| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
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.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1558.1 | Lots of solutions | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Wed Feb 05 1992 13:11 | 12 |
> 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!). | |||||
| 1558.2 | This is quite tedious! | CLT::TRACE::GILBERT | Ownership Obligates | Wed Feb 05 1992 13:41 | 36 |
f1(x) = x/(1-x)
{ 0 when x <= a
f2(x) = { c�(x-a)/(b-a) when a <= x <= b
{ c when x >= b
When is there an intersection in the first part of f2(x)?
When x <= a and x/(1-x) = 0
This happens when x = 0 <= a, so x = 0 is the point of intersection
When is there an intersection in the second part of f2(x)?
When a <= x <= b and x/(1-x) = c�(x-a)/(b-a)
a <= x <= b and cx� + (b-a-c-ac)�x + ac = 0
a <= x <= b and x = ((ac+a-b+c) � sqrt(D))/(2c),
where D = (a-1)��c� - 2�(a+1)�(b-a)�c + (b-a)�
Now D >= 0 when c <= (a+1 + 2�sqrt(a))�(b-a)/(a-1)�
or c <= (a+1 - 2�sqrt(a))�(b-a)/(a-1)�
Rephrasing in terms of a,
D >= 0 when a >= (c�+bc-c+b + 2�c�sqrt(bc-c+b))/(c+1)�
or a <= (c�+bc-c+b - 2�c�sqrt(bc-c+b))/(c+1)�
and bc-c+b >= 0 when b >= c/(c+1) or c >= -b/(b-1);
Rephrasing in terms of b,
D >= 0 when b >= ac+a+c + 2�sqrt(a)
or b <= ac+a+c - 2�sqrt(a)
So, choosing the best phrasing, they intersect when
a <= x = ((ac+a-b+c) � sqrt(D))/(2c) <= b,
where D = (a-1)��c� - 2�(a+1)�(b-a)�c + (b-a)�, and
|b - (ac+a+c)| >= 2�sqrt(a)
Note that 0 <= a <= b and c >= 0 ensures a <= ((ac+a-b+c) - sqrt(D))/(2c),
and with b >= c/(c+1) it ensures ((ac+a-b+c) + sqrt(D))/(2c) <= b.
When is there an intersection in the third part of f2(x)?
When x >= b and x/(1-x) = c
b <= x = c/(c+1)
| |||||
| 1558.3 | Thanks for the help. | 3D::ASFOUR | Thu Feb 06 1992 08:12 | 0 | |