Given f(x) = x2 + 2
⇒ f'(x) = 2x > 0 for every x > 0
⇒ f is strictly increasing function.
⇒ f is one-one function.
Also, Rf = (2, ∞) = Co-domain
⇒ f is onto function.
Thus f is a bijective function.
Therefore, the inverse of the given function is exists.
Let y = x2 + 2
⇒ x2 = y – 2
⇒ x = √(y – 2)
Hence, f–1(x) = √(x – 2)