Question

Find all the values of k that makes f(x) continuous at x = 2

Answer 1

Given:

$f(x)=\begin{cases}x^2+4x,\text{ if x<2} \\ 2k^2x-4,\text{ if x}\ge2\end{cases}$

As the given function is continuous at x = 2,

$\begin{gathered} \lim _{x\to2^-}f(x)=\lim _{x\to2^+}f(x) \\ \lim _{x\to2^-}(x^2+4x)=\lim _{x\to2^+}(2k^2x-4) \\ 2^2+4(2)=2k^2(2)-4 \\ 4+8=4k^2-4 \\ 4k^2=12+4 \\ k^2=\frac{16}{4} \\ k^2=4 \\ k=\pm2 \end{gathered}$

Answer: k = 2, -2