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Suppose that the function [tex]\( g \)[/tex] is defined, for all real numbers, as follows.
[tex]\[ g(x) = \begin{cases}
\frac{1}{4} x + 1 & \text{if } x \leq -2 \\
(x - 1)^2 & \text{if } -2 \ \textless \ x \leq 1 \\
3 & \text{if } x \ \textgreater \ 1
\end{cases} \][/tex]

Find [tex]\( g(-1) \)[/tex], [tex]\( g(1) \)[/tex], and [tex]\( g(3) \)[/tex].

[tex]\[
\begin{array}{l}
g(-1) = \square \\
g(1) = \square \\
g(3) = \square
\end{array}
\][/tex]

Sagot :

To find the values of [tex]\( g(-1) \)[/tex], [tex]\( g(1) \)[/tex], and [tex]\( g(3) \)[/tex], we can refer directly to the definition of the function [tex]\( g(x) \)[/tex] for different domains of [tex]\( x \)[/tex]. Let's go step by step:

1. Find [tex]\( g(-1) \)[/tex]:
- For [tex]\( g(-1) \)[/tex], we need to determine which part of the function [tex]\( g(x) \)[/tex] to use. Since [tex]\(-2 < -1 \leq 1\)[/tex], we use the second piece of the function: [tex]\( g(x) = (x - 1)^2 \)[/tex].
- Now, substitute [tex]\( x = -1 \)[/tex] into this piece:
[tex]\[ g(-1) = (-1 - 1)^2 = (-2)^2 = 4 \][/tex]

2. Find [tex]\( g(1) \)[/tex]:
- For [tex]\( g(1) \)[/tex], [tex]\(-2 < 1 \leq 1\)[/tex] places us in the same primary domain as before: [tex]\( g(x) = (x - 1)^2 \)[/tex].
- Now, substitute [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = (1 - 1)^2 = 0^2 = 0 \][/tex]

3. Find [tex]\( g(3) \)[/tex]:
- For [tex]\( g(3) \)[/tex], since [tex]\( 3 > 1 \)[/tex], we need to use the third part of the function: [tex]\( g(x) = 3 \)[/tex].
- Substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ g(3) = 3 \][/tex]

Summarizing our findings, we have:
[tex]\[ \begin{array}{l} g(-1) = 4 \\ g(1) = 0 \\ g(3) = 3 \end{array} \][/tex]

Thus, the values are:
[tex]\[ g(-1) = 4, \quad g(1) = 0, \quad g(3) = 3 \][/tex]