In Section FunctionsandtheirRepresentations, we defined functions as processes. In this section, we seek to reverse, or ‘undo’ those processes. As in real life, we will find that some processes (like putting on socks and shoes) are reversible while some (like baking a cake) are not.

Consider the function \(f(x) = 3x+4\). Starting with a real number input \(x\), we apply two steps in the following sequence: first we multiply the input by \(3\) and, second, we add \(4\) to the result.

To reverse this process, we seek a function \(g\) which will undo each of these steps and take the output from \(f\), \(3x+4\), and return the input \(x\). If we think of the two-step process of first putting on socks then putting on shoes, to reverse the process, we first take off the shoes and then we take off the socks. In much the same way, the function \(g\) should undo each step of \(f\) but in the opposite order. That is, the function \(g\) should first subtract \(4\) from the input \(x\) then divide the result by \(3\). This leads us to the formula \(g(x) = \frac {x-4}{3}\).

Let’s check to see if the function \(g\) does the job. If \(x=5\), then \(f(5) = 3(5)+4 = 15+4 = 19\). Taking the output \(19\) from \(f\), we substitute it into \(g\) to get \(g(19) = \frac {19-4}{3} = \frac {15}{3} = 5\), which is our original input to \(f\). To check that \(g\) does the job for all \(x\) in the domain of \(f\), we take the generic output from \(f\), \(f(x) = 3x+4\), and substitute that into \(g\). That is, we simplify \(g(f(x)) = g(3x+4) = \frac {(3x+4)-4}{3} = \frac {3x}{3} = x\), which is our original input to \(f\). If we carefully examine the arithmetic as we simplify \(g(f(x))\), we actually see \(g\) first ‘undoing’ the addition of \(4\), and then ‘undoing’ the multiplication by \(3\).

Not only does \(g\) undo \(f\), but \(f\) also undoes \(g\). That is, if we take the output from \(g\), \(g(x) = \frac {x-4}{3}\), and substitute that into \(f\), we get \(f(g(x)) = f\left (\frac {x-4}{3}\right ) = 3 \left (\frac {x-4}{3}\right ) + 4 = (x-4) + 4 = x\). Using the language of function composition developed in Section FunctionComposition, the statements \(g(f(x)) = x\) and \(f(g(x)) = x\) can be written as \((g \circ f)(x) = x\) and \((f \circ g)(x) = x\), respectively. Abstractly, we can visualize the relationship between \(f\) and \(g\) in the diagram below.

The main idea to get from the diagram is that \(g\) takes the outputs from \(f\) and returns them to their respective inputs, and conversely, \(f\) takes outputs from \(g\) and returns them to their respective inputs. We now have enough background to state the central definition of the section.

If we abstract one step further, we can express the sentiment in Definition inversefunctiondefn by saying that \(f\) and \(g\) are inverses if and only if \(g \circ f = I_{1}\) and \(f \circ g = I_{2}\) where \(I_{1}\) is the identity function restricted to the domain of \(f\) and \(I_{2}\) is the identity function restricted to the domain of \(g\).

In other words, \(I_{1}(x) = x\) for all \(x\) in the domain of \(f\) and \(I_{2}(x) = x\) for all \(x\) in the domain of \(g\). Using this description of inverses along with the properties of function composition listed in Theorem functioncompprops, we can show that function inverses are unique.

Suppose \(g\) and \(h\) are both inverses of a function \(f\). By Theorem inversefunctionprops, the domain of \(g\) is equal to the domain of \(h\), since both are the range of \(f\). This means the identity function \(I_{2}\) applies both to the domain of \(h\) and the domain of \(g\). Thus \(h = h \circ I_{2} = h \circ (f \circ g) = (h \circ f) \circ g = I_{1} \circ g = g\), as required.

We summarize the important properties of invertible functions in the following theorem. Apart from introducing notation, each of the results below are immediate consequences of the idea that inverse functions map the outputs from a function \(f\) back to their corresponding inputs.

The notation \(f^{-1}\) is an unfortunate choice since you’ve been programmed since Elementary Algebra to think of this as \(\frac {1}{f}\). This is most definitely not the case since, for instance, \(f(x) = 3x+4\) has as its inverse \(f^{-1}(x) = \frac {x-4}{3}\), which is certainly different than \(\frac {1}{f(x)} = \frac {1}{3x+4}\).

Why does this confusing notation persist? As we mentioned in Section FunctionComposition, the identity function \(I\) is to function composition what the real number \(1\) is to real number multiplication. The choice of notation \(f^{-1}\) alludes to the property that \(f^{-1} \circ f = I_{1}\) and \(f \circ f^{-1} = I_{2}\), in much the same way as \(3^{-1} \cdot 3 = 1\) and \(3 \cdot 3^{-1} = 1\).

Before we embark on an example, we demonstrate the pertinent parts of Theorem inversefunctionprops to the inverse pair \(f(x) = 3x+4\) and \(g(x) = f^{-1}(x) = \frac {x-4}{3}\). Suppose we wanted to solve \(3x+4 = 7\). Going through the usual machinations, we obtain \(x = 1\). If we view this equation as \(f(x) = 7\), however, then we are looking for the input \(x\) corresponding to the output \(f(x) = 7\). This is exactly the question \(f^{-1}\) was built to answer. In other words, the solution to \(f(x) = 7\) is \(x = f^{-1}(7) =1\). In other words, the formula \(f^{-1}(x)\) encodes all of the algebra required to ‘undo’ what the formula \(f(x)\) does to \(x\). More generally, any time you have ever solved an equation, you have really been working through an inverse problem.

We also note the graphs of \(f(x) = 3x+4\) and \(g(x) = f^{-1}(x) = \frac {x-4}{3}\) are easily seen to be reflections across the line \(y=x\) as seen below. In particular, note that the \(y\)-intercept \((0,4)\) on the graph of \(y = f(x)\) corresponds to the \(x\)-intercept on the graph of \(y = f^{-1}(x)\). Indeed, the point \((0,4)\) on the graph of \(y = f(x)\) can be interpreted as \((0,4) = (0,f(0)) = (f^{-1}(4), 4)\) just as the point \((4,0)\) on the graph of \(y = f^{-1}(x)\) can be interpreted as \((4,0) = (4, f^{-1}(4)) = (f(0), 0)\).

We now investigate under what circumstances a function is invertible. As a way to motivate the discussion, we consider \(f(x) = x^2\). A likely candidate for the inverse is the function \(g(x) = \sqrt {x}\). However, \((g\circ f)(x) = g(f(x)) = \sqrt {x^2} = |x|\), which is not equal to \(x\) unless \(x \geq 0\).

For example, when \(x=-2\), \(f(-2)= (-2)^2 = 4\), but \(g(4) = \sqrt {4}=2\). That is, \(g\) failed to return the input \(-2\) from its output \(4\). Instead, \(g\) matches the output \(4\) to a different input, namely \(2\), which satisfies \(f(2) = 4\). Schematically:

We see from the diagram that since both \(f(-2)\) and \(f(2)\) are \(4\), it is impossible to construct a function which takes \(4\) back to both \(x=2\) and \(x=-2\) since, by definition, a function can match \(4\) with only one number.

In general, in order for a function to be invertible, each output can come from only one input. Since, by definition, a function matches up each input to only one output, invertible functions have the property that they match one input to one output and vice-versa. We formalize this concept below.

Note that an equivalent way to state Definition onetoone is that a function is one-to-one if different inputs go to different outputs. That is, if \(a \neq b\), then \(f(a) \neq f(b)\).

Before we solidify the connection between invertible functions and one-to-one functions, we take a moment to see what goes wrong graphically when trying to find the inverse of \(f(x) = x^2\).

Per Theorem inversefunctionprops, the graph of \(y = f^{-1}(x)\), if it exists, is obtained from the graph of \(y=x^2\) by reflecting \(y=x^2\) about the line \(y=x\). Procedurally, this is accomplished by interchanging the \(x\) and \(y\) coordinates of each point on the graph of \(y = x^2\). Algebraically, we are swapping the variables ‘\(x\)’ and ‘\(y\)’ which results in the equation \(x = y^2\). Using desmos, we graph both \(y=x^2\) and \(x = y^2\) below.

We see immediately the graph of \(x = y^2\) fails the Vertical Line Test, Theorem VLT. In particular, the vertical line \(x=4\) intersects the graph at two points, \((4,-2)\) and \((4,2)\) meaning the relation described by \(x = y^2\) matches the \(x\)-value \(4\) with two different \(y\)-values, \(-2\) and \(2\).

Note that the vertical line \(x=4\) and the points \((4,2)\) and \((4, -2)\) on the graph of \(x=y^2\) correspond to the horizontal line \(y=4\) and the points \((2, 4)\) and \((-2,4)\) on the graph of \(y = x^2\). This brings us right back to the concept of one-to-one. The fact that both \((-2,4)\) and \((2,4)\) are on the graph of \(f\) means \(f(-2)=f(2) = 4\). Hence, \(f\) takes different inputs, \(-2\) and \(2\), to the same output, \(4\), so \(f\) is not one-to-one.

Recall the Horizontal Line Test from Exercise HLTExercise in Section FunctionsandtheirRepresentations. Applying that result to the graph of \(f\) we say the graph of \(f\) ‘fails’ the Horizontal Line Test since the horizontal line \(y=4\) intersects the graph of \(y = x^2\) more than once. This means that the equation \(y=x^2\) does not represent \(x\) is not a function of \(y\).

Said differently, the Horizontal Line Test detects when there is at least one \(y\)-value (\(4\)) which is matched to more than one \(x\)-value (\(\pm 2\)). In other words, the Horizontal Line Test can be used to detect whether or not a function is one-to-one.

So, to review, \(f(x) = x^2\) is not invertible, not one-to-one, and its graph fails the Horizontal Line Test. It turns out that these three attributes: being invertible, one-to-one, and having a graph that passes the Horizontal Line Test are mathematically equivalent. That is to say if one if these things is true about a function, then they all are; it also means that, as in this case, if one of these things isn’t true about a function, then none of them are. We summarize this result in the following theorem.

To prove Theorem inversefunctionequivalency, we first suppose \(f\) is invertible. Then there is a function \(g\) so that \(g(f(x)) = x\) for all \(x\) in the domain of \(f\). If \(f(a) = f(b)\), then \(g(f(a)) = g(f(b))\). Since \(g(f(x)) = x\), the equation \(g(f(a)) = g(f(b))\) reduces to \(a = b\). We’ve shown that if \(f(a) = f(b)\), then \(a = b\), proving \(f\) is one-to-one.

Next, assume \(f\) is one-to-one. Suppose a horizontal line \(y=c\) intersects the graph of \(y = f(x)\) at the points \((a,c)\) and \((b,c)\). This means \(f(a) = c\) and \(f(b) = c\) so \(f(a) = f(b)\). Since \(f\) is one-to-one, this means \(a=b\) so the points \((a,c)\) and \((b,c)\) are actually one in the same. This establishes that each horizontal line can intersect the graph of \(f\) at most once, so the graph of \(f\) passes the Horizontal Line Test.

Last, but not least, suppose the graph of \(f\) passes the Horizontal Line Test. Let \(c\) be a real number in the range of \(f\). Then the horizontal line \(y=c\) intersects the graph of \(y=f(x)\) just once, say at the point \((a,c) = (a, f(a))\). Define the mapping \(g\) so that \(g(c) = g(f(a)) = a\). The mapping \(g\) is a function since each horizontal line \(y=c\) where \(c\) is in the range of \(f\) intersects the graph of \(f\) only once. By construction, we have the domain of \(g\) is the range of \(f\) and that for all \(x\) in the domain of \(f\),\(g(f(x)) = x\). We leave it to the reader to show that for all \(x\) in the domain of \(g\), \(f(g(x)) = x\), too.

Hence, we’ve shown: first, if \(f\) invertible, then \(f\) is one-to-one; second, if \(f\) is one-to-one, then the graph of \(f\) passes the Horizontal Line Test; and third, if \(f\) passes the Horizontal Line Test, then \(f\) is invertible. Hence if \(f\) is satisfies any one of these three conditions, we can show \(f\) must satisfy the other two.

We put this result to work in the next example.

In Example inversefunctiononetooneex, we showed the functions \(G\) and \(g\) are invertible and graphed their inverses. While graphs are perfectly fine representations of functions, we have seen where they aren’t the most accurate. Ideally, we would like to represent \(G^{-1}\) and \(g^{-1}\) in the same manner in which \(G\) and \(g\) are presented to us. The key to doing this is to recall that inverse functions take outputs back to their associated inputs.

Consider \(G = \{ (t^3+1, 2t) \, | \, \text {$t$ is a real number.} \}\). As mentioned in Example inversefunctiononetooneex, the ordered pairs which comprise \(G\) are in the form \((\text {input}, \text {output})\). Hence to find a compatible description for \(G^{-1}\), we simply interchange the expressions in each of the coordinates to obtain \(G^{-1} = \{ (2t, t^3+1) \, | \, \text {$t$ is a real number.} \}\).

Since the function \(g\) was defined in terms of a formula we would like to find a formula representation for \(g^{-1}\). We apply the same logic as above. Here, the input, represented by the independent variable \(t\), and the output, represented by the dependent variable \(y\), are related by the equation \(y = g(t)\). Hence, to exchange inputs and outputs, we interchange the ‘\(t\)’ and ‘\(y\)’ variables. Doing so, we obtain the equation \(t = g(y)\) which is an implicit description for \(g^{-1}\). Solving for \(y\) gives an explicit formula for \(g^{-1}\), namely \(y = g^{-1}(t)\). We demonstrate this technique below.

\[ \begin{array}{rclr} y & = & g(t) & \\ y & = & \frac {2t}{1-t} & \\ t & = & \frac {2y}{1-y} & \text {interchange variables: $t$ and $y$} \\ t(1-y) & = & 2y & \\ t-ty & = & 2y & \\ t & = & ty + 2y & \\ t & = & y(t+2) & \text {factor}\\ y & = & \frac {t}{t+2} \end{array} \]

We claim \(g^{-1}(t) = \frac {t}{t+2}\), and leave the algebraic verification of this to the reader.

We generalize this approach below. As always, we resort to the default ‘\(x\)’ and ‘\(y\)’ labels for the independent and dependent variables, respectively.

Steps for finding a formula for the Inverse of a one-to-one function
  1. Write \(y=f(x)\)
  2. Interchange \(x\) and \(y\)
  3. Solve \(x = f(y)\) for \(y\) to obtain \(y=f^{-1}(x)\)

We now return to \(f(x) = x^2\). We know that \(f\) is not one-to-one, and thus, is not invertible, but our goal here is to see what way to see what goes wrong algebraically.

If we attempt to follow the algorithm above to find a formula for \(f^{-1}(x)\), we start with the equation \(y=x^2\) and interchange the variables ‘\(x\)’ and ‘\(y\)’ to produce the equation \(x = y^2\). Solving for \(y\) gives \(y = \pm \sqrt {x}\). It’s this ‘\(\pm \)’ which is causing the problem for us since this produces two \(y\)-values for any \(x>0\).

Using the language of Section Relations, the equation \(x = y^2\) implicitly defines two functions, \(g_{1}(x) = \sqrt {x}\) and \(g_{2}(x) = -\sqrt {x}\). Graphically, each of the functions \(g_{1}\) and \(g_{2}\) correspond to the top and bottom halves, respectively, of the graph of \(x = y^2\), as seen below.

Hence, in some sense, we have two partial inverses for \(f(x) = x^2\): \(g_{1}(x) = \sqrt {x}\) returns the positive inputs from \(f\) and \(g_{2}(x) = -\sqrt {x}\) returns the negative inputs to \(f\). In order to view each of these functions as strict inverses, however, we need to split \(f\) into two parts: \(f_{1}(x) = x^2\) for \(x \geq 0\) and \(f_{2}(x) = x^2\) for \(x \leq 0\). Graphically:

We claim that \(f_{1}\) and \(g_{1}\) are an inverse function pair as are \(f_{2}\) and \(g_{2}\). Indeed, we find that for \(g_{1} \circ f_{1}\):

\(\begin{array}{rcl} (g_{1} \circ f_{1})(x) & = & g_{1}(f_{1}(x)) \\ & = & g_{1}(x^2) \\ & = & \sqrt {x^2} \\ & = & |x| = x, \, \text {as $x \geq 0$.} \\ \end{array}\)

Likewise, for \(f_{1} \circ g_{1}\):

\(\begin{array}{rcl} (f_{1} \circ g_{1})(x) & = & f_{1}(g_{1}(x)) \\ & = & f_{1}(\sqrt {x}) \\ & = & (\sqrt {x})^2 \\ & = & x \\ \end{array}\)

Graphically, we see the graphs of \(f_{1}\) and \(g_{1}\) appear to be reflections across the line \(y =x\).

Likewise, for \(g_{2} \circ f_{2}\), we get:

\(\begin{array}{rcl} (g_{2} \circ f_{2})(x) & = & g_{2}(f_{2}(x)) \\ & = & g_{2}(x^2) \\ & = & -\sqrt {x^2} \\ & = & - |x| \\ & = & -(-x) = x, \, \text {as $x \leq 0$.} \end{array}\)

Lastly, we simplify \((f_{2} \circ g_{2})(x)\):

\(\begin{array}{rcl} (f_{2} \circ g_{2})(x) & = & f_{2}(g_{2}(x)) \\ & = & f_{2}(-\sqrt {x}) \\ & = & (-\sqrt {x})^2 \\ & = & (\sqrt {x})^2 \\ & = & x \\ \end{array}\)

Graphically:

Hence, by restricting the domain of \(f\) we are able to produce invertible functions. Said differently, in much the same way the equation \(x = y^2\) implicitly describes a pair of functions, the equation \(y = x^2\) implicitly describes a pair of invertible functions.

Our next example continues the theme of restricting the domain of a function to find inverse functions.

Our last example of the section gives an application of inverse functions. Recall in Example PortaBoyDemand in Section LinearFunctions, we modeled the demand for PortaBoy game systems as the price per system, \(p(x)\) as a function of the number of systems sold, \(x\). In the following example, we find \(p^{-1}(x)\) and interpret what it means.