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Finding Stationary Points

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    f(x) = x^3 +ax^2 +5x +2 and has two stationary points, one of which is x = 2. a) Find the other stationary point.
    b) classify the stationary points using the second derivative test

    3. The attempt at a solution

    f '(x) = 2ax + 3x^2 +5
    f ''(x) = 2a + 6x

    0 = (2)^3 + a(2^2) +5(2) +2
    0 = 8 + 4a + 10 +2
    0 = 20 + 4a
    -4a = 20, a = -5

    What am I supposed to do to get a stationary point?
     
    Last edited: Feb 2, 2010
  2. jcsd
  3. Feb 2, 2010 #2

    radou

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    0 = (2)^3 + a(2^2) +5(2) +2

    For a stationary point, f'(x) = 0, not f(x) = 0.
     
  4. Feb 2, 2010 #3
    f ' (x) = 0
    So that would be?
    2ax + 3x^2 + 5 = 0
    How would I find a?

    Would I use 0 = (2)^3 + a(2^2) +5(2) +2 to get a = -5

    0 = 2(-5)x + 3x^2 +5
    0 = -10x + 3x^2 +5
    0 = 3x^2 - 10x + 5
    x = (5 +sqrt(10))/3 and (5 - sqrt(10))/3 ?

    or

    0 = ( 2a(2) + 3(2)^2 +5
    0 = 4a + 12 + 5
    0 = 4a +17
    -4a = 17
    a = -17 / 4

    substitute -17/4 into 2ax + 3x^2 + 5 = 0?
    0 = 2(-17/4)x + 3x^2 + 5 = 0
    0 = 3x^2 - 8.5x + 5 = 0
    x = 2 , 5/6

    the second derivative?
     
    Last edited: Feb 2, 2010
  5. Feb 2, 2010 #4
    You have to substitute x=2 in the entire equation, not just one variable.
     
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