Consider an equation $\frac{1}{y} + a\log{y} = x$, where $a \in \mathbb{R}$ is a parameter. I want to find the value of $a$ such that maximum (minimum) of $y(x)$ is the largest (smallest).
The idea is simple, all I have to do is to find the maximum of $y(x)$ for a fixed $a$ and the maximalize over all $a$. Same for minimum However, I have some problems with this.
Using the implicit function theorem, I have to assume, $y(x)$ is not equal $\frac{1}{a}$. So assume for example that $y(x) > \frac{1}{a}$ and calculate the derivative: $y'(x) = \frac{1}{\frac{-1}{y^2}+\frac{a}{y}} = \frac{y^2}{ay - 1} > 0$. Similarly, if $y(x) < \frac{1}{a}$ then $y'(x) < 0$.
How can I interpret that? Does that mean that the minimum is equal to $\frac{1}{a}$?
As you noted, it is not possible to globally define $y$ as a function of $x$, so it is not clear what is meant as the maximum of $y(x)$. If $y(x)$ is taken as the lower part of the curve in the picture below ($y < \frac 1a$), then the maximum is precisely $\frac 1a$. Conversely, if you considerer the upper aprt ($y > \frac 1a$), the minimum will be $\frac 1a$ and there will be no maximum.
However, this will not get you any maximal or minimal $a$.