Maximizing log(x) For what functions is maximizing $\log f(x)$ the same as maximizing $f(x)$?
I've read this thread. I'm looking for a rigorous, calculus based answer.
$$\frac{d \log f(x)}{dx} = \frac{d f(x)}{dx} \frac{d \log f(x)}{d f(x)} = \frac{d f(x)}{dx} \frac{1}{f(x)}$$
How do we find the maxima from this, and what exactly implies that if and only if the function is increasing, the maxima are at the same values of $x$?
 A: If $g$ is strictly increasing, then the extrema of $f$ are at exactly the same locations as the extrema of $g\circ f$. That's just because $f(x)\leq f(x_0)$ if and only if $g(f(x))\leq g(f(x_0))$.
From the equation you wrote, note that since $\frac{1}{f(x)}$ can never be zero, one has that $\frac{d\log f(x)}{dx} = 0$ if and only if $\frac{df(x)}{dx} = 0$.
A: *

*

Your second paragraph seemed to indicate that this holds even if $f$ is decreasing, or whatever.

I think what MPW means is that 'this' holds for any possible $f$. Such $f$ must be strictly positive and all that other stuff.


*But vacuously, I guess you could say that


If $f$ is not strictly positive, then maximising $\ln f$ is equivalent to maximising $f$. $\tag{1}$



*

whether or not it's increasing,

I think there's some confusion with the preposition here. Perhaps MPW thought 'it' referred to $g$. (I think similar confusion may arise from the aforementioned 'this'.)


*Anyhoo, to be explicitly, I indeed think that


maximising $\ln f$ is equivalent to maximising $f$ $\tag{2}$

for

any such $f: A \to \mathbb R, A \subseteq \mathbb R$ such that (2) makes sense $\tag{3}$

I shall attempt to prove this using 2nd derivatives, which MPW doesn't seem to use. (Oh drat. It's just occurring to me that we might talk about maximisation even if 2nd derivative of $f$ doesn't exist.)
Proof:
Step 1: 1A. Observe that $f$ is strictly positive. Let $g=\ln(f)$, $g: A \to \mathbb R$. Observe that 1B. both $g$ and $f$ are twice differentiable, 1C. that $g'(x)=0$ iff $f'(x)=0$, 1D. that $g''(x) (f(x))^2= f(x)f''(x)-(f'(x))^2$ and 1E. that $g''(x) < 0$ iff $ f(x)f''(x) < (f'(x))^2$.
If direction:
Step 2: Let $x \in \mathbb R$ maximise $f$, i.e. $x \in A, f'(x)=0$ and $f''(x) < 0$.
Step 3: By Steps 1A, 1C, 1E and 2, $x \in A, g'(x) = 0$ and $g''(x) < 0$, i.e. $x \in \mathbb R$ maximises $g$ because we get that  $(\text{positive number}) \ f''(x) = (\text{positive number}) \ (\text{negative number})$ $= \text{negative number} < 0 = 0^2 = (f'(x))^2$.
Only if direction:
Step 4: Let $x \in \mathbb R$ maximise $g$, i.e. $x \in A, g'(x) = 0$ and $g''(x) < 0$.
Step 5: By Steps 1A, 1C, 1E and 4, $x \in A, f'(x)=0$ and $f''(x) < 0$, i.e. $x \in \mathbb R$ maximises $f$ because $\text{positive number} \ f''(x)  < 0^2 = 0$ implies $f''(x) < 0$.
QED

Edit: I honestly had no idea that 5 years ago I actually had answered the 7 year old question that OP links to. LOL. What a small world: The reason I came to OP's question is because OP commented on my questions/answers or replied to my comments on chess se.
