Examine the differentiability of ${\rm{f}}$, where ${\rm{f}}$ is defined by
$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{x[x],}&{{\rm{ if }}0 \le x < 2}\\{(x - 1)x,}&{{\rm{ if }}2 \le x < 3}\end{array}} \right.$at $x = 2$}

We have, $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{x[x],}&{{\rm{ if }}0 \le x < 2}\\{(x - 1)x,}&{{\rm{ if }}2 \le x < 3}\end{array}} \right.$at $x = 2$.

As we know that , a function $f$ is differentiable at a point a in its domain, if both $L{f^\prime }(a)$ and $R{f^\prime }(a)$ are finite and equal, where $L{f^\prime }(c) = \mathop {\lim }\limits_{h \to 0} \frac{{f(a - h) - f(a)}}{{ - h}}$ and
$R{f^\prime }(c) = \mathop {\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h}$ [ where $a$ is any positive number]

At $x = 2,$ $L{f^\prime }(2) = \mathop {\lim }\limits_{h \to 0} \frac{{f(2 - h) - f(2)}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{(2 - h)[2 - h] - (2 - 1)2}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{(2 - h)(1) - 2}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{2 - h - 2}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{ - h}} = 1$

$R{f^\prime }(2) = \mathop {\lim }\limits_{h \to 0} \frac{{f(2 + h) - f(2)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{(2 + h - 1)(2 + h) - (2 - 1) \cdot 2}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{(1 + h)(2 + h) - 2}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2 + h + 2h + {h^2} - 2}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + 3h}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h(h + 3)}}{h} = 3$

$\Rightarrow L{f^\prime }(2) \ne R{f^\prime }(2)$

Therefore we can say that $f(x)$ is not differentiable at $x = 2$.