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  4. The function f ( x )= begincases|x-3|, x ≥ 1 (x2/4)-(3 x/2)+(13/4), x < 1 endcases is

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Q. The function $f ( x )=\begin{cases}|x-3|, & x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}, & x < 1\end{cases}$ is

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Solution:

Here $f ( x )=\begin{cases}|x-3|, & x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}, & x < 1\end{cases}$ is
$\therefore RHL$ at $x=1$,
$\displaystyle\lim _{h \rightarrow 0}|1+ h -3|=2$
LHL at $x=1$.
$\displaystyle\lim _{h \rightarrow 0} \frac{(1-)^{2}}{4}-\frac{3(1-h)}{2}+\frac{13}{4}$
$=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=\frac{14}{4}-\frac{3}{2}=2$
$\therefore f(x)$ is continuous at $x=1$
Again, $f ( x )=\begin{cases}-(x-3), & 1 \leq x< 3 \\ (x-3), & x \geq 3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}, & x< 1\end{cases}$
$\therefore f'(x)= \begin{cases}-1, & 1 \leq x< 3 \\ 1, & x \geq 3 \\ \frac{x}{2}-\frac{3}{2}, & x <
1\end{cases}$
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