Question

Is the function defined by
$f\left(x\right)=$ $\{\begin{array}{c}x+5,ifx\le 1\\ x-5,ifx>1\end{array}$
a continuous function ?

Answer 1

The given function is $\{\begin{array}{c}x+5,\text{if}x\le 1\\ x-5,\text{if}x>1\end{array}$
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 $f\left(c\right)=c-5$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x-5)=c-5$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore $f$ is continuous at all points $x$, such that $x<1$
Case II
If $c=1$, then $f\left(1\right)=1+5=6$
The left hand limit of $f$ at $x=1$ is,
$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}(x+5)=1+5=6$
The right hand limit of $f$ at $x=1$ is,
$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}(x-5)=1-5=-4$
It is observed that the left and right hand limit of f at $x=1$ do not coincide.
Therefore, $f$ is not continuous at $x=1$
Case III
If $c>1$, then $f\left(c\right)=c-5$ and $f\left(x\right)=\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x-5)=c-5$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$, such that $x>1$
Thus, from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.