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

(A) $f: R_{+} \rightarrow [4, \infty)$ is given as $f(x) = x^{2} + 4$.
For one-one:
Let $f(x) = f(y)$.
$\Rightarrow x^{2} + 4 = y^{2} + 4$.
$\Rightarrow x^{2} = y^{2}$.
$\Rightarrow x = y$ (since $x, y \in R_{+}$).
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,
$(g \circ f)(x) = g(f(x)) = g(x^{2} + 4) = \sqrt{(x^{2} + 4) - 4} = \sqrt{x^{2}} = x$.
And
$(f \circ g)(y) = f(g(y)) = f(\sqrt{y - 4}) = (\sqrt{y - 4})^{2} + 4 = y - 4 + 4 = y$.
Therefore,$g \circ f = I_{R_{+}}$ and $f \circ g = I_{[4, \infty)}$.
Hence,$f$ is invertible and the inverse of $f$ is given by $f^{-1}(y) = g(y) = \sqrt{y - 4}$.