 2$ and $x > 2$ as it is polynomial function.
Now, continuity at x $=$ 2 :

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} ({x^2} + 1) = \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} {(2 + h)^2} + 1$

$= \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} (4 + 4h + {h^2}) + 1 = 5$

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} ({x^3} - 3) = \mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} {(2 - h)^3} - 3$

$\mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} (2 - 12h + 6{h^2} - {h^3}) - 3 = 5$

Also, $f(2) = 8 - 3 = 5$
therefore, $\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2)$

Hence we can say that f(x) is continuous at x $=$ 2 and at all points.
So, f has no point of discontinuity.