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Math Help - Tangent Line

  1. #1
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    Tangent Line

    thanks
    Last edited by camjenson; November 12th 2012 at 10:19 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Tangent Line

    The line tangent to a function f(x) at x=c (assuming the derivative exists at this point) is found by the point slope formula:

    y-f(c)=f'(c)(x-c)

    Writing this in slope-intercept form, we have:

    y=f'(c)x+(f(c)-cf'(c))

    Now, using f(x)=\frac{1}{\ln(bx)} and c=\frac{e}{b} can you now compute the required y-intercept?
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  3. #3
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    Re: Tangent Line

    To be honest, no, I don't know where those variables would go in the equation...
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Tangent Line

    First, compute the derivative of f(x)=\frac{1}{\ln(bx)}. Next, let x=c=\frac{e}{b} and evaluate the yintercept:

    f(c)-cf'(c)
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  5. #5
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    Re: Tangent Line

    Isn't this the derivative -1/(x*[(ln(x))^(2)])?
    So, in more simpler words, isn't it y-y1=m(x-x1)? y being the intercept y1 being the derivative after putting e/b in for x, m being the derivative, x being 0, and x1 being e/b?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Tangent Line

    Your derivative is correct, and next you correctly cite the point-slope formula for a line, but after that you contradict yourself/make wrong statements, and I am not sure what you really mean. I think what I stated is about as simple as it gets.
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