# How is the many-to-one function $f(x) = \frac{1}{x-1} + \frac{2}{x-2} + \frac{3}{x-3}$ decreasing?

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?

• Umm... "my book"? May 7, 2019 at 2:03
• @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.
– Hema
May 7, 2019 at 3:00
• You DID mention it... that's the problem! May 7, 2019 at 3:34

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.

• The function is continuous where it's defined, see math.stackexchange.com/questions/1482787/… May 7, 2019 at 10:42
• 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. May 7, 2019 at 12:07

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.

• Note that the function is continuous on the domain given and so the IVT applies. May 7, 2019 at 2:13
• The IVM demands also that the domain is an interval. May 7, 2019 at 10:43

If a $$C^1$$-function's derivative is negative then it's decreasing on any interval where it's defined.

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

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

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

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

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