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 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions.

Figure 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

and substitutes 75 for [latex]F[/latex] to calculate

Figure 2

Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast 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 [latex]F[/latex] after substituting a value for [latex]C[/latex]. For example, to convert 26 degrees Celsius, she could write

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 [latex]f\left(x\right)[/latex], we represent its inverse as [latex]{f}^{-1}\left(x\right)[/latex], read as [latex]"f[/latex] inverse of [latex]x.\text{"}[/latex] The raised [latex]-1[/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[/latex] . In other words, [latex]{f}^{-1}\left(x\right)[/latex] does not mean [latex]\frac{1}{f\left(x\right)}[/latex] because [latex]\frac{1}{f\left(x\right)}[/latex] is the reciprocal of [latex]f[/latex] and not the inverse.

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

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

and

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

Finding Domain and Range of Inverse Functions

The outputs of the function [latex]f[/latex] are the inputs to [latex]{f}^{-1}[/latex], so the range of [latex]f[/latex] is also the domain of [latex]{f}^{-1}[/latex]. Likewise, because the inputs to [latex]f[/latex] are the outputs of [latex]{f}^{-1}[/latex], the domain of [latex]f[/latex] is the range of [latex]{f}^{-1}[/latex]. We can visualize the situation.

Figure 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 [latex]f\left(x\right)=\sqrt[\leftroot{-1}\uproot{2}]{x}[/latex] is [latex]{f}^{-1}\left(x\right)={x}^{2}[/latex], because a square “undoes” a square root; but the square is only the inverse of the square root on the domain [latex]\left[0,\infty \right)[/latex], since that is the range of [latex]f\left(x\right)=\sqrt[\leftroot{-1}\uproot{2}]{x}[/latex].

We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\left(x\right)={x}^{2}[/latex]. 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 [latex]f\left(x\right)={x}^{2}[/latex] with its range limited to [latex]\left[0,\infty \right)[/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If [latex]f\left(x\right)={\left(x - 1\right)}^{2}[/latex] on [latex]\left[1,\infty \right)[/latex], then the inverse function is [latex]{f}^{-1}\left(x\right)=\sqrt[\leftroot{-1}\uproot{2}]{x}+1[/latex].

• The domain of [latex]f[/latex] = range of [latex]{f}^{-1}[/latex] = [latex]\left[1,\infty \right)[/latex].
• The domain of [latex]{f}^{-1}[/latex] = range of [latex]f[/latex] = [latex]\left[0,\infty \right)[/latex].

Example 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 below. We restrict the domain in such a fashion that the function assumes all y-values exactly once.

Constant	Identity	Quadratic	Cubic	Reciprocal
[latex]f\left(x\right)=c[/latex]	[latex]f\left(x\right)=x[/latex]	[latex]f\left(x\right)={x}^{2}[/latex]	[latex]f\left(x\right)={x}^{3}[/latex]	[latex]f\left(x\right)=\frac{1}{x}[/latex]
Reciprocal squared	Cube root	Square root	Absolute value
[latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex]	[latex]f\left(x\right)=\sqrt[\leftroot{-1}\uproot{2}3]{x}[/latex]	[latex]f\left(x\right)=\sqrt[\leftroot{-1}\uproot{2}]{x}[/latex]	[latex]f\left(x\right)=|x|[/latex]

Show 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 [latex]\left[0,\infty \right)[/latex], where it is equal to the identity function.

The reciprocal-squared function can be restricted to the domain [latex]\left(0,\infty \right)[/latex].

We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs. 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 4. (a) Absolute value (b) Reciprocal squared

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.

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.

How To: 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 6: Evaluating a Function and Its Inverse from a Graph at Specific Points

A function [latex]g\left(x\right)[/latex] is given in Figure 5. Find [latex]g\left(3\right)[/latex] and [latex]{g}^{-1}\left(3\right)[/latex].

Figure 5

Show Solution

To evaluate [latex]g\left(3\right)[/latex], we find 3 on the x-axis and find the corresponding output value on the y-axis. The point [latex]\left(3,1\right)[/latex] tells us that [latex]g\left(3\right)=1[/latex].

To evaluate [latex]{g}^{-1}\left(3\right)[/latex], recall that by definition [latex]{g}^{-1}\left(3\right)[/latex] means the value of x for which [latex]g\left(x\right)=3[/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\left(5,3\right)[/latex] on the graph, which means [latex]g\left(5\right)=3[/latex], so by definition, [latex]{g}^{-1}\left(3\right)=5[/latex].

Figure 6

Try It

Using the graph in Example 6, (a) find [latex]{g}^{-1}\left(1\right)[/latex], and (b) estimate [latex]{g}^{-1}\left(4\right)[/latex].

Show Solution

a. 3; b. 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, [latex]y[/latex] as a function of [latex]x\text{-\hspace{0.17em}}[/latex] we can often find the inverse function by solving to obtain [latex]x[/latex] as a function of [latex]y[/latex].

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

1. Make sure [latex]f[/latex] is a one-to-one function.
2. Solve for [latex]x[/latex].
3. Interchange [latex]x[/latex] and [latex]y[/latex].

Example 7: Inverting the Fahrenheit-to-Celsius Function

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

[latex]C=\frac{5}{9}\left(F - 32\right)[/latex]

Show Solution

[latex]\begin{gathered} C =\frac{5}{9}\left(F - 32\right) \\ C\cdot \frac{9}{5}=F - 32 \\ F=\frac{9}{5}C+32 \end{gathered}[/latex]

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

[latex]C=h\left(F\right)=\frac{5}{9}\left(F - 32\right)[/latex],

then

[latex]F={h}^{-1}\left(C\right)=\frac{9}{5}C+32[/latex].

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

Try It

Solve for [latex]x[/latex] in terms of [latex]y[/latex] given [latex]y=\frac{1}{3}\left(x - 5\right)[/latex]

Show Solution

[latex]x=3y+5[/latex]

Example 8: Solving to Find an Inverse Function

Find the inverse of the function [latex]f\left(x\right)=\frac{2}{x - 3}+4[/latex].

Show Solution

[latex]\begin{align}&y=\frac{2}{x - 3}+4 && \text{Set up an equation}. \\ &y - 4=\frac{2}{x - 3} && \text{Subtract 4 from both sides}. \\ &x - 3=\frac{2}{y - 4} && \text{Multiply both sides by }x - 3\text{ and divide by }y - 4. \\ &x=\frac{2}{y - 4}+3 && \text{Add 3 to both sides}. \end{align}[/latex]

So [latex]{f}^{-1}\left(y\right)=\frac{2}{y - 4}+3[/latex] or [latex]{f}^{-1}\left(x\right)=\frac{2}{x - 4}+3[/latex].

Analysis of the Solution

The domain and range of [latex]f[/latex] exclude the values 3 and 4, respectively. [latex]f[/latex] and [latex]{f}^{-1}[/latex] are equal at two points but are not the same function, as we can see by creating the table below.

[latex]x[/latex]	1	2	5	[latex]{f}^{-1}\left(y\right)[/latex]
[latex]f\left(x\right)[/latex]	3	2	5	[latex]y[/latex]

Example 9: Solving to Find an Inverse with Radicals

Find the inverse of the function [latex]f\left(x\right)=2+\sqrt[\leftroot{-1}\uproot{2}]{x - 4}[/latex].

Show Solution

[latex]\begin{align}&y=2+\sqrt{x - 4} \\ &{\left(y - 2\right)}^{2}=x - 4 \\ &x={\left(y - 2\right)}^{2}+4 \end{align}[/latex]

So [latex]{f}^{-1}\left(x\right)={\left(x - 2\right)}^{2}+4[/latex].

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

Analysis of the Solution

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

Try It

What is the inverse of the function [latex]f\left(x\right)=2-\sqrt[\leftroot{-1}\uproot{2}]{x}?[/latex] State the domains of both the function and the inverse function.

Show Solution

[latex]{f}^{-1}\left(x\right)={\left(2-x\right)}^{2}[/latex]; domain of [latex]f:\left[0,\infty \right)[/latex]; domain of [latex]{f}^{-1}:\left(-\infty ,2\right][/latex]

Try It

 

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 [latex]f\left(x\right)={x}^{2}[/latex] restricted to the domain [latex]\left[0,\infty \right)[/latex], on which this function is one-to-one, and graph it as in Figure 7.

Figure 7. Quadratic function with domain restricted to [0, ∞).

Restricting the domain to [latex]\left[0,\infty \right)[/latex] 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, [latex]{f}^{-1}\left(x\right)=\sqrt[\leftroot{-1}\uproot{2}]{x}[/latex]. What happens if we graph both [latex]f\text{ }[/latex] and [latex]{f}^{-1}[/latex] on the same set of axes, using the [latex]x\text{-}[/latex] axis for the input to both [latex]f\text{ and }{f}^{-1}?[/latex]

We notice a distinct relationship: The graph of [latex]{f}^{-1}\left(x\right)[/latex] is the graph of [latex]f\left(x\right)[/latex] reflected about the diagonal line [latex]y=x[/latex], which we will call the identity line, shown in Figure 8.

Figure 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 10: Finding the Inverse of a Function Using Reflection about the Identity Line

Given the graph of [latex]f\left(x\right)[/latex], sketch a graph of [latex]{f}^{-1}\left(x\right)[/latex].

Figure 9

Show Solution

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 [latex]\left(0,\infty \right)[/latex] and range of [latex]\left(-\infty ,\infty \right)[/latex], so the inverse will have a domain of [latex]\left(-\infty ,\infty \right)[/latex] and range of [latex]\left(0,\infty \right)[/latex].

If we reflect this graph over the line [latex]y=x[/latex], the point [latex]\left(1,0\right)[/latex] reflects to [latex]\left(0,1\right)[/latex] and the point [latex]\left(4,2\right)[/latex] reflects to [latex]\left(2,4\right)[/latex]. Sketching the inverse on the same axes as the original graph gives us the result in Figure 10.

Figure 10. The function and its inverse, showing reflection about the identity line

Try It

Q & A

Is there any function that is equal to its own inverse?

Yes. If [latex]f={f}^{-1}[/latex], then [latex]f\left(f\left(x\right)\right)=x[/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because

[latex]\frac{1}{1/x}=x[/latex]

Any function [latex]f\left(x\right)=c-x[/latex], where [latex]c[/latex] is a constant, is also equal to its own inverse.

Learning Outcomes

• Find the inverse of a polynomial function.
• Restrict the domain to find the inverse of a polynomial function.
• Find or evaluate the inverse of a function.

Figure 1

A mound of gravel is in the shape of a cone with the height equal to twice the radius.

The volume is found using a formula from elementary geometry.

[latex]\begin{align}V&=\frac{1}{3}\pi {r}^{2}h \\ &=\frac{1}{3}\pi {r}^{2}\left(2r\right) \\ &=\frac{2}{3}\pi {r}^{3} \end{align}[/latex]

We have written the volume V in terms of the radius r. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula

[latex]r=\sqrt[3]{\dfrac{3V}{2\pi }\\}[/latex]

This function is the inverse of the formula for V in terms of r.

Find the inverse of a polynomial function

Two functions f and g are inverse functions if for every coordinate pair in f, (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.

For a function to have an inverse function the function to create a new function that is one-to-one and would have an inverse function.

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Figure 2

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x measured horizontally and y measured vertically, with the origin at the vertex of the parabola.

Figure 3

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\left(x\right)=a{x}^{2}[/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a.

[latex]\begin{align} 18&=a{6}^{2} \\ a&=\frac{18}{36} \\ &=\frac{1}{2} \end{align}[/latex]

Our parabolic cross section has the equation

[latex]y\left(x\right)=\frac{1}{2}{x}^{2}[/latex]

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y the width will be given by 2x, so we need to solve the equation above for x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable:

[latex]\begin{align}y&=\frac{1}{2}{x}^{2} \\ 2y&={x}^{2} \\ x&=\pm \sqrt{2y} \end{align}[/latex]

This is not a function as written. We are limiting ourselves to positive x values, so we eliminate the negative solution, giving us the inverse function we’re looking for.

[latex]y=\frac{{x}^{2}}{2},\text{ }x>0[/latex]

Because x is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x. The trough is 3 feet (36 inches) long, so the surface area will then be:

[latex]\begin{align}\text{Area} & =l\cdot w \\ & =36\cdot 2x \\ & =72x \\ & =72\sqrt{2y} \end{align}[/latex]

This example illustrates two important points:

1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.

Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation [latex]{f}^{-1}\left(x\right)[/latex].

Warning: [latex]{f}^{-1}\left(x\right)[/latex] is not the same as the reciprocal of the function [latex]f\left(x\right)[/latex]. This use of –1 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\left(x\right)[/latex], we would need to write [latex]{\left(f\left(x\right)\right)}^{-1}=\frac{1}{f\left(x\right)}[/latex].

An important relationship between inverse functions is that they “undo” each other. If [latex]{f}^{-1}[/latex] is the inverse of a function f, then f is the inverse of the function [latex]{f}^{-1}[/latex]. In other words, whatever the function f does to x, [latex]{f}^{-1}[/latex] undoes it—and vice-versa. More formally, we write

[latex]{f}^{-1}\left(f\left(x\right)\right)=x,\text{for all }x\text{ in the domain of }f[/latex]

and

[latex]f\left({f}^{-1}\left(x\right)\right)=x,\text{for all }x\text{ in the domain of }{f}^{-1}[/latex]

A General Note: Verifying Two Functions Are Inverses of One Another

Two functions, f and g, are inverses of one another if for all x in the domain of f and g.

[latex]g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)=x[/latex]

How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.

1. Replace [latex]f\left(x\right)[/latex] with y.
2. Interchange x and y.
3. Solve for y, and rename the function [latex]{f}^{-1}\left(x\right)[/latex].

Example 1: Verifying Inverse Functions

Show that [latex]f\left(x\right)=\frac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\frac{1}{x}-1[/latex] are inverses, for [latex]x\ne 0,-1[/latex] .

Show Solution

We must show that [latex]{f}^{-1}\left(f\left(x\right)\right)=x[/latex] and [latex]f\left({f}^{-1}\left(x\right)\right)=x[/latex].

[latex]\begin{align}{f}^{-1}\left(f\left(x\right)\right)&={f}^{-1}\left(\frac{1}{x+1}\right) \\ &=\frac{1}{\frac{1}{x+1}}-1 \\ &=\left(x+1\right)-1 \\ &=x\end{align}[/latex]

[latex]\begin{align}f\left({f}^{-1}\left(x\right)\right)&=f\left(\frac{1}{x}-1\right) \\ &=\frac{1}{\left(\frac{1}{x}-1\right)+1} \\ &=\frac{1}{\frac{1}{x}} \\ &=x \end{align}[/latex]

Therefore, [latex]f\left(x\right)=\frac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\frac{1}{x}-1[/latex] are inverses.

Try It

Show that [latex]f\left(x\right)=\frac{x+5}{3}[/latex] and [latex]{f}^{-1}\left(x\right)=3x - 5[/latex] are inverses.

Show Solution

[latex]{f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(\frac{x+5}{3}\right)=3\left(\frac{x+5}{3}\right)-5=\left(x - 5\right)+5=x[/latex] and [latex]f\left({f}^{-1}\left(x\right)\right)=f\left(3x - 5\right)=\frac{\left(3x - 5\right)+5}{3}=\frac{3x}{3}=x[/latex]

Example 2: Finding the Inverse of a Cubic Function

Find the inverse of the function [latex]f\left(x\right)=5{x}^{3}+1[/latex].

Show Solution

This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x.

[latex]\begin{gathered}y=5{x}^{3}+1 \\ x=5{y}^{3}+1 \\ x - 1=5{y}^{3} \\ \frac{x - 1}{5}={y}^{3} \\ {f}^{-1}\left(x\right)=\sqrt[3]{\frac{x - 1}{5}} \end{gathered}[/latex]

Analysis of the Solution

Look at the graph of f and [latex]{f}^{-1}[/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[/latex]. This is always the case when graphing a function and its inverse function.

Also, since the method involved interchanging x and y, notice corresponding points. If [latex]\left(a,b\right)[/latex] is on the graph of f, then [latex]\left(b,a\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Since [latex]\left(0,1\right)[/latex] is on the graph of f, then [latex]\left(1,0\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Similarly, since [latex]\left(1,6\right)[/latex] is on the graph of f, then [latex]\left(6,1\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex].

Figure 4

Try It

Find the inverse function of [latex]f\left(x\right)=\sqrt[3]{x+4}[/latex].

Show Solution

[latex]{f}^{-1}\left(x\right)={x}^{3}-4[/latex]

Try It

Restrict the domain to find the inverse of a polynomial function

So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.

A General Note: Restricting the Domain

If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.

How To: Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

1. Restrict the domain by determining a domain on which the original function is one-to-one.
2. Replace f(x) with y.
3. Interchange x and y.
4. Solve for y, and rename the function or pair of function [latex]{f}^{-1}\left(x\right)[/latex].
5. Revise the formula for [latex]{f}^{-1}\left(x\right)[/latex] by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.

Example 3: Restricting the Domain to Find the Inverse of a Polynomial Function

Find the inverse function of f:

1. [latex]f\left(x\right)={\left(x - 4\right)}^{2}, x\ge 4[/latex]
2. [latex]f\left(x\right)={\left(x - 4\right)}^{2}, x\le 4[/latex]

Show Solution

The original function [latex]f\left(x\right)={\left(x - 4\right)}^{2}[/latex] is not one-to-one, but the function is restricted to a domain of [latex]x\ge 4[/latex] or [latex]x\le 4[/latex] on which it is one-to-one.

To find the inverse, start by replacing [latex]f\left(x\right)[/latex] with the simple variable y.

[latex]\begin{align}&y={\left(x - 4\right)}^{2} && \text{Interchange } x \text{ and }y. \\ &x={\left(y - 4\right)}^{2} && \text{Take the square root}. \\ &\pm \sqrt{x}=y - 4 && \text{Add } 4 \text{ to both sides}. \\ &4\pm \sqrt{x}=y \end{align}[/latex]

This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of x and y for the original f(x), we looked at the domain: the values x could assume. When we reversed the roles of x and y, this gave us the values y could assume. For this function, [latex]x\ge 4[/latex], so for the inverse, we should have [latex]y\ge 4[/latex], which is what our inverse function gives.

1. The domain of the original function was restricted to [latex]x\ge 4[/latex], so the outputs of the inverse need to be the same, [latex]f\left(x\right)\ge 4[/latex], and we must use the + case:
   [latex]{f}^{-1}\left(x\right)=4+\sqrt{x}[/latex]
2. The domain of the original function was restricted to [latex]x\le 4[/latex], so the outputs of the inverse need to be the same, [latex]f\left(x\right)\le 4[/latex], and we must use the – case:
   [latex]{f}^{-1}\left(x\right)=4-\sqrt{x}[/latex]

Analysis of the Solution

On the graphs below, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line [latex]y=x[/latex]. The coordinate pair [latex]\left(4, 0\right)[/latex] is on the graph of f and the coordinate pair [latex]\left(0, 4\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. For any coordinate pair, if (a, b) is on the graph of f, then (b, a) is on the graph of [latex]{f}^{-1}[/latex]. Finally, observe that the graph of f intersects the graph of [latex]{f}^{-1}[/latex] on the line y = x. Points of intersection for the graphs of f and [latex]{f}^{-1}[/latex] will always lie on the line y = x.

Figure 6

Example 4: Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified

Restrict the domain and then find the inverse of

[latex]f\left(x\right)={\left(x - 2\right)}^{2}-3[/latex].

Show Solution

We can see this is a parabola with vertex at [latex]\left(2, -3\right)[/latex] that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to [latex]x\ge 2[/latex].

To find the inverse, we will use the vertex form of the quadratic. We start by replacing f(x) with a simple variable, y, then solve for x.

[latex]\begin{align}&y={\left(x - 2\right)}^{2}-3 && \text{Interchange } x \text{ and } y. \\ &x={\left(y - 2\right)}^{2}-3 && \text{Add 3 to both sides}. \\ &x+3={\left(y - 2\right)}^{2} && \text{Take the square root}. \\ &\pm \sqrt{x+3}=y - 2 && \text{Add 2 to both sides}. \\ &2\pm \sqrt{x+3}=y && \text{Rename the function}. \\ &{f}^{-1}\left(x\right)=2\pm \sqrt{x+3} \end{align}[/latex]

Now we need to determine which case to use. Because we restricted our original function to a domain of [latex]x\ge 2[/latex], the outputs of the inverse should be the same, telling us to utilize the + case

[latex]{f}^{-1}\left(x\right)=2+\sqrt{x+3}[/latex]

If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

Analysis of the Solution

Notice that we arbitrarily decided to restrict the domain on [latex]x\ge 2[/latex]. We could just have easily opted to restrict the domain on [latex]x\le 2[/latex], in which case [latex]{f}^{-1}\left(x\right)=2-\sqrt{x+3}[/latex]. Observe the original function graphed on the same set of axes as its inverse function in the graph below. Notice that both graphs show symmetry about the line y = x. The coordinate pair [latex]\left(2,\text{ }-3\right)[/latex] is on the graph of f and the coordinate pair [latex]\left(-3,\text{ }2\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Observe from the graph of both functions on the same set of axes that

[latex]\text{domain of }f=\text{range of } {f}^{-1}=\left[2,\infty \right)[/latex]

and

[latex]\text{domain of }{f}^{-1}=\text{range of } f=\left[-3,\infty \right)[/latex]

Finally, observe that the graph of f intersects the graph of [latex]{f}^{-1}[/latex] along the line y = x.

Figure 7

 

Try It

Find the inverse of the function [latex]f\left(x\right)={x}^{2}+1[/latex], on the domain [latex]x\ge 0[/latex].

Show Solution

[latex]{f}^{-1}\left(x\right)=\sqrt{x - 1}[/latex]

Solving Applications of Radical Functions

Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited.

How To: Given a radical function, find the inverse.

1. Determine the range of the original function.
2. Replace f(x) with y, then solve for x.
3. If necessary, restrict the domain of the inverse function to the range of the original function.

Example 5: Finding the Inverse of a Radical Function

Restrict the domain and then find the inverse of the function [latex]f\left(x\right)=\sqrt{x - 4}[/latex].

Show Solution

Note that the original function has range [latex]f\left(x\right)\ge 0[/latex]. Replace [latex]f\left(x\right)[/latex] with y, then solve for x.

[latex]\begin{align} &y=\sqrt{x - 4} && \text{Replace}f\left(x\right)\text{with}y.\\ &x=\sqrt{y - 4} && \text{Interchange}x\text{and}y. \\ &x =\sqrt{y - 4} && \text{Square each side}. \\ &{x}^{2} =y - 4 && \text{Add 4}. \\ &{x}^{2}+4 =y && \text{Rename the function}{f}^{-1}\left(x\right). \\ &{f}^{-1}\left(x\right) ={x}^{2}+4 \end{align}[/latex]

Recall that the domain of this function must be limited to the range of the original function.

[latex]{f}^{-1}\left(x\right)={x}^{2}+4,x\ge 0[/latex]

Analysis of the Solution

Notice in the graph below that the inverse is a reflection of the original function over the line y = x. Because the original function has only positive outputs, the inverse function has only positive inputs.

Figure 8

Try It

Restrict the domain and then find the inverse of the function [latex]f\left(x\right)=\sqrt{2x+3}[/latex].

Show Solution

[latex]{f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{2},x\ge 0[/latex]

Try It

Solving Applications of Radical Functions

Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.

Example 6: Solving an Application with a Cubic Function

A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

[latex]V=\frac{2}{3}\pi {r}^{3}[/latex]

Find the inverse of the function [latex]V=\frac{2}{3}\pi {r}^{3}[/latex] that determines the volume V of a cone and is a function of the radius r. Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use [latex]\pi =3.14[/latex].

Show Solution

Start with the given function for V. Notice that the meaningful domain for the function is [latex]r\ge 0[/latex] since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is [latex]V\ge 0[/latex]. Solve for r in terms of V, using the method outlined previously.

[latex]\begin{align} V&=\frac{2}{3}\pi {r}^{3} \\ {r}^{3}&=\frac{3V}{2\pi } && \text{Solve for }{r}^{3}. \\ r&=\sqrt[3]{\frac{3V}{2\pi }} && \text{Solve for }r. \end{align}[/latex]

This is the result stated in the section opener. Now evaluate this for V = 100 and [latex]\pi =3.14[/latex].

[latex]\begin{align}r & =\sqrt[3]{\frac{3V}{2\pi }} \\ & =\sqrt[3]{\frac{3\cdot 100}{2\cdot 3.14}} \\ & \approx \sqrt[3]{47.7707} \\ & \approx 3.63 \end{align}[/latex]

Therefore, the radius is about 3.63 ft.

Determining the Domain of a Radical Function Composed with Other Functions

When radical functions are composed with other functions, determining domain can become more complicated.

Example 7: Finding the Domain of a Radical Function Composed with a Rational Function

Find the domain of the function [latex]f\left(x\right)=\sqrt{\frac{\left(x+2\right)\left(x - 3\right)}{\left(x - 1\right)}}[/latex].

Show Solution

Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where [latex]\frac{\left(x+2\right)\left(x - 3\right)}{\left(x - 1\right)}\ge 0[/latex]. The output of a rational function can change signs (change from positive to negative or vice versa) at x-intercepts and at vertical asymptotes. For this equation, the graph could change signs at x = –2, 1, and 3.

To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph.

Figure 9

This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a y-intercept at (0, 6).

From the y-intercept and x-intercept at x = –2, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.

From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function f(x) will be defined. f(x) has domain [latex]-2\le x<1\text{or}x\ge 3[/latex], or in interval notation, [latex]\left[-2,1\right)\cup \left[3,\infty \right)[/latex].

Finding Inverses of Rational Functions

As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.

Example 8: Finding the Inverse of a Rational Function

The function [latex]C=\frac{20+0.4n}{100+n}[/latex] represents the concentration C of an acid solution after n mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for n in terms of C. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.

Show Solution

We first want the inverse of the function. We will solve for n in terms of C.

[latex]\begin{gathered}C=\frac{20+0.4n}{100+n}\\ C\left(100+n\right)=20+0.4n\\ 100C+Cn=20+0.4n\\ 100C - 20=0.4n-Cn\\ 100C - 20=\left(0.4-C\right)n\\ n=\frac{100C - 20}{0.4-C}\end{gathered}[/latex]

Now evaluate this function for C=0.35 (35%).

[latex]\begin{align}n&=\frac{100\left(0.35\right)-20}{0.4 - 0.35}\\ &=\frac{15}{0.05}\\ &=300\end{align}[/latex]

We can conclude that 300 mL of the 40% solution should be added.

Try It

Find the inverse of the function [latex]f\left(x\right)=\frac{x+3}{x - 2}[/latex].

Show Solution

[latex]{f}^{-1}\left(x\right)=\frac{2x+3}{x - 1}[/latex]

Try It

Key Concepts

• If [latex]g\left(x\right)[/latex] is the inverse of [latex]f\left(x\right)[/latex], then
• [latex]g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)=x[/latex].
• 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 [latex]y=f\left(x\right)[/latex] for [latex]x[/latex] as a function of [latex]y[/latex]. Then exchange the labels [latex]x[/latex] and [latex]y[/latex].
• The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[/latex].
• The inverse of a quadratic function is a square root function.
• If [latex]{f}^{-1}[/latex] is the inverse of a function f, then f is the inverse of the function [latex]{f}^{-1}[/latex].
• While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible.
• To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.
• When finding the inverse of a radical function, we need a restriction on the domain of the answer.
• Inverse and radical and functions can be used to solve application problems.

Glossary

inverse function

for any one-to-one function [latex]f\left(x\right)[/latex], the inverse is a function [latex]{f}^{-1}\left(x\right)[/latex] such that [latex]{f}^{-1}\left(f\left(x\right)\right)=x[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex]; this also implies that [latex]f\left({f}^{-1}\left(x\right)\right)=x[/latex] for all [latex]x[/latex] in the domain of [latex]{f}^{-1}[/latex]

invertible function

any function that has an inverse function