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

[tex]\[ g(x) = \begin{cases}
x + 4, & -5 \leq x \leq -1 \\
2 - x, & -1 \ \textless \ x \leq 5
\end{cases} \][/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 function [tex]\( g(x) \)[/tex] at the given points step-by-step:

1. Evaluate [tex]\( g(-4) \)[/tex]:
- For [tex]\( x = -4 \)[/tex]:
- Since [tex]\(-5 \leq -4 \leq -1\)[/tex], we use the first piece of the function: [tex]\( g(x) = x + 4 \)[/tex].
- Substituting [tex]\( x = -4 \)[/tex] into this piece:
[tex]\[ g(-4) = -4 + 4 = 0 \][/tex]

2. Evaluate [tex]\( g(-2) \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
- Since [tex]\(-5 \leq -2 \leq -1\)[/tex], we use the first piece of the function: [tex]\( g(x) = x + 4 \)[/tex].
- Substituting [tex]\( x = -2 \)[/tex] into this piece:
[tex]\[ g(-2) = -2 + 4 = 2 \][/tex]

3. Evaluate [tex]\( g(0) \)[/tex]:
- For [tex]\( x = 0 \)[/tex]:
- Since [tex]\(-1 < 0 \leq 5\)[/tex], we use the second piece of the function: [tex]\( g(x) = 2 - x \)[/tex].
- Substituting [tex]\( x = 0 \)[/tex] into this piece:
[tex]\[ g(0) = 2 - 0 = 2 \][/tex]

4. Evaluate [tex]\( g(3) \)[/tex]:
- For [tex]\( x = 3 \)[/tex]:
- Since [tex]\(-1 < 3 \leq 5\)[/tex], we use the second piece of the function: [tex]\( g(x) = 2 - x \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] into this piece:
[tex]\[ g(3) = 2 - 3 = -1 \][/tex]

5. Evaluate [tex]\( g(4) \)[/tex]:
- For [tex]\( x = 4 \)[/tex]:
- Since [tex]\(-1 < 4 \leq 5\)[/tex], we use the second piece of the function: [tex]\( g(x) = 2 - x \)[/tex].
- Substituting [tex]\( x = 4 \)[/tex] into this piece:
[tex]\[ g(4) = 2 - 4 = -2 \][/tex]

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