Arithmetic Mean of a set of critical points

Question

Consider the following functions: $f(x) = \sin(x)$and $g(x) =x^3 + 1$. Let set $A$ include only the critical point(s) of $f$ across all reals and set $B$ include only the critical points of $g$ across all reals. Let $A\cup B$ denote the union of sets $A$ and $B$. Find the Arithmetic Mean of $A\cup B$.

Edit: So the mean of an infinite sit seems more like a measure, or so to speak, Infinite averages

Answer 1

We have $f'(x)=\cos(x),g'(x)=3x^2$ thus $f'(x)=0$ at $\frac {2n+1\pi}{2} $ where $n$ takes values of all integers. And $g'(x)=0$ only at $x=0$ hence the set $AUB={....,-\frac {3\pi}{2},-\frac {\pi}{2},0,\frac {\pi}{2},\frac {3\pi}{2},.....} $ We have 0 and on each of its side we have symmetric values with opposite sign hence they cancel out. Thus the mean of this set is $\frac {0}{n} $ hence the mean is $0$.

Answer 2

Think about the derivative of $f(x)$, which is $cos(x)$. Think about the symmetry of this function. When is it zero (i.e. where are $f(x)$'s critical points?) At $\pi/2, 3\pi/2, 5\pi/2 $, etc, but also at the negatives of those numbers, due to $cos(x)$'s symmetry. So when you sum them all up for the arithemtic mean, they all cancel. So the only contributions to the mean are from $g(x)$ (in other words, the sum of all the critical points converges to zero). It's a polynomial of degree 3 so clearly, there will be a finite number of critical points. So then you have the sum of three numbers (the critical points of $g(x)$), let's call it N, and N is a real number. Since there were infinite critical points in $f(x)$, you would be dividing this by infinity to get the mean. So essentially you have the limit of "N" divided by x, as x goes to infinity; i.e., zero. So the mean would be zero.

Answer 3

Critical points of sine function are $x=\frac{\pi}{2}+k\pi, k\in Z$. Critical point of the cubic function is $0$. So the sum of all critical points is $0$, hence mean is $0$ as well.