The domain of the derivative of the function f(x)=≤ft […] ,. is

We have,

$f(x)= \begin{cases}\frac{1}{2}(-x-1), & x<-1 \\ \tan ^{-1} x ~~~~~, &-1 \leq x \leq 1 \\ \frac{1}{2}(x-1), & x>1\end{cases}$

We observe that

$\lim\limits_{x \rightarrow-1^{-}} f(x)=\lim\limits_{x \rightarrow-1^{-}} \frac{1}{2}(-x-1)=0$

$\lim\limits_{x \rightarrow-1^{+}} f(x)=\lim\limits_{x \rightarrow-1^{+}} \tan ^{-1} x=\tan ^{-1}(-1)=-\pi / 4$

Clearly, $\lim\limits_{x \rightarrow-1^{-}} f(x) \neq \lim\limits_{x \rightarrow-1^{+}} f(x)$

So, f(x) is not continuous at $x=-1$.

Similarly, f(x) is not continuous at $x=1$.

Consequently f(x) is not differentiable at $x= \pm 1$.

At all other points f(x) is differentiable.