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Intervals where the function is increasing

  1. Jun 5, 2007 #1
    1. The problem statement, all variables and given/known data

    For the function f(x)=xln(x²), x≠0


    3. The attempt at a solution

    f'(x)=2+ln(x²)

    For the function to be increasing f'(x)>0

    2+ln(x²)>0
    ln(x²)>-2
    x²>e^(-2)

    Taking square roots of both sides

    ±x>±(1/e)

    How can I find the intervals from this?
    (Somebody has suggested to me that after taking the square root of 1/e the answer should be positive only rather than plus/minus, but I can't seem to find a reason why such would be the case)
     
  2. jcsd
  3. Jun 5, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    note that if you accept that [tex]\log (x^2) = 2 \log (x)[/tex], then now RHS implies that x should be non-negative. That's the only possible reason I can think of that may dictate that you take the +ve root. But, it is clear that for your f(x) both +/- roots are good solutions and your interval is
    [tex](-\infty,-1/e) \cup (1/e, \infty)[/tex]
     
  4. Jun 5, 2007 #3
    yeah I got it!
    If I bring the two down to the front of the natural logarithm I will get
    2ln|x|.

    With the modulus then I get |x|>(1/e) and hence the intervals (-∞, -1/e) and (1/e, ∞)

    Thanks! :tongue2:
     
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