Question

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

Explanation:

Step1: Analyze the function for \( x

eq -2 \)
The function \( g(x) = x \) when \( x
eq -2 \). This is a linear function with a slope of 1 and y - intercept 0. So we can graph the line \( y = x \) for all real numbers except \( x=-2 \). To show the exclusion at \( x = -2 \), we will draw an open circle at the point \( (-2, -2) \) (since when \( x=-2 \), the line \( y = x \) would have \( y=-2 \), but this point is not included in the \( g(x)=x \) part of the function).

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

When \( x=-2 \), \( g(x) = 2 \). So we will plot a closed circle at the point \( (-2, 2) \) to represent this part of the piece - wise function.

Step3: Graphing the line \( y = x \) (excluding \( x=-2 \))

We can find some points on the line \( y = x \): when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y = 1 \); when \( x=-1 \), \( y=-1 \); when \( x = 2 \), \( y = 2 \); when \( x=-3 \), \( y=-3 \). We draw a straight line through these points. At \( x=-2 \), we place an open circle at \( (-2,-2) \) to indicate that this point is not part of the \( y = x \) part of the function.

Step4: Plotting the point for \( x=-2 \)

We place a closed circle at \( (-2, 2) \) to show that when \( x=-2 \), the function value is 2.

(Note: Since this is a graphing problem, the final answer is the graph constructed as described above. If we were to describe the key features: The graph is the line \( y = x \) with an open circle at \( (-2,-2) \) and a closed circle at \( (-2,2) \))

Answer:

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