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

Solution:

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

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

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

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then $f(c)=f(1)=1+1=2$

 

The left hand limit of f at x = 1 is,

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

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

 

$\therefore \lim _{x \rightarrow 1} f(x)=f(1)$

Therefore, f is continuous at x = 1

Case III:

If $c>1$, then $f(c)=c+1$

 

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Hence, the given function f has no point of discontinuity.