Question

The function
$f\left(x\right)=max\{(1-x),(1+x),2\},x\in (-\infty ,\infty )$ is

• A. Continuous at all points
• B. Differentiable at all points
• C. Not differentiable at all points except at $x=1$ and $x=-1$
• D. Continuous at all points except at $x=1$ and $x=-1$ where it is discontinuous

Answer 1

The correct option is A Continuous at all points

We have, $f\left(x\right)=\text{max}\left\{\right(1-x),(1+x),2\}$
$=\{\begin{array}{c}1-x,\text{if}x\le -1\\ 2,\text{if}-1\le x\le 1\\ 1+x,\text{if}x\ge 1\end{array}$
$\therefore {lim}_{x\to -1^{-}}f\left(x\right)={lim}_{x\to -1}1-x=2$
and ${lim}_{x\to -1^{+}}f\left(x\right)={lim}_{x\to -1}2=2$
and $f(-1)=2$
$\therefore {lim}_{x\to -1^{-}}f\left(x\right)={lim}_{x\to -1^{+}}f\left(x\right)=f(-1)$
So, $f\left(x\right)$ is continuous at $x=-1$
It can be easily checked that $f\left(x\right)$ is also continuous at $x=1$
Since, $f\left(x\right)$ is a polynomial function for $x\le -1$ and $x\ge 1$ and a constant function for $-1\le x\le 1$.

Hence, $f\left(x\right)$ is continuous for all $x$.
We have, $L{f}^{\prime }(-1)=-1,R{f}^{\prime }(-1)=0,L{f}^{\prime }\left(1\right)=0,R{f}^{\prime }\left(1\right)=1$
So, $f\left(x\right)$ is not differentiable at $x=\pm 1$.