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Finding Critical numbers

  1. Jan 5, 2005 #1
    Hi,
    i was wondering if anyone could help me.
    The problem says find the critical numbers of f...
    f(x) = 2x^3 + 15x^2 - 36x +1

    ok i found f ' .....6x^2 + 30x-36,
    Now how do i start on finding the critical numbers? Do i have to take the second derivative?
     
  2. jcsd
  3. Jan 5, 2005 #2

    Tide

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    What, exactly, is the definition of "critical number?" :-)
     
  4. Jan 5, 2005 #3
    So should you take a second derivative?
     
  5. Jan 5, 2005 #4
    Right....its not, so I can just set it equal to zero, and then solve for the variable, right?
     
  6. Jan 5, 2005 #5

    dextercioby

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    Yes,your 'critical numbers' are solution of the equation
    [tex] f'(x)=0 [/tex]
    .BTW,your terminology is pretty weird.I use to call them "critical points",coz they pop up whenever i want to draw the graph of a function

    Daniel.
     
  7. Jan 5, 2005 #6

    HallsofIvy

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    "Right....its not, so I can just set it equal to zero, and then solve for the variable, right?"

    Having determined first that there are no values of x for which the derivative does not exist, yes.

    "BTW,your terminology is pretty weird.I use to call them "critical points",coz they pop up whenever i want to draw the graph of a function"

    Actually, there is a difference. Given y= f(x), the "critical numbers" are the values of x at which f ' (x) does not exist or is equal to 0. The "critical points" are the points (x,y) with x a critical number.
     
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