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

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- Nov 11th 2010, 07:16 PMchowardmaxima 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.

PLEASE HELP - Nov 11th 2010, 07:24 PMpickslides
- Nov 11th 2010, 07:39 PMchoward
i used the quotient rule and got

f '(x)= (x^4-3x^2-6)/((x^2)-2)^2 - Nov 11th 2010, 07:42 PMchoward
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

- Nov 11th 2010, 08:06 PMUnknown008
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.

- Nov 11th 2010, 08:22 PMchoward
I'm still not seeing how you got 2x^4

- Nov 11th 2010, 08:26 PMUnknown008
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.

- Nov 11th 2010, 08:47 PMchoward
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

- Nov 11th 2010, 08:52 PMUnknown008
Put x = 0 and you get the y intercept as 0, which is good.

Then, put y = 0. You get:

Which means:

Solve for the 3 solutions of x.