\begin{tabular}{l}9 Use the information given to find g . \\ In each case g is a linear function. \\ \( \mathrm{g}(x)=m x+c \), for some numbers \( m \) and \( c \). \\ \( \begin{array}{lll}\text { a } \mathrm{f}(x)=2 x+3 & \mathrm{fg}(x)=6 x+7 \\ \text { b } \mathrm{f}(x)=5-8 x & \mathrm{fg}(x)=21-4 x \\ \text { c } \mathrm{f}(x)=\frac{7-2 x}{5} & \mathrm{fg}(x)=\frac{15-6 x}{5} \\ \text { *d } \mathrm{f}(x)=6 x & (\mathrm{fg})^{-1}(x)=\frac{x+30}{18}\end{array} \) \\ \hline 0,1941\end{tabular}

Let's solve each part step-by-step to find the linear function \( \mathrm{g}(x) = m x + c \). ### **General Approach** Given: - \( \mathrm{f}(x) \) - \( \mathrm{fg}(x) = \mathrm{f}(g(x)) \) Since \( \mathrm{g}(x) = m x + c \), we substitute \( g(x) \) into \( f \): \[ \mathrm{f}(g(x)) = \mathrm{f}(m x + c) \] Set this equal to the given \( \mathrm{fg}(x) \) and solve for \( m \) and \( c \). --- ### **a.** **Given:** \[ \mathrm{f}(x) = 2x + 3 \\ \mathrm{fg}(x) = 6x + 7 \] **Solution:** \[ \mathrm{f}(g(x)) = 2(m x + c) + 3 = 2m x + 2c + 3 \] Set equal to \( \mathrm{fg}(x) \): \[ 2m x + 2c + 3 = 6x + 7 \] **Equate coefficients:** \[ 2m = 6 \implies m = 3 \\ 2c + 3 = 7 \implies 2c = 4 \implies c = 2 \] **Result:** \[ \mathrm{g}(x) = 3x + 2 \] --- ### **b.** **Given:** \[ \mathrm{f}(x) = 5 - 8x \\ \mathrm{fg}(x) = 21 - 4x \] **Solution:** \[ \mathrm{f}(g(x)) = 5 - 8(m x + c) = 5 - 8m x - 8c \] Set equal to \( \mathrm{fg}(x) \): \[ -8m x + (5 - 8c) = -4x + 21 \] **Equate coefficients:** \[ -8m = -4 \implies m = \frac{1}{2} \\ 5 - 8c = 21 \implies -8c = 16 \implies c = -2 \] **Result:** \[ \mathrm{g}(x) = \frac{1}{2}x - 2 \] --- ### **c.** **Given:** \[ \mathrm{f}(x) = \frac{7 - 2x}{5} \\ \mathrm{fg}(x) = \frac{15 - 6x}{5} \] **Solution:** \[ \mathrm{f}(g(x)) = \frac{7 - 2(m x + c)}{5} = \frac{7 - 2m x - 2c}{5} \] Set equal to \( \mathrm{fg}(x) \): \[ \frac{7 - 2m x - 2c}{5} = \frac{15 - 6x}{5} \] **Simplify and equate coefficients:** \[ -2m x + (7 - 2c) = -6x + 15 \] \[ -2m = -6 \implies m = 3 \\ 7 - 2c = 15 \implies -2c = 8 \implies c = -4 \] **Result:** \[ \mathrm{g}(x) = 3x - 4 \] --- ### **d.** **Given:** \[ \mathrm{f}(x) = 6x \\ (\mathrm{fg})^{-1}(x) = \frac{x + 30}{18} \] **Solution:** Let \( y = \mathrm{fg}(x) = \mathrm{f}(g(x)) = 6g(x) \). Given the inverse: \[ x = (\mathrm{fg})^{-1}(y) = \frac{y + 30}{18} \] Express \( y \) in terms of \( x \): \[ x = \frac{y + 30}{18} \implies y = 18x - 30 \] Since \( y = 6g(x) \): \[ 6g(x) = 18x - 30 \implies g(x) = \frac{18x - 30}{6} = 3x - 5 \] **Result:** \[ \mathrm{g}(x) = 3x - 5 \] --- ### **Final Answers:** - **a.** \( \mathrm{g}(x) = 3x + 2 \) - **b.** \( \mathrm{g}(x) = \frac{1}{2}x - 2 \) - **c.** \( \mathrm{g}(x) = 3x - 4 \) - **d.** \( \mathrm{g}(x) = 3x - 5 \)