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Math Help - show function has stationary points

  1. #1
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    show function has stationary points

    need help with this one.

    show the function below has stationary points @ x=-1 x=0 x= 2

    y=\frac{x^4}{4}-\frac{x^3}{3}-x^2

    12y=3x^4-4x^3-12x^2

     <br />
12\frac{dy}{dx}=12x^3-12x^2-24x<br />

     <br />
\frac{dy}{dx}=x^3-x^2-2x<br />

     <br />
\frac{dy}{dx}=0<br />

    correct so far?
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  2. #2
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    Quote Originally Posted by decoy808 View Post
    need help with this one.

    show the function below has stationary points @ x=-1 x=0 x= 2

    y=\frac{x^4}{4}-\frac{x^3}{3}-x^2

    12y=3x^4-4x^3-12x^2

     <br />
12\frac{dy}{dx}=12x^3-12x^2-24x<br />

     <br />
\frac{dy}{dx}=x^3-x^2-2x<br />

     <br />
\frac{dy}{dx}=0<br />

    correct so far?
    yes . Then carry on with the factorization .
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  3. #3
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    Quote Originally Posted by decoy808 View Post
    need help with this one.

    show the function below has stationary points @ x=-1 x=0 x= 2

    y=\frac{x^4}{4}-\frac{x^3}{3}-x^2

    12y=3x^4-4x^3-12x^2

     <br />
12\frac{dy}{dx}=12x^3-12x^2-24x<br />

     <br />
\frac{dy}{dx}=x^3-x^2-2x<br />

     <br />
\frac{dy}{dx}=0<br />

    correct so far?
    Yes, so solve x^3- x^2- 2x= 0 or, since you are only asked to show that the stationary points are at x= -1, x= 0, and x= 2, evaluate x^3- x^2- 2x at each of those values of x.
    Last edited by HallsofIvy; February 24th 2010 at 11:24 AM.
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