Maximum/minimum through logarithm

[Bipolarity]

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

 

[Maged Saeed]

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 ..
:)

 

[pasmith]

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

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].

 

[PeroK]

## ln (x-4)^{2} = 2 \ ln(x-4) ##

Also:

## ln (x-4)^{2} = 2 \ ln|x-4| \ \ (x \ne 4)##

 

[HallsofIvy]

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].