mohon bantu dengan caranya

               [tex]f (x) = \frac{2x + 1}{x - 3} [/tex]

     [tex] f^{- 1} (x) = \frac{ 3x + 1}{x - 2} [/tex]

[tex] f^{- 1} (x - 2) = \frac{3 (x - 2) + 1}{x - 2 - 2} [/tex]
 
             [tex]= \frac{3x - 6 + 1}{x - 4} [/tex]
 
             [tex]= \frac{3x - 5}{x - 4} ; x \neq 4 ..... jawaban : d[/tex]
     

f(x) = (2x + 1)/(x - 3)
(2x + 1)/(x - 3) = y
xy - 3y = 2x + 1
xy - 2x = 3y + 1
x(y - 2) = 3y + 1
x = (3y + 1)/(y - 2)
f^-1(x) = (3x + 1)/(x - 2)
f^-1(x - 2) = (3(x - 2) + 1)/((x - 2) - 2)
f^-1(x - 2) = (3x - 6 + 1)/(x - 4)
f^-1(x - 2) = (3x - 5)/(x - 4) ,,, x - 4 ≠ 0 => x ≠ 4