Question
Consider f : R+ → [4, ⊂) given by f(x) = x2 + 4. Show that f is invertible with the inverse f–1 of f given by 
where R+ is the set of all non-negative real numbers.
Solution
f : R+ → [4, ⊂) is given by f(x) = x2 + 4.
Let x1, x2 ∈ Df = R+ = [0, ⊂) such that f(x1) = f(x2)
∴ x12 + 4 = x 22 + 4 ⇒ x12 = x22 ⇒ | x1 | = | x2 |
⇒ x1 = x2 [∵ x1x2 ≥ 0]
∴ f(x1) = f(x2) ⇒ x1 = x2 ∴ f is one-one.
Let y ∈ Rf then y = f(x), x ∈ Df = R+
⇒ y = x2 + 4 ⇒ x2 = y – 4
Now y – 4 ≥ 0 as x ≥ 0
∴ y ≥ 4 ⇒ Rf = [4, ⊂).
∴ Rf = co-domain ⇒ f is onto.
∴ f is both one-one and onto and so invertible.
Let y = f(x)
∴ y = x2 + 4
