Question

Determine the value of the constant k so that the function
$f\left(x\right)=\left\{\begin{array}{ll}k{x}^2,\mathrm{if}& x\le 2\\ 3,\mathrm{if}& x>2\end{array}\right.\mathrm{is}\mathrm{continuous}\mathrm{at}x=2.$

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}k{x}^2,\mathrm{if}x\le 2\\ 3,\mathrm{if}x>2\end{array}\right.$
If $f\left(x\right)$ is continuous at x = 2, then
$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{\mathit{}x\to 2^{+}}{\lim }f\left(x\right)=f\left(2\right)$ ...(1)

Now,

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(2-h\right)=\underset{h\to 0}{\lim }k{\left(2-h\right)}^2=4k$



And, $f\left(2\right)=3$

From (1), we have

$$\begin{gathered} 4k=3 \\ \Rightarrow k=\frac{3}{4} \end{gathered}$$