The function f : R → R given by f(x) = -lx - 1| is

$f(x)=-|x-1|$

so $f(x)=\left\{\begin{array}{rr}-(x-1) & x \geq 1 \\ (x-1) & x<1\end{array}\right.$  since x is zero at x = 1

so  $f(x)=\left\{\begin{array}{rr}-1 & x \geq 1 \\ 1 & x<1\end{array}\right.$

So $\lim\limits_{x \rightarrow 1^{-}} f(x)=-(1-1)=0$ = LHL (left hand limit)

$\lim\limits_{x \rightarrow 1+} f(x)=(1-1) = 0$ = RHL (Right hand)

$f(1)= -|1-1|=0$

⇒  f(1) = LHL = RHL (continuous)

so  $\lim\limits_{x \rightarrow 1^{-}} f'(x) =-1$  = LHD  (left hand derivative)

$\lim\limits_{x \rightarrow 1^{+}} f(x)=1$  = RHD  (right hand derivative)

as LHD ≠ RHD

so not differentiable at x = 1