If $f(x)=\left\{\begin{array}{lr}4 & ,-3<x<-1 \\ 5+x & ,-1 \leq x<0 \\ 5-x & , 0 \leq x<2 \\ x^2+x-3 & , 2 \leq x<3\end{array}\right.$ then $f(|x|)$ is

continuous but not differentiable in (-3, 3)

We have,

$f(x)=\left\{\begin{array}{lc}4 & , -3<x<-1 \\ 5+x & , -1 \leq x<0 \\ 5-x & , 0 \leq x<2 \\ x^2+x-3 & , 2 \leq x<3\end{array}\right.$

∴  $f(|x|)= \begin{cases}5-|x|& , \text { if } 0 \leq|x|<2 \\ |x|^2+|x|-3& , \text { if } 2 \leq|x|<3\end{cases}$

$\Rightarrow f(|x|)= \begin{cases}5+x & , \text { if }-2<x<0 \\ 5-x & , \text { if } 0 \leq x<2 \\ x^2-x-3 & ,\text { if }-3<x \leq-2 \\ x^2+x-3 ! & , \text { if } 2 \leq x<3\end{cases}$

Clearly, f(x) is continuous in (-3, 3). But, it is not differentiable at x = 0, -2, 2.

Hence, f(|x|) is continuous but not differentiable in (-3, 3).