To find the inverse of a function, you switch the spots of $x$ and $y$ in the function and solve for the new $y$ output. For example:

$y = 3 + \sin\ x$

  • Switch positions of $x$ and $y$

$x = 3 + \sin\ y$

  • Solve for this new $y$

$y = \sin^{-1}(x-3)$

However, I've never wondered why this is the case. Does anyone have an explanation as to where this method came about?

In addition, how would one find the inverse of a function with 3 variables $x$, $y$, and $z$?

  • $\begingroup$ the function of your example is not invertible in all it domain because the sine function doesnt have inverse in all it domain. In this case you can choose a subset of the domain of your function where the function is strictly monotone (hence injective) to define an inverse. Many times the inverses does not have a closed (easy) form, so we must approximate them using advanced techniques of analysis. $\endgroup$ – Masacroso Jun 16 '17 at 8:04

suppose you have a function f:X->Y. If f is bijective, then we can identify a function g:Y->X which maps the images under f to its pre images. This is called the inverse function.


Elementary explanation with the example $$y(x)=3+\sin(x)$$

This means that putting a value $x$ into the above equation returns a value $y$.

To make things more simple, we ignore that this function is not bijective and should be restricted to a limited range.

Then, we look for a function $g$ such as putting $y$ into it returns $x$, that is $g(y)=x$. The function $g$ is the inverse function of $y(x)$.

$$g(3+\sin(x))=x \tag 1$$

But we are not familiar with this form of writing. We prefer to write $$Y=g(X)\tag 2$$

The use of differents symbols avoid the confusion between $x$ and $X$ and between $y$ and $Y$.

Comparing (1) and (2), we see that $\quad \begin{cases} g(X)=g(3+\sin(x)) \quad\to\quad X=3+\sin(x) \\ Y=g(X)=x \quad\to\quad Y=x \end{cases}$

Hence : $$X=3+\sin(Y)$$

So, the recipe which consists in inverting $x$ and $y$ , which gives $x=3+\sin(y)$ , is a short-cut for all the above. Except that this short-cut can make forget that the new $x$ and $y$ are not the same as the previous ones.

If fact, to be clear, we should write : $$Y=\sin^{-1}(X-3)$$ which defines a function $Y(X)$. We are allowed to chose any symbol as we want for a given function, for example $\quad f(t)=\sin^{-1}(t-3)\quad$ or any other symbols. Among the infinity of possible notations, why not this one : $$y=\sin^{-1}(x-3)$$ But, of course the $x$ and $y$ are not the same as the $x$ and $y$ at the beginning.


Consider this magic box.

$X$ => [ $F$ ] ==> $Y$

Now, by solving for $X$ in terms of $Y$, we have effectively found

$X$ <== [$F^{-1}$] <== $Y$.

Now, we just switch the names of $X$ and $Y$, because convention tell us to always express $Y$ in terms of $X$, even in inverse functions.

As far as inverses of $z=f(x,y)$, if each point $(x,y)$ is associated with exactly one $z$, and each $z$ is associated with exactly one $(x,y)$ which is almost NEVER the case, then we can find an inverse without much hassle.

In $y=f(x)$, we also need to splice the function into a sub-domain, so that on that domain each $x$ maps to one $y$, and each $y$ maps to one $x$.

Otherwise, the inverse of $y=x^2$ would be a sideways parabola.

However, we know that the usual inverse of $y=x^2$ is $y=\sqrt{x}$, because we only consider to invert the branch of $y=x^2$ that corresponds to $x \in [0,\infty)$.


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