Question

Find the value of k for which

$f\left(x\right)=kx+5,whenx\le 2$ and $x-1,whenx>2$ is continuous at $x=2$.

Answer 1

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

$li{m}_{x\to 2^{+}}f\left(x\right)=li{m}_{h\to 0}f(2+h)$

$=li{m}_{h\to 0}\left[\right(2+h)-1]=li{m}_{h\to 0}(1+h)=1$

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

$=li{m}_{h\to 0}(2k+5-kh)=2k+5$

Since, f(x) is continuous at $x=2$, therefore,

$2k+5=1$

$2k=-4$

$k=-2$