# For any continuous function f(x), how can I split up the function and restrict the domain to find an inverse?

I want to know everything there is to know about inverses for curiosity's sake. I am totally fine finding the inverse of a function where each x maps to a unique y coordinate, but when we get to quadratics and other stuff, I get really confused.

School wants me to know how to find inverse functions of quadratics, trinomials, etc, but I really want to know how to find the inverse of any function by splitting up the domain. Please place all links where you got info in the answer so I can do additional research, but if you didn't do research, seriously don't bother.

• This is a pretty broad question. You might be better off coming up with one or a few simple examples of problems you want to be able to solve and trying to work through those, maybe with help from this community if you get stuck. – crf Apr 13 '14 at 22:58
• I am trying to gain an intuition on inverse functions where, for f(x), more than 1 value for x maps to a single y by splitting it up into two functions. If you want a simple example, try $x^2 + 5x + 6$. – louie mcconnell Apr 13 '14 at 23:00
• If for a start you are happy to consider differentiable functions, Google "inverse function theorem". – David Apr 13 '14 at 23:18
• well, consider $\arcsin$ and $\arccos$ and $\arctan,$ these illustrate some of the issues en.wikipedia.org/wiki/… – Will Jagy Apr 13 '14 at 23:21

## 1 Answer

Louie, here is one way to split up the domain of a function so that it becomes invertible over each part: split the domain so that the function is strictly increasing or strictly decreasing on each part. The reason this works is because any monotone function (that is, any function which is strictly increasing or decreasing) can be inverted.

For an easy example, consider the quadratic function $f(x) = -(x-1)^2 + 2$. Since $f(x)$ is strictly increasing for $x\leq 1$ and strictly decreasing for $x\geq 1$, a natural way to divide the domain would be $D_1 = (-\infty,1]$ and $D_2=[1,\infty)$. The inverse of $f$ restricted to $D_1$ is $g_1(x)=1-\sqrt{2-x}$ and the inverse of $f$ restricted to $D_2$ is $g_2(x)=1+\sqrt{2-x}$.

For a more interesting example, consider the cosine function $\cos(x)$ which is increasing whenever $(2n-1)\pi\leq x\leq 2n\pi$ and decreasing whenever $2n\pi\leq x\leq (2n+1)\pi$, for $n$ an integer. For each of these intervals we can find a function $g_n$ which inverts $\cos$ restricted to that interval: \begin{align} g_{2n-1} &= 2n\pi-\arccos(x) \\ g_{2n} &= \arccos(x)+2n\pi. \end{align} (I think that does it. Can you check my work?)

You might find my answer unsatisfactory because I did not provide links to research articles. The reason for this omission is that such articles probably do not exist. Mathematicians understand quite well the existence of inverses to a function, although computing them is another matter. If you have questions about inverting specific functions, I recommend you ask them here; we're always happy to help!