Consider f : R → R+ → [4, ∞) given by f(x) = x2 + 4.

Solution:

Injectivity of $f$ :

Let $x$ and $y$ be two elements of the domain $(Q)$, such that

$f(x)=f(y)$

$\Rightarrow x^{2}+4=y^{2}+4$

$\Rightarrow x^{2}=y^{2}$

$\Rightarrow x=y \quad\left(\right.$ as co-domain as $\left.R^{+}\right)$

So, $f$ is one-one.

Surjectivity of $f$ :

Let $y$ be in the co-domain $(Q)$, such that $f(x)=y$

$\Rightarrow x^{2}+4=y$

$\Rightarrow x^{2}=y-4$

$\Rightarrow x=\sqrt{y-4} \in R$

$\Rightarrow f$ is onto.

So, $f$ is a bijection and, hence, it is invertible.

Finding $f^{-1}$ :

Let $f^{-1}(x)=y$

$\Rightarrow x=f(y)$                 ...(1)

$\Rightarrow x=y^{2}+4$

$\Rightarrow x-4=y^{2}$

$\Rightarrow y=\sqrt{x-4}$

So, $f^{-1}(x)=\sqrt{x-4} \quad[$ from (1) $]$