Question

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

• A. discontinuous at exactly two points
• B. differentiable $\forall x\in R$
• C. differentiable $\forall x\in R-\left\{- 1 , 1\right\}$
• D. continuous $\forall x\in R-$ $\left\{0 , 1 , - 1\right\}$

Answer 1

Answer: (C)

$f(x)=\left\{\begin{array}{lll}1-x & : & x \leq-1 \\ 2 & : & -1 < x \leq 1 \\ 1+x & : & x > 1\end{array}\right.$
Continuity at $x=-1$
$f\left(- 1\right)=1-\left(- 1\right)=2$
$f\left(- 1^{-}\right)=1-\left(- 1\right)=2$
$f\left(- 1^{+}\right)=2$
$\because f\left(- 1\right)=f\left(- 1^{-}\right)=f\left(- 1^{+}\right)$
$\therefore $ continuous at $x=-1$
$f\left(1\right)=2,f\left(1^{-}\right)=2$
$f\left(1^{+}\right)=1+1=2$
$\because f\left(1^{-}\right)=f\left(1\right)=f\left(1^{+}\right)$
$\therefore $ continuous at $x=1$
For differentiability, $f^{\prime}(x)=\left\{\begin{array}{ll}-1 & x < -1 \\ 0 & -1 < x < 1 \\ 1 & x > 1\end{array}\right.$
at $x=-1$
$f^{\prime}\left(-1^{-}\right)=-1, f\left(-1^{+}\right)=0$
$\because f^{\prime}\left(-1^{-}\right) \neq f^{\prime}\left(-1^{+}\right) \Rightarrow $ non-differentiable at $x=1$
$f^{\prime}\left(1^{-}\right)=0, f^{\prime}\left(1^{+}\right)=1$
$\because f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right) \Rightarrow $ non-differentiable