In Problems 17-22, find the function values for each function. 17. \( f(x)=8 ; f(2), f(t+8), f(-\sqrt{17}) \) 18. \( g(x)=|x-3| ; g(10), g(3), g(-3) \) 19. \( F(t)=\left\{\begin{array}{rr}2 & \text { if } t>1 \\ 0 & \text { if } t=1 \\ -1 & \text { if } t<1\end{array}\right. \) \( F(12), F(-\sqrt{3}), F(1), F\left(\frac{18}{5}\right) \)

For the first function, since \( f(x) = 8 \) is a constant function, it means no matter what value you plug in, the output will always be 8. Therefore, \( f(2) = 8 \), \( f(t+8) = 8 \), and \( f(-\sqrt{17}) = 8 \). When it comes to the second function \( g(x) = |x-3| \), you simply need to calculate the distance of x from 3. So for \( g(10) = |10-3| = 7 \), \( g(3) = |3-3| = 0 \), and \( g(-3) = |-3-3| = 6 \). For the piecewise function \( F(t) \), we check where \( t \) falls. For \( F(12) \), since \( 12 > 1 \), we have \( F(12) = 2 \). For \( F(-\sqrt{3}) \) (which is less than 1), \( F(-\sqrt{3}) = -1 \). At \( F(1) \), where \( t = 1 \), the output is 0. Finally, \( F\left(\frac{18}{5}\right) \) (which is also greater than 1) gives \( F\left(\frac{18}{5}\right) = 2 \). So in summary, \( F(12) = 2, F(-\sqrt{3}) = -1, F(1) = 0, F\left(\frac{18}{5}\right) = 2 \). Here’s a fun fact: constant functions like \( f(x) = 8 \) can serve as the trusty companions in math, providing stability without the drama of changing values! Meanwhile, absolute value functions are like that reliable friend who always provides 'straight talk'—no matter the scenario, it knows how to measure ‘distance’ perfectly!