# Finding local max, min, and points of inflection

• December 12th 2012, 08:59 PM
Chaim
Finding local max, min, and points of inflection
Find all local maximum points, local minimum points, and points of inflection on the curve y=-3x4-8x3+18x2
What I did: First derivative is y'=-12x3-24x2+36x
Then I set it equal to 0
Then I factored it out: -12x(x2+2x-3) = -12x(x-1)(x-2), so the critical numbers are x=0, x=1, and x=2

After getting the critical numbers, I did the second derivative
y''=36x2-58x+36
I used the second derivative test, and they end up all positive, so it's all min,
So I'm a bit confused.
Isn't this the way to find absolute minimum, instead of local minimum o.o

And how do I go on from there :o
• December 12th 2012, 09:03 PM
Prove It
Re: Finding local max, min, and points of inflection
Quote:

Originally Posted by Chaim
Find all local maximum points, local minimum points, and points of inflection on the curve y=-3x4-8x3+18x2
What I did: First derivative is y'=-12x3-24x2+36x
Then I set it equal to 0
Then I factored it out: -12x(x2+2x-3) = -12x(x-1)(x-2), so the critical numbers are x=0, x=1, and x=2

After getting the critical numbers, I did the second derivative
y''=36x2-58x+36
I used the second derivative test, and they end up all positive, so it's all min,
So I'm a bit confused.
Isn't this the way to find absolute minimum, instead of local minimum o.o

And how do I go on from there :o

There is no way that all three critical points can possibly be minima.

To answer your other question, absolute maxima and minima can occur EITHER at critical points OR at endpoints of your function.
• December 13th 2012, 02:24 PM
Chaim
Re: Finding local max, min, and points of inflection
Oh oops, I factored it wrong
It's actually 12x(x+3)(x-1)
So x=0, -3, and 1

So the 2nd derative would be:
-36x2-58x+36
Plugging in 0 would make it 36, (So Minimum)
Plugging in 1 would make it Negative (So Maximum)
Plugging in -3 would make it Positive (So Minimum)
Though, I'm a bit confused in the multiple choice answers, because x=-3, 1 is supposed to match up into a max or min, while x=0 is alone either a max or min.

If I use the factored one: -2(18x2+24x-18)
(Or maybe the multiple choice answers could be wrong as well)

Points of inflection would be where 2nd derivative is equal to 0 right?
So I did the quadratic formula within the paranthesis, and I got something different: (-29+ or - sqrt2137)/36

While in the multiple choice answers, it was like either (-2 + or - sqrt13)/3 or (-2 + or - sqrt19)/3
• December 13th 2012, 03:09 PM
skeeter
Re: Finding local max, min, and points of inflection
$y'' = -36x^2 - 48x + 36$

$y'' = -12(3x^2 + 4x - 3)$

$x = \frac{-4 \pm \sqrt{40}}{6} = \frac{-2 \pm \sqrt{10}}{3}$
• December 13th 2012, 04:06 PM
Chaim
Re: Finding local max, min, and points of inflection
Quote:

Originally Posted by skeeter
$y'' = -36x^2 - 48x + 36$

$y'' = -12(3x^2 + 4x - 3)$

$x = \frac{-4 \pm \sqrt{40}}{6} = \frac{-2 \pm \sqrt{10}}{3}$

Oh! No wonder!
I did 58x instead of 48x, wow xD
Thanks for pointing that out! :)