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Math Help - points where function has a horizontal tangent line

  1. #1
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    points where function has a horizontal tangent line

    Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.



    First I plugged in o to all the x's and got the point (2,0).. I don't know how to find the other two points. I want to learn the method of how to work these types of problems out, then I can do the same types of problemsusing the same method. Help would be great!
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    Quote Originally Posted by Neverending View Post
    Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.



    First I plugged in o to all the x's and got the point (2,0).. I don't know how to find the other two points. I want to learn the method of how to work these types of problems out, then I can do the same types of problemsusing the same method. Help would be great!
    Horizontal tangent means gradient of tangent is equal to zero.

    Solve dy/dx = 0 to get the x-coordinates of the required points.
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    dy/dx means find derivative of..

    so how do I find where the derivative equals 0?
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    Quote Originally Posted by Neverending View Post
    dy/dx means find derivative of..

    so how do I find where the derivative equals 0?
    You start by calculating what the derivative is. Then you set it equal to zero. Try posting what you can do and where you get stuck and it will be easier to help you.
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  5. #5
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    Quote Originally Posted by Neverending View Post
    Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.



    First I plugged in o to all the x's and got the point (2,0).. I don't know how to find the other two points. I want to learn the method of how to work these types of problems out, then I can do the same types of problemsusing the same method. Help would be great!
    If y = ax^n then \frac{dy}{dx} = nax^{n-1}.

    Using this, we get

    \frac{dy}{dx} = 4x^3 - 16x.

    We want to find the x-values where this derivative is 0. So we LET \frac{dy}{dx} = 0 and solve for x.


    0 = 4x^3 - 16x

    0 = 4x(x^2 - 4)

    0 = 4x(x + 2)(x - 2)

    0 = 4x or 0 = x + 2 or 0 = x - 2.

    So x = 0 or x = -2 or x = 2 are the points we want.
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  6. #6
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    oh so I find the derivative first.. i see.. thanks guys!
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