Question

piecewise defined functions
removing discontinuities
which statement is true for ( g(x) ) shown in the graph?
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:

A removable discontinuity (hole) exists when the limit of the function at a point exists, but the function is not defined there, or the defined value does not match the limit.

1. For $x=0$: The left-hand and right-hand limits of $g(x)$ as $x$ approaches 0 both equal 0, but the function has an open circle at $(0,2)$ (so $g(0)$ is not defined as 0). To remove this discontinuity, we need to set $g(0)$ equal to the limit, which is 0.
2. For $x=3$: The point at $x=3$ is a minimum of the curve (a closed dot), so the function is already continuous there, and there is no discontinuity to remove.
3. The option defining $g(0)=2$ is incorrect because it does not match the limit of the function at $x=0$.

Answer:

You can remove the discontinuity at $x = 0$ by defining $g(0) = 0$.