If ƒ (x) = 3x + 1 and ƒ - 1 =, then ƒ - 1 (7) =

F (x) = 3x + 1

let f (x) = y

y = 3x + 1

y - 1 = 3x

3x = y - 1

x = (y - 1) / 3

From y = f (x), x = f⁻¹ (y)

x = (y - 1) / 3

f⁻¹ (y) = (y - 1) / 3

f⁻¹ (7) = (7 - 1) / 3 = 6/3 = 2

f⁻¹ (7) = 2