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Math Help - Please help. find the coordinates of points where the graph has horizontal tanngents

  1. #1
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    Please help. find the coordinates of points where the graph has horizontal tanngents

    Hello, could somebody please help me please.
    Find the coordinates of points where the graph f(x) has horizontal tangents. As a check, graph f(x) and see whether the points you find look as though they have horizontal tangents

    F(x)= 1/3x^3-3x^2-16x+8

    my answer is
    f(x)=-141.3
    f(x)=25.3
    is that correct?
    thanks
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  2. #2
    Super Member Bacterius's Avatar
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    When the derivative of the function in a point is 0, its tangent is horizontal.

    Your function is f(x) = \frac{1}{3} x^3 - 3x^2 - 16x + 8

    Take the derivative of this function (using the sum and power differentiation principles) :

    f'(x) = x^2 - 6x - 16

    Now you are looking for x-coordinates when the tangent is horizontal. Thus, you want to solve f'(x) = 0 :

    f'(x) = 0

    x^2 - 6x - 16 = 0

    It factorizes ! -6 = 2 - 8, 16 = 2 \times (-8) :

    (x - 8)(x + 2) = 0

    Thus, f(x) has an horizontal tangent in x-coordinates 8 and -2. But you also want y-coordinates, right ? So plug this back into your first equation :

    f(8) = \frac{-424}{3}

    f(-2) = \frac{76}{3}

    Thus, the curve has horizontal tangents in coordinates (8, f(8)), and (-2, f(-2)), that is, (8, \frac{-424}{3}) and (-2, \frac{76}{3}).

    So you are right (prefer keeping fractional answer though).
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