# found a function from equations and inequality?

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 ?

EDIT 1:

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

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

$$f(x)=g(x)(x-50400)$$

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

$$f(x)=(x^2+Bx+C)(x-50400)$$

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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