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Max & Min

  1. Jan 7, 2009 #1
    1. The problem statement, all variables and given/known data

    let f(x)=(x+1)^2(x-2)
    a)Find the max and Min values of f(x)
    b)Find the inflection point of concavity

    2. Relevant equations

    using defferentiation

    3. The attempt at a solution
    for part a) differentiate (x+1)^2(x-2) we get ans=3(x-1)(x+1) Plse check for me?
    than f'(x)=0 therefore the points i got is x=1 and x=-1 respective??

    the second part not sure..plse assist.
  2. jcsd
  3. Jan 7, 2009 #2
    You are right about the local max and min. For the second part--all an inflection point is, is where the second derivative (that which we use to determine concavity) is 0. So you would have


    Can you get it from there?
  4. Jan 8, 2009 #3


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    Staff Emeritus
    Science Advisor

    Make sure you answer the question asked! Max and Min occur at x= -1 and x= 1 but you haven't yet said what the max and min are.

    Also what jeffreydk said about the inflection point is slightly misleading. An inflection point is a point where the first derivative changes sign. That means the the second derivative must be 0 there but that is not sufficient. You need to check that the first derivative really does change sign. For example if f(x)= x4, f'= 4x3 and f"= 12x2. f"(0)= 0 but f' does NOT change sign there so (0,0) is NOT an inflection point of f(x)= x4.
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