Question

Find the value of $k$ for which $f\left(x\right)=kx+5,whenx\le 2$ and $f\left(x\right)=x-1,whenx>2$ is continuous at $x=2$.

Answer 1

At $x=2,f\left(2\right)=k\left(2\right)+5=2k+5$

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

$=\underset{h\to 0}{lim}\left[\right(2+h)-1]=\underset{h\to 0}{lim}(1+h)=1$

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

$=\underset{h\to 0}{lim}\left[\right(2k+5)-kh]=2k+5$

Now, $\underset{x\to 2}{lim}f\left(x\right)$ exists only when $2k+5=1⟹k=-2$.

When $k=-2$, we have $\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)=1$

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