By Frank Wilson, Chandler-Gilbert Community College; Scott Adamson, Chandler-Gilbert Community College; Trey Cox, Chandler-Gilbert Community College; and Alan O’Bryan, Arizona State University
What would you do if you discovered a popular approach to teaching inverse functions negatively affected student understanding of the underlying ideas? Would you continue to teach the problematic procedure or would you search for a better way to help students make sense of the mathematics?
A popular approach to finding the inverse of a function is to switch the \(x\) and \( y\) variables and solve for the \(y\) variable. The strategy of swapping variables is not grounded in mathematical operations and, we will argue, is nonsensical. Nevertheless, the procedure is so ingrained in textbooks and other curricula that many teachers accept it as mathematical truth without questioning is conceptual validity. As a result, students try to memorize the strategy but struggle to “accurately carry out mathematical procedures, understand why those procedures work, and know how they might be used and their results interpreted” (NCTM, 2009; Carlson & Oehrtman, 2005). As we will illustrate, this common process for finding the inverse of a function makes it harder for students to understand fundamental inverse function concepts.
Foundational Ideas about Functions and Their Inverses
A function \(f\) describes the relationship between two covarying quantities represented by variables \(x\) and \(y\). Without loss of generality, let \(x\) be the independent variable for the function \(f\) and \(y\) be the dependent variable for the function \(f\). The inverse function \(f^{-1}\) also describes the relationship between the quantities represented by variables \(x\) and \(y\) except \(y\) is designated as the independent variable for the function \(f^{-1}\) and \(x\) is designated as the dependent variable for the function \(f^{-1}\).
The following properties hold:
Concept #1: The domain of a function \(f\) is the range of its inverse function \(f^{-1}\) and the range of the function \(f\) is the domain of its inverse function \(f^{-1}\) (Wilson, 2007).
Concept #2: \(f^{-1}(f(x))=x\). In layman’s terms, the inverse function undoes whatever the function does (Bayazit & Gray, 2004).
These two concepts form the foundational ideas of the inverse function concept and hold true for functions represented in equations, graphs, tables or words.
Problematic Conceptions Arising from the Switch x and y Approach to Finding Inverse Functions
We define a conception as “problematic” if it describes an understanding that obscures connections to related ideas, introduces mathematical inconsistencies, and/or is likely to hinder students from developing powerful meanings of future topics. There are two problematic conceptions that emerge from the switch \(x\) and \(y\) approach to finding the inverse of a function.
Problematic Conception #1: The inverse of \(y=f(x)\) is \(y=f^{-1}(x)\).
In this statement, the independent variable for both \(f\) and \(f^{-1}\) is \(x\) and the dependent variable for both functions is \(y\). This problematic conception develops out of the procedure of switching \(x\) and \(y\) to find the inverse of a function, as illustrated in the following example.
Given \(f(x)=86x+15\), find \(f^{-1}\). \[\begin{align*} f(x) &= 86x +15\\ y &= 86x +15\quad \text{since}\ y= f(x)\\ x&=86y+15\quad \textbf{switch x and y}\\ x-15 &=86y\\ y &= \frac{x-15}{86}\\ f^{-1}(x) &= \frac{x-15}{86} \end{align*}\]
To some educators, calling this statement a problematic conception may seem like heresy. However, it is easy see the conceptual problem when the variables are assigned real world meanings.
In 2016 – 2017, tuition at the Maricopa Community Colleges was \(\$86\) per credit hour. All students registering to take classes were also required to pay a \(\$15\) registration fee. The function \(y=f(x)\) where \(f(x)=86x+15\) (introduced earlier) relates the number of credit hours, \(x\), to the total tuition cost (including the registration fee), \(y\). For clarity and emphasis, we change the variables in this equation to \(c\), for the number of credit hours assigned, and to \(t\), for the total tuition cost in dollars. The resultant equation is \(t=f(c)\) where \(f(c)=86c+15\). No matter what we do to mathematically manipulate this equation, the meaning of the variables \(t\) and \(c\) will remain unchanged. Suppose we are asked to calculate and interpret the meaning of \(f^{-1} (445)\). Using the switch \(x\) and \(y\) approach, we concluded earlier that \(y=f^{-1} (x)\) where \(f^{-1}(x)=\frac{x-15}{86}\). In terms of \(c\) and \(t\) this is \(t=f^{-1} (c)\) where \( f^{-1} (c)=\frac{c-15}{86}\). So \[\begin{align*}f^{-1}(445) &= \frac{445-15}{86}\\ f^{-1}(445) &= \frac{430}{86}\\ f^{-1}(445) &= 5 \end{align*}\]
What is the meaning of the result? Since \(c\) is credits and \(t\) is tuition cost in dollars, the result must mean that \(445\) credits cost \(\$5\). This statement is false because credits cost \(\$86\) per credit hour! To make sense of \(t=f^{-1}(c)\), we would have to change the meaning of the variables \(c\) and \(t\).
The confusion is easily remedied by applying an alternate strategy to finding the inverse. The strategy of solve for the dependent variable is demonstrated in the following example. As stated earlier, \(t\) represents the total tuition cost in dollars and \(c\) represents the number of credit hours assigned. For the inverse function \(f^{-1}\), \(c\) is the dependent variable so we solve the equation for \(c\).
Given \(f(c)=86c+15\), find \(f^{-1}\).
\[\begin{align*} f(c) &= 86c+15\\ t &= 86c+15\quad \text{since}\ t=f(c)\\ t-15 &= 86c\\ c &=\frac{t-15}{86}\\ f^{-1} (t) &= \frac{t-15}{86} \end{align*}\]
Note that \(t\) is the independent variable and \(c\) is the dependent variable for the inverse function. \( f^{-1} (445)=5\) implies that when \(t=445\), \(c=5\). In other words, when the total tuition cost (including registration) is \(\$445\), then 5 credits are purchased. This statement is true.
By referring to basic inverse function concepts, we can also detect the fallacy in the statement, “The inverse of \(y=f(x)\) is \(y=f^{-1}(x)\).” Let \(x\) be the independent variable and \(y\) be the dependent variable of a function \(f\). Then \(y=f(x)\) . We know \[\begin{align*}f^{-1} (f(x)) &= x\quad \text{Concept 2}\\ f^{-1} (y) &= x\quad \text{since}\ y=f(x)\end{align*}\]
Notice that the independent variable for the inverse function \(f^{-1}\) is \(y\) and the dependent variable is \(x\). So the inverse of \(y=f(x)\) is \(x=f^{-1}(y)\) not \(y=f^{-1}(x)\).
The tuition example represents a traditional exercise where students focus only on a memorized procedure. Carlson and Oehrtman warn that “this procedural approach to determining ‘an answer’ has little or no real meaning for the student unless he or she also possesses an understanding as to why the procedure works (2005).” The conceptual weakness of the problematic approach to finding the inverse becomes clearly evident with functions representing real world contexts.
Keeping track of the meaning of variables is essential when working with exponential and logarithmic functions. Understanding that \(y=b^x\) is equivalent to \(\log_b y = x\) is key to understanding logarithms conceptually. The switch \(x \) and \(y\) approach to finding inverses obscures the inverse relationship between exponential and logarithmic functions. For example, suppose that \(f(x)=3^x\). Find \(f^{-1}\).
\[\begin{align*} \textit{Switch x}\ & \textit{and y}\ \text{approach} & \textit{Solve for the}\ & \textit{dependent variable}\ \text{approach} \\ f(x) &= 3^x & f(x) &= 3^x\\ y &= 3^x & y &= 3^x\\ x& =3^y\quad \text{switch}\ x\ \text{and}\ y & \log_3 y &= x\\ \log_3 x &= y & f^{-1}(y) &= \log_3 y \\ f^{-1}(x) &= \log_3 x & & \\ \end{align*}\]
Using the switch \(x\) and \(y\) approach, it is common for students to conclude incorrectly that \(\log_3 x=3^x\) because of the statements \(\log_3 x=y\) and \(y=3^x\) included as part of the problem solving process. No such confusion exists when the solve for the dependent variable approach is used.
Problematic Conception #2: With the horizontal axis representing the independent variable and the vertical axis representing the dependent variable, the graphs of \(f\) and \(f^{-1}\) may be drawn on the same axes. The resultant graphs are symmetric about the line \(y=x\).
It is true that the graphs of \(y=f(x)\) and \(y=f^{-1} (x)\) are symmetric about the line \(y=x\) but, as established earlier, there are inherent issues with saying that \(y=f^{-1} (x)\) is the inverse function of \(y=f(x)\). The result \(y=f^{-1} (x)\) comes from switching the \(x\) and \(y\) variables in the inverse function. In fact, switching the variables in any mathematical relation will create a graph that is symmetric about the line \(y=x\). The practice of graphing \(f(x)\) and \(f^{-1}(x)\) on the same axes should be avoided (VanDyke, 1996) because it muddles the concept of inverse. Instead \(f(x)\) and \(f^{-1}(y)\) should be graphed on separate axes labeled appropriately with \(x\) or \(y\) on the horizontal axis.
The conceptual problems which occur when graphing \(f(x)\) and \(f^{-1}(x)\) on the same axes are evident when modeling even the simplest real-world context. The weekly earnings, \(y\), of an employee earning \(\$10\) per hour who works \(x\) hours in a week is given by \(y=10x\). The independent variable for the function \(f\) is \(x\) and the dependent variable is \(y\). For the inverse function \(f^{-1}\), the dependent variable is \(x\) so we solve \(y=10x\) for \(x\) and get \(x=\frac{1}{10} y\). We have \(y=f(x)\) with \(f(x)=10x\) and \(x=f^{-1} (y)\) with \(f^{-1} (y)=\frac{1}{10} y\). If we switch the \(x\) and \(y\) variables in the inverse function equation, we get \(y=f^{-1} (x)\) with \(f^{-1} (x)=\frac{1}{10} x\) . We graph \(f(x)\) and \(f^{-1} (x)\) on the same axes and label the axes with the variables \(x\) and \(y\) as is customary. We include the units associated with the variables \(x\) and \(y\).
Graphing \(y=f(x)\) and \(y=f^{-1}(x)\) on the same set of axes.
From the graph, we see that \(f^{-1} (20)=2\). The \(x\)-axis is labeled hours worked weekly and the \(y\) axis is labeled weekly earnings (dollars) so this must mean that when the employee works \(20\) hours the employee earns \(\$2\). But this doesn’t make sense because we know the employee makes \(\$10\) per hour! We could remove the labels from the axes, but this does not help someone understand a function’s graph as a visual representation of a relationship between two quantities and is likely to make it even harder to comprehend the meaning of a point on the graph. Graphing \(y=f(x)\) and \(y=f^{-1} (x)\) on the same axes created confusion and did nothing to help us understand inverse functions.
There are two equally viable strategies for representing functions and their inverses graphically. The first strategy is to graph each function on its own pair of coordinate axes with the horizontal axis representing the independent variable and the vertical axis representing the dependent variable of the function.
Graphing \(y=f(x)\) and \(x=f^{-1}(y)\) on separate axes
From the graph of \(f\), we determine that \(f(2)=20\) means that working 2 hours weekly results in weekly earnings of \(\$20\). From the graph of \(f^{-1}\), we determine that \(f^{-1} (20)=2\) means that when weekly earnings were \(\$20\) the number of hours worked was \(2\). Both results make sense in the real-world context.
The second strategy for graphing a function and its inverse comes from changing the way we think about graphs. With this approach, we use the same graph to represent a function and its inverse but designate the horizontal axis to represent the independent variable for \(f\) and the vertical axis to represent the independent variable for \(f^{-1}\) (Moore, Liss, Silverman, Paoletti, LaForest, & Musgrave, 2013). Observe that to determine \(f(2)\) we start at \(x=2\) on the horizontal axis and move vertically until we touch the graph of \(f\). We then move horizontally until we touch the vertical axis at \(y=20\). We conclude \(f(2)=20\). To determine \(f^{-1} (20)\) we start at \(y=20\) on the vertical axis and move horizontally until we touch the graph of \(f^{-1}\). We then move vertically until we touch the horizontal axis at \(x=2\). We conclude \(f^{-1} (20)=2\).
Using the same graph to represent a function and its inverse but designating the horizontal axis to represent the independent variable for \(f\) and the vertical axis to represent the independent variable for \(f^{-1}\)
This way of thinking can be powerful for students who recognize the equation \(f(x)=30\) is equivalent to \(x=f^{-1} (30)\). The student finds \(30\) on the vertical axis and determines the corresponding value on the horizontal axis is \(3\). The student concludes that the solution to \(f(x)=30\) is \(x=3\) because \(f^{-1} (30)=3\).
Bayazit and Gray (2004) claim that learners with a conceptual understanding of inverse functions were able to deal with the inverse function concept in situations not involving formulas whereas learners limited by a procedural understanding of inverse functions (e.g. switch \(x\) and \(y\)) were less likely to be successful in a context without a formula.
A side benefit of discarding the switch \(x\) and \(y\) approach is that it frees learners from the \(x\)-addiction – the notion that only \(x\) can be the independent variable. In graphing, the \(x\)-axis becomes the horizontal axis and the \(y\)-axis becomes the vertical axis. The reality is that disciplines outside of mathematics rarely use \(x\) to represent the horizontal axis and \(y\) to represent the vertical axis. Rather, they use variable names (perhaps even complete words) that make sense in the context of the situation. Since, as we propose, the axes are no longer tied to \(x\) and \(y\), learners think more deeply about the concepts of independent and dependent variables when graphing real world data models such as \(p=f(t)\) where \(f(t)=298,213,000(1.009)^t\) and \(\textit{height}\ = f(\textit{time})\) where \(f(\textit{time})=-8.99 \cos(\frac{\pi}{6}\cdot \textit{time})+12.74.\)
When students understand the concept of inverse function in the context of a real world situation, they engage in reasoning (the process of drawing conclusions on the basis of evidence or stated assumptions (NCTM, 2009)) and sense making (developing understanding of a situation, context, or concept by connecting it with existing knowledge (NCTM, 2009)). This connects directly with the Standards for Mathematical Practices – specifically Math Practice #1 (make sense of problems and persevere in solving them) and Math Practice #2 (reason abstractly and quantitatively) (National Governors Association, 2010). The Mathematical Association of America encourages similar ways of thinking in their Committee on the Undergraduate Program in Mathematics Curriculum Guide (MAA, 2015). Cognitive Recommendation #1 states that Students should develop effective thinking and communication skills. All such connections help students understand and retain new information, something that is more challenging if students are not engaged in reasoning and sense making (Hiebert et al., 1997).
Summary
A correct understanding of inverse functions empowers learners mathematically. By eliminating the switch \(x\) and \(y\) approach and implementing the solve for the dependent variable approach, teachers can reduce confusion and enhance student understanding. By recognizing that the inverse of \(y=f(x)\) is \(x=f^{-1}(y)\), learners can make sense of inverse functions in multiple mathematical contexts including real world data analysis and modeling.
Adapted from an article by the same authors, listed in the references below.
References
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Carlson, M. & Oehrtman, M. (2005). Key aspects of knowing and learning the concept of function. Mathematical Association of America Research Sampler, No. 9, March 2005.
Hiebert, J., Carpenter, T., Fennema, E., Fuson, K., Wearne, D., Murray, H., et al. (1997). Making sense: Teaching and learning mathematics with understanding. Portsmouth, NH: Heinemann.
Mathematical Association of America (2015). 2015 CUPM Curriculum Guide to Majors in the Mathematical Sciences. Carol S. Schumacher and Martha J. Siegel, Co-Chairs, Paul Zorn Editor. Washington, DC: MAA
Moore, K. C., Liss II, D. R., Silverman, J., Paoletti, T, Laforest, K. R., and Musgrave, S. (2013). Pre-Service Teachers’ Meanings and Non-Canonical Graphs. In Martinez, M. & Castro Superfine, A. (Eds.), Proceedings of the 35th annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education (pp. 441-448). Chicago, IL: University of Illinois at Chicago.
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Van Dyke, F. (February 1996). The inverse of a function. Mathematics Teacher. 89, pp. 121 – 126.
Wilson, F. (2007). Finite mathematics and applied calculus. Boston: Houghton Mifflin Company.
Wilson, F.C., Adamson, S., Cox, T., and O’Bryan, A. (March 2011). Inverse functions: What our teachers didn’t tell us. Mathematics Teacher. 104, pp. 500-507.
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