mohon bantuanya nomer 5 dan caranya

[tex]g(x) = \frac{2x + 4}{3x - 1} \\ y = \frac{2x + 4}{3x - 1} \\ y(3x - 1) = 2x + 4 \\ 3xy - y = 2x + 4 \\ 3xy - 2x = 4 + y \\ x(3y - 2) = 4 + y \\ x = \frac{4 + y}{3y - 2} \\ g{}^{ - 1} (x) = \frac{4 + x}{3x - 2} \\ g {}^{ - 1} (3) = \frac{4 + 3}{3(3) - 2} \\ g {}^{ - 1} (
3) = \frac{7}{7} \\ g {}^{ - 1} (3)= 1[/tex]