Is the function defined by

The given function is $f(x)=\left\{\begin{array}{l}x+5, \text { if } x \leq 1 \\ x-5, \text { if } x>1\end{array}\right.$

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 $f(c)=c+5$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+5)=c+5$

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

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

Case II:

If $c=1$, then $f(1)=1+5=6$

The left hand limit of f at x = 1 is,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+5)=1+5=6$

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

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{\prime}}(x-5)=1-5=-4$

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 $f(c)=c-5$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-5)=c-5$

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

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.