Examine the continuity of function at x=2 and x =3 . f(x)=|x-2|+|… - Vidyadip

Question

Examine the continuity of function at $x=2$ and $x =3 . f(x)=|x-2|+|x-3|$

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Answer

Given function $f(x)=|x-2|+|x-3|$,
It is written as : $\begin{array}{l} \quad f(x)=\left\{\begin{array}{cc} -(x-2)-(x-3), & 0 \leq x<2 \\ x-2-(x-3), & 2 \leq x<3 \\ x-2+x-3, & x \geq 3 \end{array}\right. \\ \Rightarrow \quad f(x)=\left\{\begin{array}{cl} 5-2 x, & \text { if } 0 \leq x>2 \\ 1, & \text { if } 2 \leq x<3 \\ 2 x-5, & \text { if } x \geq 3 \end{array}\right. \end{array}$
$(i)$ Examining the continuity at $x=2$.
$ \text { R.H.L. }=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 1=1$
$ \text { L.H.L. } =\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[5-2(2-h)]$
$ =\lim _{h \rightarrow 0}(5-4+2 h)=1$ at $x=2$
$ f(x)=5-2 x$
$\therefore f(2)=5-2 \times 2=5-4=1$
hence at $x=2$
$f(2)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} f(2-h)=1$
hence function is continuous at $x=2$.
$(ii$) Examining the continuity at $x=3$
$\text{R.H.L}. \lim _{h \rightarrow 0} f(3+h) =\lim _{h \rightarrow 0}[2(3+h)-5]$
$ =\lim _{h \rightarrow 0}(6+2 h-5)=1$
$\text{L.H.L}. \lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} 1=1$
at $x=3$
$f(x)=2 x-5,$
$\therefore f(3)=2 \times 3-5=6-5=1$
$\therefore f(3)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0} f(3-h)=1$
Hence function is continuous at $x=3$.

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