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  4. The function f(x)= begincases x[x], text if 0 ≤ x<2 (x-1) x, text if 2 ≤ x < 3 endcases is is

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Q. The function $f(x)=\begin{cases} x[x], & \text { if } 0 \leq x<2 \\ (x-1) x, & \text { if } 2 \leq x < 3\end{cases}$ is is

Continuity and Differentiability

Solution:

Here, $f(x)=\begin{cases} x[x], \text { if } 0 \leq x<2 \\ (x-1) x, \text { if } 2 \leq x < 3\end{cases}$ at $x=2$
$LHD =L f^{\prime}(2)=\displaystyle\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} $
$ =\displaystyle\lim _{h \rightarrow 0} \frac{(2-h)[2-h]-(2-1) 2}{-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2-h)(1)-2}{-h} [\because(2-h)=1] $
$ =\displaystyle\lim _{h \rightarrow 0} \frac{2-h-2}{-h}=\displaystyle\lim _{h \rightarrow 0} \frac{-h}{-h}=1 $
RHD $ =R f^{\prime}(2)=\displaystyle\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} $
$ =\displaystyle\lim _{h \rightarrow 0} \frac{(2+h-1)(2+h)-(2-1)(2)}{h} $
$=\displaystyle\lim _{h \rightarrow 0} \frac{(h+1)(2+h)-2}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{h^2+3 h+2-2}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{h^2+3 h}{h}=\displaystyle\lim _{h \rightarrow 0} \frac{h(h+3)}{h}$
$=3$
$\because LHD \neq RHD$
$\therefore f(x)$ is not differentiable at $x=2$.