Function inverses

Question

From the definition of a inverse standpoint ($f^{-1}(f(x))=f(f^{-1}(x))=x$), why does interchanging variables ($x$ and $y$) work to find the inverse? It seems logical to me but I cannot come up with a coherent argument as to why.

Answer 1

When we find the inverse in this way, we are not really ‘interchanging variables’. We’re just rearranging to find the input in terms of the output (when usually we have the output defined explicitly in terms of the input), and then relabelling to make it clearer. It just happens that we usually use $y$ to refer to the output of a function and $x$ to refer to the input.

An inverse function $f^{-1}$ is just a function that ‘undoes’ the operation of $f$, so when we find the input in terms of the output we are simply finding what operations are required to turn the output of $f$ back into the input, which we often call $x$.

As a simple example, suppose $y = 3x + 2$. Then $\frac{y - 2}{3} = x$.

We didn’t interchange any variables and we have found the inverse. However, if we want to write it in the ‘standard’ $y = f(x)$ way, we can relabel to find $y = \frac{x-2}{3}$.

This is all quite loose because we are not really manipulating functions—just expressions. Really, we are taking $f(x)$ in terms of $x$ and finding $x$ in terms of $f(x)$. But you can see hopefully more intuitively why this technique works when we have relatively simple functions defined like this.

$$\rule{4cm}{0.4pt}$$

Since I feel like I have tried to fully address your concern about ‘interchanging variables’, let me try to give some extra reasoning about why we can ‘rearrange’ an equation like we did above to find the inverse.

Any function that we can write like we did above can be expressed as a composition of intermediate functions. Consider what happens to the input $x$ in our example above: first it gets multiplied by $3$ and then it gets added to $2$.

As I said, the inverse is just an ‘undoing’ function. It takes us back from the output to the input again. When we rearrange an equation, what we are doing is repeatedly applying functions to both sides.

When we find the inverse using our rearranging technique...

$$y = 3x + 2$$

$$\text{(apply $f(\lambda) = \lambda - 2$ to both sides)}$$

$$y - 2 = 3x$$

$$\text{(apply $f(\lambda) = \frac{1}{3}\lambda$ to both sides)}$$

$$\frac{y - 2}{3} = x$$

...we are just applying the inverse operations of the function, in the reverse order, which of course will undo the operation of the function.

This is also why functions that are not one-to-one do not have inverse functions—we can’t undo their operation as we did here because we don’t know which input lead to a specific output.

Answer 2

From your definition, suppose that $(x,y)$ is on that function, we have

$x=f^{-1}(f(x))=f^{-1}(y)$.

Now like you said, if we change $x$ and $y$, we obtain $y=f^{-1}(x)$, which is the inverse.