To determine the values of [tex]\( h \)[/tex] at specific points, we need to evaluate the piecewise function at those points by choosing the appropriate cases.
Step-by-Step Solution:
1. Evaluate [tex]\( h(0) \)[/tex]:
- For [tex]\( x = 0 \)[/tex], we look at the piecewise function's domain definitions:
- Since [tex]\( -2 \leq 0 \leq 1 \)[/tex], use the second case: [tex]\( h(x) = (x - 1)^2 - 2 \)[/tex].
- Plug [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[
h(0) = (0 - 1)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1
\][/tex]
- Therefore, [tex]\( h(0) = -1 \)[/tex].
2. Evaluate [tex]\( h(1) \)[/tex]:
- For [tex]\( x = 1 \)[/tex], again we check the domain definitions:
- Since [tex]\( -2 \leq 1 \leq 1 \)[/tex], use the second case: [tex]\( h(x) = (x - 1)^2 - 2 \)[/tex].
- Plug [tex]\( x = 1 \)[/tex] into the equation:
[tex]\[
h(1) = (1 - 1)^2 - 2 = 0^2 - 2 = 0 - 2 = -2
\][/tex]
- Therefore, [tex]\( h(1) = -2 \)[/tex].
3. Evaluate [tex]\( h(3) \)[/tex]:
- For [tex]\( x = 3 \)[/tex], check the domain definitions:
- Since [tex]\( 3 > 1 \)[/tex], use the third case: [tex]\( h(x) = 2 \)[/tex].
- Therefore, [tex]\( h(3) = 2 \)[/tex].
The final values for the function at the given points are:
[tex]\[
\begin{array}{l}
h(0) = -1 \\
h(1) = -2 \\
h(3) = 2
\end{array}
\][/tex]
