Answer
one-to-one function
inverse: $f^{-1}(x)=x^2-2,\text{ }x\ge0$
Work Step by Step
Some of the ordered pairs of the given function, $
f(x)=\sqrt{x+2},\text{ }x\ge-2
$, are $
\left\{(-2,0)(-1,1),(2,2),(7,3),(14,4),...\right\}
$. Note that every $y$-coordinate from this function is unique. Hence, the given function is a one-to-one function.
To find the inverse, let $y=f(x)$. Then, interchange the $x$ and $y$ variables and solve for $y$. That is,
\begin{align*}\require{cancel}
y&=\sqrt{x+2},\text{ }x\ge-2
\\&\Rightarrow
x=\sqrt{y+2},\text{ }y\ge-2
&(\text{interchange $x$ and $y$})
\\
&(\text{* Note that $y\ge-2$ implies $x\ge0$.})
\\&
(x)^2=\left(\sqrt{y+2}\right)^2,\text{ }x\ge0
&(\text{solve for $y$})
\\&
x^2=y+2
\\&
x^2-2=y
\\&
y=x^2-2
.\end{align*}
Hence, the inverse is $
f^{-1}(x)=x^2-2,\text{ }x\ge0
$.
