F(x)=9x+4/6x-5 pertanyaan= f^-1(2)=.....?

rumus singkat
f(x) = (ax+b)/(cx+d) inversnya f^-1(x) = (-dx+b)/(cx-a)

jadi f(x) =(9x+4)/(6x-5) inversnya
f^-1(x)= (5x+4)/(6x-9)
f^-1(2) = ( 5.2+4)/(6.2-9)= 14/3

[tex]y = \frac{9x + 4}{6x - 5} \\ 6xy - 5y = 9x + 4 \\ x(6y - 9) = 4 + 5y \\ x = \frac{4 + 5y}{6y - 9} \\ f { - }^{1} (2) = \frac{14}{3} [/tex]