Q. 56 - Precalculus Enhanced with Graphing Utilities

Question

The function $\mathit{f}$ is one-to-one. Find its inverse and check your answer. Graph $\mathit{f}$,${\mathit{f}}^{\mathbf{-}\mathbf{1}}$  and y = x on the same coordinate axes.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{\ge }\mathbf{0}$

Step-by-Step Solution

Answer

The inverse of the function is ${\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}}$

Step 1. Given information :

The given function is one-to-one function.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{\ge }\mathbf{0}$

Step 2. Finding the inverse of the function :

For finding the inverse of the function we need to interchange the variables.

let, $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}$

Interchanging the variables x, y,

$$\begin{gathered} \mathit{x}\mathbf{=}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{9} \\ {\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathit{x}\mathbf{-}\mathbf{9} \\ \mathit{y}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}} \end{gathered}$$

So, the inverse  ${\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}}$

Step 3. Check the solution :

To check whether the function is inverses are not,

let, $\mathit{f}\mathbf{\left(}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\right(}\mathit{x}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathit{x}\mathbf{ }\mathbf{ }$

$$\begin{gathered} \mathit{f}\mathbf{\left(}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}}\mathbf{\right)}\mathbf{=}\mathit{x}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{w}\mathit{e}\mathbf{ }\mathit{g}\mathit{o}\mathit{t}\mathbf{,}\mathbf{ }{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}} \\ {\mathbf{\left(}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{=}\mathit{x}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathbf{,}\mathbf{ }\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9} \\ \mathit{x}\mathbf{-}\mathbf{9}\mathbf{+}\mathbf{9}\mathbf{=}\mathit{x} \\ \mathit{x}\mathbf{=}\mathit{x} \end{gathered}$$

Similarly, ${\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{f}\left(x\right)\mathbf{\right)}\mathbf{=}\mathit{x}$

$$\begin{gathered} {\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathbf{,}\mathbf{ }\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9} \\ \sqrt{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{)}\mathbf{-}\mathbf{9}}\mathbf{=}\mathit{x}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{,}\mathbf{ }{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{9}} \\ \sqrt{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathit{x} \\ \mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{=}\mathit{x}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{,}\mathbf{ }{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)} \\ \mathit{h}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathbf{ }\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}\mathbf{.} \end{gathered}$$

Step 4. Graph representation of f , f - 1 and y = x

Graph ;

Domain of $\mathit{f}$ = range of ${\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{[}\mathbf{9}\mathbf{,}\mathbf{\infty }\mathbf{)}$ 

Domain of  ${\mathit{f}}^{\mathbf{-}\mathbf{1}}$= range of $\mathit{f}\mathbf{=}\mathbf{[}\mathbf{9}\mathbf{,}\mathbf{\infty }\mathbf{)}$