<h2>Answer and
Step-by-step explanation:</h2>
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
approaches some value
as
approaches
from both the right and the left, then the limit of
exists and equals
. Here we won't calculate the limit, but apply some concepts of it. So:
a. ![as \ x \rightarrow +\infty, \ k(x) \rightarrow +\infty](https://tex.z-dn.net/?f=as%20%5C%20x%20%5Crightarrow%20%2B%5Cinfty%2C%20%5C%20k%28x%29%20%5Crightarrow%20%2B%5Cinfty)
Move on the x-axis from the left to the right and you realize that as x increases y also increases without bound.
b. ![as \ x \rightarrow -\infty, \ k(x) \rightarrow 0](https://tex.z-dn.net/?f=as%20%5C%20x%20%5Crightarrow%20-%5Cinfty%2C%20%5C%20k%28x%29%20%5Crightarrow%200)
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. ![as \ x \rightarrow 2, \ k(x) \rightarrow 0](https://tex.z-dn.net/?f=as%20%5C%20x%20%5Crightarrow%202%2C%20%5C%20k%28x%29%20%5Crightarrow%200)
Since the function is continuous here, we can say that ![k(2)=0](https://tex.z-dn.net/?f=k%282%29%3D0)
d. ![as \ x \rightarrow -2, \ k(x) \rightarrow 0](https://tex.z-dn.net/?f=as%20%5C%20x%20%5Crightarrow%20-2%2C%20%5C%20k%28x%29%20%5Crightarrow%200)
The function is discontinuous here, but
exists and equals 0 as the black hole indicates at
.
e. ![as \ x \rightarrow -4, \ k(x) \rightarrow 2](https://tex.z-dn.net/?f=as%20%5C%20x%20%5Crightarrow%20-4%2C%20%5C%20k%28x%29%20%5Crightarrow%202)
The function is also discontinuous here, but the black hole indicates that this exists at
, so ![k(-4)=2](https://tex.z-dn.net/?f=k%28-4%29%3D2)
f. ![as \ x \rightarrow 0, \ k(x) \rightarrow 4](https://tex.z-dn.net/?f=as%20%5C%20x%20%5Crightarrow%200%2C%20%5C%20k%28x%29%20%5Crightarrow%204)
Since the function is continuous here, we can say that ![k(0)=4](https://tex.z-dn.net/?f=k%280%29%3D4)