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.
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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!
i used the quotient rule and got
f '(x)= (x^4-3x^2-6)/((x^2)-2)^2
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
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.
Last edited by Unknown008; November 11th 2010 at 09:25 PM.
I'm still not seeing how you got 2x^4
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.
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
Put x = 0 and you get the y intercept as 0, which is good.
Then, put y = 0. You get:
Solve for the 3 solutions of x.
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