Need Help with a mathematical modeling question

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Homework Help Overview

This discussion revolves around a mathematical modeling question involving a 1-parameter equation, specifically focusing on finding the positive fixed point of the function fc(x) = (4/x) + (x/2) - c and determining its interval of existence for values of c between 0 and 2.3.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the quadratic formula to find the fixed point p(c) and question the correctness of their derived expressions. There is also exploration of the first derivative to establish the interval of existence and stability for c, with some participants expressing uncertainty about their calculations and the implications of their findings.

Discussion Status

The conversation includes attempts to clarify the fixed point and its stability conditions, with some participants providing alternative perspectives on the existence of solutions. There is an ongoing exploration of the mathematical reasoning involved, but no consensus has been reached regarding the correctness of the approaches taken.

Contextual Notes

Participants are working under the constraints of the problem's parameters, specifically the range of c from 0 to 2.3, and are grappling with the implications of their algebraic manipulations and the behavior of the function.

shadedude123
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This question deals with a 1-parameter equation:

fc(x)= (4/x)+(x/2)-c for 0≤c≤ 2.3

a.) Use algebra to find the positive fixed point p(c) of fc(x) (in terms of c) and identify its exact interval of existence

a.) Ok so for the fixed point I got p(c)= -c+√(c2+8) I'm not sure if this is correct though. I used the quadratic formula to find this after setting the initial equation equal to x and solving in terms of c. Now for the interval of existence I need to take the first derivative of the initial equation which I have as -4x-2+(1/2). Is this right? And I think I'm supposed to plug in the fixed point i have into this first derivative to get the interval of stability for c by setting an inequality up like: -1<c<1. This would be the interval of existence for the parameter c right? Please correct me if I'm making any mistakes. I can't do this last step because i end up getting ridiculous numbers.
 
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Hi,
I do not know if you still need the help, but i found it interesting to see the answer :)

so from
f(x) = x
or
4/x+x/2-c = x
we have
x^2+2cx-8 = 0
or
x = p(c) = -c+-sqrt{c^2+8}
and since we are only looking for the positive solution we choose
p(c) = -c+sqrt{c^2+8}for the interval we choose the c such that the solution of x=f(x) exists
or in other words that
-1 < f'(x) < 1 , where x = p(c)

Since
f'(x) = -4/(x^2)+1/2
we have

-1 < -4/(2 c^2+8-2c sqrt{c^2+8})+1/2 < 1

from where we obtain
0 < c < 2 sqrt{2/3}

I think this is correct, but if you stil need it I could check and make a better calculation.
 
Last edited:
shadedude123 said:
This question deals with a 1-parameter equation:

fc(x)= (4/x)+(x/2)-c for 0≤c≤ 2.3

a.) Use algebra to find the positive fixed point p(c) of fc(x) (in terms of c) and identify its exact interval of existence

a.) Ok so for the fixed point I got p(c)= -c+√(c2+8) I'm not sure if this is correct though. I used the quadratic formula to find this after setting the initial equation equal to x and solving in terms of c. Now for the interval of existence I need to take the first derivative of the initial equation which I have as -4x-2+(1/2). Is this right? And I think I'm supposed to plug in the fixed point i have into this first derivative to get the interval of stability for c by setting an inequality up like: -1<c<1. This would be the interval of existence for the parameter c right? Please correct me if I'm making any mistakes. I can't do this last step because i end up getting ridiculous numbers.

I don't understand the question. You have found a solution of x = f(x) in terms of c, and it exists and is positive for all real values of c.

If you don't trust algebra and logic, try instead to use plotting: you can plot f(x) for any given value of c and see right away that the line y = x cuts the graph y = f(x) in just one point x > 0.

RGV
 

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