$f: R_{+} \rightarrow[4, \infty)$ is given as $f(x)=x^{2}+4$

For one – one

Let $f ( x )= f ( y )$

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

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

$\Rightarrow x = y$                $[$ as $x=y \in R_+]$

$\therefore f$ is a one $-$ one function.

For onto

For $y \in[4, \infty),$ let $y=x^{2}+4$

$\Rightarrow x^{2}=y-4 \geq 0$   $[$ as  $y \geq 4]$

$\Rightarrow x=\sqrt{Y-4} \geq 0$

Therefore, for any $y \in[4, \infty)$, there exists $x=\sqrt{Y-4} \in R_+$, such that

$f(x)=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=y-4+4=y$

$\therefore$ $f$ is onto.

Thus, $f$ is one – one and onto and therefore, $f^{-1}$ exists.

Let us define $g:$  $[4, \infty) \rightarrow R +$ by $g ( y )=\sqrt{Y-4}$

Now,

$(gof)(x)=g(f(x))=g\left(x^{2}+4\right)=\sqrt{\left(x^{2}+4\right)-4}=\sqrt{x^{2}}=x$

and

$(f o g)(y)=f(g(y))=f(\sqrt{Y-4})=(\sqrt{Y-4})^{2}+4=y-4+4=y$

$\therefore $ $gof= fog = I _{ R }$

Hence, $f$ is invertible and the inverse of $f$ is given by $f^{1}(y)=g(y)=\sqrt{y-4}$