Create a function where $f(1000)=1.99$ and where $f(400000)=0.49$ As per object I need to create a continuous and decreasing function with one variable where result is as above, but not only.
Function should also be easy to modify to get 
$1<f(1000)<10$ and $0.1<f(400000)<1.$
From what I remember it should be a $f(1/x)$ function but...I got no result. The best match is
$$r=\frac{1}{10^{-12}x^2+0.00000334 x+0.5413},$$
but I was wondering if there is a more elegant solution.
I think have to give you some clarifications:
this formula "should" be used to calculate price for a service according to the quantity purchased, for this reason there are some other condition to be considered:

1) $y$ (result/price) should reduce quickly with low values of $x$ and
  then become more or less plain: for this purpose, straight line is not
  acceptable
2) clearly $y$ (price) cannot be negative
3) it is strictly required that $x\cdot y(x) < (x+1)\cdot y(x+1)$ since total amount cannot reduce
4) it would be nice that once reached a level around $0.49$, $y$ should not reduce much further: just for example: $f(10.000.000)=0.40$
$x\;y(x) < (x+1)\;y(x+1)$

means that the result of function $y$ multiplied by the parameter $x$ shall only increase:
in other words: since $y$ is the price and $x$ is the quantity, the total amount has to increase only: in fact it is not reasonable for example that

$f(10000)=1.5\;\;\; and\;\;\; f(20000)=0.6$

since this means that for 10,000 pieces I will pay 15,000.00 and that for 20.000 pieces will pay only 12,000.00
Thanks for your time!
joe
 A: EDIT: After the question was updated this answer no longer works. Lulu seems to have a good answer that meets the additional criteria.
Why not just a straight line that passes through those two points?
$f(x) = mx+b$, such that $f(1000)=1.99$ and $f(400000)=0.49$. This easily generalizes to the more general requirement as well.
A: I still don't understand the constraints.  What does "$yx<y(x+1)$" mean? I don't think it means $y(x)<y(x+1)$ since you previously specified that the function should be decreasing.
Just to take a stab at it, why not stick with your first guess and adapt $\frac 1x$?  Specifically, look at $$f(x) = \frac ax+b$$
Using your data we easily specify the parameters and get:
$$f(x) = \frac {1503.7594}{x}+.4862406$$
This certainly passes your limit test (the function decays slowly to .486, just below the upper value you mentioned).
I see from your comments that the constraint I didn't understand means that you require $xf(x)<(x+1)f(x+1)$.  I note that for functions of my form we have:$$xf(x)=a+bx$$ and as $bx<b(x+1)$, so long as $b$ is positive, my function passes that test as well.
A: Assume  
$y=f(x)=ae^{-bx},\ a,b>0$  
Or equivalently  
$log(y)=log(a)-bx$  
Substituting the numerical data for x and y and solving the associated linear system yields:  
$f(x)=1.997002e^{-0.351249\times10^{-5}x}$
