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Math Help - help with critical points....

  1. #1
    Newbie
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    help with critical points....

    hi!
    ok... i was given this function:
    f(x) = (x+2)^3 (1-x) (3-x)
    and i have some trouble getting it's critical points...
    I understand i get those by finding the points in which the derivative is 0.... right?
    I got the derivative....
    f'(x) = 5x^4 -8x^3 -27x^2 -44x +4

    so... i equaled that to zero... and i got the four points, which are:

    x1 = 2.314
    x2 = -2.005
    x3 = -1.995
    x4 = 0.086


    so the question is... does that even make any sense? I checked those points on the graph, and they dont mark the minimum values or anything... am i doing something wrong??
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  2. #2
    Senior Member
    Joined
    Nov 2008
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    Hello buddy.

    Quote Originally Posted by c.park View Post
    hi!
    ok... i was given this function:
    f(x) = (x+2)^3 (1-x) (3-x)
    and i have some trouble getting it's critical points...
    I understand i get those by finding the points in which the derivative is 0.... right?
    Yep, thats how it works.

    Quote Originally Posted by c.park View Post
    I got the derivative....
    f'(x) = 5x^4 -8x^3 -27x^2 -44x +4

    Uhhhhh, so close.
    it's more like
    5x^4 + 8x^3 - 27x^2 - 44x + 4

    did you noticed the +8x^3, not -8x^3

    Quote Originally Posted by c.park View Post
    so... i equaled that to zero... and i got the four points, which are:

    x1 = 2.314
    x2 = -2.005
    x3 = -1.995
    x4 = 0.086


    so the question is... does that even make any sense? I checked those points on the graph, and they dont mark the minimum values or anything... am i doing something wrong??
    The solution should be

    x=0.0864, x = 2.314, x = -2

    Alright?

    Yours
    Rapha
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  3. #3
    MHF Contributor

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    By the way, those are not points, they are values of x. The "critical points" are (x, y) points with y calculated for each value of x.
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