\[ h(x)=\left\{\begin{array}{ll} \frac{1}{2} x+1 & \text { if } x<-2 \\ (x-1)^{2}-1 & \text { if }-2 \leq x<1 \\ -\frac{1}{2} x+2 & \text { if } x \geq 1 \end{array}\right. \] 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: \[\begin{array}{c} h(-5)= \boxed{-1.5} \\ h(-1)= \boxed{3} \\ h(1)= \boxed{1.5} \end{array}\]