Consider $f: R \rightarrow R_{+} \rightarrow[4, \infty)$ given by $f(x)=x^{2}+4$. Show that $f$ is invertible with inverse $f^{-1}$ of $f$ given by $f^{-1}(x)=\sqrt{x-4}$, where $R^{+}$is the set of all non-negative real numbers.
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) $]$