Let $q:[1, + \infty) \subset \mathbb{R} \longrightarrow \mathbb{R}$ be a function defined as
$ \qquad \qquad \qquad \qquad \qquad \quad q(x) = \left \{ \begin{array}{lcl} \delta_{1} & \text{ if } & 1 \leq x < \epsilon_{2} \\ \vdots & & \\ \delta_{n-1} & \text{ if } & \epsilon_{n-1} \leq x < \epsilon_{n} \\ \delta_{n} & \text{ if } & \epsilon_{n} \geq x \end{array} \right . $
for a given partition $[1, \epsilon_{2}),\dots ,[\epsilon_{n-1}, \epsilon_{n}), [\epsilon_{n}, +\infty)$ of its domain, and $1 >\delta_{1} \geq \delta_{2}\geq ... \geq \delta_{n} > 0$. Fix an integer $t\in \mathbb{Z}_{>0}$, and consider the function
$\qquad \qquad \qquad \qquad l : [1, t]\subset \mathbb{R}\longrightarrow \mathbb{R} : x \longmapsto \frac{1}{1-q^{x}\left (\frac{t}{x} \right)}\left ( 1 + \sum_{j = 1}^{\lfloor \frac{t}{x} \rfloor - 1} q^{x}(j) \right)$
with the convention that the summatory cancels out for $\lfloor\frac{t}{x}\rfloor - 1 < 1$. My problem is to study the trend and find global minima on the domain $[1, t]$. The function is clearly not globally continuous, in its generic form. Would it make sense to study the derivative in this context?
This is a complicated problem, and I do not expect a full answer. However, any hint on how to proceed analytically would be greatly appreciated. In the context I a working, a good approximation or reduction to a simpler similar problem would be ok.
The derivative can help you see what is happening in each interval. Is the function increasing or decreasing there? Does it have a local minimum on the interval? If so, this might be a candidate for the global minimum.
Then you also want to see what happens at the discontinuities. Does the function jump up or down?