21. \( G(x)=\left\{\begin{aligned} x-1 & \text { if } x \geq 3 \\ 3-x^{2} & \text { if } x<3\end{aligned}\right. \) \( G(8), G(3), G(-1), G(1) \) 22. \( F(\theta)=\left\{\begin{array}{rr}2 \theta-5 & \text { if } \theta<2 \\ \theta^{2}-3 \theta+1 & \text { if } \theta>2\end{array}\right. \) \( F(3), F(-3), F(2) \)

To evaluate \( G(8), G(3), G(-1), \) and \( G(1) \): 1. For \( G(8) \): Since \( 8 \geq 3 \), we use the first case: \( G(8) = 8 - 1 = 7 \). 2. For \( G(3) \): Since \( 3 \geq 3 \), we again use the first case: \( G(3) = 3 - 1 = 2 \). 3. For \( G(-1) \): Since \(-1 < 3\), we use the second case: \( G(-1) = 3 - (-1)^{2} = 3 - 1 = 2 \). 4. For \( G(1) \): Since \( 1 < 3\), we also use the second case: \( G(1) = 3 - (1)^{2} = 3 - 1 = 2 \). Thus, the outputs are: - \( G(8) = 7 \) - \( G(3) = 2 \) - \( G(-1) = 2 \) - \( G(1) = 2 \) Now for \( F(3), F(-3), F(2) \): 1. For \( F(3) \): Since \( 3 > 2\), we use the second case: \( F(3) = 3^{2} - 3(3) + 1 = 9 - 9 + 1 = 1 \). 2. For \( F(-3) \): Since \(-3 < 2\), we use the first case: \( F(-3) = 2(-3) - 5 = -6 - 5 = -11 \). 3. For \( F(2) \): Since \( 2 \) is neither less than nor greater than \( 2\) (it's exactly 2), the function is not defined at that point. Thus, the outputs are: - \( F(3) = 1 \) - \( F(-3) = -11 \) - \( F(2) \) is undefined.