# Intervals of increase: -3arccot(x)

1. Apr 5, 2016

### Cosmophile

1. The problem statement, all variables and given/known data
Find the regions of increase and decrease for $f(x) = -3 \cot^{-1}(x)$

2. The attempt at a solution

My instructor and I disagree. The answer he gave was $(-\infty, \infty)$, while my answer was $(-\infty, 0) \cup (0, \infty)$.

My reasoning is this: $-3\cot ^{-1}(x)$ is not defined for $x=0$. So, we cannot say what the behavior of the function is (increasing, decreasing, constant) at $x=0$.

If my reasoning is wrong, I would like to understand why.

2. Apr 5, 2016

### LCKurtz

$\cot(\frac \pi 2) = 0$ so $\cot^{-1}(0) = \frac \pi 2$.

3. Apr 5, 2016

### Cosmophile

This is the graph I'm seeing.

4. Apr 5, 2016

### LCKurtz

The graph I gave you doesn't have the $-3$ in front of it. But the problem you are seeing is in the definition of the principle range for the arccotangent. Most texts use the branch of $\cot x$ on $(0,\pi)$ when calculating its inverse. This preserves continuity of the inverse function and the choice also makes the derivative formulas work. Whether or not your calculator or graphing program uses the same interval, who knows. Apparently your professor uses the standard one.

5. Apr 5, 2016

### Cosmophile

So is my answer wrong?

Actually, wait. I'm aware of what you're referring to, but the answer he provided was $(-\infty, \infty)$. Clearly, that is not correct, as he is including $0$ in the interval.

6. Apr 5, 2016

### LCKurtz

Assuming your professor is using the standard definition for the principle value of the arccot function, which is the graph I showed you, I would say your answer is not completely correct. The arccot function is continuous at and decreasing at $x=0$, and you have excluded that point. Of course, your problem has that -3 in front of the function so yours is increasing. In your first picture in post #3, if you moved the right half of the graph up until it connected with the left half, it would be correct.

[Edit, added:] By "correct", I mean qualitatively. I am referring to the continuity, not necessarily the vertical placement of the graph of your given function.

Last edited: Apr 5, 2016
7. Apr 5, 2016

### Cosmophile

@LCKurtz wasn't sure if you saw my edit.

8. Apr 5, 2016

### LCKurtz

I hadn't seen your edit, but I have now. Your professor is correct. What I said in post #6 should explain it for you. Note the comment about the graph at the end of post #6.

9. Apr 5, 2016

### SammyS

Staff Emeritus
Even going with the version from WolframAlpha, it says there that the domain is $(-\infty, \infty)$ .

10. Apr 5, 2016

### Cosmophile

Hmm. That's so odd; why does the graph show a disconnect at $x = 0$ then?

11. Apr 5, 2016

### SammyS

Staff Emeritus
There is a discontinuity, a jump discontinuity, but $\ \cot^{-1}(0)=\pi/2\$.

12. Apr 5, 2016

### LCKurtz

There is no discontinuity in the graph of $-3\cot^{-1}x$ if you use the standard principle value.

13. Apr 6, 2016

### Cosmophile

Could you (or anyone else who sees this) elaborate on the "standard principle value?" I don't understand how we can just shift the right-hand side up.

Edit: I've done a bit of research and have found that there's no real sense of consensus! This surprised me. Not even the creators of top maths softwares can come to an agreement!

Last edited: Apr 6, 2016
14. Apr 6, 2016

### LCKurtz

I'm going to ignore the -3 in your problem for now. Since the cotangent has many different segments that map onto the y axis, the "inverse" would be a multi-valued function. A partial graph of that is:

If you want a single valued definition (that is, a function), you have pick what part of this graph to use. It doesn't matter which part is chosen as long as everyone knows which it is. That's why you need a standard choice. If you want the arccotangent to be continuous, the standard choice would be the blue one. Many texts choose this branch and then call it Arccot, with a capital A. With this choice, Arccot(x) is always an angle in $(0,\pi)$.

An alternate, less common, choice is to choose the left part of the green and right part of the blue graphs, which would put the standard angle in $(-\frac \pi 2, \frac \pi 2)$. This gives a jump at $x=0$, but you still must choose either $\frac \pi 2$ or $-\frac \pi 2$ for its value at 0.

So, no matter which way you do it, it is incorrect to say arccot(0) is undefined. In any case, as I pointed out in post #4, it is apparent that your professor uses Arccot(x) as above as his standard one. So would I, and I would suggest you do so too. That being said, I think you should move on from this problem. I think you understand the issue and that's all that is needed.