Finding Local Maxima and Minima Points for a Cubic Function

  • Thread starter Thread starter prophet05
  • Start date Start date
  • Tags Tags
    Amp Local Max
Click For Summary

Homework Help Overview

The discussion revolves around finding local maxima and minima points for the cubic function g(x) = 1 + 4x - 10x^2 + x^4. Participants are exploring methods to determine the critical points of the function through calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods, including factoring the derivative and using Newton's method to find zeros. There is uncertainty about the applicability of the cubic formula and the challenges of solving the derivative algebraically.

Discussion Status

The conversation includes attempts to apply Newton's method and references to the cubic formula, with some participants expressing difficulty in memorizing or applying these methods. There is a light-hearted exchange among participants, but no consensus on a specific approach has been reached.

Contextual Notes

One participant notes that the professor intended for the problem to be solved using a calculator, which may influence the methods discussed.

prophet05
Messages
12
Reaction score
0
[SOLVED] Local Max & Min problem

I'm having a real tough time finishing this problem. I have to find the local maxima and minima points for g(x).

g(x) = 1 + 4x - 10x^2 + x^4

(dy/dx) = 4 - 20x + 4x^3

I've been trying to factor it to get the max and min points, but I find it impossible to simplify. I've tried factoring and just can't seem to find a way. And I can't use quadratic formula since it's to the 3rd degree, right?
 
Physics news on Phys.org
The first thing I thought of was Newton's method because I'm not great when it comes to solving the zeros for that with pure algebra.
 
Well here's the cubic formula:

cubic.gif


For a cubic of the form ax^3+bx^2+cx+d

The roots are: 0.2016396757234, 2.12841906384458, -2.33005873956798

So Newton's method seems like that way to go.
 
Yea, the cubic formula is out, I am not going to mesmerize that. I'll try and read up on Newton's method.
 
Newton's method is a powerful tool for finding zeros. Take a guess, x_n, at what x-value the zero could be at:

Then [tex]x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}[/tex]

Then take the value of [tex]x_{n+1}[/tex] that you got and do the above calculation over and over again (around four or five times) until the number that you obtain settles down.

The only problem is that it doesn't tell you how many zeros the function has.
 
"Mesmerize" it? Now why didn't I think of that! I'll just hynotize formulas into doing my bidding!
 
HallsofIvy said:
"Mesmerize" it? Now why didn't I think of that! I'll just hynotize formulas into doing my bidding!

It's all part of the plan for world domination mwhahahaha!11!oneone!1!1
 
Feldoh said:
It's all part of the plan for world domination mwhahahaha!11!oneone!1!1

I don't usually jump on the bandwagon of a string of jokes, but that 'mwahahaha!11!oneone... had me rofl-ing!
 
Haha. Thanks for all the help. Turns out the professor wanted us to solve it with calculator. Mesmerize would have been a better way to go. =P

[tex]g(x) = 1 + 4x - 10x^2 + x^4[/tex]
[tex](dy/dx) = 4 - 20x + 4x^3[/tex]
 
Last edited:

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K