just finding inverse of a function basically

[sour rain]

Find a constant so that a function is a self inverse

[wintry tartan]

(a) y = 1/x
xy = 1
x = 1/y
f^-1(x) = 1/x
(b) y = (3x - 5)/(x + k)
y(x + k) = 3x - 5
xy + ky = 3x - 5
ky + 5 = 3x + xy
3x + xy = ky + 5
x(3 + y) = ky + 5
x = (ky + 5)/(3 + y)
x = (ky + 5)/(y + 3)
f^-1(x) = (kx + 5)/(x + 3)

   f(x) = f^-1(x)
   (3x - 5)/(x + k) = (kx + 5)/(x + 3)
   (3x - 5)(x + 3) = (x + k)(kx + 5)
   3x² + 4x - 15 = kx² + (5 + k²)x + 5k
  k = 3, but doesn't suits the other 2 coefficients, 
  i.e. 5+k²=4 and 5k = -15, so no solution or k = ø

[sour rain]

Alright, thank you!

[compact drift]

@sour rain has given 1 rep to @wintry tartan