Section 2.9 Inverse Functions

Objective 1: Inverses

Before defining inverse functions, it is essential to understand a special type of function known as a one-to-one function.

One-to-One Function

• Definition: A function $f$ is one-to-one if different elements in its domain are assigned to different values in its range.

  • That is, if $x<em>1 \neq x</em>2$, then $f(x<em>1) \neq f(x</em>2)$.

  • Equivalently, if $f(x<em>1) = f(x</em>2)$, then $x<em>1 = x</em>2$.

  • In simpler terms: Each distinct x-value corresponds to a distinct y-value.

• Conditions for not being one-to-one: If there are two distinct elements in the domain of $f$ that are assigned the same value in the range, then the function $f$ is not one-to-one.

• **Conceptual Representation (similar to Figure 2.100):

  • A one-to-one function:** Each y-value in the range corresponds to only one x-value in the domain.

  • A function that is not one-to-one: At least one y-value in the range corresponds to multiple x-values from the domain.

  • A relation that is not a function: At least one x-value in the domain maps to more than one y-value in the range.

Inverse Function

Let $f$ represent a one-to-one function. If $y$ is in the range of $f$, there is only one value of $x$ in the domain of $f$ such that $f(x)=y$. We define the inverse of $f$, called the inverse function of $f$, denoted $f^{-1}$, by $f^{-1}(y)=x$ if and only if $y=f(x)$.

From this definition, we have the following:

• Domain of $f$ = Range of $f^{-1}$

• Range of $f$ = Domain of $f^{-1}$

Side Note: The reason only a one-to-one function can have an inverse function is that if $x<em>1 \neq x</em>2$ but $f(x<em>1)=y$ and $f(x</em>2)=y$, then $f^{-1}(y)$ would have to be both $x<em>1$ and $x</em>2$. This is not possible because $x<em>1 \neq x</em>2$, which violates the definition of a function (one input, one output).

One interpretation of the equation $f^{-1}(f(x))=x$ (and also $f(f^{-1}(x))=x$) is that $f^{-1}$ reverses the effect that $f$ has on $x$. For example, let $f(x)=x+2$. $f$ adds 2 to any input $x$. To undo what $f$ does to $x$, we should subtract 2 from $x$. That is, the inverse of $f$ should be $g(x)=x-2$. $g$ subtracts 2 from $x$.

Let's verify that $g(x)=x-2$ is indeed the inverse function of $f(x)=x+2$:

$f(g(x)) = f(x-2)$

$= (x-2)+2$

$= x$

We leave it for you to check that $g(f(x))=x$.

Example 3: Verifying Inverse Functions

Verify that the following pairs of functions are inverses of each other:

$f(x) = 2x+3$

$g(x) = \frac{x-3}{2}$

Solution:

First, evaluate $f(g(x))$:

$f(g(x)) = f\left(\frac{x-3}{2}\right)$

$= 2\left(\frac{x-3}{2}\right) + 3$

$= (x-3) + 3$

$= x$

Thus, $f(g(x)) = x$.

Next, evaluate $g(f(x))$:

$g(f(x)) = g(2x+3)$

$= \frac{(2x+3)-3}{2}$

$= \frac{2x}{2}$

$= x$

Thus, $g(f(x)) = x$. Because $f(g(x))=x$ and $g(f(x))=x$, $f$ and $g$ are inverses of each other.

Practice Problem 3

Verify that $f(x) = 3x-1$ and $g(x) = \frac{x+1}{3}$ are inverses of each other.

In Example 3, notice how the functions $f$ and $g$ reverse (undo) the effect of each other. The function $f$ takes an input $x$, multiplies it by 2, and adds 3; $g$ reverses (or undoes) this effect by subtracting 3 and dividing by 2. This process is illustrated in Figure 2.104. Notice that $g$ reverses the operations performed by $f$ and the order in which they are done.

Important Note: The notation $f^{-1}(x)$ does not mean $\frac{1}{f(x)}$. The expression $\frac{1}{f(x)}$ represents the reciprocal of $f(x)$ and is sometimes written as $(f(x))^{-1}$.

Example 2: Relating the Values of a Function and Its Inverse

Assume that $f$ is a one-to-one function.

• If $f(3)=5$, find $f^{-1}(5)$.

• If $f^{-1}(-1)=7$, find $f(7)$.

Solution:

By definition, $f^{-1}(y)=x$ if and only if $y=f(x)$.

• Let $x=3$ and $y=5$. Reading the definition from right to left, $5=f(3)$ if and only if $f^{-1}(5)=3$. So, $f^{-1}(5)=3$.

• Let $y=-1$ and $x=7$. We are given $f^{-1}(-1)=7$. From the definition, this means $f(7)=-1$. So, $f(7)=-1$.

Practice Problem 2

Assume that $f$ is a one-to-one function.

• If $f(-3)=12$, find $f^{-1}(12)$.

• If $f^{-1}(4)=9$, find $f(9)$.

Objective 2: Finding the Inverse Function

Let $y=f(x)$ be a one-to-one function; then $f$ has an inverse function. Suppose $(a, b)$ is a point on the graph of $f$. Then $b=f(a)$. This means that $a=f^{-1}(b)$; so $(b,a)$ is a point on the graph of $f^{-1}$. The points $(a, b)$ and $(b, a)$ are symmetric about the line $y=x$, as shown in Figure 2.105. That is, if the graph paper is folded along the line $y=x$, the points $(a, b)$ and $(b,a)$ will coincide. Therefore, we have the following property.

Symmetry Property of the Graphs of $f$ and $f^{-1}$:

The graph of a one-to-one function $f$ and the graph of $f^{-1}$ are symmetric about the line $y=x$.

Example 4: Finding the Graph of $f^{-1}$ from the Graph of $f$

The graph of a function $f$ is shown in Figure 2.106. Sketch the graph of $f^{-1}$. Figure 2.106 is a 'Graph of f'. The symmetry between the graphs of $f$ and $f^{-1}$ about the line $y=x$ tells us that we can find an equation for the inverse function $y=f^{-1}(x)$ from the equation of a one-to-one function $y=f(x)$ by interchanging the roles of $x$ and $y$ in the equation $y=f(x)$. This results in the equation $x=f(y)$. Then we solve the equation $x=f(y)$ for $y$ in terms of $x$ to get $y=f^{-1}(x)$.

Procedure in Action: Example 5 Finding an Equation for $f^{-1}$

OBJECTIVE

Find the inverse of a one-to-one function $f$.

EXAMPLE

Find the inverse of $f(x) = 3x-4$.

• Step 1: Replace $f(x)$ with $y$ in the equation defining $f(x)$.:
  

  $y=3x-4$ (Replace $f(x)$ with $y$)

• Step 2: Interchange $x$ and $y$.
  

  $x=3y-4$ (Interchange $x$ and $y$)

• Step 3: Solve the equation in Step 2 for $y$.
  

  $x+4=3y$ (Add 4 to both sides.)
  

  $\frac{x+4}{3}=y$ (Divide both sides by 3.)

• Step 4: Write $f^{-1}(x)$ for $y$.
  

  $f^{-1}(x) = \frac{x+4}{3}$ (We usually end with $f^{-1}(x)$ on the left.)

Practice Problem 5

Find the inverse of $f(x) = -2x+3$.

Example 6: Finding the Inverse Function

Find the inverse of the one-to-one function $f(x) = \frac{x+1}{x-2}$, $x \neq 2$.

Solution:

• Step 1:
  

  $y=\frac{x+1}{x-2}$ (Replace $f(x)$ with $y$)

• Step 2:
  

  $x=\frac{y+1}{y-2}$ (Interchange $x$ and $y$)

• Step 3: Solve $x=\frac{y+1}{y-2}$ for $y$. This is the most challenging step.
  

  $x(y-2)=y+1$ (Multiply both sides by $y-2$)
  

  $xy-2x=y+1$ (Distributive property)
  

  $xy-2x+2x-y=y+1+2x-y$ (Add $2x-y$ to both sides.)
  

  $xy-y=2x+1$ (Simplify.)
  

  $y(x-1)=2x+1$ (Factor out $y$.)
  

  $y=\frac{2x+1}{x-1}$ (Divide both sides by $x-1$, assuming that $x \neq 1$.)

Objective 3: Finding the Range of a One-to-One Function

It is not always easy to determine the range of a function that is defined by an equation. However, for a one-to-one function, we can find the range of $f$ by finding the domain of $f^{-1}$. Recall that the domain of a function given by a formula is the largest set of real numbers for which the outputs are real numbers.

Example 7: Finding the Domain and Range of a One-to-One Function

Find the domain and the range of the function $f(x) = \frac{x+1}{x-2}$ of Example 6.

Solution:

The domain of $f(x) = \frac{x+1}{x-2}$ is the set of all real numbers $x$ such that $x \neq 2$.

In interval notation, the domain of $f$ is $(-\infty, 2) \cup (2, \infty)$.

From Example 6, $f^{-1}(x) = \frac{2x+1}{x-1}$, $x \neq 1$; therefore,

range of $f$ = domain of $f^{-1}$ = \left{x \mid x \neq 1\right}.

In interval notation, the range of $f$ is $(-\infty, 1) \cup (1, \infty)$.

Practice Problem 7

Find the domain and the range of the function $f(x)=\frac{x}{x+3}$.

If a function $f$ is not one-to-one, then it does not have an inverse function. Sometimes by changing its domain, we can produce an interesting function that does have an inverse. (This technique is frequently used in trigonometry.) We saw in Example 1(b) that $g(x) = x^2-1$ is not a one-to-one function; so $g$ does not have an inverse function. However, the horizontal-line test shows that the function

$G(x) = x^2-1$, $x \geq 0$

with domain $[0, \infty)$ is one-to-one. See Figure 2.109. Therefore, $G$ has an inverse function $G^{-1}$. Figure 2.109 is titled 'The function G has an inverse'. The horizontal axis is