Given f(x) = 2x(x–1).
⇒ f'(x) = 2x(x– 1) × (2x – 1) × loge 2 > 0 for all x in [1, ∞)
⇒ f is strictly increasing function.
⇒ f is one-one function.
Also, Rf = [1, ∞)
⇒ Rf = [1, ∞) = Co-domain
⇒ f is onto function.
Thus, f is a bijective function.
So its inverse is exists.
Let y = 2x(x – 1)
⇒ y = 2x2 – x