jika fungsi f dn g adl f: x -> 2x^3 g: x-> x^3/2 maka (gf)^-1 (√2) =

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f(x) = 2x³ ⇒ f⁻¹(x) = (x/2)^(1/3)
g(x) = x^(3/2) ⇒ g⁻¹(x) = x^(2/3)
(g o f)⁻¹(√2) = f⁻¹ (√2)^2/3
= (1/2)(√2)^(2/9)

^ = pangkat