2. Unit 3: Inverses of Functions
  3. Lesson 2: Determining Inverses of Linear Functions Algebraically

Alternative Format — Lesson 2: Determining Inverses of Linear Functions Algebraically

Let's Start Thinking

Order of Operations and Inverse Functions

Recall

The inverse of a function is the relation which undoes the work of the function.

Here's a helpful way of thinking of a function and its inverse.

If the function is tie shoes, then the inverse is untie shoes.

A tied pair of shoes undergoes function f inverse: untie shoes to become a pair of untied shoes.

Source: Shoes - VectorUp/iStock/GettyImages

We can think of this function and its inverse within a larger context too.

Let's add socks to the picture.

When I get ready to leave for the day among other things I normally follow this procedure.

Leaving

A cycle. First put on socks. Put on shoes. Apply function f, tie shoes. Tied shoes. Leave the house.

I put on socks.

I put on shoes.

I apply the tie shoes function.

Then I leave my house.

Arriving

A cycle. Arrive home. Tied shoes. Apply inverse function f, Untie shoes. Untied Shoes. Take off socks.

Sources: Man Leaving - Serge-Maksimov/iStock/GettyImages Shoes - VectorUp/iStock/GettyImages;
Socks - Nikiteev_Konstantin/iStock/GettyImages;

At the end of the day, I arrive home and enter my house.

I apply the untie shoes function.

I take off my shoes and I take off my socks.

Notice that at each stage the opposite thing is happening.

For example, exiting my house or entering my house or putting socks on versus taking socks off.

In addition, the stages are performed in opposite order. So the operations themselves and the order of operations are reversed when an inverse function is applied.

We'll see more of this throughout this lesson.


Lesson Goals

  • Determine the inverse of a linear function algebraically.
  • Determine the domain and range of the inverse of a function.

Try This

Typically, a person's height is approximately related to the length of their femur (or thighbone) by 

\[g(x)=\dfrac{5}{13}x-25\]

where \(x\) is the person's height, in centimetres, and \(g(x)\) is the femur length, in centimetres.

Lea is an anthropologist. She would like to use this function in the opposite way: given a person's femur length, she would like to predict the height of the person. Determine the function that Lea should use.


Determining Inverses of Linear Functions


Let's look at determining the inverse of a linear function algebraically.

Example 1

The inverse of \(f(x)=5x+2\) is a function. Determine \(f^{-1}(x)\).

Solution

Recall

To determine the inverse of \(f(x)\), interchange the independent and dependent variables.

This is true regardless of the way the function is represented.

In a table, we interchange the \(x\) and \(y\) columns.

In a graph, we interchange the \(x\) and \(y\)-coordinates.

Here we will interchange \(x\) and \(y\) as well.

Except right now the function doesn't have a \(y\).

So the first order of business is to write this function with a \(y\).

We replace \(f(x)\) with \(y\),

\(\begin{aligned} f(x) &=5 x+2 \\ y &=5 x+2 \end{aligned}\)

Here, \(y=f(x)\).

To determine \(f^{-1}(x)\), interchange \(x\) and \(y\).

\( x =5 y+2 \)

But this new equation no longer describes the original function.

From this point forward, \(y=f^{-1}(x)\).

The change in the meaning of \(y\) happened when we interchange \(x\) and \(y\).

To determine an expression for the inverse, we rearranged the equation to isolate for \(y\).

\( x = 5 y+2 \)

First, we subtract \(2\) from both sides.

\( 5 y =x-2 \)

Then divide both sides by \(5\).

\( y =\dfrac{x-2}{5} \)

In this part of the solution, \(y = f^{-1}(x)\).

Therefore, \(f^{-1}(x) = \dfrac{x-2}{5}\).

Compare \(f(x)\) and \(f^{-1}(x)\)

So we've completed this question already, but let's take a moment to compare \(f(x)\) and \(f^{-1}(x)\).

First, let's compare their operations and their order of operations.

Operations to Evaluate \(f(x)\) 

\( f(x)= 5x+2 \)

  1. Multiply by \(5\). ​​​​
  2. Add \(2\).

Operations to Evaluate \(f^{-1}(x)\)

\(f^{-1}(x) = \dfrac{x-2}{5}\)

  1. Subtract \(2\).
  2. Divide by \(5\).

The operations and the order in which they are applied are opposites.

Where we used to have a multiply by \(5\), we now have a divide by \(5\). Where we used to have add \(2\), we now have subtract \(2\) and whereas the multiply by \(5\) used to come first, now it's the subtract \(2\) that comes first.

Compare the Graphs of \(y=f(x)\) and \(y=f^{-1}(x)\)

First, we'll graph \(y=f(x)\). 

It is a linear function with slope \(5\) and \(y\)-intercept \(2\).

The linear function passes through the point (0, 2) with slope 5.

We can write \(f^{-1}(x)\) in \(y = mx + b\) form as well.

\(\begin{aligned} f^{-1}(x) &=\frac{x-2}{5} \\ &=\frac{1}{5} x-\frac{2}{5} \end{aligned}\)

\(f^{-1}(x)\) is a linear function with slope \(\dfrac{1}{5}\) and \(y\)-intercept \(-\dfrac{2}{5}\).

Here is its graph.

Graph of y = 5x plus 2 and it's inverse y= one-fifth x minus 2 over 5.

Now if we add the line \(y = x\) to this graph, we see something that you might recall from a previous lesson, the graph of \(y = f(x)\) and the graph of its inverse are reflections in the line \(y = x\).

The function and its inverse are reflections in the line y = x.

We also see that the inverse is a function as indicated in the question.

So the use of function notation for the inverse throughout this question was indeed justified.

Solution Revisited

Let's make one final comment about this example, this time about the solution itself.

Here is the solution again as presented earlier.

Here, \(y=f(x)\).

\(\begin{aligned} f(x) &=5 x+2 \\ y &=5 x+2 \end{aligned}\)

To determine \(f^{-1}(x)\), interchange \(x\) and \(y\).

\( x =5 y+2 \)

From this point forward, \(y=f^{-1}(x)\).

\(\begin{align*} 5 y &=x-2 \\ y &=\frac{x-2}{5} \end{align*}\)

Therefore, \(f^{-1}(x) = \dfrac{x-2}{5}\).

Notice the phrase to determine \(f^{-1}(x)\) interchange \(x\) and \(y\) included here in the solution.

This phrase is necessary for the solution.

Interchanging \(x\) and \(y\) changes the function.

The algebra shown on this page is about two different functions.

The first lines are about \(f(x)\) and the remainder is about \(f^{-1}(x)\).


Steps to Determine the Inverse of \(f(x)\) Algebraically

When \(f(x)\) is described algebraically, determine the inverse by following these steps.

  1. Rewrite \(f(x)\) as \(y\).
  2. Write "To determine the inverse, interchange \(x\) and \(y\)."
  3. Interchange \(x \) and \(y\).
  4. Rearrange the equation to isolate for \(y\).
  5. If the inverse is a function, rewrite \(y\) as \(f^{-1}(x)\).

In some instances, it is not possible to complete step 4. However, in this unit, for the most part, only quadratic and non-horizontal linear functions are considered, so rearrangement will always be possible. 


Check Your Understanding 1

Question

The inverse of \(f(x)=-6x-5\)  is a function. Determine \(f^{-1}(x)\).

Answer

\(f^{-1}(x) = -\dfrac{x + 5}{6}\)

Feedback

To determine \(f^{-1}(x)\), rewrite \(f(x)\) as \(y\), interchange \(x\) and \(y\) and then isolate for \(y\).

  1. Replace \(f(x)\) with \(y\).

    \(y= -6x-5\)

  2. Interchange \(x\) and \(y\)

    \(x= -6y-5\)

  3. Solve for \(y\)

    \(\begin{align*} -6y &= x + 5 \\ y &= \dfrac{x + 5}{-6} \end{align*}\)

     

  4. Replace \(y\) with \(f^{-1}(x)\).

    \(f^{-1}(x) = -\dfrac{x + 5}{6}\)


Try This Revisited

Typically, a person's height is approximately related to the length of their femur (or thighbone) by 

\[g(x)=\dfrac{5}{13}x-25\]

where \(x\) is the person's height, in centimetres, and \(g(x)\) is the femur length, in centimetres.

Lea is an anthropologist. She would like to use this function in the opposite way: given a person's femur length, she would like to predict the height of the person. Determine the function that Lea should use.

Solution

The function \(g(x)\) and its inverse operate as shown:

In a cycle, the femur length is input into function g, which outputs the height. The height is input into the inverse of g and outputs the femur length.

Thus, the function Lea needs to use is \(g^{-1}(x)\).  We find the inverse as follows.

\(\begin{align*} g(x)&=\dfrac{5}{13}x-25\\ y&=\dfrac{5}{13}x-25\\ \end{align*}\)

To determine \(g^{-1}(x)\), interchange \(x\) and \(y\):

\(\begin{align*} x&=\dfrac{5}{13}y-25\\ \dfrac{5}{13}y&=x+25\\ 5y&=13x+325\\ y &= \dfrac{13}{5}x+65\\ g^{-1}(x) &= \dfrac{13}{5}x+65\\ \end{align*}\)

Therefore, Lea should use the function \( g^{-1}(x)= \dfrac{13}{5}x+65\\ \).

Compare \(g(x)\) and \(g^{-1}(x)\)

It is not immediately obvious that \(g(x)=\dfrac{5}{13}x-25\) and \( g^{-1}(x)= \dfrac{13}{5}x+65\\ \) are inverses of each other: it is not clear that opposite operations are performed in reverse order.

Recall, though, that the same function can have different representations. In this case, we can rewrite each function with common denominators:

\(\begin{align*} g(x) &=\dfrac{5}{13}x-25\\ &= \dfrac{5x}{13}-\dfrac{325}{13}\\ &= \dfrac{5x-325}{13} \end{align*}\)

\(\begin{align*} g^{-1}(x) &=\dfrac{13}{5}x+65\\ &= \dfrac{13x}{5}+\dfrac{325}{5}\\ &= \dfrac{13x+325}{5} \end{align*}\)

And now if we compare operations and the order in which they are performed:

Operations to Evaluate \(g(x)\)

  1. Multiply by \(5\)
  2. Subtract \(325\)
  3. Divide by \(13\)

Operations to Evaluate \(g^{-1}(x)\)

  1. Multiply by \(13\)
  2. Add \(325\)
  3. Divide by \(5\)

As anticipated, to evaluate \(g(x)\) and \(g^{-1}(x)\), opposite operations are performed in reverse order.


Check Your Understanding 2

Question

The inverse of \(f(x)= \dfrac{8}{5}x + 7\) is a function. Determine \(f^{-1}(x)\).

Answer

\( f^{-1}(x) = \dfrac{5}{8}x - \dfrac{35}{8}\)

Feedback

We first replace \(f(x)\) with \(y\):

\(\begin{align*} f(x)&=\dfrac{8}{5}x + 7 \\ y&= \dfrac{8}{5}x + 7 \\ \end{align*}\)

To determine \(f^{-1}(x),\) interchange \(x\) and \(y\):

\(\begin{align*} \dfrac{8}{5}y + 7 &= x \\ \dfrac{8}{5}y &= x - 7 \\ 8y &= 5x - 35\\ y &= \dfrac{5}{8}x - \dfrac{35}{8} \\ f^{-1}(x) &= \dfrac{5}{8}x - \dfrac{35}{8} \end{align*}\)

Therefore, \( f^{-1}(x) = \dfrac{5}{8}x - \dfrac{35}{8}\).


Example 2

Determine the inverse of each relation.

  1. \(f(x)=4\)
  2. \(x=-2\)

Solution — Part A

To determine the inverse of \(f(x)\):

\(\begin{align*} f(x) &= 4\\ y &= 4\\ \end{align*}\)

In determining the inverse we interchange \(x\) and \(y\). In this case, this involves replacing \(y\) with \(x\) only.

 \(\begin{align*} x &= 4\\ \end{align*}\)

What results is an equation of a vertical line. Here is a graph of \(f(x)=4\) and its inverse \(x=4\). Notice that \(x\)- and \(y\)-coordinates have been interchanged and that the graphs are reflections in the line \(y=x\).

In this case, the inverse of \(f(x)\) is not a function. When \(f(x)\) is linear, this can only happen when the graph of \(f(x)\) is a horizontal line.

Solution — Part B

To determine the inverse of \(x=-2\), we interchange \(x\) and \(y\). In this case, this involves replacing \(x\) with \(y\) only. Thus, the inverse is \(y=-2\).

Once again, the graphs of the original relation, \(x=-2\), and its inverse, \(y=-2\), are reflections of each other in the line \(y=x\).

The inverse of the vertical line \(x=a\) is the horizontal line \(y=a\).

The inverse of the horizontal line \(y=a\) is the vertical line \(x=a\). In this case, the inverse is not a function.

 Example 3

Mylene works as a wedding planner, assisting couples to make arrangements for their wedding ceremony and reception. One venue that Mylene works with charges \($300\) to rent the hall and an additional \($79\) per person who attends the wedding.

  1. Determine a function, \(f(x)\), to calculate the total cost of the hall rental for \(x\) guests.
  2. Determine the inverse of \(f(x)\).
  3. When would Mylene use \(f(x)\) and when would she use \(f^{-1}(x)\)?

Solution — Part A

The function \(f(x)\) is linear and so is of the form \(f(x)=mx+b\) with

  • \(m=79\) since the cost per guest is \($79\), and
  • \(b=300\) since the fixed cost is \($300\).

Therefore, \(f(x)=79x+300\). This function calculates the cost to rent the hall for a wedding reception with \(x\) guests.

Solution — Part B 

Since \(f(x)\) is linear (and not horizontal), the inverse is a function. Therefore, we can represent the inverse of \(f(x)\) as \(f^{-1}(x)\).

\(\begin{align*} f(x)&=79x+300\\ y &= 79x+300 \end{align*}\)

To determine \(f^{-1}(x)\), interchange \(x\) and \(y\):

\(\begin{align*} x &= 79y+300\\ 79y &= x-300\\ y &= \dfrac{x-300}{79}\\ f^{-1}(x)&=\dfrac{x-300}{79} \end{align*}\)

Therefore, \(f^{-1}(x)=\dfrac{x-300}{79}\). This function calculates the number of guests that can be invited to a wedding reception costing \($x\).

Using the same letter to indicated the independent variable in \(f(x)\) and \(f^{-1}(x)\) can be confusing since the independent variables represent different quantities in the two functions. If we let \(n\) be the number of guests and we let \(C\) be the cost of the wedding reception, in dollars, then we can also write:

  • \(C=f(n)=79n+300\), and
  • \(n=f^{-1}(C)=\dfrac{C-300}{79}\).

Solution — Part C

The functions \(f(x)\) and \(f^{-1}(x)\) operate as shown:

In a cycle, the number of guests is input into f of x which outputs the cost. Cost is input into the inverse of f of x and outputs the number of guests.

Some couples may have a fixed guest list when they meet with Mylene. In this situation, Mylene knows the number of guests and needs to calculate the cost; she would use \(f(x)\).

Other couples may have a fixed budget when they meet with Mylene. In this situation, Mylene knows the cost and needs to calculate how many guests can be invited; she would use \(f^{-1}(x)\).


Domain and Range of Inverses of Functions


Explore This 1

Description — Version 1

What is the relationship between the domain and range of a function and its inverse?

\(f(x) = 2x +1, -6 \le x \le 5\)

The graph of f of x and its inverse.

  \(f(x)\) \(f^{-1}(x)\)
Domain \(-6 \le x \le 5\) \(-11 \le x \le 11\)
Range \(-11 \le x \le 11\) \(-6 \le x \le 5\)

\(f(x) = 2x +1, -5 \le x \le 2\)

The graph of f of x and its inverse.

  \(f(x)\) \(f^{-1}(x)\)
Domain \(-5 \le x \le 2\) \(-9 \le x \le 5\)
Range \(-9 \le x \le 5\) \(-5 \le x \le 2\)

Description — Version 2

What is the relationship between the domain and range of a function and its inverse?

\(f(x) = \sqrt{x + 3.9} - 7.7, -3.9 \le x \le 10\)

The graph of f of x and its inverse.

  \(f(x)\) \(f^{-1}(x)\)
Domain \(-3.9 \le x \le 10\) \(-7.7 \le x \le -4\)
Range \(-7.7 \le x \le -4\) \(-3.9 \le x \le 10\)

\(f(x) = \sqrt{x + 3.9} - 7.7, -3.4 \le x \le 5\)

The graph of f of x and its inverse.

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

  \(f(x)\) \(f^{-1}(x)\)
Domain \(-3.4 \le x \le 5\) \(-7 \le x \le -4.7\)
Range \(-7 \le x \le -4.7\) \(-3.4 \le x \le 5\)

Interactive Version

Domain and Range of Functions and their Inverses


Explore This 1 Summary

Perhaps you noticed the following things in Explore This 1.

  • The domain of \(f(x)\) is the same as the range of the inverse of \(f(x)\).
  • The range of \(f(x)\) is the same as the domain of the inverse of \(f(x)\).

Example 4

For the given functions, determine the following:

  1. The inverse of the function.
  2. The domain and range of the function.
  3. The domain and range of the inverse of the function.

Function 1

x values map to unique values from f of x. Specifically, negative 2 maps to negative 8, negative 1 to negative 1, 0 to 0, 1 to 1, and 2 to 8.

Function 2

\(x\) \(g(x)\)
\(-2\) \(8\)
\(-1\) \(2\)
\(0\) \(0\)
\(1\) \(2\)
\(2\) \(8\)

Solution — Function 1

x values map to unique values from f of x. Specifically, negative 2 maps to negative 8, negative 1 to negative 1, 0 to 0, 1 to 1, and 2 to 8.

  1. Interchanging the independent and dependent variables, the inverse of \(f(x)\) is:

    x values map to unique values from f inverse of x. Specifically, negative 8 maps to negative 2, negative 1 to negative 1, 0 to 0, 1 to 1, and 8 to 2.

    For the inverse of \(f(x)\), each input has a unique output; therefore, the inverse is a function. The notation \(f^{-1}(x)\) can be used.
  2. The domain is the set of all possible values of the independent variable. Therefore, the domain of \(f(x)\) is \(D_f=\{-2, -1, 0, 1, 2\}\).

    The range is the set of all possible values of the dependent variable. Therefore, the range of \(f(x)\) is \(R_f=\{-8, -1, 0, 1, 8\}\).

  3. The domain of \(f^{-1}(x)\) is \(D_{f^{-1}}=\{-8,-1,0,1,8\}\).

    The range of \(f^{-1}(x)\) is \(R_{f^{-1}}=\{-2,-1,0,1,2\}\).

    Notice that:

    • the domain of \(f\) is the range of \(f^{-1}\) , and
    • the range of \(f\) is the domain of \(f^{-1}\).

Solution — Function 2

\(x\) \(g(x)\)
\(-2\) \(8\)
\(-1\) \(2\)
\(0\) \(0\)
\(1\) \(2\)
\(2\) \(8\)

  1. Interchanging the independent and dependent variables, the inverse of \(g(x) \) is:

    \(x\) Inverse of \(g(x)\)
    \(8\) \(-2\)
    \(2\) \(-1\)
    \(0\) \(0\)
    \(2\) \(1\)
    \(8\) \(2\)

    Since the inverse of \(g(x)\) is not a function, the notation \(g^{-1}(x)\) will not be used in this solution.

  2. The domain is the set of all possible values of the independent variable. Therefore, the domain of \(g(x)\) is \(D_g=\{-2, -1, 0, 1, 2\}\).

    The range is the set of all possible values of the dependent variable. Therefore, the range of \(g(x)\) is \(R_g=\{0, 2, 8\}\).

  3. The domain of the inverse of \(g(x)\) is \(D_{\text{inv of }g}=\{0,2,8\}\).

    The range of the inverse of \(g(x)\) is \(R_{\text{inv of }g}=\{-2,-1,0,1,2\}\).

    Notice again that: 

    • the domain of \(g\) is the range of the inverse of \(g\) , and
    • the range of \(g\) is the domain of the inverse of \(g\).

Domain and Range of Inverse Functions

Recall

To determine the inverse of a function, interchange the independent and dependent variables of the original function.

Consequentially, we have the following facts about the domain and range of a function and its inverse.

The domain of \(f(x)\) is the range of the inverse of \(f(x)\).

The range of \(f(x)\) is the domain of the inverse of \(f(x)\).


Check Your Understanding 3

Question — Version 1

The table shows all values of the function \(f(x)\). Determine the domain and range of the inverse of \(f(x)\).

\(x\) \(f(x)\)
\(0\) \(3\)
\(3\) \(10\)
\(6\) \(13\)
\(9\) \(21\)
\(12\) \(25\)

Answer — Version 1

\(D_{f^{-1}}=R_f=\{3,~10,~13,~21,~25\}\)

\(R_{f^{-1}}=D_f=\{0,~3,~6,~9,~12\}\)

Feedback — Version 1

Since the domain of the inverse of \(f\) is the range of \(f\):

\(D_{f^{-1}}=R_f=\{3,~10,~13,~21,~25\}\)

Since the range of the inverse of \(f\) is the domain of \(f\):

\(R_{f^{-1}}=D_f=\{0,~3,~6,~9,~12\}\)

Question — Version 2

The table shows all values of the function \(f(x)\). Determine the domain and range of the inverse of \(f(x)\).

\(x\) \(f(x)\)
\(0\) \(4\)
\(4\) \(13\)
\(8\) \(14\)
\(12\) \(15\)

Answer — Version 2

\(D_{f^{-1}}=R_f=\{4,~13,~14,~15\}\)

\(R_{f^{-1}}=D_f=\{0,~4,~8,~12\}\)

Feedback — Version 2

Since the domain of the inverse of \(f\) is the range of \(f\):

\(D_{f^{-1}}=R_f=\{4,~13,~14,~15\}\)

Since the range of the inverse of \(f\) is the domain of \(f\):

\(R_{f^{-1}}=D_f=\{0,~4,~8,~12\}\)


Let's continue exploring the domain in range of a function and its inverse.

Example 5

Consider \(f(x)=3x-5\) defined on \(\{x \in \mathbb{R} \mid -1 \le x \le 4\}\). Determine the domain and range of \(f(x)\) and \(f^{-1}(x)\).

Solution

Let's begin with a graph.

\(f(x)=3x-5\) is a linear function with

  • slope \(=3\), and

  • \(y\)-intercept \(=-5\).

We first plot the \(y\)-intercept.

The y-intercept is at the point (0, negative 5).

Then we can use the slope to find a second point of the line by going up \(3\) and over \(1\).

The second point we find on the graph is (1, negative 2)

 Next, we connect the dots with a straight line, \((-1, -8)\) to \((0, -5)\) to \((1, -2)\) to \((4, 7)\).

The graph of f of x = 3x minus 5, in the domain between negative 1 and 4.

We need to be mindful of the domain in the problem. The line extends only from \(x = -1\) to \(x = 4\).

Closed dots are used since the endpoints are included.

Let's also graph the inverse function.

To graph \(y=f^{-1}(x)\):

  • Interchange \(x\)- and \(y\)-coordinates.
  • When we do that, we will see \(y=f^{-1}(x)\) is a reflection of \(y=f(x)\) in the line \(y=x\).

So here's the line \(y = x\).

We'll take the point \((-1, -8)\), which is on \(f(x)\). We will interchange \(x\) and \(y\).

So that gives us the new point, \((-8, -1)\).

Let's do the same for the other endpoint.

We'll interchange \(x\) and \(y\) for the point \((4, 7)\) to get \((7, 4)\).

Since the image of a line under a reflection will still be a line, we can connect these two points to get the graph of \(y = f^{-1}(x)\).

Connecting the points (negative 8, negative 1) and (7, 4) we find plot the inverse function of f of x.

We are now ready to state the domain and range of \(f(x)\) and \(f^{-1}(x)\).

  • \(D_f=\{x \in \mathbb{R}\mid-1 \le x \le 4\}\)

  • \(R_f=\{y \in \mathbb{R}\mid-8 \le y \le 7\}\)

  • \(D_{f^{-1}}= \{x \in \mathbb{R}\mid-8 \le x \le 7\}\)

  • \(R_{f^{-1}}=\{y \in \mathbb{R}\mid -1 \le y \le 4\}\)

Notice that we see here something said earlier

Recall

The domain of \(f(x)\) is the range of \(f^{-1}(x)\).

The ​​​​range of \(f(x)\) is the domain of \(f^{-1}(x)\). 


Check Your Understanding 4

Question

Determine the domain and range of \(f^{-1}(x)\) given \(f(x)= -2x+3\) defined on \(\{x\in\mathbb{R}\mid -2 \le x \le 3\}\).

Answer

\(D_{f^{-1}}=\{x\in \mathbb{R} | {-3} \le x \le 7\}\)

\(R_{f^{-1}}=\{y\in\mathbb{R}\mid {-2} \le y \le 3\}\)

Feedback

The graph of \(y=f(x)\) is a line

  • with slope \(-2\),
  • with \(y\)-intercept \(3\), and
  • defined on the domain \(\{x\in\mathbb{R}\mid -2 \le x \le 3\}\).

Further, the endpoints have \(y\)-coordinates \(f(-2)=-2(-2)+3=7\) and \(f(3)=-2(3)+3=-3\).

Thus, the graphs of \(y=f(x)\) and \(y=f^{-1}(x)\) are as shown.

Notice that the graph of \(g(x)=f^{-1}(x)\) was obtained by interchanging the \(x\)- and \(y\)-coordinates of \(y=f(x)\).

Therefore, for the function \(y=f^{-1}(x)\):

  • \(D_{f^{-1}}=\{x\in \mathbb{R} | {-3} \le x \le 7\}\), and
  • \(R_{f^{-1}}=\{y\in\mathbb{R}\mid {-2} \le y \le 3\}\).

Wrap-Up


Lesson Summary

We saw that when \(f(x)\) is defined algebraically, the steps to determine its inverse are:

  1. Rewrite \(f(x)\) as \(y\).
  2. Write "To determine the inverse, interchange \(x\) and \(y\)."
  3. Interchange \(x \) and \(y\).
  4. Rearrange the equation to isolate for \(y\).
  5. If the inverse is a function, rewrite \(y\) as \(f^{-1}(x)\).

In some instances, it is not possible to complete step 4. However, in this unit, for the most part only quadratic and non-horizontal linear functions were considered, so rearrangement was always possible. 

We saw that the domain and range of a function and its inverse are related:

  • The domain of \(f(x)\) is the range of the inverse of \(f(x)\).
  • The range of \(f(x)\) is the domain of the inverse of \(f(x)\).

Take It With You

In this lesson, we studied inverses of linear functions. In a future lesson, we will look at inverses of quadratic functions. Is the inverse of \(f(x)=x^2\) a function? How do you know?