# 1.7 Inverse functions  (Page 2/10)

 Page 2 / 10
$\left({f}^{-1}\circ f\right)\left(x\right)={f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(y\right)=x$

This holds for all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal , some functions do not have inverses.

Given a function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ we can verify whether some other function $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ is the inverse of $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ by checking whether either $\text{\hspace{0.17em}}g\left(f\left(x\right)\right)=x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}f\left(g\left(x\right)\right)=x\text{\hspace{0.17em}}$ is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)

For example, $\text{\hspace{0.17em}}y=4x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\frac{1}{4}x\text{\hspace{0.17em}}$ are inverse functions.

$\left({f}^{-1}\circ f\right)\left(x\right)={f}^{-1}\left(4x\right)=\frac{1}{4}\left(4x\right)=x$

and

$\left({f}^{}\circ {f}^{-1}\right)\left(x\right)=f\left(\frac{1}{4}x\right)=4\left(\frac{1}{4}x\right)=x$

A few coordinate pairs from the graph of the function $\text{\hspace{0.17em}}y=4x\text{\hspace{0.17em}}$ are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function $\text{\hspace{0.17em}}y=\frac{1}{4}x\text{\hspace{0.17em}}$ are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.

## Inverse function

For any one-to-one function     $\text{\hspace{0.17em}}f\left(x\right)=y,\text{\hspace{0.17em}}$ a function $\text{\hspace{0.17em}}{f}^{-1}\left(x\right)\text{\hspace{0.17em}}$ is an inverse function    of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}{f}^{-1}\left(y\right)=x.\text{\hspace{0.17em}}$ This can also be written as $\text{\hspace{0.17em}}{f}^{-1}\left(f\left(x\right)\right)=x\text{\hspace{0.17em}}$ for all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ It also follows that $\text{\hspace{0.17em}}f\left({f}^{-1}\left(x\right)\right)=x\text{\hspace{0.17em}}$ for all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}{f}^{-1}\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}{f}^{-1}\text{\hspace{0.17em}}$ is the inverse of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$

The notation ${f}^{-1}$ is read $\text{“}f$ inverse.” Like any other function, we can use any variable name as the input for ${f}^{-1},$ so we will often write $\text{\hspace{0.17em}}{f}^{-1}\left(x\right),$ which we read as $“f$ inverse of $x.”$ Keep in mind that

${f}^{-1}\left(x\right)\ne \frac{1}{f\left(x\right)}$

and not all functions have inverses.

## Identifying an inverse function for a given input-output pair

If for a particular one-to-one function $\text{\hspace{0.17em}}f\left(2\right)=4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(5\right)=12,\text{\hspace{0.17em}}$ what are the corresponding input and output values for the inverse function?

The inverse function reverses the input and output quantities, so if

Alternatively, if we want to name the inverse function $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}g\left(4\right)=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(12\right)=5.$

Given that $\text{\hspace{0.17em}}{h}^{-1}\left(6\right)=2,\text{\hspace{0.17em}}$ what are the corresponding input and output values of the original function $\text{\hspace{0.17em}}h?\text{\hspace{0.17em}}$

$h\left(2\right)=6$

Given two functions $\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right),\text{\hspace{0.17em}}$ test whether the functions are inverses of each other.

1. Determine whether $\text{\hspace{0.17em}}f\left(g\left(x\right)\right)=x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}g\left(f\left(x\right)\right)=x.$
2. If both statements are true, then $g={f}^{-1}$ and $f={g}^{-1}.\text{\hspace{0.17em}}$ If either statement is false, then both are false, and $\text{\hspace{0.17em}}g\ne {f}^{-1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\ne {g}^{-1}.$

## Testing inverse relationships algebraically

If $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x+2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)=\frac{1}{x}-2,\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}g={f}^{-1}?$

$\begin{array}{l}\begin{array}{l}\hfill \\ g\left(f\left(x\right)\right)=\frac{1}{\left(\frac{1}{x+2}\right)}-2\hfill \end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=x+2-2\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=x\hfill \end{array}$

so

This is enough to answer yes to the question, but we can also verify the other formula.

$\begin{array}{l}\begin{array}{l}\\ f\left(g\left(x\right)\right)=\frac{1}{\frac{1}{x}-2+2}\end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{\frac{1}{x}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=x\hfill \end{array}$

If $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}-4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)=\sqrt[\text{\hspace{0.17em}}3]{x+4},\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}g={f}^{-1}?$

Yes

## Determining inverse relationships for power functions

If $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}\text{\hspace{0.17em}}$ (the cube function) and $\text{\hspace{0.17em}}g\left(x\right)=\frac{1}{3}x,\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}g={f}^{-1}?$

$f\left(g\left(x\right)\right)=\frac{{x}^{3}}{27}\ne x$

No, the functions are not inverses.

If $\text{\hspace{0.17em}}f\left(x\right)={\left(x-1\right)}^{3}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}g\left(x\right)=\sqrt[3]{x}+1,\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}g={f}^{-1}?$

Yes

## Finding domain and range of inverse functions

The outputs of the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ are the inputs to $\text{\hspace{0.17em}}{f}^{-1},\text{\hspace{0.17em}}$ so the range of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is also the domain of $\text{\hspace{0.17em}}{f}^{-1}.\text{\hspace{0.17em}}$ Likewise, because the inputs to $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ are the outputs of $\text{\hspace{0.17em}}{f}^{-1},\text{\hspace{0.17em}}$ the domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is the range of $\text{\hspace{0.17em}}{f}^{-1}.\text{\hspace{0.17em}}$ We can visualize the situation as in [link] .

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