Is $1/x$ a descending function? It is descending on all its connected components, but not as a function, right?
But on the other hand the $f'(x) = -x^{-2} < 0$, then it should be descending, what am I missing?
 A: The theorem about the relationship between the monotony  of $f$ and the sign of $f'$ has to be applied to an interval (a connected set in $\mathbb{R}$). Indeed, reading its proof, we have that for $x_1<x_2$, by the Mean Value Theorem, there is some $t\in (x_1,x_2)$ such that
$$f(x_2)-f(x_1)=f'(t)(x_2-x_1)$$
So $f(x_2)\geq f(x_1)$ if and only if $f'(t)\geq 0$.
Here, in order to apply the Mean Value Theorem, we need $f$ to be a continuous function on the closed interval $[x_1,x_2]$  and differentiable on the open interval $(x_1,x_2)$.
A: The theorem which says that $f'(x)<0$ for all $x \in I$ implies $f$ is decreasing on $I$ is applicable only when the function is defined and differentiable on that interval. Since your function is not defined at $0$ you cannot apply that theorem to any interval containing $0$. Of course, the function is decreasing in $(0,\infty)$ as well as $(-\infty,0)$. 
A: $\frac{1}{x}$ is not defined when $x=0$, so when $x > 0$ it might not be decreasing.
A: Note that $$f(-1)=-1<1=f(1).$$ Thus $f$ is not decreasing. Is there a contradiction with $f'(x)<0?$ No, since $f$ is not defined at $x=0.$
The domain of $f$ is $(-\infty,0)\cup (0,\infty).$ Since $f'<0$ all we can say is that $f$ is decreasing on $(-\infty,0)$ and on $(0,\infty).$ But this doesn't imply that $f$ is decreasing on $(-\infty,0)\cup (0,\infty).$
A: graph of 1/x
As you can see from the graph of 1/x , its not defined at 0 but decreasing for other values of x since y'=-(1/x^2)
A: "Graphical approach"
When you look at graph of $1/x$ you can mark the point for which function not decrease?

