\[ g(x)\left\{\begin{array}{ll} 1 & \text { if } x<-2 \\ -(x-1)^{2}+1 & \text { if }-2 \leq x \leq 1 \\ -3 & \text { if } x>1 \end{array}\right. \] find $g(-2), g(0)$, and $g(2)$

Step 1 ：The function g(x) is defined piecewise, so we need to determine which piece of the function applies for each value of x.

Step 2 ：For x = -2, we use the second piece of the function, because -2 is included in the interval [-2, 1]. So, we substitute x = -2 into the function to get \(g(-2) = -(-2-1)^{2}+1 = -8\).

Step 3 ：For x = 0, we also use the second piece of the function, because 0 is in the interval [-2, 1]. So, we substitute x = 0 into the function to get \(g(0) = - (0-1)^{2}+1 = 0\).

Step 4 ：For x = 2, we use the third piece of the function, because 2 is greater than 1. So, we substitute x = 2 into the function to get \(g(2) = -3\).

Step 5 ：The values of the function g(x) at x = -2, 0, and 2 are \(g(-2) = -8\), \(g(0) = 0\), and \(g(2) = -3\).

Step 6 ：So, the final answer is \(\boxed{g(-2) = -8, g(0) = 0, g(2) = -3}\)