Question

Find the value of k that makes f(x) continuous at x = -2

Answer 1

The given function is:

$f(x)=\begin{cases}-2x^2-5kx;x\leq-2 \\ -x^2+k;x>-2 \\ \square\end{cases}$

The function is continuous at x=-2 if:

$\begin{gathered} \lim _{x\to-2^-}f(x)=\lim _{x\to-2^+}f(x) \\ \lim _{x\to-2}(-2x^2-5kx)=\lim _{x\to-2}(-x^2+k) \\ -2(-2)^2-5k(-2)=-(-2)^2+k \\ -8+10k=-4+k \\ 9k=-4+8 \\ k=\frac{4}{9} \end{gathered}$

So the value of k is 4/9.