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

  • $\begingroup$ It looks like they are two separate problems, so you should solve each one separately using one equation. $\endgroup$ Oct 22, 2012 at 14:10
  • 1
    $\begingroup$ Is the "algebra-precalculus" tag intended - because the obvious way of dealing with this (per answer) uses calculus. $\endgroup$ Oct 22, 2012 at 15:29
  • $\begingroup$ No, because i didn't found cubic-equations. So I use algebra. $\endgroup$
    – lambda23
    Oct 23, 2012 at 9:48

1 Answer 1


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:

enter image description here

  • $\begingroup$ eye-catching!! ;-) +1 $\endgroup$
    – amWhy
    Apr 1, 2013 at 0:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .