# 1.4 Inverse functions  (Page 2/11)

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## Rule: horizontal line test

A function $f$ is one-to-one if and only if every horizontal line intersects the graph of $f$ no more than once. (a) The function f ( x ) = x 2 is not one-to-one because it fails the horizontal line test. (b) The function f ( x ) = x 3 is one-to-one because it passes the horizontal line test.

## Determining whether a function is one-to-one

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

1. 2. 1. Since the horizontal line $y=n$ for any integer $n\ge 0$ intersects the graph more than once, this function is not one-to-one. 2. Since every horizontal line intersects the graph once (at most), this function is one-to-one. Is the function $f$ graphed in the following image one-to-one? No.

## Finding a function’s inverse

We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of $f$ to elements in the range of $f.$ The inverse function maps each element from the range of $f$ back to its corresponding element from the domain of $f.$ Therefore, to find the inverse function of a one-to-one function $f,$ given any $y$ in the range of $f,$ we need to determine which $x$ in the domain of $f$ satisfies $f\left(x\right)=y.$ Since $f$ is one-to-one, there is exactly one such value $x.$ We can find that value $x$ by solving the equation $f\left(x\right)=y$ for $x.$ Doing so, we are able to write $x$ as a function of $y$ where the domain of this function is the range of $f$ and the range of this new function is the domain of $f.$ Consequently, this function is the inverse of $f,$ and we write $x={f}^{-1}\left(y\right).$ Since we typically use the variable $x$ to denote the independent variable and $y$ to denote the dependent variable, we often interchange the roles of $x$ and $y,$ and write $y={f}^{-1}\left(x\right).$ Representing the inverse function in this way is also helpful later when we graph a function $f$ and its inverse ${f}^{-1}$ on the same axes.

## Problem-solving strategy: finding an inverse function

1. Solve the equation $y=f\left(x\right)$ for $x.$
2. Interchange the variables $x$ and $y$ and write $y={f}^{-1}\left(x\right).$

## Finding an inverse function

Find the inverse for the function $f\left(x\right)=3x-4.$ State the domain and range of the inverse function. Verify that ${f}^{-1}\left(f\left(x\right)\right)=x.$

Follow the steps outlined in the strategy.

Step 1. If $y=3x-4,$ then $3x=y+4$ and $x=\frac{1}{3}y+\frac{4}{3}.$

Step 2. Rewrite as $y=\frac{1}{3}x+\frac{4}{3}$ and let $y={f}^{-1}\left(x\right).$

Therefore, ${f}^{-1}\left(x\right)=\frac{1}{3}x+\frac{4}{3}.$

Since the domain of $f$ is $\left(\text{−}\infty ,\infty \right),$ the range of ${f}^{-1}$ is $\left(\text{−}\infty ,\infty \right).$ Since the range of $f$ is $\left(\text{−}\infty ,\infty \right),$ the domain of ${f}^{-1}$ is $\left(\text{−}\infty ,\infty \right).$

You can verify that ${f}^{-1}\left(f\left(x\right)\right)=x$ by writing

${f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(3x-4\right)=\frac{1}{3}\left(3x-4\right)+\frac{4}{3}=x-\frac{4}{3}+\frac{4}{3}=x.$

Note that for ${f}^{-1}\left(x\right)$ to be the inverse of $f\left(x\right),$ both ${f}^{-1}\left(f\left(x\right)\right)=x$ and $f\left({f}^{-1}\left(x\right)\right)=x$ for all x in the domain of the inside function.

Find the inverse of the function $f\left(x\right)=3x\text{/}\left(x-2\right).$ State the domain and range of the inverse function.

${f}^{-1}\left(x\right)=\frac{2x}{x-3}.$ The domain of ${f}^{-1}$ is $\left\{x|x\ne 3\right\}.$ The range of ${f}^{-1}$ is $\left\{y|y\ne 2\right\}.$

## Graphing inverse functions

Let’s consider the relationship between the graph of a function $f$ and the graph of its inverse. Consider the graph of $f$ shown in [link] and a point $\left(a,b\right)$ on the graph. Since $b=f\left(a\right),$ then ${f}^{-1}\left(b\right)=a.$ Therefore, when we graph ${f}^{-1},$ the point $\left(b,a\right)$ is on the graph. As a result, the graph of ${f}^{-1}$ is a reflection of the graph of $f$ about the line $y=x.$

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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