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Critical points

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data
    I need to find the critical points of
    f(x) = (x^3 - 2x)e^x

    I found the derivative, and set it equal to zero

    ended up with e^x (x^3 +3x^2 -2x -2) = 0

    I am having trouble factoring the second factor, any suggestions?
  2. jcsd
  3. Nov 22, 2007 #2


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    let f(x)=x^3 +3x^2 -2x -2
    try f(1) and see what happens...
  4. Nov 22, 2007 #3
    yeah I found f(1) just by looking at it, but how can I solve it without guessing?
    + there might be more solutions?
  5. Nov 22, 2007 #4


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    Well if f(1)=0 then it means that (x-1) is a factor of f(x) just divide the polynomial by that linear factor and you'll get the other quadratic factor and then you can solve....Remember the remainder and factor theorem?
  6. Nov 22, 2007 #5
    lol, wow I guess that was a while ago

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