# 1.7: Inverse Functions - Mathematics

Learning Objectives

• Verify inverse functions.
• Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.
• Find or evaluate the inverse of a function.
• Use the graph of a one-to-one function to graph its inverse function on the same axes.

A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.

If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure (PageIndex{1}) provides a visual representation of this question. In this section, we will consider the reverse nature of functions. Figure (PageIndex{1}): Can a function “machine” operate in reverse?

## Verifying That Two Functions Are Inverse Functions

Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula

[C=dfrac{5}{9}(F−32)]

and substitutes 75 for (F) to calculate

[dfrac{5}{9}(75−32)approx24^{circ}]

Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure (PageIndex{2}) for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for (F) after substituting a value for (C). For example, to convert 26 degrees Celsius, she could write

[egin{align} 26&=dfrac{5}{9}(F-32) 26⋅dfrac{9}{5}&=F−32 F&=26⋅dfrac{9}{5}+32approx79end{align}]

After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.

The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.

Given a function (f(x)), we represent its inverse as (f^{−1}(x)), read as “(f) inverse of (x).” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1 . In other words, (f^{−1}(x)) does not mean (frac{1}{f(x)}) because (frac{1}{f(x)}) is the reciprocal of (f) and not the inverse.

The “exponent-like” notation comes from an analogy between function composition and multiplication: just as (a^{−1}a=1) (1 is the identity element for multiplication) for any nonzero number (a), so (f^{−1}{circ}f) equals the identity function, that is,

[(f^{−1}{circ}f)(x)=f^{−1}(f(x))=f^{−1}(y)=x]

This holds for all (x) in the domain of (f). 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 (f(x)), we can verify whether some other function (g(x)) is the inverse of (f(x)) by checking whether either (g(f(x))=x) or (f(g(x))=x) 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, (y=4x) and (y=frac{1}{4}x) are inverse functions.

[(f^{−1}{circ}f)(x)=f^{-1}(4x)=dfrac{1}{4}(4x)=x]

and

[(f{circ}f^{−1})(x)=fBig(dfrac{1}{4}xBig)=4Big(dfrac{1}{4}xBig)=x]

A few coordinate pairs from the graph of the function (y=4x) are ((−2, −8)), ((0, 0)), and ((2, 8)). A few coordinate pairs from the graph of the function (y=frac{1}{4}x) 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.

Definition: Inverse Function

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

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

[f^{−1}(x) eqdfrac{1}{f(x)}]

and not all functions have inverses.

Example (PageIndex{1}): Identifying an Inverse Function for a Given Input-Output Pair

If for a particular one-to-one function (f(2)=4) and (f(5)=12), what are the corresponding input and output values for the inverse function?

Solution

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

[f(2)=4, ext{ then } f^{-1}(4)=2 ; f(5)=12, ext{ then }f^{-1}(12)=5].

Alternatively, if we want to name the inverse function (g), then (g(4)=2) and (g(12)=5).

Analysis

Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table (PageIndex{1}).

Table (PageIndex{1})
((x,f(x)))((x,g(x)))
((2,4))((4,2))
((5,12))((12,5))

Exercise (PageIndex{1})

Given that (h^{-1}(6)=2), what are the corresponding input and output values of the original function (h)?

(h(2)=6)

How To: Given two functions (f(x)) and (g(x)), test whether the functions are inverses of each other.

1. Determine whether (f(g(x))=x) or (g(f(x))=x).
2. If both statements are true, then (g=f^{-1}) and (f=g^{-1}). If either statement is false, then both are false, and (g{ eq}f^{-1}) and (f{ eq}g^{-1}).

Example (PageIndex{2}): Testing Inverse Relationships Algebraically

If (f(x)=frac{1}{x+2}) and (g(x)=frac{1}{x}−2), is (g=f^{-1})?

Solution

[egin{align} g(f(x))&=dfrac{1}{(frac{1}{x+2})−2} &=x+2−2 &=x end{align}]

so

[g=f^{-1} ext{ and } f=g^{-1}]

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

[egin{align} f(g(x))&=dfrac{1}{frac{1}{x}-2+2} &= dfrac{1}{frac{1}{x}} &=x end{align}]

Analysis

Notice the inverse operations are in reverse order of the operations from the original function.

Exercise (PageIndex{2})

If (f(x)=x^3−4) and (g(x)=sqrt{x+4}), is (g=f^{-1})?

Yes

Example (PageIndex{3}): Determining Inverse Relationships for Power Functions

If (f(x)=x^3) (the cube function) and (g(x)=frac{1}{3}x), is (g=f^{-1})?

Solution

[f(g(x))=dfrac{x^3}{27}{ eq}x]

No, the functions are not inverses.
Analysis

The correct inverse to the cube is, of course, the cube root (sqrt{x}=x^{frac{1}{3}}), that is, the one-third is an exponent, not a multiplier.

Exercise (PageIndex{3})

If (f(x)=(x−1)^3) and (g(x)=sqrt{x}+1), is (g=f^{-1})?

Yes

## Finding Domain and Range of Inverse Functions

The outputs of the function (f) are the inputs to (f^{-1}), so the range of (f) is also the domain of (f^{-1}). Likewise, because the inputs to (f) are the outputs of (f^{-1}), the domain of (f) is the range of (f^{-1}). We can visualize the situation as in Figure (PageIndex{3}). Figure (PageIndex{3}): Domain and range of a function and its inverse.

When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of (f(x)=sqrt{x}) is (f^{-1}(x)=x^2), because a square “undoes” a square root; but the square is only the inverse of the square root on the domain (left[0,infty ight)), since that is the range of (f(x)=sqrt{x}).

We can look at this problem from the other side, starting with the square (toolkit quadratic) function (f(x)=x^2). If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.

In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function (f(x)=x^2) with its range limited to (left[0,infty ight)), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If (f(x)=(x−1)^2) on ([1,∞)), then the inverse function is (f^{-1}(x)=sqrt{x}+1).

• The domain of (f) = range of (f^{-1} = left[1,infty ight)).
• The domain of (f^{-1}) = range of (f = left[0,infty ight)). Is it possible for a function to have more than one inverse?

No. If two supposedly different functions, say, (g) and h, both meet the definition of being inverses of another function (f), then you can prove that (g=h). We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.

Note: Domain and Range of Inverse Functions

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

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

How To: Given a function, find the domain and range of its inverse.

1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.
2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.

Example (PageIndex{4}): Finding the Inverses of Toolkit Functions

Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table (PageIndex{2}). We restrict the domain in such a fashion that the function assumes all y-values exactly once.

Table (PageIndex{2})
(f(x)=c)(f(x)=x)(f(x)=x^2)(f(x)=x^3)(f(x)=frac{1}{x})
Reciprocal squaredCube RootSquare RootAbsolute Value
(f(x)=frac{1}{x^2})(f(x)=sqrt{x})(f(x)=sqrt{x})(f(x)=|x|)

Solution

The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.

The absolute value function can be restricted to the domain (left[0,infty ight)),where it is equal to the identity function.

The reciprocal-squared function can be restricted to the domain ((0,infty)).

Analysis

We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure (PageIndex{4}). They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse. Figure (PageIndex{4}): (a) Absolute value (b) Reciprocal squared (PageIndex{4}): The domain of function (f) is ((1,infty)) and the range of function (f) is ((−infty,−2)). Find the domain and range of the inverse function.

Solution

The domain of function (f^{-1}) is ((−infty,−2)) and the range of function (f^{-1}) is ((1,infty)).

## Finding and Evaluating Inverse Functions

Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.

### Inverting Tabular Functions

Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.

Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.

Example (PageIndex{5}): Interpreting the Inverse of a Tabular Function

A function (f(t)) is given in Table (PageIndex{3}), showing distance in miles that a car has traveled in (t) minutes. Find and interpret (f^{-1}(70))

 (t) (minutes) (f(t)) (miles) 30 50 70 90 20 40 60 70

The inverse function takes an output of (f) and returns an input for (f). So in the expression (f^{-1}(70)), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function (f), 90 minutes, so (f^{-1}(70)=90). The interpretation of this is that, to drive 70 miles, it took 90 minutes.

Alternatively, recall that the definition of the inverse was that if (f(a)=b), then (f^{-1}(b)=a). By this definition, if we are given (f^{-1}(70)=a), then we are looking for a value (a) so that (f(a)=70). In this case, we are looking for a (t) so that (f(t)=70), which is when (t=90).

Exercise (PageIndex{5})

Using Table (PageIndex{4}), find and interpret (a) (f(60)),and (b) (f^{-1}(60)).

 (t) (minutes) (f(t)) (miles) 30 50 60 70 90 20 40 50 60 70

(f(60)=50). In 60 minutes, 50 miles are traveled.
(f^{-1}(60)=70). To travel 60 miles, it will take 70 minutes.

### Evaluating the Inverse of a Function, Given a Graph of the Original Function

We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. Given the graph of a function, evaluate its inverse at specific points.

1. Find the desired input on the y-axis of the given graph.
2. Read the inverse function’s output from the x-axis of the given graph.

Example (PageIndex{6}): Evaluating a Function and Its Inverse from a Graph at Specific Points

A function (g(x)) is given in Figure (PageIndex{5}). Find (g(3)) and (g^{-1}(3)).
. Solution

To evaluate (g(3)), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point ((3,1)) tells us that (g(3)=1).

To evaluate (g^{-1}(3)), recall that by definition (g^{-1}(3)) means the value of (x) for which (g(x)=3). By looking for the output value 3 on the vertical axis, we find the point ((5,3)) on the graph, which means (g(5)=3), so by definition, (g^{-1}(3)=5.) See Figure (PageIndex{6}). Exercise (PageIndex{6})

Using the graph in Figure (PageIndex{6}), (a) find (g^{-1}(1)),and (b) estimate (g^{-1}(4)).

3

5.6

### Finding Inverses of Functions Represented by Formulas

Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula— for example, (y) as a function of (x)— we can often find the inverse function by solving to obtain (x) as a function of (y).

How To: Given a function represented by a formula, find the inverse.

1. Make sure (f) is a one-to-one function.
2. Solve for (x)
3. Interchange (x) and (y).

Example (PageIndex{7}): Inverting the Fahrenheit-to-Celsius Function

Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.

[C=dfrac{5}{9}(F−32)]

Solution

[egin{align} C&=frac{5}{9}(F-32) C{cdot}frac{9}{5}&=F−32 F&=frac{9}{5}C+32end{align}]

By solving in general, we have uncovered the inverse function. If

[C=h(F)=dfrac{5}{9}(F−32)],

then

[F=h^{-1}(C)=dfrac{9}{5}C+32.]

In this case, we introduced a function (h) to represent the conversion because the input and output variables are descriptive, and writing (C^{-1}) could get confusing.

Exercise (PageIndex{7})

Solve for (x) in terms of (y) given (y=frac{1}{3}(x−5))

(x=3y+5)

Example (PageIndex{8}): Solving to Find an Inverse Function

Find the inverse of the function (f(x)=frac{2}{x−3}+4).

Solution

[egin{align} y&=dfrac{2}{x−3+4} & ext{Set up an equation.} y−4&=dfrac{2}{x−3} & ext{Subtract 4 from both sides.} x−3&=dfrac{2}{y−4} & ext{Multiply both sides by x−3 and divide by y−4.} x&=dfrac{2}{y−4}+3 & ext{Add 3 to both sides.} end{align}]

So (f^{-1}(y)=frac{2}{y−4}+3) or (f^{-1}(x)=frac{2}{x−4}+3).

Analysis

The domain and range of (f) exclude the values 3 and 4, respectively. (f) and (f^{-1}) are equal at two points but are not the same function, as we can see by creating Table (PageIndex{5}).

 (x) (f(x)) 1 2 5 (f^{-1}(y)) 3 2 5 (y)

Example (PageIndex{9}): Solving to Find an Inverse with Radicals

Find the inverse of the function (f(x)=2+sqrt{x−4}).

Solution

[ egin{align} y&=2+sqrt{x-4} (y-2)^2&=x-4 x&=(y-2)^2+4 end{align}]

So (f^{-1}(x)=(x−2)^2+4).

The domain of (f) is (left[4,infty ight)). Notice that the range of (f) is (left[2,infty ight)), so this means that the domain of the inverse function (f^{-1}) is also (left[2,infty ight))

Analysis

The formula we found for (f^{-1}(x)) looks like it would be valid for all real (x). However, (f^{-1}) itself must have an inverse (namely, (f) ) so we have to restrict the domain of (f^{-1}) to (left[2,infty ight)) in order to make (f^{-1}) a one-to-one function. This domain of (f^{-1}) is exactly the range of (f).

Exercise (PageIndex{8})

What is the inverse of the function (f(x)=2-sqrt{x})? State the domains of both the function and the inverse function.

(f^{-1}(x)=(2−x)^2); domain of (f): (left[0,infty ight)); domain of (f^{-1}): (left(−infty,2 ight])

## Finding Inverse Functions and Their Graphs

Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function (f(x)=x^2) restricted to the domain (left[0,infty ight)), on which this function is one-to-one, and graph it as in Figure (PageIndex{7}). Figure (PageIndex{7}): Quadratic function with domain restricted to ([0, infty)).

Restricting the domain to (left[0,infty ight)) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.

We already know that the inverse of the toolkit quadratic function is the square root function, that is, (f^{-1}(x)=sqrt{x}). What happens if we graph both (f) and (f^{-1}) on the same set of axes, using the x-axis for the input to both (f) and (f^{-1})?

We notice a distinct relationship: The graph of (f^{-1}(x)) is the graph of (f(x)) reflected about the diagonal line (y=x), which we will call the identity line, shown in Figure (PageIndex{8}). .
Figure (PageIndex{8}): Square and square-root functions on the non-negative domain

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.

Example (PageIndex{10}): Finding the Inverse of a Function Using Reflection about the Identity Line

Given the graph of (f(x)) in Figure (PageIndex{9}), sketch a graph of (f^{-1}(x)). This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of ((0,infty)) and range of ((−infty,infty)), so the inverse will have a domain of ((−infty,infty)) and range of ((0,infty)).

If we reflect this graph over the line (y=x), the point ((1,0)) reflects to ((0,1)) and the point ((4,2)) reflects to ((2,4)). Sketching the inverse on the same axes as the original graph gives Figure (PageIndex{10}). Exercise (PageIndex{1})

Draw graphs of the functions (f) and (f^{-1}) from Example (PageIndex{8}).  Is there any function that is equal to its own inverse?

Yes. If (f=f^{-1}), then (f(f(x))=x), and we can think of several functions that have this property. The identity function

does, and so does the reciprocal function, because

[dfrac{1}{frac{1}{x}}=x]

Any function (f(x)=c−x), where (c) is a constant, is also equal to its own inverse.

## Key Concepts

• If (g(x)) is the inverse of (f(x)), then (g(f(x))=f(g(x))=x).
• Each of the toolkit functions has an inverse.
• For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
• A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
• For a tabular function, exchange the input and output rows to obtain the inverse.
• The inverse of a function can be determined at specific points on its graph.
• To find the inverse of a formula, solve the equation (y=f(x)) for (x) as a function of (y). Then exchange the labels (x) and (y).
• The graph of an inverse function is the reflection of the graph of the original function across the line (y=x).

## 1.7 Integrals Resulting in Inverse Trigonometric Functions

In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.

### Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

### Rule: Integration Formulas Resulting in Inverse Trigonometric Functions

The following integration formulas yield inverse trigonometric functions. Assume a > 0 a > 0 :

## 1.7: Inverse Functions - Mathematics

In the last example from the previous section we looked at the two functions (fleft( x ight) = 3x - 2) and (gleft( x ight) = frac <3>+ frac<2><3>) and saw that

[left( ight)left( x ight) = left( ight)left( x ight) = x]

and as noted in that section this means that these are very special functions. Let’s see just what makes them so special. Consider the following evaluations.

In the first case we plugged (x = - 1) into (fleft( x ight)) and got a value of -5. We then turned around and plugged (x = - 5) into (gleft( x ight)) and got a value of -1, the number that we started off with.

In the second case we did something similar. Here we plugged (x = 2) into (gleft( x ight)) and got a value of(frac<4><3>), we turned around and plugged this into (fleft( x ight)) and got a value of 2, which is again the number that we started with.

Note that we really are doing some function composition here. The first case is really,

[left( ight)left( < - 1> ight) = gleft[ ight)> ight] = gleft[ < - 5> ight] = - 1]

and the second case is really,

Note as well that these both agree with the formula for the compositions that we found in the previous section. We get back out of the function evaluation the number that we originally plugged into the composition.

So, just what is going on here? In some way we can think of these two functions as undoing what the other did to a number. In the first case we plugged (x = - 1) into (fleft( x ight)) and then plugged the result from this function evaluation back into (gleft( x ight)) and in some way (gleft( x ight)) undid what (fleft( x ight)) had done to (x = - 1) and gave us back the original (x) that we started with.

Function pairs that exhibit this behavior are called inverse functions. Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way.

A function is called one-to-one if no two values of (x) produce the same (y). This is a fairly simple definition of one-to-one but it takes an example of a function that isn’t one-to-one to show just what it means. Before doing that however we should note that this definition of one-to-one is not really the mathematically correct definition of one-to-one. It is identical to the mathematically correct definition it just doesn’t use all the notation from the formal definition.

Now, let’s see an example of a function that isn’t one-to-one. The function (fleft( x ight) = ) is not one-to-one because both (fleft( < - 2> ight) = 4) and (fleft( 2 ight) = 4). In other words, there are two different values of (x) that produce the same value of (y). Note that we can turn (fleft( x ight) = ) into a one-to-one function if we restrict ourselves to (0 le x < infty ). This can sometimes be done with functions.

Showing that a function is one-to-one is often a tedious and difficult process. For the most part we are going to assume that the functions that we’re going to be dealing with in this section are one-to-one. We did need to talk about one-to-one functions however since only one-to-one functions can be inverse functions.

Now, let’s formally define just what inverse functions are.

#### Inverse Functions

Given two one-to-one functions (fleft( x ight)) and (gleft( x ight)) if

then we say that (fleft( x ight)) and (gleft( x ight)) are inverses of each other. More specifically we will say that (gleft( x ight)) is the inverse of (fleft( x ight)) and denote it by

[gleft( x ight) = >left( x ight)]

Likewise, we could also say that (fleft( x ight)) is the inverse of (gleft( x ight)) and denote it by

[fleft( x ight) = >left( x ight)]

The notation that we use really depends upon the problem. In most cases either is acceptable.

For the two functions that we started off this section with we could write either of the following two sets of notation.

Now, be careful with the notation for inverses. The “-1” is NOT an exponent despite the fact that it sure does look like one! When dealing with inverse functions we’ve got to remember that

This is one of the more common mistakes that students make when first studying inverse functions.

The process for finding the inverse of a function is a fairly simple one although there is a couple of steps that can on occasion be somewhat messy. Here is the process

#### Finding the Inverse of a Function

Given the function (fleft( x ight)) we want to find the inverse function, (>left( x ight)).

1. First, replace (fleft( x ight)) with (y). This is done to make the rest of the process easier.
2. Replace every (x) with a (y) and replace every (y) with an (x).
3. Solve the equation from Step 2 for (y). This is the step where mistakes are most often made so be careful with this step.
4. Replace (y) with (>left( x ight)). In other words, we’ve managed to find the inverse at this point!
5. Verify your work by checking that (left( >> ight)left( x ight) = x) and (left( <> circ f> ight)left( x ight) = x) are both true. This work can sometimes be messy making it easy to make mistakes so again be careful.

That’s the process. Most of the steps are not all that bad but as mentioned in the process there are a couple of steps that we really need to be careful with.

In the verification step we technically really do need to check that both (left( >> ight)left( x ight) = x) and (left( <> circ f> ight)left( x ight) = x) are true. For all the functions that we are going to be looking at in this section if one is true then the other will also be true. However, there are functions (they are far beyond the scope of this course however) for which it is possible for only one of these to be true. This is brought up because in all the problems here we will be just checking one of them. We just need to always remember that technically we should check both.

Now, we already know what the inverse to this function is as we’ve already done some work with it. However, it would be nice to actually start with this since we know what we should get. This will work as a nice verification of the process.

So, let’s get started. We’ll first replace (fleft( x ight)) with (y).

Next, replace all (x)’s with (y) and all y’s with (x).

Finally replace (y) with (>left( x ight)).

Now, we need to verify the results. We already took care of this in the previous section, however, we really should follow the process so we’ll do that here. It doesn’t matter which of the two that we check we just need to check one of them. This time we’ll check that (left( >> ight)left( x ight) = x) is true.

Now the fact that we’re now using (gleft( x ight)) instead of (fleft( x ight)) doesn’t change how the process works. Here are the first few steps.

Now, to solve for (y) we will need to first square both sides and then proceed as normal.

Finally let’s verify and this time we’ll use the other one just so we can say that we’ve gotten both down somewhere in an example.

So, we did the work correctly and we do indeed have the inverse.

Before we move on we should also acknowledge the restrictions of (x ge 0) that we gave in the problem statement but never apparently did anything with. Note that this restriction is required to make sure that the inverse, (>left( x ight)) given above is in fact one-to-one.

Without this restriction the inverse would not be one-to-one as is easily seen by a couple of quick evaluations.

Therefore, the restriction is required in order to make sure the inverse is one-to-one.

The next example can be a little messy so be careful with the work here.

The first couple of steps are pretty much the same as the previous examples so here they are,

Now, be careful with the solution step. With this kind of problem it is very easy to make a mistake here.

[eginxleft( <2y - 5> ight) & = y + 4 2xy - 5x & = y + 4 2xy - y & = 4 + 5x left( <2x - 1> ight)y & = 4 + 5x y & = frac<<4 + 5x>><<2x - 1>>end]

So, if we’ve done all of our work correctly the inverse should be,

Finally, we’ll need to do the verification. This is also a fairly messy process and it doesn’t really matter which one we work with.

Okay, this is a mess. Let’s simplify things up a little bit by multiplying the numerator and denominator by (2x - 1).

Wow. That was a lot of work, but it all worked out in the end. We did all of our work correctly and we do in fact have the inverse.

There is one final topic that we need to address quickly before we leave this section. There is an interesting relationship between the graph of a function and its inverse.

Here is the graph of the function and inverse from the first two examples. We’ll not deal with the final example since that is a function that we haven’t really talked about graphing yet. In both cases we can see that the graph of the inverse is a reflection of the actual function about the line (y = x). This will always be the case with the graphs of a function and its inverse.

An inverse function is a function which does the “reverse” of a given function. More formally, if (f) is a function with domain (X), then (^<-1>) is its inverse function if and only if (^<-1>left(fleft(x ight) ight)=x) for every (x in X).

A function must be a one-to-one relation if its inverse is to be a function. If a function (f) has an inverse function (f^<-1>), then (f) is said to be invertible.

Given the function (f(x)), we determine the inverse (f^<-1>(x)) by:

• interchanging (x) and (y) in the equation
• making (y) the subject of the equation
• expressing the new equation in function notation.

Note: if the inverse is not a function then it cannot be written in function notation. For example, the inverse of (f(x) = 3x^2) cannot be written as (f^<-1>(x) = pm sqrt<3>x>) as it is not a function. We write the inverse as (y = pm sqrt<3>x>) and conclude that (f) is not invertible.

If we represent the function (f) and the inverse function (^<-1>) graphically, the two graphs are reflected about the line (y=x). Any point on the line (y = x) has (x)- and (y)-coordinates with the same numerical value, for example ((-3-3)) and (left( frac<4><5> frac<4> <5> ight)). Therefore interchanging the (x)- and (y)-values makes no difference. This diagram shows an exponential function (black graph) and its inverse (blue graph) reflected about the line (y = x) (grey line).

Important: for (^<-1>), the superscript (- ext<1>) is not an exponent. It is the notation for indicating the inverse of a function. Do not confuse this with exponents, such as (left( frac<1> <2> ight)^<-1>) or (3 + x^<-1>).

Be careful not to confuse the inverse of a function and the reciprocal of a function:

## Inverse Functions Contents: This page corresponds to § 1.7 (p. 150) of the text. Suggested Problems from Text p.158 #1-4, 5, 8, 9, 12, 13, 15, 18, 21, 22, 27, 31, 34, 37, 46, 48, 51, 71, 74, 83 Definition of Inverse Function

Before defining the inverse of a function we need to have the right mental image of function.

Consider the function f(x) = 2x + 1. We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. In this section it helps to think of f as transforming a 3 into a 7, and f transforms a 5 into an 11, etc.

Now that we think of f as "acting on" numbers and transforming them, we can define the inverse of f as the function that "undoes" what f did. In other words, the inverse of f needs to take 7 back to 3, and take -3 back to -2, etc.

Let g(x) = (x - 1)/2. Then g(7) = 3, g(-3) = -2, and g(11) = 5, so g seems to be undoing what f did, at least for these three values. To prove that g is the inverse of f we must show that this is true for any value of x in the domain of f. In other words, g must take f(x) back to x for all values of x in the domain of f. So, g(f(x)) = x must hold for all x in the domain of f. The way to check this condition is to see that the formula for g(f(x)) simplifies to x.

g(f(x)) = g(2x + 1) = (2x + 1 -1)/2 = 2x/2 = x.

This simplification shows that if we choose any number and let f act it, then applying g to the result recovers our original number. We also need to see that this process works in reverse, or that f also undoes what g does.

f(g(x)) = f((x - 1)/2) = 2(x - 1)/2 + 1 = x - 1 + 1 = x.

Letting f -1 denote the inverse of f, we have just shown that g = f -1 .

Let f and g be two functions. If

f(g(x)) = x and g(f(x)) = x,

then g is the inverse of f and f is the inverse of g.

(a) Open the Java Calculator and enter the formulas for f and g. Note that you take a cube root by raising to the (1/3), and you do need to enter the exponent as (1/3), and not a decimal approximation. So the text for the g box will be

(x - 2)^(1/3)

Use the calculator to evaluate f(g(4)) and g(f(-3)). g is the inverse of f, but due to round off error, the calculator may not return the exact value that you start with. Try f(g(-2)). The answers will vary for different computers. However, on our test machine f(g(4)) returned 4 g(f(-3)) returned 3 but, f(g(-2)) returned -1.9999999999999991, which is pretty close to -2.

The calculator can give us a good indication that g is the inverse of f, but we cannot check all possible values of x.

(b) Prove that g is the inverse of f by simplifying the formulas for f(g(x) and g(f(x)).

## Graphs of Inverse Functions

We have seen examples of reflections in the plane. The reflection of a point (a,b) about the x-axis is (a,-b), and the reflection of (a,b) about the y-axis is (-a,b). Now we want to reflect about the line y = x.

The reflection of the point (a,b) about the line y = x is the point (b,a) .

Let f(x) = x 3 + 2. Then f(2) = 10 and the point (2,10) is on the graph of f. The inverse of f must take 10 back to 2, i.e. f -1 (10)=2, so the point (10,2) is on the graph of f -1 . The point (10,2) is the reflection in the line y = x of the point (2,10). The same argument can be made for all points on the graphs of f and f -1 .

The graph of f -1 is the reflection about the line y = x of the graph of f.

## Existence of an Inverse

Some functions do not have inverse functions. For example, consider f(x) = x 2 . There are two numbers that f takes to 4, f(2) = 4 and f(-2) = 4. If f had an inverse, then the fact that f(2) = 4 would imply that the inverse of f takes 4 back to 2. On the other hand, since f(-2) = 4, the inverse of f would have to take 4 to -2. Therefore, there is no function that is the inverse of f.

Look at the same problem in terms of graphs. If f had an inverse, then its graph would be the reflection of the graph of f about the line y = x. The graph of f and its reflection about y = x are drawn below.

Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function.

This generalizes as follows: A function f has an inverse if and only if when its graph is reflected about the line y = x, the result is the graph of a function (passes the vertical line test). But this can be simplified. We can tell before we reflect the graph whether or not any vertical line will intersect more than once by looking at how horizontal lines intersect the original graph!

### Horizontal Line Test

Let f be a function.

If any horizontal line intersects the graph of f more than once, then f does not have an inverse.

If no horizontal line intersects the graph of f more than once, then f does have an inverse.

The property of having an inverse is very important in mathematics, and it has a name.

Definition : A function f is one-to-one if and only if f has an inverse.

The following definition is equivalent, and it is the one most commonly given for one-to-one.

Alternate Definition : A function f is one-to-one if, for every a and b in its domain, f(a) = f(b) implies a = b.

Graph the following functions and determine whether or not they have inverses.

(a) f(x) = (x - 3) x 2 . Answer

(b) f(x) = x 3 + 3x 2 +3x. Answer

(c) f(x) = x ^(1/3) ( the cube root of x). Answer

## Finding Inverses

Example 1. First consider a simple example f(x) = 3x + 2 .

The graph of f is a line with slope 3, so it passes the horizontal line test and does have an inverse.

There are two steps required to evaluate f at a number x. First we multiply x by 3, then we add 2.

Thinking of the inverse function as undoing what f did, we must undo these steps in reverse order.

The steps required to evaluate f -1 are to first undo the adding of 2 by subtracting 2. Then we undo multiplication by 3 by dividing by 3.

Therefore, f -1 (x) = (x - 2)/3.

Steps for finding the inverse of a function f.

1. Replace f(x) by y in the equation describing the function.
2. Interchange x and y. In other words, replace every x by a y and vice versa.
3. Solve for y.
4. Replace y by f -1 (x).

 Step 1 y = 6 - x/2. Step 2 x = 6 - y/2. Step 3 x = 6 - y/2. y = 12 - 2x. Step 4 f -1 (x) = 12 - 2x.

Step 2 often confuses students. We could omit step 2, and solve for x instead of y, but then we would end up with a formula in y instead of x. The formula would be the same, but the variable would be different. To avoid this we simply interchange the roles of x and y before we solve.

This is the function we worked with in Exercise 1. From its graph (shown above) we see that it does have an inverse. (In fact, its inverse was given in Exercise 1.)

 Step 1 y = x 3 + 2. Step 2 x = y 3 + 2. Step 3 x - 2 = y 3 .

(x - 2)^(1/3) = y. Step 4 f -1 (x) = (x - 2)^(1/3).

Graph f(x) = 1 - 2x 3 to see that it does have an inverse. Find f -1 (x). Answer

## MAT 112 Ancient and Contemporary Mathematics

Sometimes it is possible to “undo” the effect of a function. In these cases, we can define a function that reverses the effects of another function. A function that we can “undo” is called invertible.

In the video in Figure 7.5.1 we introduce inverse functions and give examples. For details and examples, see our treatment of inverse

We give the formal definition of an invertible function and of the inverse of an invertible function.

###### Definition 7.5.2 .

Let (A) and (B) be non-empty sets. We say that a function (f:A o B) is if for every (bin B) there is exactly one (ain A) such that (f(a)=b ext<.>) The of an invertible function (f:A o B ext<,>) denoted by (f^<-1> ext<,>) is the function (f^<-1>:B o A) that assigns to each element (b in B) the unique element (a in A) such that (f(a) = b ext<.>)

In other words, a function (f:A o B) is invertible if every (bin B) has exactly one preimage (ain A ext<.>) So if (f(a) = b ext<,>) then (f^<-1>(b) = a ext<.>)

Confirm your understanding of the definition by completing it in the exercise.

###### Checkpoint 7.5.3 . Definition of invertible function.

We investigate whether the two functions (mathrm ) and ( ext) are invertible.

###### Example 7.5.5 . Are (mathrm ) and ( ext) invertible ?

We use the functions (mathrm colon N o I) and (mathrm colon I o G) from Example 7.1.5 and Example 7.1.7 given by the tables in Figure 7.1.4 and Figure 7.1.6 respectively.

The function (mathrm colon N o I) where (I) is the set of student identification numbers and (N) is the set of student names is invertible as long as every student has a different name. The function (mathrm^<-1>: I o N) is the function that tells us the student's name for a given identification number. With the table in Figure 7.1.4 we get

Recall the function grade from Example 7.1.7. The function (mathrm colon I o G) where (I) is the set of identification numbers and (G) is the set of grades is not invertible since many students may earn the same grade in a class. Both the students with the identification number 1007 and 1008 earn an A in MAT 112. We have (mathrm ( ext < 1007 >) = ext) and ( ext ( ext < 1008 >) = ext ext<,>) and we would not be able to uniquely define (mathrm^<-1>(A) ext<.>)

In Figure 7.5.6 we give an example of an invertible function from (_5) to (_5) and its inverse. The function (e) in Figure 7.5.9 illustrates that for an invertible function the domain and codomain do not have to be the same.

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## Math: How to Find the Inverse of a Function The inverse function of a function f is mostly denoted as f -1 . A function f has an input variable x and gives then an output f(x). The inverse of a function f does exactly the opposite. Instead it uses as input f(x) and then as output it gives the x that when you would fill it in in f will give you f(x). To be more clear:

If f(x) = y then f -1 (y) = x. So the output of the inverse is indeed the value that you should fill in in f to get y. So f(f -1 (x)) = x.

Not every function has an inverse. A function that does have an inverse is called invertible. Only if f is bijective an inverse of f will exist. But what does this mean?

The easy explanation of a function that is bijective is a function that is both injective and surjective. However, for most of you this will not make it any clearer.

A function is injective if there are no two inputs that map to the same output. Or said differently: every output is reached by at most one input.

An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. If we fill in -2 and 2 both give the same output, namely 4. So x 2 is not injective and therefore also not bijective and hence it won&apost have an inverse.

A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. So f(x)= x 2 is also not surjective if you take as range all real numbers, since for example -2 cannot be reached since a square is always positive.

So while you might think that the inverse of f(x) = x 2 would be f -1 (y) = sqrt(y) this is only true when we treat f as a function from the nonnegative numbers to the nonnegative numbers, since only then it is a bijection.

This does show that the inverse of a function is unique, meaning that every function has only one inverse. ## More Questions with Solutions

Use the table below to find the following if possible:
1) g -1 (0) , b) g -1 (-10) , c) g -1 (- 5) , d) g -1 (-7) , e) g -1 (3)

Solution
a) According to the the definition of the inverse function:
a = g -1 (0) if and only if g(a) = 0
Which means that a is the value of x such g(x) = 0.
Using the table above for x = 11, g(x) = 0. Hence a = 11 and therefore g -1 (0) = 11
b) a = g - 1 (- 5) if and only if g(a) = - 5
The value of x for which g(x) = - 5 is equal to 0 and therefore g -1 ( - 5) = 0
c) a = g -1 (-10) if and only if g(a) = - 10
There is no value of x for which g(x) = -10 and therefore g -1 (-10) is undefined.
d) a = g -1 (- 7) if and only if g(a) = - 7
There no value of x for which g(x) = - 7 and therefore g -1 (- 7) is undefined.
e) a = g -1 (3) if and only if g(a) = 3
The value of x for which g(x) = 3 is equal to - 2 and therefore g -1 (3) = - 2

## Questions on Inverse Functions with Solutions and Answers

Analytical and graphing methods are used to solve maths problems and questions related to inverse functions. Detailed solutions are also presented. Several questions involve the use of the property that the graphs of a function and the graph of its inverse are reflection of each other on the line y = x.

1) Sketch the graph of the inverse of f in the same system of axes.
2) Find the inverse of and check your answer using some points.
Solution
1) Locate few points on the graph of f. Here is a list of points whose coordinates (a , b) can easily be determined from the graph:
(1 , 1) , (0 , -1) , (-1 , -3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(1 , 1) , (-1 , 0) , (-3 , -1)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.

1) Sketch the inverse of f in the same graph.
2) Find the inverse of and check your answer using some points.
Solution
1) Locate few points on the graph of f. A possible list of points whose coordinates (a , b) is as follows:
(1.5 , 0) , (2 , 1) , (6 , 3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(0 , 1.5) , (1 , 2) , (3 , 6)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.

2) Write the given function f(x) = √(2 x - 3) as an equation in two unknowns.
y = √(2 x - 3)
Solve the above for x. First square both sides
2 x - 3 = y 2
2 x = y 2 + 3
x = (y 2 + 3) / 2
Interchange x and y and write the equation of the inverse function f -1 and write the domain of the inverse.
y = (x 2 + 3) / 2
f -1 (x) = (x 2 + 3) / 2 , x ≥ 0 (domain which is the range of f from its graph above)
We now verify that the points (0 , 1.5) , (1 , 2) and (3 , 6) used to sketch the graph of the inverse function are on the graph of f -1 .
f -1 (0) = (0 2 + 3) / 2 = 1.5
f -1 (1) = (1 2 + 3) / 2 = 2
f -1 (3) = (3 2 + 3) / 2 = 6

Solution
1) Use the graph to find points on the graph of f. A possible list of points whose coordinates (a , b) is as follows:
(0 , 3) , (2 , -1) , (5 , - 3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(3 , 0) , (- 1 , 2) , (- 3 , 5)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.

2) We now determine f -1 (x). For -3 ≤ x ≤ - 1 , f -1 (x) has a linear expression with slope m1 through the points (- 1 , 2) , (- 3 , 5) given by
m1 = (5 - 2) / (-3 - (-1)) = - 3 / 2
For -3 ≤ x ≤ - 1, f -1 (x) is given by:
f -1 (x) = - (3 / 2)(x - (-1)) + 2 = - (3 / 2)(x + 1) + 2
For - 1 < x ≤ 3 , f -1 (x) has a linear expression with slope through the points (- 1 , 2) , (3 , 0) given by
m2 = (0 - 2) / (3 - (-1)) = - 1 / 2
For - 1 < x ≤ 3, f -1 (x) is given by:
f -1 (x) = - (1 / 2)(x - (-1)) + 2 = - (1 / 2)(x + 1) + 2

1) What is the domain and range of f?
2) Sketch the graph of f -1 .
3) Find f -1 (x) (include domain).
Solution
1) f(x) is defined as a real number if the radicand 2 / x - 1 is greater than or equal to 0. Hence we need to solve the inequality:
2 / x - 1 ≥ 0
(2 - x) / x ≥ 0
The expression on the left of the inequality changes sign at the zeros of the numerator and denominator which are x = 2 and x = 0. See table below.

Domain: (0 , 2]
Range: (-∞ , 0]
2) Points on the graph of f
(2 , 0) , (1 , -1)
The above points on the graph of the inverse function, will have coordinates (b , a) as follows:
(0 , 2) , (- 1 , 1)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.

3) Write f(x) as an equation in y and x.
( y = -sqrt-1> )
Solve the above equation for x. Square both sides of the above equation
( y^2 = dfrac<2>-1 )
( dfrac<2> = y^2 + 1 )
( x = dfrac<2> )
Interchange x and y and write the inverse function
( y = dfrac<2> )
( f^<-1>(x) = dfrac<2> )
Domain and range of f -1 are the range and domain of f . Hence
Domain of f -1 : (-∞ , 0]
Range of f -1 : (0 , 2]

Solution
For each graph, select points whose coordinates are easy to determine. Use these points and also the reflection of the graph of function f and its inverse on the line y = x to skectch to sketch the inverse functions as shown below