Is $\frac{1}{x}$ a monotonic-decreasing function? The derivative of $f(x)=\frac{1}{x}$ is always negative for all $x$ except $x=0$ (where it is not defined). But $f(-1)<f(1)$ , so the statement that $f(x)$ either decreases or remains constant as $x$ increases for all $x$ does not seem to be holding good here. So , can I say that $f(x)=\frac{1}{x}$ a monotonic-decreasing function?
Thanks !
 A: The answer is No, the inverse function is not monotone on $\Bbb R\setminus\{0\}$.
I will gather the information from the previous comments to be more precise. As you said yourself, the derivative is strictly negative everywhere on the domain, but that doesn't imply that the function is monotonely decreasing. This kind of implication is only true on intervals. Since the domain is not an interval, such an implication is false in general, and in particular in this case. Indeed, since $f(-1)=-1\leq1=f(1)$, the function cannot be montonely decreasing. As it has been said previously, you can at most say that the function is monotonely decreasing on both intervals $(-\infty,0)$ and $(0,+\infty)$.
This doesn't mean, that a function cannot be monotone on $\Bbb R\setminus\{0\}$. Indeed, consider the function $f$ on this set defined by $f(x)=x$. This is not a very sophisticated answer, but it will serve the purpose, because it is clearly monotonic.
A: $$f'(x)=-\frac{1}{x^2}<0, x\ne 0.$$ So f(x) is decreasing function for $x\in (-\infty, 0)$ and for $x\in (0, \infty).$
