Given function is
[tex]f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.[/tex]
now we need to find the value of k such that function f(x) continuous everywhere.
We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.
So we just need to find both left and right hand limits then set equal to each other to find the value of k
To find the left hand limit (LHD) we plug x=-4 into 3x+k
so LHD= 3(-4)+k
To find the Right hand limit (RHD) we plug x=-4 into
[tex]kx^2-5[/tex]
so RHD= [tex]k(-4)^2-5[/tex]
Now set both equal
[tex]k(-4)^2-5=3(-4)+k[/tex]
[tex]16k-5=-12+k[/tex]
[tex]16k-k=-12+5[/tex]
[tex]15k=-7[/tex]
[tex]k=-\frac{7}{15}[/tex]
k=-0.47
Hence final answer is -0.47.
