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Thread: stationary points?

  1. #1
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    stationary points?

    1. For the following function:


    y= x^3/3+ 2x^2 + 3x - 1



    a. Find the stationary points.
    b. Use the second-order derivatives to test whether the stationary points are at a
    maximum or minimum value of.

    thank you for your help


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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    a. find f''(x), then x1 and x2 of f'(x)=0
    b. find f''(x), if f''(xt)>0 then xt min. if f''(xt),0 xt max. if f''(xt)=0 and f'''(xt)!=0 then it is nor max/min.
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  3. #3
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    thanks however i am clueless to finding f''(x), then x1 and x2 of f'(x)=0
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  4. #4
    Senior Member yeKciM's Avatar
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    $\displaystyle f'_(x)=x^2+4x+3$

    when u put $\displaystyle f'(x)=0$ means that $\displaystyle x^2+4x+3=0$

    $\displaystyle x_1=-1 , x_2=-3$ There u have stationary points...

    $\displaystyle f''(x)=(f'(x))'=2x+4$

    lol... u can go from here ?

    just put stationary points in $\displaystyle f''_{(x)}$

    $\displaystyle f''_{(x_1)}=2 $ so it's min because it's $\displaystyle f''_{(x_1)}>0$
    $\displaystyle f''_{(x_2)}=-2}$ so it's max because it's $\displaystyle f''_{(x_2)}<0$
    Last edited by yeKciM; Jul 20th 2010 at 07:35 AM.
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