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:
Find a function $f(x)$ that satisfies the equation $5f(x)+x = 4$. [Hint: Use basic algebra to isolate $f(x)$.]
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:
- $P(Q(0))$
- $P(Q(4))$
- $Q(P(2.5))$
- $Q(P(14.8))$
- What is the relationship between the functions $P$ and $Q$?
- Simplify $P(Q(x))$.
- 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:
- 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.