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  1. #1
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    differentiation problem

    Given that [latex] y= \frac{1}{3}x^3-9x [/latex]

    This part asks me to differentiate the equation and find the minimum and maximums, but I did all that, byt the next part got me:

    Given that [latex] 24x+3y+2=0 [/latex] is the equation of the tangent to the curve at the point (p,q) , find p and q.


    Now I got the answer to be (1,-22/3) and (-1, 26/3)

    But theres only one answer which is (1,-22/3) and the other one isn't mentioned in the answer book. I know that the tangent is only hitting the curve at one point, but what can help me decide between the two? (by the way I used the method where I found what values of x made the derivative = -8, rather than solving the 2 simultaneously because Im bad at factorising cubics.)
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  2. #2
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    Quote Originally Posted by david18 View Post
    Given that  y= \frac{1}{3}x^{3}-9x

    This part asks me to differentiate the equation and find the minimum and maximums, but I did all that, byt the next part got me:

    Given that  24x+3y+2=0 is the equation of the tangent to the curve at the point (p,q) , find p and q.


    Now I got the answer to be (1,-22/3) and (-1, 26/3)

    But theres only one answer which is (1,-22/3) and the other one isn't mentioned in the answer book. I know that the tangent is only hitting the curve at one point, but what can help me decide between the two? (by the way I used the method where I found what values of x made the derivative = -8, rather than solving the 2 simultaneously because Im bad at factorising cubics.)

    The tags are 'math', not 'latex'.

    The given line equation, when solved for y, yields y=-8x-\frac{2}{3}

    \frac{1}{3}x^{3}-9x = -8x-\frac{2}{3}

    Solving this gives us x=1 or x=-2.

    But one of these may just be an intersection point. The slope of said line is -8.

    So, where does the curve have slope -8?.

    x^{2}-9=-8

    x=1
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by david18 View Post
    Now I got the answer to be (1,-22/3) and (-1, 26/3)

    I get (1; - \frac{26}{3} ) \ and \ (-1 ; \frac{22}{3})

    Secondly, if x^2 = 1, you are not allowed to say x = \sqrt{1} = \pm 1 (That's what i was taught anyway).
    Last edited by janvdl; January 5th 2008 at 10:45 AM. Reason: Sorry, forgot a negative sign
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    \frac{1}{3}x^{3}-9x = -8x-\frac{2}{3}
    Galactus shouldn't we be using x^2 - 9 ?

    The tangent to the derivative of the original equation? Isn't that what he's asking?
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  5. #5
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    Quote Originally Posted by janvdl View Post
    Galactus shouldn't we be using x^2 - 9 ?

    The tangent to the derivative of the original equation? Isn't that what he's asking?
    Absolutely, janvdl. That's why I amended my post.
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