# How to search extreme relative on cubic equation.

I have a cubic equation:

$y = 2x^3-12x^2+18x$

I want to search its extreme relative (extreme point). What should i do? With which process should i use?

And can I do this equation with that process?
$y = x^3+3x^2$

Note: I'm beginner

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It looks like they are two separate problems, so you should solve each one separately using one equation. – Ross Millikan Oct 22 '12 at 14:10
Is the "algebra-precalculus" tag intended - because the obvious way of dealing with this (per answer) uses calculus. – Mark Bennet Oct 22 '12 at 15:29
No, because i didn't found cubic-equations. So I use algebra. – lambda23 Oct 23 '12 at 9:48

I intend to search for one of your functions, say the first one: $$y = 2x^3-12x^2+18x$$ The method says do the first derivative of $y$ and put it to zero because at relative extreme points $y'$ exists and is equal to $0$. You get $$y'=6x^2-24x+18=0\longrightarrow 6(x-1)(x-3)=0$$ and then $x=1,x=3$. These values can lead us to the relative extreme points, called critical points. Now you can use another method to find out which one of above $x$ is related to $max$ and which one is related to $min$. It says if some $x$ is caused by solving $y'=0$ such that $y''$ exists, then if $y''(x)>0$ then that $x$ lead us to $min$ point and if $y''(x)<0$ then that $x$ lead us to $max$. Here, we have $y''=12x-24$ and so $$x=1\longrightarrow y''\big|_{x=1}=12-24=-12<0$$ so $y(1)=2-12+18=8$ is the relative maximum. And so $$x=1\longrightarrow y''\big|_{x=3}=36-24=+12>0$$ so $y(3)=0$ is the relative minimum. Here a plot of the first function:

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eye-catching!! ;-) +1 – amWhy Apr 1 '13 at 0:09