Should we stop teaching “interchange $x$ and $y$” when finding the inverse function?

Question

In one textbook I use for College Algebra, the author teaches that one should interchange $x$ and $y$ when looking for inverse functions. For example, the inverse function of $$y=2x+2$$ is $$y=0.5x-1.$$

In a calculus textbook the author does not teach interchanging variables. For example, the inverse function of $$y=2x+2$$ is $$x=0.5y-1.$$

Surely there is no substantial difference between using $x$ as the variable and using $y$ as the variable for a function. I am a bit concerned about the two different teachings my students received from two different courses as students may get confused. Should we stop teaching "interchange $x$ and $y$" when finding the inverse function?

p.s. If we do teach "interchange $x$ and $y$" when finding the inverse function", the inverse function of

$$C=\frac{5}{9}(F-32)$$

should be $$C=\frac{9}{5}F+32,$$where in the inverse function, $F$ stands for the temperature in Celsius while $C$ stands for the temperature in Fahrenheit. Isn't this confusing?

Answer 1

It's important to clarify the difference between the way mathematicians do things, and the way scientists do things. In math, we typically reserve $x$ for the independent variable (aka input) and the horizontal axis, while $y$ is reserved for the dependent variable (aka output) and the vertical axis. With these variable choices, we need to swap variables in the process of finding the inverse.

In science, variables generally stand for something meaningful. $F$ means Fahrenheit, $C$ means Celsius, and the relationship can be written with either one as input. So we can think of $F$ as a function of $c$: $$F(c) = \frac{9}{5}\cdot c + 32, $$ or of $C$ as a function of $f$: $$ C(f) = \frac{5}{9}\cdot(f-32). $$ In this case, there is of course no swapping, in going from the first equation to its inverse, meaning is properly preserved by writing $f$ instead of $F(c)$, and then solving for $c$.

I think it's important for algebra students to see the science way, which is more concretely meaningful. Eventually, they need to understand inverse functions done with $x$ and $y$, also.

Answer 2

Since $y=2x+2$ and $x=0.5y-1$ have the same graph on the x-y plane, I am hesitant to call them inverse functions.

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Clearly, it's just a matter of making a distinction between independent variables and dependent variables. It is both traditional and sensible to treat $x$ as the default independent variable and $y$ as the default dependent variable. Based on that, I don't have a problem teaching that an inverse function switches the roles of the independent and dependent variables and that is why we exchange the variable names. If an author had a bee in their bonnet that we shouldn't assume the dependence of a variable they used to describe it, I suppose they're welcome to fight for that hill.

It is also the tradition to use $x$ as the dummy independent variable in function notation. So I have no difficulty in saying that $f(x)=e^x$ and $f(x)=\ln x$ are inverse functions, and would be unnecessarily confused if an author felt compelled to switch one of those dummy variables.

Answer 3

I usually introduce the idea of inverse functions by linking it to the idea of basic composite functions

For your example, I'd ask: ``What are the (intermediary) functions done to $x$ to get $2x+2$?"

• Given $x$, we'd multiply by $2$ and then we'd add $2$
• So it's like we have $y=f\left(g\left(x\right)\right)=2x+2$, where $f(x)=x+2$ and $g(x)=2x$

Then to get $x$ from $y=f\left(g\left(x\right)\right)$, we'd have to undo those $f(x)$ and $g(x)$ functions.

• The inverse of $f(x)$ is what undoes $x+2$, so $f^{-1}(x)=x-2$
• Likewise, $g^{-1}(x)=\frac{x}{2}$

And introduce that $f^{-1}\left(f(x)\right)=x$, which is basically "whatever is done to $x$ by the function $f(x)$ is undone by $f^{-1}(x)$"

• So if $f^{-1}(x) $ was applied to $y$, we'd have $f^{-1}\left(f\left(g\left(x\right)\right)\right)=g\left(x\right)$
• Algebraically, $[2x+2]-2=2x$
• Then if we apply $g^{-1}(x)$, we'd get $g^{-1}\left(g(x)\right)=x$
• Algebraically, $\frac{2x}{2}=x$

If you're teaching younger students (like middle school grade?), I might use a sort of "last in; first out" analogy. Like "when you are getting ready to walk out the door, first you'd put your sock on and, lastly, put on your shoe. Coming home, you'd need to take off your shoe, and then take off the sock"

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Obviously a big downside is that it only works for very simple functions but I do like it since it always keeps the idea of an inverse function linked to the function itself, rather than just swapping variables and solving.

I've found it most helpful when teaching students new to trigonometry since, for example, $\sin(\theta)=\frac{O}{H}$ and you want to solve for $\theta$

• To get $\theta$ alone, you need to "undo" that $\sin$ function that is done to $\theta$.
• To undo $\sin$ you need $\sin^{-1}$, and since it's an equation what is done to one side you need to do it to the other.
• Algebraically, $\sin^{-1}\left(\sin(\theta)\right)=\sin^{-1}\left(\frac{O}{H}\right)$
• $\longrightarrow \theta = \sin^{-1}\left(\frac{O}{H}\right)$

Answer 4

I can't spot in any of the other answers what I think is the main point.

Neither teaching students to swap variables or teaching them not to switch variables is really the solution, as both of these simply train students to carry out a mechanical process without understanding what is going on.

Either way, for the student the 'inverse' of a function remains 'the formula I get at the end of carrying out these steps'.

What is lacking is the more basic idea of what a function is. While a function is a formula written on a page, the concept of an inverse doesn't really mean much. I'd say what needs adding is:

• a function is a rule for taking an input number and giving an output number (this definition should be sufficient in the circumstances);
• and the same function can be written in more than one way.

I was really hit by how little understanding students have of functions when I wrote down $f(x)=x^2+2$ and asked them to write down what $f(y)$ was. Most of them got there after some thought, but really weren't sure they had the right answer (which explains why they were finding linear functions so hard).

Once you have those ideas, you are in a better position to teach that:

Sometimes is it right to change the variables. Sometimes it's not. You have to make a decision based on the meaning you are attaching to the symbols you have written down.

Answer 5

Let me take a moment to look at an example which involves more variables. Consider the map defined by: $$f(x,y) = \left( \frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right)$$ for $(x,y) \neq (0,0)$. Let us study how to find the inverse map. We seek $f^{-1}$ such that $$ f(f^{-1}(u,v)) = (u,v) \qquad \& \qquad f^{-1}(f(x,y))=(x,y) $$ for all appropriate $(u,v)$ and $(x,y)$. If $f: dom(f) \rightarrow range(f)$ then $f^{-1}: range(f) \rightarrow dom(f)$. In this example we should eventually be convinced that $dom(f)=range(f) = \mathbb{R}^2-\{ (0,0)\}$.

Pragmatically, to calculate the formula for the inverse we can set $f(x,y) = (u,v)$ and solve the following for $x$ and $y$. $$ \left( \frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right) = (u,v) \ \ (*) $$ Remember, we assume $(x,y) \neq (0,0)$ hence multiplication by $x^2+y^2$ is a reasonable step. $$(x,-y) = (u(x^2+y^2), v(x^2+y^2)) $$ which means $ x = u(x^2+y^2)$ and $-y = v(x^2+y^2)$. Suppose $u,v \neq 0$, we can return to the cases $u=0, v\neq 0$ and $u \neq 0, v=0$ after we've finished the interesting case. We already can rule out the case $u=v=0$ since that forces $x=y=0$ in view of the above equations. Now solve both equations for $x^2+y^2$, $$ x^2+y^2 = \frac{x}{u} = \frac{-y}{v} \ \ (**)$$ Next, consider $$ u^2+v^2 = \left(\frac{x}{x^2+y^2} \right)^2+ \left(\frac{-y}{x^2+y^2} \right)^2 = \frac{x^2+y^2}{(x^2+y^2)^2} = \frac{1}{x^2+y^2} \ \ \Rightarrow x^2+y^2 = \frac{1}{u^2+v^2}. $$ Great, now with the above and (**) we derive: $$ \frac{x}{u} = \frac{1}{u^2+v^2} \qquad \& \qquad \frac{-y}{v} = \frac{1}{u^2+v^2} $$ Thus, $$ x = \frac{u}{u^2+v^2} \qquad \& \qquad y = \frac{-v}{u^2+v^2} $$ We find, $$ f^{-1}(u,v) = \left( \frac{u}{u^2+v^2}, \frac{-v}{u^2+v^2} \right) $$ But, what about $u=0, v\neq 0$ ? Let's investigate if the formula above works in such a context, \begin{align} f(f^{-1}(0,v)) &= f\left( \frac{0}{0^2+v^2}, \frac{-v}{0^2+v^2} \right) \\ &= f(0, -1/v) \\ &= \left(\frac{0}{0^2+(-1/v)^2}, \frac{-(-1/v)}{0^2+(-1/v)^2}\right) \\ &= \left(0, \frac{1/v}{1/v^2}\right) \\ &= \left(0,v \right). \end{align} Similar calculations show that $f(f^{-1}(u,0)) = (u,0)$. Indeed, you can check, $f(f^{-1}(u,v) = (u,v)$ for all $(u,v) \neq (0,0)$. Amusingly, pretty much identical calculations reveal that $f^{-1}(f(x,y)) = (x,y)$ for $(x,y) \neq 0$. In fact, since $dom(f) = range(f)$ we might as well say $f=f^{-1}$ in this case. But, if I intend to keep distinct the copies of $\mathbb{R}^2$ in which $dom(f)$ and $range(f)$ reside then writing $f=f^{-1}$ is not a good idea. I suspect a notation such as: $$ f: dom(f) \subset \mathbb{R}^2_{xy} \rightarrow range(f) \subset \mathbb{R}^2_{uv} $$ and $$ f^{-1}: range(f) \subset \mathbb{R}^2_{uv} \rightarrow dom(f) \subset \mathbb{R}^2_{xy} $$ is helpful.

Fine, full disclosure, $f(z) = 1/z$ hence $w=1/z$ yielding $z =1/w$ gives $f^{-1}(w) = 1/w$ is the better way to see that the reciprocal map on the punctured plane is it's own inverse.

If students in the precalculus are taught to solve for the independent variable in terms of the dependent variable then it is a fairly easy transition to higher dimensions where we simply solve for the independent variables in terms of the dependent variables.

Answer 6

Consider the equation in its two different arrangement of variables: $y=x+5$ and $x=y-5$;

for $x$ (independent variable value) as input, I am getting a set of $y$ (values dependent on input '$x$' variable values). Plotting them on a graph(where $x$ is independent and $y$ is the dependent variable) I get:(Line in red)

for $y$ (independent variable value) as input, I am getting a set of $x$ (values dependent on input '$y$' variable values). Plotting them on a graph (where $y$ is independent and $x$ is the dependent variable) I get:(Line in blue(overlapping the red line made by the first equation)

which is the same thing (since it's the same equation), but in this case the axes aren't the same, they are in a way "interchanged", so in order to represent the the inverse function (which is a function, lets say, $g(x)$ i.e a function where $x$ is dependent and $y$ is independent) we swap the variables so they can be plotted to get the inverse function in terms of $x$ as independent variable.

The inverse function is the line in green (with $x$ as dependent variable)

So, in conclusion, this is how I understand it and how I explain the "interchanging of x and y" to myself. It ultimately translates to undoing of a function that is done on independent variable x. Maybe you could explain it in this way while teaching.