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Easier way to solve cubics

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Just wondering if anyone can suggest methods for solving cubic functions that are easy to remember.

    For example, I am solving for a minimum value and I have the following equation:
    x+4x^3-1=0
     
  2. jcsd
  3. Oct 15, 2011 #2
    also should mention I only need real roots
     
  4. Oct 15, 2011 #3
    Local min? Because global min is negative infinity...

    Take the derivative of the function and find the 2 points where the slope is zero. One will be a local min, the other a local max. Plug in numbers on either side of your answers to get the signs (tells you if the slope is positive or negative. Positive on the left side and negative on the right side means it's a local max. Negative on the left side and positive on the right side means it's a local min.
     
  5. Oct 15, 2011 #4
    Okay.. Doesnt seem to work for this equation.

    4x^3+x-1=0
    12x^2+1=0
    x^2=-1/12
    ??

    The function I provided you with was already a derivative. I just needed to solve it for the root to know what the min was.
     
  6. Oct 15, 2011 #5

    SammyS

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    Write the terms with powers in decreasing order.

    Use the rational root theorem, the remainder theorem, and long or synthetic division.
     
  7. Oct 15, 2011 #6
    Are the roots going to be the same for the derivative of a cubic and the actual cubic function?
     
  8. Oct 15, 2011 #7

    SammyS

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    The derivative will not be a big help !
     
  9. Oct 15, 2011 #8
    I already know what the answer is, I just needed a little refresher as I will probably need to be solving these quite a bit without electronic devices.
     
  10. Oct 15, 2011 #9

    SammyS

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    Do you know the rational root theorem, & the remainder theorem?
     
  11. Oct 15, 2011 #10
    This is not true. The op is saying they need to solve the algebraic equation:
    [tex]
    x + 4 x^3 - 1 = 0
    [/tex]
    which means they already had found the first derivative and equated it to zero. A primitive function of the above l.h.s. is:
    [tex]
    F(x) = \frac{x^2}{2} + x^4 - x + C
    [/tex]
    which tends to positive infinity at both [itex] x \rightarrow \pm \infty[/itex] because the fourth power is the most dominant term and has a possitive coefficient. This means that the function is bounded from below and has a well defined global minimum, which is a local minimum at the same time.
     
  12. Oct 15, 2011 #11
    The only possible rational zeros of:
    [tex]
    4 x^3 + x - 1= 0
    [/tex]
    have a numerator that is a divisor of 1, i.e. +1 or -1, and a denominator that is a divisor of 4, i.e. 1, 2, 4. Thus, the only possible rational zeros are:
    [tex]
    \pm \frac{1}{4}, \pm \frac{1}{2}, \pm 1
    [/tex]
    You should use Horner scheme to see if any of these are roots of your cubic equation, and, if so, to get the remaining quadratic equation which can be solved easily.
     
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