If f(x) = x |x|,

Solution:

$|x|= \begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}$

$\therefore f(x)=x|x|= \begin{cases}x^{2}, & x \geq 0 \\ -x^{2}, & x<0\end{cases}$

Now,

$L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(2-h)^{2}-2^{2}}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{4-4 h+h^{2}-4}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(-4+h) h}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0}(4-h)$

$\Rightarrow L f^{\prime}(2)=4-0=4$

Also,

$R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(2+h)^{2}-2^{2}}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{4+4 h+h^{2}-4}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(4+h) h}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0}(4+h)$

$\Rightarrow R f^{\prime}(2)=4+0=4$

So, $L f^{\prime}(2)=R f^{\prime}(2)=4$

$\therefore f^{\prime}(2)=4$

If $f(x)=x|x|$, then $f(2)=$ ___4____.