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Math Help - [SOLVED] Finding the equation of a function

  1. #1
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    [SOLVED] Finding the equation of a function

    find the function f(x) the graph passes through (1,2) and whose tangent line at the point (x,F(x)) has a slope of 1/x3
    Last edited by mr fantastic; March 11th 2010 at 03:43 AM.
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  2. #2
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    Quote Originally Posted by cole123 View Post
    find the function f(x) the graph passes through (1,2) and whose tangent line at the point (x,F(x)) has a slope of 1/x3
    Dear cole,

    What is the slope of the graph? Is it \frac{1}{3} ? There is a typo.
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  3. #3
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    sorry it is 1/x^3
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  4. #4
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    Dear cole123,

    Suppose (x,y) is any point on this curve. Then, since (1,2) is also on this curve, the tangent of the curve is,

    \frac{y-2}{x-1}=\frac{1}{x^{3}}

    You could simplify the expression as you wish(eg; take y=f(x) if you want) to get the equation of the curve.

    Hope this will help you.
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  5. #5
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    Im having trouble with that part I know the next step leads to:

    f(x)= -(1/2)x^-2 + c but I have no idea how to get this!
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  6. #6
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    Hi cole123

    \frac{d}{dx}\left(-\frac{1}{2x^2}+C\right)=\frac{d}{dx}\left(-\frac{x^{-2}}{2}+C\right)=x^{-3}

    If f(1)=2

    -\frac{1}{2}+C=2\ \Rightarrow\ C=2.5

    f(x)=2.5-\frac{1}{2x^2}

    If we integrate \frac{1}{x^3}

    we get \frac{x^{-2}}{-2}+C
    Last edited by Archie Meade; March 11th 2010 at 04:52 AM. Reason: quoted the wrong posting accidentally
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  7. #7
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    I dont understand why it is not -(1/3) x^-2?
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  8. #8
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    Quote Originally Posted by cole123 View Post
    Im having trouble with that part I know the next step leads to:

    f(x)= -(1/2)x^-2 + c but I have no idea how to get this!
    Dear cole123,

    My method is incorrect. I had done it assuming the curve is a straight line which is of course incorrect. But Archie Meade's answer is correct.

    Dear Archie Meade,

    Thank you for giving the correct answer. Now I know what I did wrong.
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  9. #9
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    Hi cole123,

    It's because, when we integrate, we must calculate the new power
    before dividing by it.

    \int{x^n}dx=\frac{x^{n+1}}{n+1}+C

    and when we differentiate, we multiply by the original power
    before reducing the power by 1.

    \frac{d}{dx}\left(x^n+C\right)=nx^{n-1}

    Hi Sudharaka,
    yes, with the curve, only the point of tangency is on the line
    in that vicinity.
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  10. #10
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    Thank you very much!
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