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Disclaimer: I'm a beginner with inverse functions.

Can anyone explain what I'm doing wrong here? I'd like to avoid using "y" -- that is, I want to show everything in terms of x and f(x).

enter image description here

Thanks!

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  • $\begingroup$ If you are looking for the inverse function of $f(x) = \frac1x$ for $x\ne 0$, then you have found it. Fun facts: There are more functions with the property $f^{-1} = f$. They are called involutions. $\endgroup$
    – user251257
    Jul 30, 2015 at 0:00
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    $\begingroup$ I would recommend using $y$ cause too many $x$'s just complicate the calculation when you take a bit harder function. $\endgroup$
    – A.Γ.
    Jul 30, 2015 at 0:27

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This is a very good question!

One of the main viewpoints, when it comes to inverse functions, is that they make "input" and "output" switch places - if $\big(x, f(x) \big)$ is a point on the graph of $y = f(x)$, then the point $\big( f(x), x \big)$ is a point on the graph of $y = f^{-1}(x)$.

Algebraically, we can think of this as replacing $x$ (our old input) with $f^{-1}(x)$ (our new output), and $f(x)$ (our old output) with $x$ (our new input): We start thinking of $f(x)$ as our "input", and $x$ as the "output", $f^{-1}(x)$. What makes things confusing is that we agree to always use "$x$" for our input, which means $x$ gets a little overused here.

Since that reads clunkier than I'd intended, how about a table?

\begin{array}{|r|c|c|}\hline&\rm{Input} & \rm{Output} \\\hline\rm{Old}& x & f(x) \\ \\ \rm{New}& f(x) & x \\ \\ \rm{New\ Renamed} & x & f^{-1}(x) \\\hline\end{array}

Going from Old to New, we're making $x$ and $f(x)$ switch places (since we start thinking about the inverse function). Then we rename things by calling our input $x$ and our output $f^{-1}(x)$. You could even think of this renaming in an algebraic way, by applying $f^{-1}$ to what we have:

$$\begin{align*}f(x) &\mapsto f^{-1}(f(x)) = x \\ x &\mapsto f^{-1}(x).\end{align*}$$

So, when you reach the step with $$x = \frac{1}{f(x)},$$ you've already effectively switched input and output (by solving for the old output, $f(x)$, which becomes the new input). Now you just need to rename:

$$x = \frac{1}{f(x)} \overset{\rm{rename}}{\longrightarrow} f^{-1}(x) = \frac{1}{x}.$$

As you can see, the study of functions requires new ways of thinking; things are more abstract. While in the end we can reduce this process to a sequence steps to be followed, it's more subtle than, for example, the "just multiply both sides by 2" sort of algebra you have more practice with.

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    $\begingroup$ Nice work! The only thing I am worried about here is too much symmetry in the example. Especially when there is a trend among beginners to think of the inverse of $f(x)$ as $\frac{1}{f(x)}$, i.e. mixing up the multiplicative inverse of the value with the inverse mapping. $\endgroup$
    – A.Γ.
    Jul 30, 2015 at 0:53
  • $\begingroup$ Thanks, @A.G.! That's a very good point I hadn't considered. This would indeed not be my preferred function to explain inverse functions... $\endgroup$
    – pjs36
    Jul 30, 2015 at 1:12
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I believe I figured this out. I may be wrong though!

enter image description here

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