Let [tex]$g$[/tex] be the piecewise defined function shown.

[tex]\[ g(x)=\left\{\begin{array}{ll}
x+4, & -5 \leq x \leq -1 \\
2-x, & -1 \ \textless \ x \leq 5
\end{array}\right. \][/tex]

Evaluate [tex]$g$[/tex] at different values in its domain.

[tex]\[
\begin{array}{l}
g(-4)=\square \\
g(-2)=\square \\
g(0)=\square \\
g(3)=\square \\
g(4)=\square
\end{array}
\][/tex]



Answer :

Let's evaluate the piecewise function [tex]\( g(x) \)[/tex] at the specified values.

The function [tex]\( g(x) \)[/tex] is defined as follows:
[tex]\[ g(x) = \begin{cases} x + 4, & \text{if } -5 \leq x \leq -1 \\ 2 - x, & \text{if } -1 < x \leq 5 \end{cases} \][/tex]

### Evaluating [tex]\( g(-4) \)[/tex]:
For [tex]\( x = -4 \)[/tex]:
[tex]\[ -5 \leq -4 \leq -1 \quad \text{(so we use the first piece of the function)} \][/tex]
[tex]\[ g(-4) = -4 + 4 = 0 \][/tex]

### Evaluating [tex]\( g(-2) \)[/tex]:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ -5 \leq -2 \leq -1 \quad \text{(so we use the first piece of the function)} \][/tex]
[tex]\[ g(-2) = -2 + 4 = 2 \][/tex]

### Evaluating [tex]\( g(0) \)[/tex]:
For [tex]\( x = 0 \)[/tex]:
[tex]\[ -1 < 0 \leq 5 \quad \text{(so we use the second piece of the function)} \][/tex]
[tex]\[ g(0) = 2 - 0 = 2 \][/tex]

### Evaluating [tex]\( g(3) \)[/tex]:
For [tex]\( x = 3 \)[/tex]:
[tex]\[ -1 < 3 \leq 5 \quad \text{(so we use the second piece of the function)} \][/tex]
[tex]\[ g(3) = 2 - 3 = -1 \][/tex]

### Evaluating [tex]\( g(4) \)[/tex]:
For [tex]\( x = 4 \)[/tex]:
[tex]\[ -1 < 4 \leq 5 \quad \text{(so we use the second piece of the function)} \][/tex]
[tex]\[ g(4) = 2 - 4 = -2 \][/tex]

Thus, the evaluated values are:
[tex]\[ \begin{array}{l} g(-4) = 0 \\ g(-2) = 2 \\ g(0) = 2 \\ g(3) = -1 \\ g(4) = -2 \end{array} \][/tex]