Which values of x are point(s) of discontinuity for this function? Please explain how you did it.
g(x)= {-4, x<=-2} {-x^2, -2<x<0} {x^2, 0<x<2} {2, x>=2}

Answer Choices (Check All That Apply):
x=-4
x=-2
x=0
x=2
x=4
HURRY! WILL MARK BRAINLIEST! 

Respuesta :

The correct answers are:
x = 0
x = 2

Explanation:
When x < 0, the value of y = -
[tex]x^2[/tex].
When x > 0, the value of y = 
[tex]x^2[/tex].
When x = 0, there is no value of y mentioned in the conditions; it means that there is discontinuity at x = 0.

When x < 2, the value of y = 
[tex]x^2[/tex].
When x >= 2, the value of y = 2.

It means that, at x = 2, there is a discontinuity because the graph of y = 
[tex]x^2[/tex] is not touching the graph of y = 2 at x = 2.

Answer:

x = -2

x = 0

x = 4

x = 0 and x = 4 (for the 2nd part)

Step-by-step explanation: