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Thread: Tangent Line to y=f(x)

  1. #1
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    Tangent Line to y=f(x)

    How do I go about finding all the values of x where the tangent line to y = f(x) is horizontal?

    f(x) = x^3(2x^2 + x - 3)^2

    I assume I would set f(x) and f '(x) equal to 0 and solve for x?
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  2. #2
    MHF Contributor Quick's Avatar
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    don't quote me

    I can't help you because I don't know derivatives

    however, as I recall from previous posts, you need to find the value of x when the f '(x)=0

    finding when f(x)=0 does nothing.
    Last edited by Quick; Aug 26th 2006 at 06:27 AM.
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  3. #3
    Member Glaysher's Avatar
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    Quick is right the question is asking to find all the points where f '(x) = 0

    $\displaystyle f(x) = x^3(2x^2 + x - 3)^2$

    $\displaystyle u = x^3$ $\displaystyle v = (2x^2 + x - 3)^2$

    Use product rule (with chain rule needed for differentiating v)

    $\displaystyle u' = 3x^2$ $\displaystyle v' = 2(4x + 1)(2x^2 + x - 3)$

    $\displaystyle f '(x) = 3x^2(2x^2 + x - 3)^2 + 2x^3(4x + 1)(2x^2 + x - 3)$

    $\displaystyle 3x^2(2x^2 + x - 3)^2 + 2x^3(4x + 1)(2x^2 + x - 3) = 0$

    Factorising

    $\displaystyle x^2(2x^2 + x - 3)(3(2x^2 + x - 3) + 2x(4x + 1)) = 0$

    $\displaystyle x^2(2x + 3)(x - 1)(14x^2 + 5x - 9) = 0$

    $\displaystyle x^2 = 0$ gives $\displaystyle x = 0$

    $\displaystyle 2x + 3 = 0$ gives $\displaystyle x = -\frac{3}{2}$

    $\displaystyle x - 1 = 0$ gives $\displaystyle x = 1$

    $\displaystyle 14x^2 + 5x - 9 = 0$ gives $\displaystyle (14x - 9)(x + 1) = 0$

    Gives $\displaystyle x = \frac{9}{14}$ and $\displaystyle x = -1$
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