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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 !

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    $\begingroup$ $f$ is not decreasing on $\mathbb{R} \setminus \lbrace 0 \rbrace$. It is decreasing on $\mathbb{R}_+ \setminus \lbrace 0 \rbrace$ and on $\mathbb{R}_- \setminus \lbrace 0 \rbrace$. The implication between the sign of the derivative and the monotony of the function is true only on intervals. $\endgroup$ – TheSilverDoe Aug 31 '20 at 10:12
  • $\begingroup$ @TheSilverDoe If I got it right , you meant that calling a function monotonic-decreasing (or increasing ) is sensible only when I mention what interval I am talking about . And in case of discontinuous functions , relation between values of function in two different intervals would not matter for the function to be monotonic. It should just be decreasing within that interval . Right ? $\endgroup$ – ARROW Aug 31 '20 at 12:41
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    $\begingroup$ It may look counterintuitive but this function is continuous. Its domain $\mathbb{R}\setminus\{0\}$ is disconnected and that why its graph has two parts, but the function is continuous. The theorem that says that if the derivative is negative then function is descreasing requieres the domain to be an interval. $\endgroup$ – jjagmath Aug 31 '20 at 13:42
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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.

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$$f'(x)=-\frac{1}{x^2}<0, x\ne 0.$$ So f(x) is decreasing function for $x\in (-\infty, 0)$ and for $x\in (0, \infty).$

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  • $\begingroup$ I get your point , but could you please clarify just once that if it was a yes or no type question : Is $1/x$ a monotonic-decreasing function ? , what should be the answer , yes or no ? $\endgroup$ – ARROW Aug 31 '20 at 12:35

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