# Draw the graph of arctan((x-1)/(x+1))

Tags:
1. Jan 29, 2016

### kolua

I am so far able to find the domain, intercepts, concavity and increase/decrease. But I am stuck at finding the asymptotes for the graph.

I think there is no vertical asymptote or is there one at x=-1?
I think the horizontal asymptote is y=pi/4

2. Jan 30, 2016

### Staff: Mentor

$\frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$
What happens to $arctan(1 - \frac{2}{x + 1})$ as x gets very large or very negative?

3. Jan 31, 2016

### Isaac0427

For the vertical asymptote, ask yourself this: can you take the arctan of an undefined value? When will $\frac{x-1}{x+1}$ be undefined?

4. Jan 31, 2016

### kolua

I think the limit of arctan((x-1)/(x+1)) as x approaches -1 from the right is pi/2 and as x approaches -1 from the left is -pi/2, which means there is no vertical asymptote here but a jump discontinuity here right?

5. Jan 31, 2016

### kolua

when x=-1 arctan is undefined but as x approaches x=-1 (x-1)/(x+1) approaches negative infinity and positive infinity from left and right. Let (x-1)/(x+1) = t, then arctan(t) approaches pi/2 and -pi/2 as t approaches the two infinity is this correct? then there is no vertical asymptote?

6. Jan 31, 2016

### kolua

is the horizontal asymptote correct? is there any other horizontal asymptote that I am missing? I got a graph like this, but I'm not sure if it is correct

#### Attached Files:

• ###### Screen Shot 2016-01-31 at 3.57.11 PM.png
File size:
98.7 KB
Views:
117
7. Jan 31, 2016

### Isaac0427

I don't believe that is so. It is undefined at x=-1, but I don't believe that as x approaches -1 that the function approaches infinity with either sign.
No, x=-1 is the vertical asymptote.

8. Jan 31, 2016

### Isaac0427

The graph is good.

9. Jan 31, 2016

### Samy_A

That's not correct.

10. Jan 31, 2016

### Isaac0427

Ok, I am probably thinking of an incorrect definition of asymptote. The function does get infinitely close to but never reaches π/4 as x approaches -1. Does this not make it an asymptote?

11. Jan 31, 2016

### kolua

x approaches infinity and function goes to pi/4 thats the definition of a horizontal asymptote, correct?

12. Jan 31, 2016

### Isaac0427

Yes that it definitely an asymptote. I'm just not sure about x=-1.

13. Jan 31, 2016

### Samy_A

No. A function f has a vertical asymptote at x=a if $\displaystyle\lim_{x\rightarrow a^+}f(x)=\pm \infty$ or $\displaystyle\lim_{x\rightarrow a^-}f(x)=\pm \infty$.
That's definitely not the case with an arctan.

14. Jan 31, 2016

### Isaac0427

15. Jan 31, 2016

### kolua

I'm not sure if the graph looks like that. Because the calculator gives a different image.

#### Attached Files:

• ###### Screen Shot 2016-01-31 at 3.57.11 PM.png
File size:
98.7 KB
Views:
69
16. Jan 31, 2016

### Isaac0427

That's the graph my calculator gives.

17. Feb 1, 2016

### Samy_A

You misunderstood me. I was answering @Isaac0427 's post concerning vertical asymptotes. The definition of a vertical asymptote makes it evident that your function doesn't have vertical asymptotes. At x=-1 it has a jump discontinuity, exactly as you wrote.

Last edited: Feb 1, 2016
18. Feb 2, 2016

### Svein

I used Atan2, which takes the x and y values separately.

19. Feb 2, 2016

### zinq

The trickiest part is figuring out what happens as x → -1 from above and from below. It appears from your graph that you've already done that.

But " The function does get infinitely close to but never reaches π/4 as x approaches -1." is not true, because of a mistake in this statement. It's also important to consider the cases x < -1 and x > -1 separately when x → -1.

Last edited: Feb 2, 2016