Question

Consider f : R+ → [4, ∝) given by f(x) = x² + 4. Show that f is invertible with the inverse of f-1 of f given by f-1(y) = \(\sqrt{y-4}\), where R+ is the set of all non-negative real numbers.

Answer 1

f(x1) = x²1 + 4 and f(x2) = x²2 + 4

So, f(x1) = f(x2) ⇒ x²1 + 4 = x²2 + 4

or x²1 = x²2 ⇒ x1 = x2

As x ∈ R, ∴ x > 0,

x²1 = x²2 ⇒ x1 = x2

⇒ f is one-one.

Let y = x² + 4 or x² = y – 4 or x = ± \(\sqrt{y-4}\).

x being > 0, -vc sign not to be taken.

∴ = \(\sqrt{y-4}\)

∴ f-1(y) = g(y) = \(\sqrt{y-4}\), y ≥ 4.

For every y ≥ 4, g(y) has real positive value.

∴ The inverse of f is f-1(y) = \(\sqrt{y-4}\).