Find functions \( f \) and \( g \) so that \( f \circ g=H \). \( H(x)=|2 x+9| \) Choose the correct pair of functions. A. \( f(x)=|-x|, g(x)=\frac{x-9}{2} \) B. \( f(x)=2 x+9, g(x)=|x| \) C. \( f(x)=\frac{x-9}{2}, g(x)=|-x| \) D. \( f(x)=|x|, g(x)=2 x+9 \)

To find functions \( f \) and \( g \) such that \( f \circ g = H \), we want \( f(g(x)) = H(x) = |2x + 9| \). Let's analyze each option: A. \( f(x) = |-x|, g(x) = \frac{x - 9}{2} \) Calculating \( f(g(x)) \): \[ g(x) = \frac{x - 9}{2} \Rightarrow f(g(x)) = f\left(\frac{x - 9}{2}\right) = \left| -\frac{x - 9}{2} \right| = \frac{x - 9}{2} \] This does not equal \( |2x + 9| \), so this option is incorrect. B. \( f(x) = 2x + 9, g(x) = |x| \) Calculating \( f(g(x)) \): \[ g(x) = |x| \Rightarrow f(g(x)) = f(|x|) = 2|x| + 9 \] This is incorrect since it does not match \( |2x + 9| \). C. \( f(x) = \frac{x - 9}{2}, g(x) = |-x| \) Calculating \( f(g(x)) \): \[ g(x) = |-x| \Rightarrow f(g(x)) = f(|x|) = \frac{|x| - 9}{2} \] This does not yield \( |2x + 9| \). D. \( f(x) = |x|, g(x) = 2x + 9 \) Calculating \( f(g(x)) \): \[ g(x) = 2x + 9 \Rightarrow f(g(x)) = f(2x + 9) = |2x + 9| \] This correctly equals \( H(x) \). Thus, the correct pair of functions is **D**: \( f(x) = |x|, g(x) = 2x + 9 \).