Locating Local Extrema in a Polynomial Equation

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Discussion Overview

The discussion revolves around finding the local extrema of the polynomial function f(x) = x^4 - 3x^2 + 2x. Participants explore methods for calculating the derivative, setting it to zero, and solving for x-coordinates of local extrema.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant suggests taking the first derivative of the function and setting it equal to zero to find critical points.
  • Another participant expresses confusion about how to handle the derivative, specifically the term 4x^3.
  • A different participant advises that there is no need to remove the derivative and hints at using polynomial division after finding an obvious solution.
  • One participant presents their solution involving the derivative 4x^3 - 6x + 2 and attempts to solve it, but expresses uncertainty about its correctness.
  • Another participant points out an error in the previous solution, questioning the manipulation of the equation and suggesting that an obvious solution can simplify the problem.
  • There is a request for a general formula to solve cubic equations due to time constraints from a participant.
  • One participant reiterates their solution and emphasizes the need to verify the roots by substituting them back into the original equation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct roots of the polynomial equation, and there are multiple competing views on the steps needed to find the local extrema. The discussion remains unresolved regarding the correctness of the proposed solutions.

Contextual Notes

There are limitations in the participants' understanding of polynomial manipulation and solving cubic equations, with some steps remaining unclear or incorrect. The discussion reflects varying levels of confidence in the mathematical processes involved.

alex123aaa
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how can i find the local extrema of x-coordinates in this equation?

f(x)=x^4-3x^2+2x
 
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Take a first derivative and set it equal to 0. Solve for x.
 
yeah i have to solve for x but how wil i get rid of the derivative 4x^3? i don't know how to remove it
 
You don't need to remove it.

Hint: There's a pretty obvious solution, find it and use polynomial division
 
here's my solution i really don't know if this is correct but please correct me so

take the derivative = 4x^3-6x+2

Set it equal to zero and begin solving. 4x^3-6x+2 = 0
4x^3-6x = -2
2x(2x^2-3) = -2
2x^2 - 3 = 0 | 2x = 0
2x^2 = 3 | x = 0
x^2 = 3/2
x = plus or minus the square root of 3/2 and x = 0
 
Well, you have a wrong step

how would 2x(2x^2-3)=-2
Suggests 2x^2-3=0 or 2x=0? If 2x^2-3=0 or 2x=0 then 2x(2x^2-3) would equal 0, certainly not -2.

Look at 4x^3-6x+2=0.
There's a complicated general formula for this. But I say you can at least guess one obvious solution, then your problem reduces to solving a second degree equation.
 
could you tell me at least what is that formula please I am running out of time i have to turn this paper in tomorrow
 
alex123aaa said:
here's my solution i really don't know if this is correct but please correct me so

take the derivative = 4x^3-6x+2

Set it equal to zero and begin solving. 4x^3-6x+2 = 0
4x^3-6x = -2
2x(2x^2-3) = -2
2x^2 - 3 = 0 | 2x = 0
"If ab= 0 then either a= 0 or b= 0". Not ab equal to anything other than 0!

2x^2 = 3 | x = 0
x^2 = 3/2
x = plus or minus the square root of 3/2 and x = 0
Did you notice that [itex]4(1)^3- 6(1)+ 2= 0[/itex]? Once you know that 1 is a root, you know that x- 1 is a factor. Divide [itex]4x^2- 6x+ 2[/itex] by x- 1 to reduce to a quadratic equation for the other roots.
 
alex123aaa said:
x = 0
x = plus or minus the square root of 3/2 and x = 0

Your roots are wrong, but once you get them correctly (plug them back into the equation to make sure it checks out). Then plug them back one by one in the original equation ( x^4 - 3x + 2x) and see which one gives you the highest value - that would be your maximum.
 
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