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The function $f(x) = \frac{1}{x-1} + \frac{2}{x-2} + \frac{3}{x-3}$ is many-to-one, despite it having a strictly negative derivative (the domain being $\mathbb{R} - \{1,2,3\}$). Why is this so? Is there any way of knowing this without actually graphing $f(x)$, which seems rather difficult?

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    $\begingroup$ Umm... "my book"? $\endgroup$ May 7, 2019 at 2:03
  • $\begingroup$ @DavidG.Stork Resonance rank booster, it is a preparatory book for joint entrance advanced examination in India, I didn't think it would be well known so I didn't mention it. $\endgroup$
    – Hema
    May 7, 2019 at 3:00
  • $\begingroup$ You DID mention it... that's the problem! $\endgroup$ May 7, 2019 at 3:34

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When we approach $x=1$ very close from the negative direction, $x-1$ will be a very small negative value and so $\frac{1}{x-1}$ will be a very large negative value. Approaching from the other side we have $\frac{1}{x-1}$ becoming a very positive value. In fact, it can become as large or as small as you like, provided you get as close to $x=1$ as necessary.

By looking at the other terms, we see that the same thing happens at $x=2$ and $x=3$ and we see that, near each of these points, the function can attain any value we like. Therefore, there exist distinct $x$ values for which $f(x)$ is the same, and so the function is not one-to-one.


By the way, the reason we can have a function with a strictly negative derivative that isn’t one-to-one is because it isn’t continuous. Specifically, this function isn’t continuous at the points $x=1, x=2, x=3$. Even if we remove these points from the domain as we have, we still find that our function can attain a particular value more than once without the derivative ever becoming zero or switching sign. This is because, at a certain point, the function ‘blows up’ to negative infinity and then magically returns again from positive infinity, so on the domain we have defined, the derivative is always negative.

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  • $\begingroup$ The function is continuous where it's defined, see math.stackexchange.com/questions/1482787/… $\endgroup$ May 7, 2019 at 10:42
  • $\begingroup$ You are right—of course. But I don’t think raising that issue here is going to do anyone any favours. My answer serves as a short, intuitive picture. $\endgroup$ May 7, 2019 at 12:07
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It's because it goes from plus infinity to minus infinity on both $(1,2)$ and $(2,3)$ (so by Intermediate value theorem it assumes every real value at least twice) which you can see by computing the appropriate limits.

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    $\begingroup$ Note that the function is continuous on the domain given and so the IVT applies. $\endgroup$ May 7, 2019 at 2:13
  • $\begingroup$ The IVM demands also that the domain is an interval. $\endgroup$ May 7, 2019 at 10:43
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If a $C^1$-function's derivative is negative then it's decreasing on any interval where it's defined.

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$f(x)=\displaystyle\sum_{k=1}^{n}f_{k}(x)$

$f_{n}(x)=\frac{n}{x-n}:n\in N^{*},x\in R-\{ n \}$

$\forall n\in N^{*},x\in R-\{ n \} :f_{n}'(x)=\frac{-n}{(x-n)^{2}}\lt 0$

$\Rightarrow \displaystyle\sum_{k=1}^{n}f_{k}(x)\lt 0$

$\forall n\in N^{*},x\in R-\{ n \} : f'(x)\lt 0$

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