Answer and Step-by-step explanation:
This is a piecewise function because it is defined by more than two functions. Basically, we want to take the limit here. Recall that if a function [tex]f(x)[/tex] approaches some value [tex]L[/tex] as [tex]x[/tex] approaches [tex]a[/tex] from both the right and the left, then the limit of [tex]f(x)[/tex] exists and equals [tex]L[/tex]. Here we won't calculate the limit, but apply some concepts of it. So:
a. [tex]as \ x \rightarrow +\infty, \ k(x) \rightarrow +\infty[/tex]
Move on the x-axis from the left to the right and you realize that as x increases y also increases without bound.
b. [tex]as \ x \rightarrow -\infty, \ k(x) \rightarrow 0[/tex]
Move on the x-axis from the right to the left and you realize that as x decreases to negative values y approaches zero.
c. [tex]as \ x \rightarrow 2, \ k(x) \rightarrow 0[/tex]
Since the function is continuous here, we can say that [tex]k(2)=0[/tex]
d. [tex]as \ x \rightarrow -2, \ k(x) \rightarrow 0[/tex]
The function is discontinuous here, but [tex]k(-2)[/tex] exists and equals 0 as the black hole indicates at [tex]x=-2[/tex].
e. [tex]as \ x \rightarrow -4, \ k(x) \rightarrow 2[/tex]
The function is also discontinuous here, but the black hole indicates that this exists at [tex]x=-4[/tex], so [tex]k(-4)=2[/tex]
f. [tex]as \ x \rightarrow 0, \ k(x) \rightarrow 4[/tex]
Since the function is continuous here, we can say that [tex]k(0)=4[/tex]
