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=(16)/(4) \\ k^2=4 \\ k=\pm2 \end{gathered}`

Answer: k = 2, -2