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Analysing graph of a logarithmic function

  1. Dec 12, 2013 #1
    Hello everyone! I'm stuck on a small detail in a math task, and would really appreciate some help!

    1. The problem statement, all variables and given/known data
    Determine all local extreme points and possible max/min values for the function f(x) = x*lnx+(x*lnx)^2 where 0<x≤1/2


    2. Relevant equations



    3. The attempt at a solution
    The first thing I did was to differentiate f, giving me f'(x)=(lnx+1)(1+2x*lnx)
    Then, in order to find extreme points, I tried to solve f'(x)=0
    So, either (lnx+1)=0 [itex]\rightarrow[/itex] x=1/e
    or (1+2x*lnx) = 0 [itex]\rightarrow[/itex] x*lnx = -1/2

    And this is where I get stuck. How do I solve that equation?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 12, 2013 #2

    Simon Bridge

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    intersection of y=lnx and y=-1/2x ... sketch out the curves and see what occurs to you.
     
  4. Dec 12, 2013 #3
    Thank you for your response!

    I did as you suggested, and quickly realized that there simply are no real solutions. However, in the course I'm attending we are not supposed to use any calculators or graphers, so how would I, without a grapher, realize that these two curves never intersect?
     
  5. Dec 12, 2013 #4

    Simon Bridge

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    That would involve demonstrating that lnx > -1/2x

    You would probably get away with just using the properties of the two functions.

    let f=ln(x) and g=-1/2x

    Assume that f=g for some value of x - and set out to find it by process of elimination:

    * ln(x) is only defined for x>0, so that rules out all x<0
    * f=g => ln(x)<0 so that rules out x>1.
    * f(1) > g(1) so... if f'<g' (i.e. g is steeper than f) in 0<x<1 then f>g everywhere.

    Thing is that you'd have to suspect that this is the case to start out with ... which you do by having experience with these functions. Which is why you are given a lot of exercises like this one.

    Aside:
    if you put g=-1/kx ... can k take a value that would make f=g for some x?
    what would have happened if we'd said f=ln|x| instead of ln(x)?

    Note:
    You don't need calculators of "graphers" to make a sketch of two functions.
     
    Last edited: Dec 12, 2013
  6. Dec 12, 2013 #5
    Thanks for another great response. I'm glad to have this sorted out now!
     
  7. Dec 12, 2013 #6

    Ray Vickson

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    You could solve the problem ##\min g(x) = \ln(x) + 1/(2x)## on ##x > 0##, and show that the minimum value is > 0. Minimizing g is an easy problem.
     
  8. Dec 12, 2013 #7

    Simon Bridge

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    Or, in terms of post #4 ... minimize f-g ... means finding x: f'-g'=0 ...
    discovering that this is only the case for x<0 is pretty much what the process of elimination in post #4 does ... but thought about a different way.

    Again, to realize you need do that test, you'd have to suspect that this is the case already right?
    You wouldn't "know" until you did the test. So it seems to be begging the question:
    ... i.e. how would you know to apply the test?

    The method I demonstrated to you in #4 was less direct that was needed - I chose that approach for a special reason. I wanted to show you how to think about a difficult problem.

    The first step was to assume the problem had a solution.

    I rewrote the relationship in a way that made it easier to investigate, and set about finding out where the solution may lie.

    It may have been that there was a solution - in which case I'd know where to find it, and there may be some specific method that may help me there. In this case it turned out that there wasn't a solution.

    In general - if finding a solution looks like it is going to be a great deal of hard work ... you are well advised to, first, check that the solution exists. Don't wait until you realize there may not be a solution to check for the existence of a solution.

    The bottom line. though, is that it takes experience to arrive at these conclusions quickly - there is no shortcut - you just have to remember your lessons.
     
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