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Finding stationary points. no x in second derivative

  1. Mar 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Just started getting introduced to calculus and a couple applications. After i've found the stationary point i understand that i can put the x value into the second derivative to find if its a maximum or minimum point. i.e

    12x-2x2
    ƒ'(x)= 12-4x
    12-4x=0
    x=3
    so 12(3)-2(3)2 = 18
    So if i've worked it out correct then the stationary point is 3,18


    3. The attempt at a solution
    So if i take the second derivative i have

    ƒ''(x)= -4

    if theres no x in the second derivative then how do I find out whether it's a maximum or minimum value?

    Thanks
     
  2. jcsd
  3. Mar 27, 2014 #2

    SteamKing

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    You can always plot the original function.
     
  4. Mar 27, 2014 #3

    Mark44

    Staff: Mentor

    You can use the 2nd derivative to determine if a critical point is a local maximum or minimum. Your textbook should have this test and some examples. Also, y = 12x - 2x2 is a very simple function. A quick sketch of its graph, as SteamKing suggests, will show immediately what's going on.
     
  5. Mar 27, 2014 #4

    LCKurtz

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    You ask yourself whether ##-4## is positive or negative for concavity.
     
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