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Thread: Math homework help (should be easy for you cos I'm noob)

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    goodieboy's Avatar
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    Math homework help (should be easy for you cos I'm noob)

    It's the basics of Calculus, I think, need to use some differentiation

    Find the range of values of 'a' such that the gradient of the tangent to the curve y = (2x + a)/(4 - x^2) is always positive. ('a' is a constant here)

    I could get up to the point of dy/dx = (2x^2 + 2ax + 8)/((4 - x^2)^2), but can't go on after that.

    All help appreciated!
    Last edited by goodieboy; 12-10-2012 at 08:20 PM.

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    Bump, still need help

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    @Reese, help me? Thanks! Will + rep.

    Still need this done

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    The derivative you found is the slope. The question is asking about what ranges the slope is positive.




    Is it from -8 to positive infinity?

    y = (2x + a)/(4 - x)

    y prime= [(4 - x)(2) -(2x + a)(-1)] / (4 - x)^2

    y prime = 8-2x+2x+a/(4 - x)^2

    y prime = 8+a/(4 - x)^2 //denominator doesn't matter because it is squared. always positive except for a undefined at x=4

    8+a>0

    a>-8 <<---that's my conclusion
    Last edited by Graff; 12-09-2012 at 10:52 PM.

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    Reese's Avatar
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    Heh, I started writing it out but Graff beat me to it . I can go and doublecheck, but everything looks good ^_^
    I can try and explain everything if you'd like(gradient of tan=y2-21/x2-x1, there's lots of ways to look at what values are positive and what are negative, and that's usually just what makes sense to you)
    Last edited by Reese; 12-09-2012 at 11:04 PM.

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    Dammit I made a friggin mistake when typing the question. Will + full rep to both of you if you guys can help me solve it again. I've edited the question, help me again please?


    @Graff @Reese, the answer should be -4 < a < 4

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    y = (2x + a)/(4 - x^2)

    y prime= [(4 - x^2)(2) -(2x + a)(-2x)] / (4 - x^2)^2

    y prime= 2(x^2 + xa + 4)/(4-x^2)^2

    2(x^2 + xa + 4)/(4-x^2)^2>0

    [2(x^2 + xa + 4)/(4-x^2)^2>0]*(4-x^2)^2
    [2(x^2 + xa + 4)>0]/2

    (x^2 + xa + 4)>0

    use quad formula (to find critical points)

    {-a(+/-)sqrt[a^2-(4*1*4)] }/2

    (a^2 -16)^.5 --->> [(a+4)(a-4)]^.5

    CPs @ a=-4 and 4



    yprime = 2(x^2 + xa + 4)/(4-x^2)^2

    plug in 3 values that are in between +/- infinity and the CP's to make sure that it's positive in that range only.



    Struggled a bit, sorry. This is my final answer, I'll let you do the algebra.
    Last edited by Graff; 12-10-2012 at 09:56 PM.

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