# maxima minima problem HELP!

• November 11th 2010, 07:16 PM
choward
maxima minima problem HELP!
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.
• November 11th 2010, 07:24 PM
pickslides
Lets not worry too much, should be quite simple

Quote:

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?

Quote:

Originally Posted by choward
Then i tried to find the derivative but i don't think it was right because it's rly long.

What rule did you use? Show me your attempt, it might be correct!
• November 11th 2010, 07:39 PM
choward
i used the quotient rule and got
f '(x)= (x^4-3x^2-6)/((x^2)-2)^2
• November 11th 2010, 07:42 PM
choward
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
• November 11th 2010, 08:06 PM
Unknown008
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.

$f'(x) = \dfrac{(x^2 - 2)(3x^2-3) - (x^3-3x)(2x)}{(x^2 - 2 )^2} = \dfrac{x^4 -3x^2 + 6}{(x^2 -2)^2}$
• November 11th 2010, 08:22 PM
choward
I'm still not seeing how you got 2x^4
• November 11th 2010, 08:26 PM
Unknown008
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.

$\begin{array}{cl}
f'(x) = \dfrac{(x^2 - 2)(3x^2-3) - (x^3-3x)(2x)}{(x^2 - 2 )^2} & = \dfrac{3x^4 - 3x^2 -6x^2 + 6 - (2x^4-6x^2)}{(x^2 - 2 )^2} \\
& = \dfrac{x^4 -3x^2 + 6}{(x^2 -2)^2}\end{array}$
• November 11th 2010, 08:47 PM
choward
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
• November 11th 2010, 08:52 PM
Unknown008
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.