maxima minima problem HELP!

**choward** · November 11th 2010, 07:16 PM

the problem is f(x)=(x^3-3x)/(x^2-2)
I don't even know how to start I have to find the y and x intercepts first can someone tell me how i do that?
Then i tried to find the derivative but i don't think it was right because it's rly long.
PLEASE HELP

**pickslides** · November 11th 2010, 07:24 PM

Lets not worry too much, should be quite simple

Originally Posted by choward

I don't even know how to start I have to find the y and x intercepts first can someone tell me how i do that?

To find the y-intercept make x=0, what do you get?

To find the x-intercept make y=0, what do you get?

Originally Posted by choward

Then i tried to find the derivative but i don't think it was right because it's rly long.
PLEASE HELP

What rule did you use? Show me your attempt, it might be correct!

**choward** · November 11th 2010, 07:39 PM

i used the quotient rule and got
f '(x)= (x^4-3x^2-6)/((x^2)-2)^2

**choward** · November 11th 2010, 07:42 PM

to find the x and y intercepts am I supposed to get the derivative first and then plug in 0 for x and y because when you plug 0 in for x in the original eqn you get 0/-2 which isnt right

**Unknown008** · November 11th 2010, 08:06 PM

No, oyu plug in y = 0 and x = 0 in the equation of the curve, not it's derivative.

And your derivative has a mistake.

$f'(x) = \dfrac{(x^2 - 2)(3x^2-3) - (x^3-3x)(2x)}{(x^2 - 2 )^2} = \dfrac{x^4 -3x^2 + 6}{(x^2 -2)^2}$

**choward** · November 11th 2010, 08:22 PM

I'm still not seeing how you got 2x^4

**Unknown008** · November 11th 2010, 08:26 PM

Oh, drat... This is a typo from me, sorry for that. I was pointing to the +6 which you put as -6 in your post.

$\begin{array}{cl}<br /> f'(x) = \dfrac{(x^2 - 2)(3x^2-3) - (x^3-3x)(2x)}{(x^2 - 2 )^2} & = \dfrac{3x^4 - 3x^2 -6x^2 + 6 - (2x^4-6x^2)}{(x^2 - 2 )^2} \\<br /> & = \dfrac{x^4 -3x^2 + 6}{(x^2 -2)^2}\end{array}$

**choward** · November 11th 2010, 08:47 PM

when I plugged in 0 for x and y into the orignial equ'n i got things like 0/-2 which isnt right What am I doing wrong? the equ'n is y=f(x)=x^3-3x/x^2-2 and i have to find all x and y intercepts

**Unknown008** · November 11th 2010, 08:52 PM

Put x = 0 and you get the y intercept as 0, which is good.

Then, put y = 0. You get:

$\dfrac{x^3 - 3x}{x^2-2} = 0$

Which means:

$x^3 - 3x = 0$

Solve for the 3 solutions of x.

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