f(x) = 3 · √x + 12
Let f(x) = y
y = 3 · √x + 12
y - 12 = 3 · √x .............. (Subtract 12 from both sides)
(y - 12)/3 = √x .............. (Divide both sides by 3)
(y/3) - (12/3) = √x
(y/3) - 4 = √x
[(y/3) - 4]² = x ............... (Square both sides)
Therefore, f⁻¹(y) = [(y/3) - 4]²
f⁻¹(x) = [(x/3) - 4]²
