Efficient Methods for Solving Cubic Functions

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Homework Help Overview

The discussion revolves around methods for solving cubic functions, specifically focusing on finding real roots and determining minimum values. The original poster presents the equation x + 4x^3 - 1 = 0 and seeks easy-to-remember techniques for solving such equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods, including the use of derivatives to find local minima and maxima, the rational root theorem, and polynomial division techniques. There are questions about the relationship between the roots of the derivative and the original cubic function.

Discussion Status

The conversation includes attempts to clarify the original poster's needs and the effectiveness of suggested methods. Some participants express skepticism about the utility of derivatives in this context, while others provide insights into potential rational roots and techniques for solving cubic equations.

Contextual Notes

There is mention of the need for real roots and a focus on solving without electronic devices. The original poster indicates familiarity with the material but seeks a refresher on solving cubic functions.

sandy.bridge
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Homework Statement


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
 
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also should mention I only need real roots
 
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.
 
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.
 
Write the terms with powers in decreasing order.

Use the rational root theorem, the remainder theorem, and long or synthetic division.
 
sandy.bridge said:
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.

Are the roots going to be the same for the derivative of a cubic and the actual cubic function?
 
The derivative will not be a big help !
 
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.
 
Do you know the rational root theorem, & the remainder theorem?
 
  • #10
Allenman said:
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.

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.
 
  • #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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