Consider $f:R^{+} \rightarrow\left[4, \infty\right.$ ) given by $f(x)=x^2+4$ (where $R^{+}$is set of non negative real numbers) then $f^{-1}(x)$ is :

$\sqrt{x-4}$

$f: R^{+} \rightarrow {[4, \infty)}$

$f(x)= x^2+4$

let $f(x)=y$

$y= x^2+4$

so $x^2 =y-4$

$\Rightarrow x =\sqrt{y-4}$       since $f: R^{+} \rightarrow[4, \infty)$

hence only positive value of x considered

so $f^{-1}(x)=\sqrt{x-4}$