To find the values of the piecewise function [tex]\( h(x) \)[/tex] for the given inputs, we need to evaluate the function according to the defined conditions:
[tex]\[ h(x) = \begin{cases}
-3x - 14 & \text{for } x < -8 \\
1 & \text{for } -8 \leq x < 1 \\
x + 5 & \text{for } x \geq 1
\end{cases} \][/tex]
1. Evaluating [tex]\( h(-10) \)[/tex]:
- Since [tex]\(-10 < -8\)[/tex], we use the first piece of the function: [tex]\( h(x) = -3x - 14 \)[/tex].
- Substituting [tex]\(-10\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
h(-10) = -3(-10) - 14 = 30 - 14 = 16
\][/tex]
- Therefore, [tex]\( h(-10) = 16 \)[/tex].
2. Evaluating [tex]\( h(-1) \)[/tex]:
- Since [tex]\(-8 \leq -1 < 1\)[/tex], we use the second piece of the function: [tex]\( h(x) = 1 \)[/tex].
- Therefore, [tex]\( h(-1) = 1 \)[/tex].
3. Evaluating [tex]\( h(1) \)[/tex]:
- Since [tex]\(1 \geq 1\)[/tex], we use the third piece of the function: [tex]\( h(x) = x + 5 \)[/tex].
- Substituting [tex]\(1\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
h(1) = 1 + 5 = 6
\][/tex]
- Therefore, [tex]\( h(1) = 6 \)[/tex].
4. Evaluating [tex]\( h(8) \)[/tex]:
- Since [tex]\(8 \geq 1\)[/tex], we use the third piece of the function: [tex]\( h(x) = x + 5 \)[/tex].
- Substituting [tex]\(8\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
h(8) = 8 + 5 = 13
\][/tex]
- Therefore, [tex]\( h(8) = 13 \)[/tex].
So, the values are:
[tex]\[ h(-10) = 16, \quad h(-1) = 1, \quad h(1) = 6, \quad h(8) = 13 \][/tex]
