Ok, so you’re at that point in your (Algebra II / Precalculus) course where you’re either teaching (or learning) about inverse functions. And then it happens, there is an open revolt about $$f^{-1}$$

Students have learned exponent notation by now and have just painfully gotten the hang of negative exponents. We have that $$x^{-1} = \frac{1}{x}$$ and in fact, textbooks give the rule that $$x^{-k} = \frac{1}{x^{k}}$$ when describing the laws of exponent. So then, it makes perfect sense that there is a lot of confusion with the notation of function inverse as \(f^{-1}\).

Here’s why the notation is the way it is for function inverse and not “one over”. Let’s see what the difference is between (a) \(f^{-1}(x)\) and (b) \([f(x)]^{-1}\) and parse the notation. Notice the placement of the exponent. In (a) the \(-1\) is associated with \(f\) whereas in (b) the \(-1\) is associated with the result of \(f(x)\).

It’s an unfortunate overuse of notation, but generally consistent. So how do we get students to see the difference? Part of it will be a simple matter of getting used to it. But another part of it is parsing the notation of functions.

Mathematics is filled with overused notation in the same way that English is filled with exceptions [laughter, slaughter, daughter, taught; read, red, reed, read; food, good, blood;]. But we’ve gotten used to the wacky “rules”. And those who don’t speak English haven’t gotten used to the wacky rules in the language they speak. That’s why we have puns.

Back to math …

We can write \(4x\) to mean “4 times \(x\)”, but why isn’t \(41\) “4 times 1”? What about \(x(y+1)\)? That’s “\(x\) times \(y + 1\)”. But then what about \(f(x)\)? Why isn’t that “\(f\) times \(x\)”? Herein lies the conundrum of notational context, convention, and culture. If we accept, by convention, that in a contextless setting \(f(x)\) is to mean “the function \(f\) applied to \(x\)”, then it’s a matter of understanding what the pieces are — \(f\) and \(x\) — and how they interact.

The name of the function is \(f\). The name of the function is not \(f(x)\). The latter is the result of \(f\) applied to \(x\). Thus, if we want to talk about the inverse of our function, we mean to say that we want the inverse of \(f\) and the way we notate that is \(f^{-1}\) and then we will apply this inverse function to \(x\). On the other hand, if we write \([f(x)]^{-1}\), we are saying “apply \(f\) to \(x\) and then give the multiplicative inverse”, or \(\frac{1}{f(x)}\).

You may have noticed that I write \([f(x)]^{-1}\) for \(\frac{1}{f(x)}\). I do this on purpose when introducing the notation of the inverse of a function and disambiguating it from the reciprocal of \(f(x)\). This is the “getting used to it” part. For students, this excessive notation is somewhat necessary because \(f^{-1}(x)\) and \(f(x)^{-1}\) are too similar for the detail of the placement of the \(-1\) to be noticed. However, using the extra brackets, calls to attention the difference in notation and hence (and hopefully) the difference in meaning.

Next, we can start to play with the notation. I would suggest that you ask your students what the following mean, assuming that \(f\) is a function and its inverse exists. Note, here I drop the extra brackets.

$$
\begin{equation}
f^{-1}(x) \mbox{ vs } f(x^{-1})\\
f(f^{-1}(x)) \mbox{ vs } f(f(x)^{-1})\\
f^{-1}(f^{-1}(x)) \mbox { vs } f^{-1}(x)f^{-1}(x) \mbox{ vs } f^{-1}(x)f(x)^{-1}\\
\end{equation}
$$

Another example of where this type of functional notation shows up is in Calculus. Take for example the derivative of a function \(f\). Sometimes it is written \(f^{\prime}(x)\) (pronounced “\(f\) prime of \(x\)” to mean “the derivative of \(f\) applied to \(x\)”).

Maybe this can help?

[further edit]
Of course, this is a wacky rule since, other exponents on \(f\) don’t work the same way, but sometimes they do. For example, \(f^{2}(x)\) isn’t \(f(f(x))\) — case in point, \(\sin^{2}(x) = \sin(x)\sin(x)\). However, the second derivative of \(f\) can be written as \(f^{\prime\prime}(x)\) or \(f^{(2)}(x)\) or \(\frac{d^{2}f}{dx^{2}}\) (but then where does the “of \(x\)” go and notice that the \(dx^{2}\) part really is \(dxdx\)?). Notice that \(\frac{d^{2}f}{dx^{2}} = \frac{d}{dx}\frac{df}{dx}\) and hence the exponents. It’s confusing and not internally consistent! Hence, we just have to get used to the overused notation.

[further further edit]
Ah, the perils of being your own proofreader … the previous paragraph is poorly worded as readers have pointed out [thank you!]. It is indeed true that the notation of \(f^{2}(x)\) can mean \(f(f(x)\) [often it does]. I was trying to highlight the general inconsistency of use when comparing with the notation used with the trig functions — \(\sin^{2}(x)\) unfortunately does not mean \(\sin(\sin(x))\) rather it means \(\sin(x)\sin(x)\). Further, omitting the argument and writing \(f^{2}\) can be ambiguous, but it is often used in the context of \(f\cdot f\) rather than \(f(f)\). The latter is written as \(f \circ f\) when arguments are not present (but also when arguments are present), where \(\circ\) symbol is to denote composition.

Hopefully, that’s less confusing!