Let \[f(x) = \frac{2x + 3}{kx - 2}.\]Find all real numbers $k$ so that $f^{-1}(x) = f(x).$
+128460 y = (2x + 3) / (kx - 2) get x by itself
y ( kx - 2) = 2x + 3
ykx - 2y = 2x + 3
x (ky) - 2y = 2x + 3
-2y - 3 = 2x - x (ky)
-2y - 3 = x ( 2 - ky)
-(2y + 3) / ( 2 - ky ) = x swap x and y
(2x + 3) / ( kx - 2) = y = the inverse
The function is its own inverse
We just can't have kx - 2 = 0
So.... kx = 2 will make the denominator = 0
k = 2/x
k can be any value except 2/x
