2.7: Inverse Functions

Note to the Instructor (click to expand) We begin this chapter with defining and finding the inverse of a function. This is a traditional treatment of inverses, including how to graph them and how to find their equations given the original function. We also include domain restriction (as we are about to encounter radical functions). To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics. Prerequisite Skills and Support Topics (click to expand) Arithmetic Mathematical Operations - Higher-Index Roots: Finding inverses of functions like \(f(x)=5x^3+1\) involves using cube roots, as seen in Example 2 where \(r = \sqrt[3]{\frac{3V}{2\pi}}\) and \(f^{-1}(x) = \sqrt[3]{\frac{x-1}{5}}\). Understanding higher-index roots is crucial. Graphing Functions Graphs of Functions: The ability to interpret points on a graph and to understand that the graph of an inverse function is a reflection of the original function's graph across the line \(y=x\) is key (Example 6, Example 10, and several homework questions). The following is a list of learning objectives for this section. Learning Objectives (click to expand) 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. Restrict the domain to find the inverse of a polynomial function. Find the inverse of a rational function.

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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\[ C = \dfrac{5}{9} (F−32) \nonumber \]and substitutes 75 for \(F\) to calculate\[ \dfrac{5}{9} (75−32) \approx 24^{\circ} \, \text{C}. \nonumber \]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.

Figure \( \PageIndex{ 2 } \)
 

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\[ \begin{array}{rrcl}
& 26 & = & \dfrac{5}{9} (F−32) \\[6pt]
\implies & 26 \cdot \dfrac{9}{5} & = & F−32 \\[6pt]
\implies & F & = & 26 \cdot \dfrac{9}{5} +32 \approx 79 \\[6pt]
\end{array} \nonumber \]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.

Definition: Inverse Function

For any one-to-one function \(y = f(x)\), 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 \(f\). The notation \(f^{-1}\) is read "\(f\) inverse."

Caution

It is important to note that\[f^{−1}(x) \neq \dfrac{1}{f(x)}, \nonumber \]and not all functions have inverses.

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,\[( f^{-1} \circ f )(x) = f^{-1} \left( f(x) \right) =  x. \nonumber \]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.

Inverse Function Notation

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 \(x\)."

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.)

Theorem

If \(f^{-1}\) is the inverse of \(f\), then \(f( f^{-1}(x))=x\) for all \(x\) in the domain of \(f^{-1}\).

For example, \(y=4x\) and \(y= \frac{1}{4} x\) are inverse functions.\[( f^{-1} \circ f )(x)= f^{-1} ( 4x )= \dfrac{1}{4} ( 4x )=x \nonumber \]and\[( f \circ f^{-1} )(x)=f\left(\dfrac{1}{4}x \right)=4\left( \dfrac{1}{4}\right)=x. \nonumber \]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.

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 \( f(2)=4\), then \(f^{-1} (4)=2\); If \(f( 5 )=12\), then \(f^(-1) ( 12 )=5\). Alternatively, if we want to name the inverse function \(g\), then \(g(4)=2\) and \(g(12)=5\).

If we show the coordinate pairs from Example \( \PageIndex{ 1 } \) in a table form, the input and output are clearly reversed. See Table \( \PageIndex{ 1 } \).

Table \( \PageIndex{ 1 } \)
 

\(( x,f(x) )\)	\(( x,g(x) )\)
\(( 2,4 )\)	\(( 4,2 )\)
\(( 5,12 )\)	\(( 12,5 )\)

Checkpoint \( \PageIndex{ 1 } \)

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

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

To determine if \( g \) is the inverse function of \( f \), we just need to check if \( g \) "undoes" \( f \).\[ \begin{array}{rclcl}
g(f(x)) & = & g\left( \dfrac{1}{x + 2} \right) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{1}{\frac{1}{x + 2}} - 2 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & x + 2 - 2 & \quad & \left( \text{recognizing that }\dfrac{1}{1/a} = a \right) \\[6pt]
& = & x & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]so\[g= f^{-1} \text{ and }f= g^{-1} \nonumber \]This is enough to answer yes to the question, but we can also verify the other formula.\[ \begin{array}{rclcl}
f(g(x)) & = & f\left( \dfrac{1}{x} - 2 \right) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{1}{\left(\frac{1}{x} + 2\right) - 2} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{1}{\frac{1}{x}} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & x & \quad & \left( \text{recognizing that }\dfrac{1}{1/a} = a \right) \\[6pt]
\end{array} \nonumber \]

Checkpoint \( \PageIndex{ 2 } \)

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

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

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

Solution

\[f( g( x ) )= f\left( \dfrac{x}{3} \right) = \left( \dfrac{x}{3} \right)^3 = \dfrac{x^3}{27} \neq x. \nonumber \]No, the functions are not inverses. The correct inverse to the cube is, of course, the cube root \(\sqrt[3]{x} = x^{1/3} \), that is, the one-third is an exponent, not a multiplier.

Checkpoint \( \PageIndex{ 3 } \)

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

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
 

Theorem: Domain and Range of Inverse Functions

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

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 \([ 0, \infty )\), 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 domain limited to \([ 0, \infty )\), 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 \([ 1, \infty )\), then the inverse function is \(f^{-1} (x)=\sqrt{ x } +1\).

• The domain of \(f\) = range of \(f^{-1}\) = \([ 1, \infty )\).
• The domain of \(f^{-1}\) = range of \(f\) = \([ 0, \infty )\).

Q&A

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.

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 \( \PageIndex{ 2 } \)
 

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

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 \([ 0, \infty )\), where it is equal to the identity function.

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

We can see that the absolute value and reciprocal-squared 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
 

Checkpoint \( \PageIndex{ 4 } \)

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.

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 \(f^{-1} (70)\).

Table \( \PageIndex{ 3 } \)
 

\(t\text{ (minutes)}\)	30	50	70	90
\(f( t )\text{ (miles)}\)	20	40	60	70

Solution

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 \(t=90\).

Checkpoint \( \PageIndex{ 5 } \)

Using Table \( \PageIndex{ 4 } \), find and interpret each of the following:

1. \(f(60)\)
2. \(f^{-1} (60)\)

Table \( \PageIndex{ 4 } \)
 

\(t\text{ (minutes)}\)	30	50	60	70	90
\(f( t )\text{ (miles)}\)	20	40	50	60	70

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

We know 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.

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 \(g^{-1} (3)\).

Figure \( \PageIndex{ 5 } \)
 

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 \( \PageIndex{ 6 } \).

Figure \( \PageIndex{ 6 } \)
 

Checkpoint \( \PageIndex{ 6 } \)

Using the graph in Figure \( \PageIndex{ 6 } \) to answer the following questions.

1. Find \(g^{-1} (1)\).
2. Estimate \(g^{-1} (4)\).

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: Find the Equation of an Invertible Function

Given a function, \( f(x) \), we find the inverse (if it exists) by doing the following:

1. Let \( y = f(x) \).
2. Solve for \( x \).
3. Interchange \( x \) and \( y \).

The result of these operations, if it is a function, is \( f^{-1}(x) \).

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,\[C= \dfrac{5}{9} (F−32). \nonumber \]

Solution

\[ \begin{array}{rrclcl}
& C & = & \dfrac{5}{9}(F - 32) & & \\[6pt]
\implies & \dfrac{9}{5} C & = & F - 32 & \quad & \left( \text{multiplying both sides by }\dfrac{9}{5} \right) \\[6pt]
\implies & \dfrac{9}{5} C + 32 & = & F & \quad & \left( \text{adding }32 \text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]By solving in general, we have uncovered the inverse function. If\[C=h(F)= \dfrac{5}{9} (F−32), \nonumber \]then\[F= h^{-1} (C)= \dfrac{9}{5} C+32. \nonumber \]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.

Checkpoint \( \PageIndex{ 7 } \)

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

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

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

Solution

\[ \begin{array}{rrclcl}
& y & = & \dfrac{2}{x - 3} + 4 & \quad & \left( \text{starting with the original equation} \right) \\[6pt]
\implies & y - 4 & = & \dfrac{2}{x - 3} & \quad & \left( \text{subtracting }4\text{ from both sides} \right) \\[6pt]
\implies & (x - 3)(y - 4) & = & 2 & \quad & \left( \text{multiplying both sides by }(x - 3) \right) \\[6pt]
\implies & x - 3 & = & \dfrac{2}{y - 4} & \quad & \left( \text{dividing both sides by }(y - 4) \right) \\[6pt]
\implies & x & = & \dfrac{2}{y - 4} + 3 & \quad & \left( \text{adding }3\text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]So \(f^{-1} ( y )= \frac{2}{y−4} +3\) or \(f^{-1} ( x )= \frac{2}{x−4} +3\).

Note that the domain and range of \(f\) in Example \( \PageIndex{ 8 } \) 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 \( \PageIndex{ 5 } \).

Table \( \PageIndex{ 5 } \)
 

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

\[ \begin{array}{rrclcl}
& y & = & 2 + \sqrt{x - 4} & \quad & \left( \text{starting with the original equation} \right) \\[6pt]
\implies & y - 2 & = & \sqrt{x - 4} & \quad & \left( \text{subtracting }2\text{ from both sides} \right) \\[6pt]
\implies & \left(y - 2\right)^2 & = & x - 4 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & \left(y - 2\right)^2 + 4 & = & x & \quad & \left( \text{adding }4\text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]So \(f^{-1} ( x )= ( x−2 )^2 +4\).

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

The formula we found for \(f^{-1}( x )\) in Example \( \PageIndex{ 9 } \) 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 \([2, \infty )\) in order to make \(f^{-1}\) a one-to-one function. This domain of \(f^{-1}\) is exactly the range of \(f\).

Checkpoint \( \PageIndex{ 9 } \)

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

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 \([0, \infty )\), 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 \([0, \infty )\) 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 \(f^{-1} (x)\).

Figure \( \PageIndex{ 9 } \)
 

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 \(( 0, \infty )\) 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 \( \PageIndex{ 10 } \).

Figure \( \PageIndex{ 10 } \)

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

Checkpoint \( \PageIndex{ 10 } \)

Draw graphs of the functions \(f\) and \(f^{-1}\) from Example \( \PageIndex{ 8 } \).

Q&A

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\[\dfrac{1}{\frac{1}{x}} =x. \nonumber \]Any function \(f( x )=c−x\), where \(c\) is a constant, is also equal to its own inverse.

Restricting 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.

Theorem: Forcing Invertibility by Restricting Domain

If we restrict the domain of a non-invertible function so that the resulting function becomes one-to-one, then this new function will have an inverse.

How To: Restrict Domain and Force Invertibility

Given a function that is not one-to-one,

1. Restrict the domain by determining a domain on which the original function is one-to-one (i.e., a domain where the function passes the Horizontal Line Test).
2. Replace \(f(x)\) with \(y\).
3. Solve for \(x\) interchange \( x \) and \( y\).
4. Rename the function \(f^{-1} (x)\).
5. Revise the formula for \(f^{-1} (x)\) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.

Example \( \PageIndex{ 11} \): Restricting the Domain to Find the Inverse of a Polynomial Function

Find the inverse function of \(f:\)

1. \(f(x)= (x-4)^2, \, x \geq 4\)
2. \(f(x)= (x-4)^2, \, x \leq 4\)

Solution

The original function \(f(x)= (x-4)^2\) is not one-to-one, but the function is restricted to a domain of \(x \geq 4\) or \(x \leq 4\) on which it is one-to-one. See Figure \( \PageIndex{ 11 } \).

Figure \( \PageIndex{ 11 } \)
 

To find the inverse, start by replacing \(f(x)\) with the simple variable \(y\).\[ \begin{array}{rrclcl} & y & = & (x-4)^2 & \quad & \left( \text{letting }y = f(x) \right) \\[6pt] \implies & \pm \sqrt{ y } & = & x-4 & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \implies & 4 \pm \sqrt{ y } & = & x & \quad & \left( \text{adding }4 \text{ to both sides} \right) \\[6pt] \implies & 4 \pm \sqrt{x} & = & y & \quad & \left( \text{exchanging }y\text{ and }x \right) \\[6pt] \end{array} \nonumber \]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 need to look at the domain of the original function. When we reversed the roles of \(x\) and \(y\), this gave us the values \(y\) could assume. For this function, \(x \geq 4\), so for the inverse, we should have \(y \geq 4\), which is what our inverse function gives.

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

On the graphs in Figure \( \PageIndex{ 12 } \), we see the original function from Example \( \PageIndex{ 11} \) graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line \(y=x\). The coordinate pair \((4,0)\) is on the graph of \(f\) and the coordinate pair \((0,4)\) is on the graph of \(f^{-1} \). For any coordinate pair, if \(( a,b )\) is on the graph of \(f\), then \(( b,a )\) is on the graph of \(f^{-1} \). Finally, observe that the graph of \(f\) intersects the graph of \(f^{-1}\) on the line \(y=x\). Points of intersection for the graphs of \(f\) and \(f^{-1}\) will always lie on the line \(y=x\).

Figure \( \PageIndex{ 12 } \)
 

Example \( \PageIndex{ 12} \): Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified

Restrict the domain and then find the inverse of\[f(x)= (x-2)^2 -3. \nonumber \]

Solution

We can see this is a parabola with vertex at \((2,–3)\) 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 \(x \geq 2\).

To find the inverse, we repeat what we did in Example \( \PageIndex{ 11} \).\[ \begin{array}{rrclcl} & y & = & (x - 2)^2 - 3 & \quad & \left( \text{substituting }y = f(x) \right) \\[6pt] \implies & y + 3 & = & (x - 2)^2 & \quad & \left( \text{adding }3\text{ to both sides} \right) \\[6pt] \implies & \pm \sqrt{y + 3} & = & x - 2 & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \implies & 2 \pm \sqrt{y + 3} & = & x & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt] \implies & 2 \pm \sqrt{x + 3} & = & y & \quad & \left( \text{interchanging }x\text{ and }y \right) \\[6pt] \end{array} \nonumber \]Now we need to determine which case to use. Because we restricted our original function to a domain of \(x \geq 2\), the outputs of the inverse should be the same, telling us to utilize the "+" case. Therefore,\[f^{-1}(x)=2+\sqrt{ x+3 }. \nonumber \]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.

Notice that we arbitrarily decided to restrict the domain on \(x \geq 2\). We could just have easily opted to restrict the domain on \(x \leq 2\), in which case \(f^{-1}(x)=2-\sqrt{ x+3 } \). Observe the original function graphed on the same set of axes as its inverse function in Figure \( \PageIndex{ 13 } \). Notice that both graphs show symmetry about the line \(y=x\). The coordinate pair \(( 2,-3 )\) is on the graph of \(f\) and the coordinate pair \(( -3,2 )\) is on the graph of \(f^{-1} \). Observe from the graph of both functions on the same set of axes that\[\text{domain of }f=\text{range of } f^{-1} =[ 2, \infty ) \nonumber \]and\[\text{domain of } f^{-1} =\text{range of }f=[ –3, \infty ). \nonumber \]Finally, observe that the graph of \(f\) intersects the graph of \(f^{-1}\) along the line \(y=x\).

Figure \( \PageIndex{ 13 } \)
 

Checkpoint \( \PageIndex{ 12} \)

Find the inverse of the function \(f(x)= x^2 +1\), on the domain \(x \geq 0\).

Solving Applications Involving the Inverse of Polynomial 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 \( \PageIndex{ 13} \): 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\[V= \dfrac{2}{3} \pi r^3. \nonumber \]Find the inverse of this function. Then use the inverse function to calculate the radius of such a mound of gravel measuring \( 100 \) cubic feet. Use \(\pi \approx 3.14\).

Solution

Start with the given function for \(V\). Notice that the meaningful domain for the function is \(r \geq 0\) 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 \(V \geq 0\). Solve for \(r\) in terms of \(V\), using the method outlined previously.\[ \begin{array}{rrcl} & V & = & \dfrac{2}{3} \pi r^3 \\[6pt] \implies & \dfrac{3}{2} V & = & \pi r^3 \\[6pt] \implies & \dfrac{3}{2 \pi} V & = & r^3 \\[6pt] \implies & \sqrt[3]{\dfrac{3}{2 \pi} V} & = & r \\[6pt] \end{array} \nonumber \]This is the result stated in the section opener. Now evaluate this for \(V=100\) and \(\pi \approx 3.14\).\[ \begin{array}{rcl} r & = & \sqrt[3]{\dfrac{3V}{2\pi}} \\[6pt] & \approx & \sqrt[3]{\dfrac{3(100)}{2 (3.14)}} \\[6pt] & \approx & \sqrt[3]{47.7707} \\[6pt] & \approx & 3.63 \\[6pt] \end{array} \nonumber \]Therefore, the radius is about \( 3.63 \) ft.

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 \( \PageIndex{ 14} \): Finding the Inverse of a Rational Function

The function \(C = \frac{20+0.4n}{100+n}\) 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.

Solution

We first want the inverse of the function.\[ \begin{array}{rrcl} & C & = & \dfrac{20 + 0.4n}{100 + n} \\[6pt] \implies & C(100 + n) & = & 20 + 0.4n \\[6pt] \implies & 100C + Cn & = & 20 + 0.4n \\[6pt] \implies & 100C - 20 & = & 0.4n - Cn \\[6pt] \implies & 100C - 20 & = & (0.4 - C)n \\[6pt] \implies & \dfrac{100C - 20}{0.4 - C} & = & n \\[6pt] \end{array} \nonumber \]Now evaluate this function for \(C= 0.35\) (35%).\[ n = \dfrac{100(0.35)-20}{0.4-0.35} = \dfrac{15}{0.05} = 300. \nonumber \]We can conclude that 300 mL of the 40% solution should be added.

Checkpoint \( \PageIndex{ 14} \)

Find the inverse of the function \(f(x)= \dfrac{x+3}{x-2} \).