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)

$f(x)=\{\begin{array}{lll}1-x& :& x\le -1\\ 2& :& -1<x\le 1\\ 1+x& :& x>1\end{array}$
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, $f^{\prime }(x)=\{\begin{array}{ll}-1& x<-1\\ 0& -1<x<1\\ 1& x>1\end{array}$
at $x=-1$
$f^{\prime }\left(-1^{-}\right)=-1,f\left(-1^{+}\right)=0$
$\because f^{\prime }\left(-1^{-}\right)\ne f^{\prime }\left(-1^{+}\right)\Rightarrow$ non-differentiable at $x=1$
$f^{\prime }\left(1^{-}\right)=0,f^{\prime }\left(1^{+}\right)=1$
$\because f^{\prime }\left(1^{-}\right)\ne f^{\prime }\left(1^{+}\right)\Rightarrow$ non-differentiable