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Math Help - Horizontal Tangent Lines

  1. #1
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    Horizontal Tangent Lines

    f(x)= xe^(x)/(9x+b)


    How would you find the coordinates of f where the tangent lines are horizonal when b=-7?
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  2. #2
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    Quote Originally Posted by ctran View Post
    f(x)= xe^(x)/(9x+b)


    How would you find the coordinates of f where the tangent lines are horizonal when b=-7?
    If b = -7, then

    f(x) = \frac{xe^x}{9x - 7}.


    Tangent lines are horizontal when the derivative is 0.


    So f'(x) = \frac{(9x - 7)\,\frac{d}{dx}(xe^x) - xe^x\,\frac{d}{dx}(9x - 7)}{(9x - 7)^2}

     = \frac{(9x - 7)(xe^x + e^x) - 9xe^x}{(9x - 7)^2}


    Now let f'(x) = 0 and solve for x.


    \frac{(9x - 7)(xe^x + e^x) - 9xe^x}{(9x - 7)^2} = 0

    e^x(9x - 7)(x + 1) - 9xe^x = 0

    e^x[(9x - 7)(x + 1) - 9x] = 0


    Since e^x \neq 0 for any x, we have

    (9x - 7)(x + 1) - 9x = 0

    9x^2 +9x - 7x - 7 - 9x = 0

    9x^2 - 7x - 7 = 0

    x = \frac{7 \pm \sqrt{301}}{18}.


    Now, substitute these values of x back into f(x).
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