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Another critical numbers problem

  1. Mar 29, 2007 #1
    g(T)=5^2/3+ t^5/3

    g't= 10/3T^-1/3+ 5/3t^2/3

    ok do i factor this one out, this one looks confusing
     
  2. jcsd
  3. Mar 29, 2007 #2

    Dick

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    Are you trying to solve g'(t)=0? Then sure, factor out a t^(-1/3).
     
  4. Mar 29, 2007 #3
    remember that when it says ^-1/3
    its not saying (5t^-1)/3
     
  5. Mar 29, 2007 #4

    Dick

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    That's why I put parentheses around the (-1/3). You should too.
     
  6. Mar 30, 2007 #5

    HallsofIvy

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    I assume you meant g(T)= 6T^(2/3)+ T^(5/3).

    After you got g'(T)= (10/3)T^(-1/3)+ (5/3)T^(2/3)= 0 you can multiply the entire equation by T^(1/3). (Obviously, the derivative does not exist at T= 0 so that is also a critical point.)

    It's a lot easier to read with parentheses! Also, please do not use "T" and "t" to mean the same thing.
     
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