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 :
Answer & explanation
Correct answer: option 3
$f: R^{+} \rightarrow {[4, \infty)}$
$f(x)= x^2+4$
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 $f^{-1}(x)=\sqrt{x-4}$