Alright! Let's determine the values for the piecewise-defined function [tex]\( f(x) \)[/tex] at the given points.
The piecewise function [tex]\( f(x) \)[/tex] is defined as follows:
[tex]\[
f(x) = \begin{cases}
3 - 5x & \text{if } x \leq 1 \\
3x & \text{if } 1 < x < 6 \\
5x + 2 & \text{if } x \geq 6
\end{cases}
\][/tex]
### Finding [tex]\( f(-4) \)[/tex]:
Since [tex]\( -4 \leq 1 \)[/tex], we use the first case of the function:
[tex]\[
f(-4) = 3 - 5(-4)
\][/tex]
[tex]\[
f(-4) = 3 + 20
\][/tex]
[tex]\[
f(-4) = 23
\][/tex]
### Finding [tex]\( f(0) \)[/tex]:
Since [tex]\( 0 \leq 1 \)[/tex], we use the first case of the function:
[tex]\[
f(0) = 3 - 5(0)
\][/tex]
[tex]\[
f(0) = 3
\][/tex]
### Finding [tex]\( f(3) \)[/tex]:
Since [tex]\( 1 < 3 < 6 \)[/tex], we use the second case of the function:
[tex]\[
f(3) = 3(3)
\][/tex]
[tex]\[
f(3) = 9
\][/tex]
### Finding [tex]\( f(4) \)[/tex]:
Since [tex]\( 1 < 4 < 6 \)[/tex], we use the second case of the function:
[tex]\[
f(4) = 3(4)
\][/tex]
[tex]\[
f(4) = 12
\][/tex]
### Finding [tex]\( f(8) \)[/tex]:
Since [tex]\( 8 \geq 6 \)[/tex], we use the third case of the function:
[tex]\[
f(8) = 5(8) + 2
\][/tex]
[tex]\[
f(8) = 40 + 2
\][/tex]
[tex]\[
f(8) = 42
\][/tex]
Thus, the values are:
[tex]\[
\begin{aligned}
f(-4) & = 23 \\
f(0) & = 3 \\
f(3) & = 9 \\
f(4) & = 12 \\
f(8) & = 42 \\
\end{aligned}
\][/tex]
