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elis-cool

Cubics...

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Ok, I have a test soon and I need to know some stuff How do you find the turning points of a cubic in the form of: ax^3+bx^2+cx+d [edit]I think you have to differentiate or something then the second derivitive tests for the max or min...[/edit] And how do you find the roots using the Bisection method and the Newton-Raphson method? CEO Plunder Studios [edited by - elis-cool on June 24, 2002 12:24:34 AM]

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Want to find turning points of f(x) = ax^3 + bx^2 + cx + d, ie, want to find where the first derivative is zero.

f'(x) = 3ax^2 + 2bx + c
solve f'(x) = 0, use quadratic formula
x = (-2b +- sqrt(4b^2 - 12ac)) / (6a)
x = (-b +- sqrt(b^2 - 3ac)) / (3a)


EDIT: you want to find the roots, too.

x^3 + ax + b = 0 (all cubics can be simplified to this)

Suppose x = s - t

(s - t)^3 + a(s - t) + b = 0
s^3 - 3s^2t + 3st^2 - t^3 + as - at + b = 0
s^3 - t^3 + b + (s - t)(a - 3st) = 0


We have some flexibility over the values of s and t. Suppose s^3 - t^3 + b = 0 and a - 3st = 0. Then the above equation certainly holds true. Now:

s = a/(3t)
a^3/(27t^3) - t^3 + b = 0
-t^6 + bt^3 + a^3/27 = 0


The rest shouldn't be too hard

[edited by - Beer Hunter on June 25, 2002 6:57:03 PM]

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