Expert: Paul Klarreich - 10/20/2010

Question
I am trying to solve for the five variables in this equation.  However, I have only three points.  I tried solving for the variables using the method explained below:  

My equation is
F(x) = qx^4-rx^3-sx^2+tx+u

local minimums
(-.398238, -.457621)
(1.20556, -3.46602)

local maximum
(.130182, .069458)

I know that minima and maxima are equal to zero at f'(x).  So that means I have six equations f(x) and f'(x) at each of the above points.  I then solved for the variables.  However, my answer checker shows that I should have whole numbers (or close to whole numbers) for all five variables.  What I came up with was small decimal numbers (all less than 1.0).  I rechecked my addition/subtraction, but I did not see any obvious mistakes.  Am I approaching this problem the right way or do I need to look at this differently?  If I am incorrect, can you push me in the right direction?

Thanks

Answer
Questioner:   David
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Calculus Problem
Question:  I am trying to solve for the five variables in this equation.  However, I have only three points.  I tried solving for the variables using the method explained below:  

F(x) = qx^4-rx^3-sx^2+tx+u

local minimums
(-.398238, -.457621)
(1.20556, -3.46602)

local maximum
(.130182, .069458)

I know that minima and maxima are equal to zero at f'(x).  So that means I have six equations f(x) and f'(x) at each of the above points.  I then solved for the variables.  However, my answer checker shows that I should have whole numbers (or close to whole numbers) for all five variables.  What I came up with was small decimal numbers (all less than 1.0).  I rechecked my addition/subtraction, but I did not see any obvious mistakes.  Am I approaching this problem the right way or do I need to look at this differently?  If I am incorrect, can you push me in the right direction?

Thanks
......................................
I think you have these six equations:

If (-.398238, -.457621) is a local minimum, then

F(-.398238) = -.457621
and
F'(-.398238) = 0

Likewise you have two at:
(1.20556, -3.46602)

and at your local maximum
(.130182, .069458)

Since you have only five variables, you should be able to discard one of the equations.  The remaining five should be solvable, but I am not in a hurry to do the arithmetic.  Probably a small spreadsheet, like in Excel, will give you the actual coefficients, then you need some simultaneous equation-solver application.