Maximum/minimum through logarithm

1. Nov 26, 2014

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

2. Nov 26, 2014

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 ..
:)

3. Nov 26, 2014

pasmith

Your problem is that $\ln x$ is not defined for $x \leq 0$. Thus the critical point of $(x - 4)^2$ is not in the domain of $\ln(x - 4)$.

4. Nov 26, 2014

PeroK

Also:

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

5. Dec 1, 2014

HallsofIvy

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