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Math Help - Tangent line, math wrong somewhere

  1. #1
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    Tangent line, math wrong somewhere

    I have an equation  f(x) = \frac{5x}{(3-5x)^5} and I want the equation of the tangent line at x = 2. I know what to do, but my math is wrong somewhere down the line.

    Subbing in 2  f(2) = \frac{5(2)}{(3-5(2))^5} = -0.000595

    Taking derivative  f'(x) = \frac {100x+15}{(3-5x)^6} and then plugging in 2  f'(2) = \frac {100(2)+15}{(3-5(2))^6} = 0.00183

    So, I use equation of a line now that I have the solutions for f(x) and m....  -0.000595 = 0.00183(2) + b and I wind up with  b = 0.0043

    I get the equation of the tangent line should be  y = 0.00183x + 0.0043 ,but this is incorrect, and I was hoping somebody could eyeball this if they get a chance because I think I'm going in circles trying to fix it.
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    MHF Contributor MarkFL's Avatar
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    Re: Tangent line, math wrong somewhere

    Your y-intercept is wrong. Using your decimal approximations, you should have:

    b = (-2)0.00183 + 0.000595 = -0.003065
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    Re: Tangent line, math wrong somewhere

    Of course, there's no reason why you should be using decimal approximations anyway...
    Thanks from AZach
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    Re: Tangent line, math wrong somewhere

    I see how your math works, but why did you change 2 to a -2?
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    Re: Tangent line, math wrong somewhere

    Quote Originally Posted by Prove It View Post
    Of course, there's no reason why you should be using decimal approximations anyway...
    True.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Tangent line, math wrong somewhere

    Quote Originally Posted by AZach View Post
    I see how your math works, but why did you change 2 to a -2?
    Think of the point-slope formula for a line.

    Also, I agree with the above, I would use the true rational values...because we can.
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