Given:
[tex]f(x)=\begin{cases}-3x^2+3x,\text{ if x<-1} \\ -5kx+k^2,\text{ if x}\ge-1\end{cases}[/tex]
As the given function is continuous at x=-1,
[tex]\begin{gathered} \lim _{x\to-1^-}f(x)=\lim _{x\to-1^+}f(x) \\ \lim _{x\to-1^-}(-3x^2+3x)=\lim _{x\to-1^+}(-5kx+k^2) \\ -3(-1)^2+3(-1)=-5k(-1)+k^2 \\ -3-3=5k+k^2 \\ k^2+5k+6=0 \end{gathered}[/tex]
Solve this quadratic equation,
[tex]\begin{gathered} k^2+5k+6=0 \\ k^2+2k+3k+6=0 \\ k(k+2)+3(k+2)=0 \\ (k+2)(k+3)=0 \\ \Rightarrow k+2=0,k+3=0 \\ k=-2,k=-3 \end{gathered}[/tex]
Answer: k= -2 , -3
