Question

Prove that the function f given by $f(x)=|x+1|$, x∈R is not differentiable at x=1.

Answer 1

The given function is f(x) = |x-1|, x∈R
It is known that a function f is differentiable at a point x=c in its domain if both
$\lim\limits_{h \to 0^-}$ $\frac {f(c+h)-f(c)}{h}$ and \lim\limits_{h \to 0^+}$$\frac {f(c+h)-f(c)}{h} are finite and equal
To check the differentiability of the given function at x=1,
consider the left hand limit of f at x=1
$\lim\limits_{h \to 0^-}$ $\frac {f(1+h)-f(1)}{h}$ = $\lim\limits_{h \to 0^-}$ $\frac {|1+h-1|-|1-1|}{h}$
=$\lim\limits_{h \to 0^-}$ $\frac {|h|-0}{h}$ = $\lim\limits_{h \to 0^-}$ $\frac {-h}{h}$ (h<0 $\implies$ |h|=-h)

Consider the right hand limit of f at x=1
$\lim\limits_{h \to 0^+}$ $\frac {f(c+h)-f(c)}{h}$ = $\lim\limits_{h \to 0^+}$ $\frac {f(1+h)-f(1)}{h}$
=l$\lim\limits_{h \to 0^+}$ $\frac {|1+h-1|-|1-1|}{h}$
=$\lim\limits_{h \to 0^+}$ $\frac {|h|-0}{h}$ = $\lim\limits_{h \to 0^+}$ h/h (h>0$\implies$|h|=h)

Since the left and right hand limits of f at x=1 are not equal, f is not differentiable at x=1