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Expert: - 10/16/2010


Question
I am trying to solve for the five variables in this equation.  However, I have only three points.  I am a litte lost on how to solve for five variables with only three points.  

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)


Thanks for the help

Answer
There are more equations here than is first seen.
To be a local minimum, it must solve F'(x) = 0 and
F(x) = y value given.

The equations are
F(-0.398238) = -.457621, F'(-0.398238) = 0,
F(1.20556) = -3.46602, F'(1.20556) = 0,
F(0.130182) = 0.069458, and F'(0.130182) = 0.

Here we have three zeros of F'(x), and it is known that
F'(x) = 4qx^3 + 3rx^2 + 2sx + t.
Multiply (x-r1)(x-r2)(x-r3) where r1, r2, and r3, are, respectively,
r1 = -0.398238, r2 = 1.20556, and  r3 = 0.130182.

Note this gives
x^3 - (r1+r2+r3)x^2 + 2(r1*r2 + r1*r3 + f2*r3)x - r1*r2*r3.

I would put the values in a spreadsheet and solve from there.

If the value of the funtion at r1, r2, and r3 is f1, f2, and f3, respectively, try f(x) = f1(x-r2)(x-r3)/[(r1-r2)(r1-r3)] +
f2(x-r1)(x-r3)/[(r2-r1)(r2-r3)] + f3(x-r1)(x-r2)/[(r3-r1)(r3-r2)].

Does this help?

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