Question

Find the value of $k$ so $f\left(x\right)$ is continuous at $x=2.$
$f\left(x\right)=\{\begin{array}{c}2x+1;ifx<2\\ k;x=2\\ 3x-1;x>2\end{array}$.

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}2x+1,& \text{if}x<2\\ k,& x=2\\ 3x-1,& x>2\end{array}$

LHL at $x=2$,

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2-h)=\underset{h\to 0}{lim}2(2-h)+1=\underset{h\to 0}{lim}(5-2h)=5$

RHL at $x=2$,

$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2+h)=\underset{h\to 0}{lim}2(2+h)+1=\underset{h\to 0}{lim}(5+2h)=5$

Also,

$f\left(2\right)=k$

Since $f\left(x\right)$ is continuous at $x=2$.

Therefore,

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=f\left(2\right)$

$\Rightarrow 5=5=k$

Hence for $k=5,f\left(x\right)$ is continuous at $x=2$.