At which point does f(x) = |x – 1| has itslocal minimum?

The given function is f(x) = ?x ? 1?, x ∈ R.
It is known that a function f is differentiable at point x = c in its domain if both
\(\lim\limits_{h \rightarrow 0^-}\) hf(c + h) – f( c )
And
\(\lim\limits_{h \rightarrow 0^+}\) hf(c + h) – f( c ) are finite and equal.
To check the differentiability of the function at x = 1,
LHS,
Consider the left hand limit of f at x=1
\(\lim\limits_{h \rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h}\)
= \(\lim\limits_{h \rightarrow 0^-}\frac{|h|}{h}\)
= \(\lim\limits_{h \rightarrow 0^-}\frac{-h}{h}\)
= ?1
RHS,
Consider the right hand limit of f at x ? 1
\(\lim\limits_{h \rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h}\)
= \(\lim\limits_{h \rightarrow 0^+}\frac{|h|}{h}\)
= 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.
As, LHS = -1 and RHS = 1, it is clear that, f?(1) < 0 on the left of x = 1 and f?(x) > 0 on the right of the point x = 1.
Hence, f?(x) changes sign, from negative on the left to positive on the right of the point x = 1.
Therefore, f(x) has a local minima at x = 1.