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Math Help - Calculus Problem

  1. #1
    Affinity
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    Smile Calculus Problem

    Find all x-coordinates of points (x,y) on the curve of where the tangent line is horizontal. (Thanks!!)
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Affinity View Post
    Find all x-coordinates of points (x,y) on the curve of where the tangent line is horizontal. (Thanks!!)
    The tangent line will be horizontal when y'(x) = 0. So:
    y(x) = \frac{(x-2)^5}{(x-4)^3}

    Using the quotient rule:
    y'(x) = \frac{5(x-2)^4(x-4)^3 - (x-2)^53(x-4)^2}{(x-4)^6}

    Doing a little factoring:
    y'(x) = \frac{(x-2)^4(x-4)^2}{(x-4)^6} \cdot (5(x-4) - 3(x-2))

    y'(x) = \frac{(x-2)^4}{(x-4)^4} \cdot (5x - 20 - 3x + 6)

    y'(x) = \frac{(x-2)^4}{(x-4)^4} \cdot (2x - 14)

    y'(x) = \frac{2(x-2)^4(x - 7)}{(x-4)^4}

    We need to solve this for y'(x) = 0.

    As long as x \neq 4 may simply set the numerator equal to 0:

    0 = 2(x-2)^4(x - 7)

    This will obviously happen at x = 2 and x = 7.

    -Dan
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