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Max or min of xlnx^2

  1. Oct 30, 2012 #1

    939

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    1. The problem statement, all variables and given/known data

    Please check my answer. I am weak at math.

    2. Relevant equations

    Y = xlnx^2

    3. The attempt at a solution

    Y' = (lnx^2) + (x)(2x/x^2)

    For the max/min, I would just set Y' = 0 by trial and error and then do y", but I don't want to do that until I know y' is correct.

    Please help =(
     
  2. jcsd
  3. Oct 30, 2012 #2

    Ray Vickson

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    Your Y' is OK. Solving the equation Y' = 0 should be easy, and does not need "trial and error". Hint: look at the form of Y'; simplify it as much as possible.

    RGV
     
  4. Oct 30, 2012 #3

    Dick

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    Looks ok so far if you mean x*ln(x^2). You could simplify the second term quite a bit.
     
  5. Oct 30, 2012 #4

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    Thanks for such quick anwers.

    So I simplify y' a little bit and have:

    y ' = (lnx^2) + (x)(2x/x^2)
    y ' = (lnx^2) + (2x^2/x^2)
    y ' = (2lnx) + (2x/x)
    y = (2lnx) + (2)

    But how can I set y ' = 0? 2lnx would have to be - 2, no? It would be invalid, wouldn't it? :-/
     
  6. Oct 30, 2012 #5

    Dick

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    Not at all. Some numbers have negative logs. Think about that a bit more.
     
  7. Oct 30, 2012 #6

    939

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    Got it. Thanks a lot for the help! =)
     
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