A) continuous at all points
B) differentiable at all points
C) differentiable at all points except at \[x=1\] and \[x=-1\]
D) none of the above
Correct Answer: C
Solution :
We have, \[f(x)=\max .[(1-x),(1+x),2]\]for \[x\in (-\infty ,\infty )\] or \[fx=\left\{ \begin{matrix} 1+x, & x>1 \\ 2, & -1\le x\le 1 \\ 1-x, & x<-1 \\ \end{matrix} \right.\] Since,\[f(x)\] is a polynomial and constant function which is defined for every values of\[x,\] therefore \[f(x)\]is continuous for all values of\[x\] \[\therefore \]\[f(x)\] is differentiable for all values of \[x\] except at \[x=1\]and \[-1.\] Alternate Solution: We have, \[f(x)=\left\{ \begin{matrix} 1+x, & x>1 \\ 2, & -1\le x\le 1 \\ 1-x, & x<-1 \\ \end{matrix} \right.\] It is clear from the figure that \[f(x)\] is continuous everywhere and \[f(x)\] is differentiable everywhere except at \[x=1,-1.\] Note: Every differentiable function is continuous but every continuous function is not differentiable.You need to login to perform this action.
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