Conference rusure::math

1558.0. "Help needed in choosing parameters to produce a solution" by 3D::ASFOUR () Wed Feb 05 1992 13:33

Given that

	f1(x) = x/(1-x) 


and                  |
	f2(x) =     c+............+-----------
                     |           /.
                     |          / .
                     |         /  .
                     +--------+---+---------- 
                              a   b

and given that   0 <= x < 1, a>=0,b>=0,c>=0, and a<=b.


When do the two functions intersect? (Alternatively, when is f2(x)=f1(x)) ?


My intuition tells me there are three cases:

1-  intersection at (0,0) only. 

2 - intersection at (0,0) and one other point.  

3-  intersection at (0,0) and two other points.

But I can't figure out what choices of a,b, and c produce each of the cases.
(all choices of a,b,c produce a solution at (0,0))

Some trivial cases are:
 a > 1, guarantees case 1. 
 a choice of c=0 will also produce case 1.
 if 0 < a < 1 and a=b (infinite slope), then case 3 holds as long as 
  0<c is finite. 

 What are the other solutions? 
 For example,is there a choice of b and c where  0 < a < 1 that satisfies 
  case 2? How about case 1 when 0 < a < 1 ?

Tbanks for any help.

	Yousif.

>	f1(x) = x/(1-x)

The derivative (or slope) of f1 is 
	    1
	--------
	(1-x)**2

Single-point intersections occur either when the curve attains c at x=b, or 
when the ac line also has this slope at the point of interesction. For
example, for x = .5 the slope is 4; and f1(.5) = 1. So a single-point
intersection with f2 occurs at b = .5, c = f1(b) = 1, a = b-c/4 = .475 (for
both reasons!).