Question

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

Answer 1

When $x=1$
Given function is
$f\left(x\right)=\{\begin{array}{cc}x+5,& \text{if}x\le 1\\ x-5,& \text{if}x>1\end{array}$

If $f\left(x\right)$ is continuous at $x=1$ then
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$

Finding L.H.L.

$\underset{x\to 1^{-}}{lim}x+5$
$\underset{h\to 0}{lim}(1-h)+5$

Putting $h=0$ then we get,
$(1-0)+5=1+5=6$

Finding R.H.L.

$\underset{x\to 1^{+}}{lim}x-5$
$\underset{h\to 0}{lim}(1+h)-5$

Putting $h=0$ then we get,
$(1+0)-5=1-5=-4$

Find $f\left(x\right)$ at $x=1$
$f\left(x\right)=x+5$ at $x=1$
$f\left(1\right)=1+5=6$

Hence, $\underset{x\to 1^{-}}{lim}f\left(x\right)\ne \underset{x\to 1^{+}}{lim}f\left(x\right)$

Therefore, the function
$f\left(x\right)=\{\begin{array}{cc}x+5,& \text{if}x\le 1\\ x-5,& \text{if}x>1\end{array}$ is discontinuous at $x=1$

When $x<1$
For $x<1,f\left(x\right)=x+5$
Since the function $f\left(x\right)=x+5$ is a polynomial, so it is continuous.
$\therefore f\left(x\right)$ is continuous for $x<1$

When $x>1$
For $x>1,f\left(x\right)=x-5$
Since the function $f\left(x\right)=x-5$ is a polynomial so it is continuous.
$\therefore f\left(x\right)$ is continuous for $x>1$

Therefore, the function
$f\left(x\right)=\{\begin{array}{cc}{x}^3-3,& \text{if}x\le 2\\ x^2+1,& \text{if}x>1\end{array}$. So, only one point of discontinuity i.e., $x=1$