When discussing inverse functions, how can our notation and methods reinforce student understanding?

Question

Yesterday in my precalculus class, I decided to teach students how to find the formula for an inverse function in a new way (to me).

In this curriculum, they have already used the notation $f^{-1}(x)$ to evaluate numerically, with the understanding that $x$ is the output of the function $f$, so evaluating $f^{-1}(3)$ means "find the input value for $f$ if the output is $3$. They have practiced this with a function's formula by setting it equal to $3$ and solving; they have also practiced this graphically and interpreted it in the context of verbally-defined functions.

Also, they have used $f(g(x))$ numerically, graphically and verbally, and recently they have composed two formulas together and simplified. They have numerically composed two inverse functions together, understanding that what one function does, the other undoes.

In an attempt to build on this function composition notation, I have strayed from the typical "Switch" method (exchange $x$ and $y$, solve for $y$, and then call this $f^{-1}(x)$), instead giving students a sequence of problems, such as:

1. Find a function $f(x)$ that satisfies the equation $5f(x)+x = 4$. [Hint: Use basic algebra to isolate $f(x)$.]

2. Let $g(x) = x^3-5$. Find a function $h(x)$ so that $g(h(x)) = x+2$. [Hint: Simplify the left-hand side of this equation by using the function $g$, where the unknown function $h(x)$ is the input.]

Then, returning to numerical understanding:

Let $P(x) = 4x-3$ and $Q(x)=0.25x+0.75$, compute the following:

3. $P(Q(0))$
4. $P(Q(4))$
5. $Q(P(2.5))$
6. $Q(P(14.8))$
7. What is the relationship between the functions $P$ and $Q$?
8. Simplify $P(Q(x))$.
9. Simplify $Q(P(x))$.

With the stage now set, and after a brief conversation about what the result should be when composing a function and its inverse, they try the following problem:

10. Find the inverse of the function $f(x) = 6-8x$. [Hint: Remember what the composition $f(f^{-1}(x))$ should equal? Write this as an equation, and then simplify the left-hand-side as in problem 2. Use algebra to solve for the function $f^{-1}(x)$.]

Now, this is perhaps a longer introduction than the Switch method found in many books, but it seems worth it in a few respects. First, it reinforces the notion and notation of composition -- students are taught to recite this each time they begin solving for an inverse. Second, they are getting a first glimpse that functions are objects which may be manipulated with algebra, something they will see again in differential equations. Finally, and not the most compelling reason, it causes students to use the new notation we've been working with all term -- for most of the functions we're dealing with (until logarithms), just switching $x$ and $y$ and solving is something we could have asked them to do in an intermediate algebra class, without any thought of functions or inverse processes.

However, this method may downplay the notion that a function and its inverse have switched roles in terms of input and output. The Switch method reinforces that idea each time, where students literally exchange input for output between the functions.

Question 1: Are my proposed benefits worth it for students, either in the long or short run? For those who have tried this method, did your students come away better off in any way?

Question 2: Am I missing any other ways in which the "switch $x$ and $y$ and solve" method is preferable?

With regard to Question 2, one situation I am anticipating for my class is that, given a formula for $f(x)$, they may write $f^{-1}(f(x)) = x$ instead of $f(f^{-1}(x)) = x$ and then have no idea what to do. The Switch method imposes no order of composition on the process.

Thanks for taking the time with this long question.

update: Removed a third part of this question as it was too broad in scope.

Answer 1

In classes I've taught (at least since the mid 1980s), I always put the "switch $x$ and $y$" part where I think it belongs --- at the end of the calculation. For example, $f(x) = 2x+1$ becomes $y=2x+1$ and then $x=\frac{1}{2}(y-1)$ and then $f^{-1}(y)=\frac{1}{2}(y-1),$ at which point we CHANGE TO USUAL FUNCTION NOTATION (i.e. use $x$'s for the independent variable) and write $f^{-1}(x)=\frac{1}{2}(x-1).$ Note that the two function-inverse function identities, $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$ are easy to prove using $y=f(x)$ and $x=f^{-1}(y)$ (i.e. when you don't switch $x$'s and $y$'s).

In motivating the change to usual function notation at the end, I remind students that $f^{-1}$ is the name of the rule whereas $f^{-1}(y)=\frac{1}{2}(y-1)$ is simply one (algebraic) way to describe the rule. The same rule can be described by $f^{-1}(u)=\frac{1}{2}(u-1)$ and $f^{-1}(P)=\frac{1}{2}(P-1)$ and $f^{-1}(z)=\frac{1}{2}(z-1).$ Thus, if the question asks you to give $f^{-1}(x),$ then you answer with $\frac{1}{2}(x-1)$ rather than with $\frac{1}{2}(y-1).$

Personally, I never understood why books, and teachers who have taught this material more than a few times, put the "switch $x$ and $y$" part at the beginning. Doing it that way seems to me to reinforce the belief in some students that mathematics consists of ad hoc procedures that one has to memorize.

Answer 2

I'll answer question 2, providing an opportunity you will miss by avoiding the "switch $x$ and $y$" approach. My goal is the same as Dave L Renfro's: try to minimize the number of ad hoc procedures that students are expected to memorize.

If the students internalize the fact that the inverse function switches inputs with outputs in every case, then it provides a mental framework for each of the following questions, tying them together into a neat conceptual package:

1. Find a formula for the inverse of the function $f(x) = 2x + 3$.

2. Find a table for the inverse of the following function:

   x   f(x)
   2   6
   3   8
   4   13
   

3. Find a graph of the inverse of the following function:

The graphical example is I think the best argument in favor of "switch the inputs and outputs," because it frees the students from their destructive instinct to find an equation for everything first.

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Finally, I have an in-class demonstration I do that helps with the graphical form of "switch inputs and outputs" that I think is the real payoff. It goes like this:

• Using a thick marker of one color, draw the x-axis along the center of a square piece of paper on both sides of the paper. Label it "$x$" on both sides.
• Using a thick marker of another color, draw the y-axis along the center of the square piece of paper on both sides of the paper. Label it "$y$" on both sides.
• Fasten a dowel to the paper with tape along the line $y=x$.
• Display the paper to the class holding it by the line $y=x$, and ask what you do to find the inverse function ("switch inputs and outputs," or "switch $x$ and $y$" as the target answers).
• Twirl the paper quickly 180 degrees, keeping the dowel in the same spot. It makes a satisfying snap sound, and reflects the paper over the line $y=x$. The $x$ and $y$ axes trade places!

This demonstrates that the transformation "reflect over the line $y=x$" is not another separate ad hoc magic rule to memorize, but is actually just another form of the overarching concept "switch the inputs and outputs to get the inverse function."

Answer 3

This idea is a little unusual, but you might consider making the inverse function part of a bigger picture.

For example...

If $$f(x)=2x+1$$ then two applications of the function gives us $$f^{(2)}(x)=f(f(x))=2(2x+1)+1=4x+3$$ and similarly three applications $$f^{(3)}(x)=f(f(f(x)))=2(4x+3)+1=8x+7$$ Once that has been mastered, then mention that no applications of the function gives us $$f^{(0)}(x)=x$$ and then you can introduce the idea that we can do the reverse. $$f^{-1}(x)=f^{(-1)}(x)=\frac{x-1}{2}$$

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If you wanted, you could even find general formulas for the applications that are consistent with the inverse. $$f^{(n)}=2^n (x+1)-1$$

And then, if you have students with a passion for math, from there you could even introduce the concept that a function could be applied a fractional number of times. $$f^{(\frac{3}{2})}(x)=2^{\frac{3}{2}} (x+1)-1=2x \sqrt 2+2 \sqrt 2-1$$