Question

Find all points of discontinuity of $f$, where $f$ is defined by
f(x) = $\{\begin{array}{c}x+1,\text{if}x\ge 1\\ x^2+1,\text{if}x<1\end{array}$

Answer 1

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

The given function is defined at all the points of the real line.
Let $c$ be a point on the real line.

Case I: $c<1$, then $f\left(c\right)=c^2+1$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x^2+1)=c^2+1$

$\therefore$ $\underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$

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

Case II : $c=1$, then $f\left(c\right)=f\left(1\right)=1+1=2$

The left hand limit of $f$ at $x=1$is,

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

The right hand limit of $f$ at $x=1$ is,

$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}(x+1)=1+1=2$

$\therefore$ $\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$

Therefore, $f$ is continuous at $x=1$

Case III : $c>1$, then $f\left(c\right)=c+1$

$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x+1)=c+1$

$\therefore$ $\underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$
Hence, the given function $f$ has no point of discontinuity.