If ƒ(x ) = 3x + 1 and ƒ -1 = , then ƒ -1(7) =
f(x) = 3x + 1let f(x) = yy = 3x + 1y - 1 = 3x3x = y - 1x = (y - 1)/3From y = f(x), x = f⁻¹(y) x = (y - 1)/3 f⁻¹(y) = (y - 1)/3f⁻¹(7) = (7 - 1)/3 = 6/3 = 2f⁻¹(7) = 2