# Local Max &amp; Min problem

[SOLVED] Local Max &amp; 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?

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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:

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 $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

Then take the value of $$x_{n+1}$$ 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.

HallsofIvy
Homework Helper
"Mesmerize" it? Now why didn't I think of that! I'll just hynotize formulas into doing my bidding!

"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

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

$$g(x) = 1 + 4x - 10x^2 + x^4$$
$$(dy/dx) = 4 - 20x + 4x^3$$

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