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Take some analytic function, $f(x)$, that goes from $-\infty$ to $\infty$, with a finite number of points such that $\frac{df}{dx}=0$. You can divide the y axis into intervals, where the boundary between each interval is the y value at a critical points (see graph).

Within each interval, there are an odd number of real inverse functions, $g_n(y)$. Number them in ascending order. Then, for every function I've tried, $\sum_n (-1)^n g_n(y)=ag_m(y)+b$ for some a,b and m that is constant across the interval. Does anyone know why this would be the case?

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It's not true in general. What is true is this. If $f(x) = a_n x^n + \ldots + a_0$ is a polynomial with $n$ odd, then in any interval where there are $n$ real inverse functions, the sum of those is constant, namely $-a_{n-1}/a_n$. So if you happened to take $n=3$, you would get $-g_1(y) + g_2(y) - g_3(y) = a_2/a_3 + 2 g_2(y)$.

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    Nevermind, I think that my function looked linear, but actually wasn't. Thanks for your help.2012-08-04