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I'm not quite sure what kind of problem this is called. I guess it may be some kind of solving of a system of nonlinear equations. My math is a little rusty.

Suppose I have a function $f(x)$ like the one displayed in the figure:

Unfortunately I don't have enough reputation to include the image. So instead, picture the following function:

$$f(x) = (1+(1-x)^2)^{-1}(1+(5-x)^2)^{-1}$$

I wish to find the values of $x$ at which the horizontal line $k$ intersects the function, but I wish to find the points in such a way that if I integrate over all the intervals they equal a specific value.

E.g. suppose like in the graph there are four points of intersection $\{x_1,x_2,x_3,x_4\}$. Due to their nature, $f(x_1)=...=f(x_4)$. I wish to find a constant function $g(x)=k$ such that

$$\int_{x_1}^{x_2} f(x) dx + \int_{x_3}^{x_4} f(x) dx = c$$

where $\{x_1,x_2,x_3,x_4\}$ are the values of $x$ at which $f(x)=k$.

However, it may be the case that to achieve a value of $c$, there will only be two intersection points. So, then it would be

$$\int_{x_1}^{x_2} f(x) dx = c$$

How could one solve this numerically? In a simple analytic case I imagine one would be able to find what each $\{x_1,x_2,x_3,x_4\}$ is in terms of the variable $k$, and then substitute this into the integral equation and solve so that $k=h(c)$, i.e. $k$ is some function of $c$, but what if this can not be done analytically?

I suppose that the main question is: what kind of problem is this? What method might one use to solve it?

Thanks for any help.

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If $f$ has a closed-form antiderivative, $F(x)=\int f(x)\,dx$, then (at least in the two-intersection case) it reduces to solving the pair of equations, $f(x_1)=f(x_2)$, $F(x_2)-F(x_1)=c$. – Gerry Myerson Apr 16 '13 at 8:58
Your $f(x)$ does not depend on $x$, only on $\theta$? And why don't you first simplify $(1+(1-\theta))^{-1}(1+(5-\theta))^{-1}=(2-\theta)^{-1}(6-\theta)^{-1}$? – Hagen von Eitzen Apr 16 '13 at 9:00
You have to be careful on which interval you want to work because your function is only defined on the interval $I=]-\infty,2[\space\cup\space]2,6[\space\cup\space]6,+\infty[$ – Dolma Apr 16 '13 at 9:14
Apologies, I had incorrectly specified the equation. It has been corrected. – jatotterdell Apr 16 '13 at 9:24

You might be mistaken at some point because the function you have written looks like this:

enter image description here

So you can see on the graph that it's impossible for you to have 4 points of intersection if you draw a horizontal line on your graph ... Think about it, the points of intersection are actually the roots of $f(x)=k$.

So on the interval $I=]-\infty,2[\space\cup\space]2,6[\space\cup\space]6,+\infty[$ since $(2-x)(6-x)\neq0$, the equation becomes $\large (2-x)(6-x)-\frac{1}{k}=0$. This is a second order polynomial and can't have more than 2 distinct roots ...

The other possibility is that you made a mistake in your formula for $f$. Maybe you misplaced the $(-1)$ exponents and it is really $\large f:x\mapsto (1+\frac{1}{1-x})(1+\frac{1}{5-x})$ ?

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Yes, I had typed the equation incorrectly. I have corrected this. – jatotterdell Apr 16 '13 at 9:31
Well now in the same way as what I told you, you will have a 4-th degree polynomial and you want to find its roots which shouldn't be so hard to do ;) – Dolma Apr 16 '13 at 9:40

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