Suppose that the function \( g \) is defined, for all real numbers, as follows. \[ \begin{array}{ll}-2 & \text { if } x<-2 \\ (x+1)^{2}-3 & \text { if }-2 \leq x<1 \\ \frac{1}{2} x+1 & \text { if } x \geq 1\end{array} \] Find \( g(-4), g(0) \), and \( g(1) \). \( g(-4)=\square \)

We are given the piecewise function \[ g(x)= \begin{cases} -2 & \text{if } x<-2, \\ (x+1)^2-3 & \text{if } -2 \leq x<1, \\ \frac{1}{2}x+1 & \text{if } x \geq 1. \end{cases} \] **Step 1:** Find \( g(-4) \). Since \(-4 < -2\), we use the first rule: \[ g(-4) = -2. \] **Step 2:** Find \( g(0) \). Since \(-2 \leq 0 < 1\), we use the second rule: \[ g(0) = (0+1)^2-3 = 1^2-3 = 1-3 = -2. \] **Step 3:** Find \( g(1) \). Since \(1 \geq 1\), we use the third rule: \[ g(1) = \frac{1}{2}(1) + 1 = \frac{1}{2} + 1 = \frac{3}{2} = 1.5. \] Thus, the answers are: \[ g(-4) = -2,\quad g(0) = -2,\quad g(1) = 1.5. \] The value of \( g(-4) \) is \( -2 \).