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I'm Software engineer and I'm having little issue solving this problem let's called H.

Well I'm looking for the mathematical expression of the function f(x) based on 3 equations and one inequality. f(x) passes through 3 points and f(x) is a decreasing function when X€[0,END]

H = {f(0)=1, f(t)=α, f(50400)=0, f '(x)<0} and 0< t < END=50400, 0< a <1.

what i tried: I supposed the solution is polynomial of order 3 and can be expressed this way:

f(x)= ax(x-t)(x-END) + bx(x-t)+ cx+ d I found d,c,b still a !

  1. Q1: Well am I doing this right ?
  2. Q2: is there any theory I can use to find the solution ?


I know this have infinit solution so a solution with a lesser degree is preferable.

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I don't understand. You function is constant equal to $\alpha$, for $t\neq 0,END$? – Sigur Aug 27 '12 at 22:41
yes, my function is equal to α, for t≠0,END ? where α and t are const defined in ]0,1[ and ]0,END[ – M'hamed Aug 27 '12 at 22:46
@Sigur so what you think ? – M'hamed Aug 27 '12 at 23:01
If you fit a cubic, there are four parameters. If you only have three points there will be one free parameter. You have been clever to write it in a way that isolates a as the free parameter, but need one more data point or one less parameter. The constraint on f' doesn't give you a specific value. – Ross Millikan Aug 27 '12 at 23:48
up vote 0 down vote accepted

$$f(50400)=0$$ Leads to:


if you want or know that $f(x)$ is of degree $3$, then $g(x)$ is of degree $2$ and $f(x)$ is:


Since $f(0)=1$, we have:

$$f(0)=(c)(-50400)=1 $$

Now $f(x)$ is:

$$f(x)= (x^2+Bx-\frac{1}{50400})(x-50400)$$

You say that $f(t)=a$, using this fact, you get the value of $B$ in terms of $a$.

Note: I don't know what you mean by "I found d,c,b still a" - Maybe you could apply the above steps on your function and get a.

f(x)= ax(x-t)(x-END) + bx(x-t)+ cx+ d I found d,c,b still a !

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Quadratic is one option. There are infinitely many other options. Some examples: Scaled & shifted exponential function, $\frac{a}{x + b} - c$, etc.

Anyway, you can only have up to 3 degrees of freedom.

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