\[
h(x)=\left\{\begin{array}{ll}
\frac{1}{2} x+1 & \text { if } x< -2 \\
(x-1)^{2}-1 & \text { if }-2 \leq x< 1 \\
-\frac{1}{2} x+2 & \text { if } x \geq 1
\end{array}\right.
\]
Find $h(-5), h(-1)$, and $h(1)$
\[
\begin{array}{c}
h(-5)= \\
h(-1)= \\
h(1)=
\end{array}
\]

Step 1 ：The function h(x) is a piecewise function, which means it is defined by different expressions depending on the value of x. To find the values of h(-5), h(-1), and h(1), we need to substitute these values into the appropriate part of the function definition.

Step 2 ：For h(-5), since -5 < -2, we use the first part of the function definition: \(\frac{1}{2} x + 1\).

Step 3 ：For h(-1), since -2 <= -1 < 1, we use the second part of the function definition: \((x - 1)^{2} - 1\).

Step 4 ：For h(1), since 1 >= 1, we use the third part of the function definition: \(-\frac{1}{2} x + 2\).

Step 5 ：Substituting the values, we get h(-5) = -1.5, h(-1) = 3, and h(1) = 1.5.

Step 6 ：Final Answer: \[\begin{array}{c} h(-5)= \boxed{-1.5} \\ h(-1)= \boxed{3} \\ h(1)= \boxed{1.5} \end{array}\]