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Maximum/minimum through logarithm

  1. Nov 26, 2014 #1
    Hi I seem to have run into a strange problem.
    Suppose one wishes to maximize/minimize the function ## f(x) = (x-4)^{2} ##. Clearly, this function has a minimum at x = 4. One could find the extremum by taking the derivative and setting to zero.

    One could also compute the logarithm of this function, i.e. ## ln (x-4)^{2} = 2 \ ln(x-4) ##, and then find the corresponding extremum of the resulting function, and since ## ln(x )## is monotonic over the region on which it is defined, the maximum/minimum of ##f(x)## and the maximum/minimum of ## f(ln(x))## should coincide.

    This however is not the case, since ## 2 \ ln(x-4) ## seems not to have a maximum or a minimum. What am I missing here?

    Thanks.

    BiP
     
  2. jcsd
  3. Nov 26, 2014 #2
    I hope I understand your point ...

    Actually the two function would have the same critical point
    see:
    $$y=(x-4)^2$$
    $$y'=2(x-4)$$
    when you equate by zero your critical point is 4.

    Also
    $$lny=2ln(x-4)$$
    $$\frac{y'}{y}=\frac{2}{x-4}$$
    $$y'=y\times\frac{2}{x-4}$$
    $$y'=(x-4)^2\times\frac{2}{x-4}$$
    $$y'=2(x-4)$$
    Again your critical point is 4.

    is that what you mean?

    Am I right ..
    :)
     
  4. Nov 26, 2014 #3

    pasmith

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    Your problem is that [itex]\ln x[/itex] is not defined for [itex]x \leq 0[/itex]. Thus the critical point of [itex](x - 4)^2[/itex] is not in the domain of [itex]\ln(x - 4)[/itex].
     
  5. Nov 26, 2014 #4

    PeroK

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    Also:

    ## ln (x-4)^{2} = 2 \ ln|x-4| \ \ (x \ne 4)##
     
  6. Dec 1, 2014 #5

    HallsofIvy

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    Personally, I wouldn't use "logarithms" or "derivatives". A [tex]x^2[/tex] is never negative and is 0 only for x= 0, so that [tex](x- 4)^2[/tex] is never negative and is 0 only for [tex]x- 4= 0[/tex] or [tex]x= 4[/tex].
     
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