Question
For what value of k is the following function continuous at x = 2?
$\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$

Answer

Given, $\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
We have,
$(\text{LHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2(2-\text{h})+1)=5$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})=\lim_\limits{\text{h}\rightarrow0}3(2+\text{h})-1=5$
Also, $\text{f}(2)=\text{k}$
If f(x) is continuous at x = 2, then
$=\lim_\limits{\text{x}\rightarrow2^-}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow2^+}\text{f}(\text{x})=\text{f}(2)$
$\Rightarrow5=5=\text{k}$
Hence, for k = 5, (fx) is continuous at x = 2

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