We have, $\text{f}(\text{x})=\begin{cases}\text{x}-3&,\text{x}\geq3\\3-\text{x}&,1\leq\text{x}<3\\\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}&,\text{x}<1\end{cases}$ - (b) -1
Solution:
$\text{Rf}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0}\frac{3-(1+\text{h})-2}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}-\frac{\text{h}}{\text{h}}=-1$
- (b) -1
Solution:
$\text{Lf}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big[\frac{(1-\text{h})^2}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big(\frac{1+\text{h}^2-2\text{h}-6+6\text{h}+13-8}{-4\text{h}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}^2+4\text{h}}{-4\text{h}}\Big)=-1$
- (c) x = 3
Solution:
Since, R.H.D. at x = 3 is 1
and L.H.D. at x = 3 is - 1
$\therefore$ f(x) is non-differentiable at x = 3.
- (d) -1
- (c) -2
Solution:
From above, we have
$\text{f}'(\text{x})=\frac{\text{x}}{2}-\frac{3}{2},\text{x}<1$
$\therefore\text{f}'(-1)=\frac{-1}{2}-\frac{3}{2}=-2$