# Prove the following

Question:

$f(x)=\left\{\begin{array}{l}x[x], \quad \text {, if } 0 \leq x<2 \\ (x-1) x, \quad \text { if } 2 \leq x<3\end{array}\right.$ at $\mathrm{x}=2$

Solution:

We know that, a function f is differentiable at a point ‘a’ in its domain if

Lf’(c) = Rf’(c)

where L $f^{\prime}(c)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$ and

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

Here, $f(x)=\left\{\begin{array}{r}x[x], \text { if } 0 \leq x<2 \\ (x-1) x, \text { if } 2 \leq x<2\end{array}\right.$ at $x=2$ at $x=2$.

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

$=\lim _{h \rightarrow 0} \frac{(2-h) \cdot 1-2}{-h} \quad[\because[2-h]=1]$

$=\lim _{h \rightarrow 0} \frac{2-h-2}{-h}=1$

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

$=\lim _{h \rightarrow 0} \frac{(2+h-1)(2+h)-(2-1) \cdot 2}{h}$

$=\lim _{h \rightarrow 0} \frac{(1+h)(2+h)-2}{h}=\lim _{h \rightarrow 0} \frac{2+h+2 h+h^{2}-2}{h}$

$=\lim _{h \rightarrow 0} \frac{3 h+h^{2}}{h}=\lim _{h \rightarrow 0} \frac{h(3+h)}{h}=3$

$\mathrm{L} f^{\prime}(2) \neq \mathrm{R} f^{\prime}(2)$

Therefore, f(x) is not differentiable at x = 2.