. $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{x + 1,}&{ifx \ge 1}\\{{x^2} + 1,}&{ifx < 1}\end{array}} \right.$ }

We observe that f(x) is continuous at all real numbers $x < 1$ and $x > 1$ as it is polynomial function.

Now, continuity at x $=$ 1 :

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

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

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

$\therefore$ $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1)$

Hence we can say that, f(x) is continuous at x $=$ 1 and at all points.

So, f(x) has no point of discontinuity.