If [tex]f^{-1}(x)[/tex] is the inverse of
[tex]f(x) = \dfrac23x+9[/tex]
then
[tex]f\left(f^{-1}(x)\right) = \dfrac23f^{-1}(x) + 9 = x[/tex]
Solve for [tex]f^{-1}(x)[/tex] :
[tex]\dfrac23f^{-1}(x)+9 = x \\\\ \dfrac23f^{-1}(x) = x-9 \\\\ f^{-1}(x) = \dfrac32(x-9) = \dfrac32x-\dfrac{27}2[/tex]
which makes D the correct answer.
