To ensure continuity of the function [tex]\( f(x) \)[/tex] at [tex]\( x = -2 \)[/tex], we must ensure that the left-hand limit as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the left ([tex]\(x \le -2\)[/tex]) is equal to the right-hand limit as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the right ([tex]\(x > -2\)[/tex]). In other words, we need
[tex]\[
\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x).
\][/tex]
First, let's evaluate the right-hand limit, where [tex]\( f(x) \)[/tex] is defined as [tex]\( k^2 + 40x \)[/tex]:
[tex]\[
\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} (k^2 + 40x).
\][/tex]
Substitute [tex]\( x = -2 \)[/tex] into this expression:
[tex]\[
\lim_{x \to -2^+} (k^2 + 40x) = k^2 + 40(-2) = k^2 - 80.
\][/tex]
Next, we evaluate the left-hand limit, where [tex]\( f(x) \)[/tex] is defined as [tex]\( kx \)[/tex]:
[tex]\[
\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (kx).
\][/tex]
Substitute [tex]\( x = -2 \)[/tex] into this expression:
[tex]\[
\lim_{x \to -2^-} (kx) = k(-2) = -2k.
\][/tex]
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = -2 \)[/tex], these two limits must be equal:
[tex]\[
k^2 - 80 = -2k.
\][/tex]
This gives us a quadratic equation to solve for [tex]\( k \)[/tex]:
[tex]\[
k^2 + 2k - 80 = 0.
\][/tex]
We can solve this quadratic equation using the quadratic formula, [tex]\( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -80 \)[/tex]:
[tex]\[
k = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-80)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 320}}{2} = \frac{-2 \pm \sqrt{324}}{2} = \frac{-2 \pm 18}{2}.
\][/tex]
This gives us two possible values for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{-2 + 18}{2} = \frac{16}{2} = 8,
\][/tex]
[tex]\[
k = \frac{-2 - 18}{2} = \frac{-20}{2} = -10.
\][/tex]
Thus, the function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = -2 \)[/tex] for [tex]\( k = 8 \)[/tex] and [tex]\( k = -10 \)[/tex].
