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,

$L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{f(-1-h)-f(-1)}{-h}$

$\Rightarrow L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-(-1-h)^{2}-\left[-(-1)^{2}\right]}{-h}$

$\Rightarrow L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-\left(1+2 h+h^{2}\right)+1}{-h}$

$\Rightarrow L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-\left(2 h+h^{2}\right)}{-h}$

$\Rightarrow L f^{\prime}(-1)=\lim _{h \rightarrow 0}(2+h)$

$\Rightarrow L f^{\prime}(-1)=2+0=2$

Also,

$R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}$

$\Rightarrow R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-(-1+h)^{2}-\left[-(-1)^{2}\right]}{h}$

$\Rightarrow R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-\left(1-2 h+h^{2}\right)+1}{h}$

$\Rightarrow R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-\left(-2 h+h^{2}\right)}{h}$

$\Rightarrow R f^{\prime}(-1)=\lim _{h \rightarrow 0}(2-h)$

 

$\Rightarrow R f^{\prime}(-1)=2-0=2$

So, $L f^{\prime}(-1)=R f^{\prime}(-1)=2$

$\therefore f^{\prime}(-1)=2$

If $f(x)=x|x|$, then $f(-1)=$ __2___.