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Suppose that the function [tex]h[/tex] is defined, for all real numbers, as follows.

[tex]\[
h(x)=\left\{\begin{array}{ll}
-\frac{1}{4} x-1 & \text{if } x\ \textless \ -2 \\
(x-1)^2-2 & \text{if } -2 \leq x \leq 1 \\
2 & \text{if } x\ \textgreater \ 1
\end{array}\right.
\][/tex]

Find [tex]h(0)[/tex], [tex]h(1)[/tex], and [tex]h(3)[/tex].

[tex]\[
\begin{array}{l}
h(0)= \\
h(1)= \\
h(3)=
\end{array}
\][/tex]



Answer :

GinnyAnswer
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]

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