Sure! Let's evaluate the piecewise function at the given points step-by-step.
The piecewise function in question is defined as:
[tex]\[
f(x) =
\begin{cases}
5 & \text{if } x \leq -1 \\
2x + 1 & \text{if } 1 < x < 3 \\
7 & \text{if } x \geq 3
\end{cases}
\][/tex]
Step 1: Evaluating [tex]\( f(-3) \)[/tex]
Since [tex]\( -3 \leq -1 \)[/tex], we use the first part of the piecewise function:
[tex]\[
f(-3) = 5
\][/tex]
Step 2: Evaluating [tex]\( f(-1) \)[/tex]
Since [tex]\( -1 \leq -1 \)[/tex], it falls under the same case as the previous step:
[tex]\[
f(-1) = 5
\][/tex]
Step 3: Evaluating [tex]\( f(3) \)[/tex]
Since [tex]\( 3 \geq 3 \)[/tex], we use the last part of the piecewise function:
[tex]\[
f(3) = 7
\][/tex]
Summarizing the evaluations:
[tex]\[
\begin{align*}
f(-3) &= 5, \\
f(-1) &= 5, \\
f(3) &= 7.
\end{align*}
\][/tex]
Thus, the values are:
[tex]\[
\begin{array}{l}
f(-3) = 5 \\
f(-1) = 5 \\
f(3) = 7
\end{array}
\][/tex]
