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

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?)