Let f(x) ≡ Ax+B Then f(g(x)) ≡ A(2x-4)+B ≡ 2Ax-4A+B 2Ax-4A+B ≡ x Therefore equating coefficients of x: 2A = 1 A = 1/2 Equating constant terms: -4A+B = 0 -4(1/2)+B = 0 -2+B = 0 B = 2 Therefore f(x) ≡ Ax+B ≡ (1/2)x+2 Edwin