Question

you can remove the discontinuity at ( x = 3 ) by defining ( g(3) = -1 ). you can remove the discontinuity at ( x = 0 ) by defining ( g(0) = 0 ). you can remove the discontinuity at ( x = 0 ) by defining ( g(0) = 2 ).

Explanation:

Step1: Identify removable discontinuities

Removable discontinuities occur at $x$-values where the graph has a hole, meaning the left-hand and right-hand limits exist and are equal, but the function is not defined (or defined incorrectly) at that point.
From the graph:

• At $x=0$: The left and right sides of the graph approach the point $(0, 2)$ (the hole is at $x=0$, $y=2$).
• At $x=3$: The left and right sides of the graph approach the point $(3, -1)$ (the hole is at $x=3$, $y=-1$).

Step2: Match discontinuities to correct definitions

For $x=0$, to remove the discontinuity, we define $g(0)$ equal to the limit value, which is 2.
For $x=3$, to remove the discontinuity, we define $g(3)$ equal to the limit value, which is -1.

Answer:

• You can remove the discontinuity at $x=3$ by defining $g(3)=-1$.
• You can remove the discontinuity at $x=0$ by defining $g(0)=2$.