Say, $f(x)$ is increasing and $g(x)$ is decreasing.
Based on (provable) different speed of increasing/decreasing, can we formulate rules like:
And how to prove such "different speed of increasing/decreasing"?
For example, can we somehow use infinitesimals with a higher order of smallness / different orders of smallnes?
On
$f + g$ can be increasing or decreasing or neither (e.g. constant).
$f - g$ -increasing since $-g$ is increasing.
$f \circ g$ is decreasing : take $x \leqslant y$, then $g(y) \leqslant g(x)$ hence $f(g(y)) \leqslant f(g(x))$.
$f \cdot g$ can be increasing or decreasing or neither (e.g. constant).
And how to prove such "different speed of increasing/decreasing"?
The tool you are looking for is differentiation. If a function $f : \mathbb{R} \mapsto \mathbb{R}$ is differentiable, say with derivative $f'$, we can mesure how much the function $f$ increases/decreases on a certain interval $I$ :
$f'(x) = 0$ for almost all $x \in I$ iff $f$ is constant on $I$.
$f'(x) > 0$ for almost all $x \in I$ iff $f$ is strictly increasing on $I$.
$f'(x) < 0$ for almost all $x \in I$ iff $f$ is strictly decreasing on $I$.
This gives you a criterion for question 1. : Assume $|f'(x)| > |g'(x)|$ for almost all $x \in I$, then $f + g$ is increasing on $I$ since $(f + g)' = f' + g'$ and $f'(x) + g'(x) > 0$.
But beware for question 4. ! In general, $(f\cdot g)' \neq f'\cdot g'$. Luckily, we have the Leibnitz formula : $\boxed{(f \cdot g)' = f' \cdot g + f \cdot g'}$
On
In general, only $f-g$ is sure to be increasing and $f\circ g$ is sure to be decreasing. Indeed, $f$ increasing means that
for every $x_1$ and $x_2$ in the domain of $f$, if $x_1<x_2$, then $f(x_1)<f(x_2)$
Similarly, $g$ decreasing means that
for every $x_1$ and $x_2$ in the domain of $g$, if $x_1<x_2$, then $g(x_1)>g(x_2)$.
$f-g$ is increasing
Assume that $x_1<x_2$ in the domain of $f-g$ (that is, the intersection of the domains of $f$ and $g$).
By definition, $f(x_1)<f(x_2)$ and $g(x_1)>g(x_2)$. The second inequality implies that $-g(x_1)<-g(x_2)$, so $(f-g)(x_1)<(f-g)(x_2)$.
By definition, $f-g$ is increasing.
$f\circ g$ is decreasing
Assume that $x_1<x_2$ in the domain of $f\circ g$.
By definition, $g(x_1)>g(x_2)$. Since $f$ is increasing, we have $f(g(x_1))>f(g(x_2))$, which means $f\circ g(x_1)>f\circ g(x_2)$.
By definition, $f\circ g$ is decreasing.
Cases of $f+g$ and $fg$
Consider $f(x)=mx$ and $g(x)=nx$ with $m>0$ (so that $f$ is increasing) and $n<0$ (so that $g$ is decreasing).
Then $(f+g)(x)=(m+n)x$ and $f+g$ can be increasing, decreasing or constant depending on the choice of the values of $m$ and $n$.
Also, $fg(x)=mnx^2$ is quadratic, so not increasing and not decreasing.
I suppose $f$ and $g$ are differentiable, then $f'\geq0$ and $g'\leq 0$.
By this assumption, we can only say surely that $f-g$ is increasing and $f\circ g$ is decreasing because $$(f(x)-g(x))'=f'(x)-g'(x)\geq0$$ and $$f(g(x))'=f'(g(x))g'(x)\leq0.$$ For other cases we can show by examples that the functions $f+g$ and $fg$ may be non-monotonic.