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Math Help - maxima minima problem HELP!

  1. #1
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    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.
    PLEASE HELP
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  2. #2
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    Lets not worry too much, should be quite simple

    Quote Originally Posted by choward View Post
    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 View Post
    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!
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  3. #3
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    i used the quotient rule and got
    f '(x)= (x^4-3x^2-6)/((x^2)-2)^2
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  4. #4
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    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
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  5. #5
    MHF Contributor Unknown008's Avatar
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    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}
    Last edited by Unknown008; November 11th 2010 at 09:25 PM. Reason: typo
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  6. #6
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    I'm still not seeing how you got 2x^4
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  7. #7
    MHF Contributor Unknown008's Avatar
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    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}
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  8. #8
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    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
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  9. #9
    MHF Contributor Unknown008's Avatar
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    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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