Question

suppose that the function g is defined, for all real numbers, as follows.
( g(x) = \begin{cases} -x & \text{if } x
eq -2 \\ -3 & \text{if } x = -2 end{cases} )
graph the function g.

Explanation:

Step1: Analyze the function for \( x

eq -2 \)
The function is \( g(x) = -x \) when \( x
eq -2 \). This is a linear function with a slope of \(-1\) and y-intercept at \( (0,0) \). We can plot some points: when \( x = 0 \), \( g(0) = 0 \); when \( x = 1 \), \( g(1) = -1 \); when \( x = -1 \), \( g(-1) = 1 \); when \( x = 2 \), \( g(2) = -2 \); when \( x = -3 \), \( g(-3) = 3 \), etc. But we need to leave a hole (or an open circle) at \( x = -2 \) for this part of the function because \( x
eq -2 \) here. Let's find the value at \( x = -2 \) for \( g(x) = -x \): \( g(-2) = -(-2) = 2 \), so we draw an open circle at \( (-2, 2) \).

Step2: Analyze the function for \( x = -2 \)

When \( x = -2 \), \( g(x) = -3 \). So we plot a closed circle (a filled dot) at the point \( (-2, -3) \).

Step3: Draw the graph

• For the line \( g(x) = -x \) (where \( x

eq -2 \)): Draw the line with slope \(-1\) passing through the origin, and mark an open circle at \( (-2, 2) \) to indicate that the function does not take the value \( 2 \) at \( x = -2 \) for this part.

• For the point \( x = -2 \): Mark a closed circle at \( (-2, -3) \) to show that the function takes the value \( -3 \) at \( x = -2 \).

Answer:

The graph consists of the line \( y = -x \) with an open circle at \( (-2, 2) \) and a closed circle at \( (-2, -3) \). To sketch it:

1. Draw the line \( y = -x \) (passing through points like \( (0,0) \), \( (1, -1) \), \( (-1, 1) \), etc.).
2. At \( x = -2 \), place an open circle at \( (-2, 2) \) (since \( g(x)

eq 2 \) when \( x = -2 \) for the \( y = -x \) part) and a closed circle at \( (-2, -3) \) (since \( g(-2) = -3 \)).