Question

Consider f: R+ → [4, ∞) given by f(x) = x^2 + 4. Show that f is invertible with the inverse f−1 (f-inverse) of given f by f-1(y) = (y - 4)^1/2, where R+ is the set of all non-negative real numbers.

Answer 1

Given, f : R⁺→ [4, ∞) given by f (x) = x² + 4.
Now, Let x, y ∈ R⁺ such that f(x) = f(y)
⇒ x² + 4 = y² + 4
⇒ x² = y²
⇒ x = y
therefore, f is one-one function.

Now, for y ∈ [4, ∞) , assume y = x² + 4
⇒x² = y - 4 ≥ 0 [ because squaring of any function is always positive.]
⇒x = $\sqrt{y-4}$ ≥ 0
⇒ for any y ∈ R, there exists x =$\sqrt{y-4}$ ∈ R such that
$f(\sqrt{y-4})=(\sqrt{y-4})^2+4=y-4+4=y$
therefore, f is onto .
since f is one - one and onto.
so, f is inversible function.

Let's find inverse of f(x),
f(x) = x² + 4
⇒y = x² + 4
⇒y - 4 = x²
taking square root both sides,
x = ±$\sqrt{y-4}$
because x ∈ R⁺ , x = $\sqrt{y-4}$
we can also write x = f(y) = $\sqrt{y-4}$
we know, f(y) will be inverse of f(x) when we put x in place of y.
so, $\displaystyle{f^{-1}(x)=\sqrt{x-4}}$