maximization of a particular ratio

Question

We are given a ratio:

$$\frac{g(x)}{f(x)}$$

where:

$$g(x) \in \mathbb{R}^{+}$$ $$f(x) \in \mathbb{N}\: \cap f(x)\ge 2$$

So $g(x)$ returns values in $[0,+\infty]$ while $f(x)$ returns values in $\{2,3,4,\ldots\}$.

I am looking for a confirmation about a very simple question: if I maximize $\dfrac{g(x)}{f(x)}$, do I also maximize $\dfrac{g(x)}{f(x)-1}$ in this very particular case?

Answer 1

Of course you do not.

An example is $g(x) = x^2 - 100x + 10000$, and $f(x) = \text{max} \{ \lceil \sqrt{x} \rceil, 2 \}. $ [Sorry for not finding a nice example]

In one case you have minimum is somewhere around 76-77, in the other one 79-80. You can play with it here.

EDIT: I see you wanted to maximize, I did minimize. But the reason why it doesn't hold is the same.

Answer 2

If I interpret your question correctly, this depends on $f(x)$ and $g(x)$, e.g.:

if $f(x)$ is constant then $\frac{g(x)}{f(x)}=c\cdot g(x)$ (c being a constant), so it is maximized at the same $x$ (obviously).

if e.g. $g(x)$ is constant and $f(x)$ is not and there is a $x_0$ such that $f(x_0)-1=f(x)$, then clearly(well, by substitution) this $x_0$ is the maximum. Example: $\frac1{\lceil x+2\rceil}$ is maximal for $x=-1$ whereas $\frac1{\lceil x+2\rceil-1}$ is maximal for $x=0$. ($\lceil\cdot\rceil$ is the ceiling function that returns the next larger integer for any real number)

Also, $\frac{g(x)}{f(x)-1}$ is not defined if $f(x)=1$.