Calculas.

Iwasfrozen

Hey, I was wondering if anybody could give me a hand with this problem I'm stuck on.
y=3x^4-2x^3-9x^2+8
Find the co-ordinates of the two local minimum points.
I know that dy/dx=o and d^2y/dx^2>0 but it doesen't workout right.
thanks in advance.

Des23

when you differentiate you get 12x^3-6x^2-18x
take 6x outside the bracket and you left inside with a quadratic. 2x^2-x-3
solve quadratic to get x=3/2 and x=-1 and you already have x=0 from taking 6x out side.
Fill these in to the second differential. the two non 0 terms are >0 fill these back into f(x) to get your y values and voila! you done!.... I think..

Iwasfrozen

So lets see, that would be.
y=3x^4-2x^3-9x^2+8
dy/dx=12x^3-6x^2-18x => 6x[2x^2-x-3] => 6x[2X^2-3x+2x-3] => 6x[x(2x-3)+1(2x-3)]
x=3/2 x=-1
Add to y:
12(3/2)^3-6(3/2)^2-18(3/2) => 81/2-27/2-27=0
12(-1)^3-6(-1)^2-18(-1) => -12-6-18=-36
Does that seem right ? How can y1=o ?

Des23

Iwasfrozen wrote: »

Does that seem right ? How can y1=o ?

It touches the x axis? ... maybe I didn't read the question right or something.

Iwasfrozen

him ok, I suppose that makes sense.

x = 3/2 and x = -1 are correct.

http://www46.wolframalpha.com/input/?i=2x%5E2+-+3x+%2B+2x+-+3+%3D+0

Fremen

If you're spelling it like that, you might be in more trouble than you think.

Iwasfrozen

If you're spelling it like that, you might be in more trouble than you think.

Yes, because spelling is very important when studying maths. :rolleyes:

LeixlipRed

Ladies and gents leave it at that please.