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## Manipulating Radical Functions 1.3 - Solution

We want to determine whether the given pair of functions are inverse functions. $\begin{gathered} f(x)=(x+6)^2 \quad \text{and} \quad g(x)=\sqrt{x}-6 \end{gathered}$ To do so, we need to verify that the compositions of $f(x)$ and $g(x)$ are the identity function. Since the domain of $g(x)$ is all real numbers greater than or equal to zero, we will only consider nonnegative values for the $x\text{-}$variable.

### Calculating $f(g(x))$

To find the expression for $f(g(x))$ we will start by substituting $\sqrt{x}-6$ for $g(x).$ $f(g(x)) \quad \Rightarrow \quad f \left( {\color{#0000FF}{\sqrt{x}-6}} \right)$ Now we apply the definition of $f(x).$ $\begin{gathered} f(x)=(x+6)^2 \\ \Downarrow \\ f\left({\color{#0000FF}{\sqrt{x}-6}} \right) = \left( {\color{#0000FF}{\sqrt{x}-6}}+6 \right)^2 \end{gathered}$ Finally, let's simplify and see if the function is the identity function.
$f(g(x)) = \left( \sqrt{x}-6+6 \right) ^2$
$f(g(x)) = \left( \sqrt{x} \right) ^2$
$f(g(x)) = x\ {\color{#009600}{\huge{\checkmark}}}$
We found that $f(g(x))$ is the identity function.

### Calculating $g(f(x))$

We will now investigate the expression $g(f(x)).$ To find this we will start by substituting $(x+6)^2$ for $f(x).$ $g(f(x)) \quad \Rightarrow \quad g \left( {\color{#0000FF}{(x+6)^2}} \right)$ Now we apply the definition of $g(x).$ $\begin{gathered} g(x)=\sqrt{x}-6 \\ \Downarrow \\ g\left( {\color{#0000FF}{(x+6)^2}} \right) = \sqrt{{\color{#0000FF}{(x+6)^2}}}-6 \end{gathered}$ Finally, let's simplify and see if the function is the identity function.
$g(f(x)) = \sqrt{(x+6)^2}-6$
$g(f(x)) = |x+6|-6$
As we said at the beginning of the exercise, we are only considering nonnegative values for the $x\text{-}$variable. Thus, $|x+6|$ $=$ $x+6.$
$g(f(x)) = |x+6|-6$
$g(f(x)) = x+6-6$
$g(f(x)) = x\ {\color{#009600}{\huge{\checkmark}}}$
We found that $g(f(x))$ is also the identity function. Thus, $f(x)$ and $g(x)$ are inverse functions.