how does a kink in the graph of the function represents the point of non-differentiability? explain in details with examples

A kink in a graph means a sharp, pointed point in a graph. If the point is smooth or rounded, and not pointed or sharp, it is not called a kink. To understand why the presence of a kink implies non-differentiability of the graph at that point, let us take an example and study it.

Take $f\left(x\right)=\left|x\right|$

This is defined as $f\left(x\right)=\left\{\begin{array}{l}x, x\ge 0\\ -x, x<0\end{array}\right.$

So, we see that f(0) is defined as f(0)=0.

But, when x is even ​slightly negative (i.e., x approaches 0 from the left side) the slope if -1, and if x is even slightly positive (i.e., x approaches 0 from the right side) the slope is +1.

So, you see, at x=0, we don't have one unique slope of f(x). That is why f(x)=|x| is not differentiable at x=0, which is a kink on it. Hope this helps.