Asymptotes of y = x - arctanx

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August 2nd 2010, 01:08 PM
SyNtHeSiS

Asymptotes of y = x - arctanx

Sketch the graph of y = x - arctanx, showing its asymptotes.

Now I got the correct shape of the graph, but my asymptotes were wrong.

I dont understand how you would get the asymptote of y = x + π/2 and y = x - π/2. Could someone please explain?
August 2nd 2010, 01:18 PM
Math Major

Remember that tangent is not a one-to-one function (ie, it sends multiple points in its domain to the same point in its range). Thus, in order for arctangent to be a function, its range has to be restricted to $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , otherwise it would be multiply defined (without the restriction, where, for instance, would arctan(0) go?). So, to find the asymptotes, we need to find where the graph does not pass.

We know that $|arctan(x)| < \frac{\pi}{2}$ , so where are those asymptotes?
August 3rd 2010, 04:03 AM
SyNtHeSiS

Quote:

Originally Posted by Math Major

We know that $|arctan(x)| < \frac{\pi}{2}$ , so where are those asymptotes?

I know that arctanx has horizontal asymptotes of y = π/2 and y = -π/2 (inversion of tanx's undefined x values) but I dont see any x values in y = x - arctanx which would make it undefined?
August 3rd 2010, 04:14 AM
HallsofIvy

If you are talking about vertical asymptotes, it has no vertical asymptotes.
August 3rd 2010, 07:10 AM
SyNtHeSiS

Quote:

Originally Posted by HallsofIvy

If you are talking about vertical asymptotes, it has no vertical asymptotes.

Yeah thats what I meant, that it has no vertical/horizontal asymptotes, which is why I wanted to know, how can you tell that y = x - arctanx has oblique asymptotes?
August 3rd 2010, 08:26 AM
skeeter

Quote:

Originally Posted by SyNtHeSiS

Yeah thats what I meant, that it has no vertical/horizontal asymptotes, which is why I wanted to know, how can you tell that y = x - arctanx has oblique asymptotes?

as $x \to \pm \infty$ , $\arctan{x} \to \pm \frac{\pi}{2}$ .

so ... as $x \to \pm \infty$ , $y = x - \arctan{x}$ approaches the line $y = x$
August 4th 2010, 05:50 AM
SyNtHeSiS

Quote:

Originally Posted by skeeter

as $x \to \pm \infty$ , $\arctan{x} \to \pm \frac{\pi}{2}$ .

so ... as $x \to \pm \infty$ , $y = x - \arctan{x}$ approaches the line $y = x$

Sorry but I am a bit confused. I understand the first line, but for the second line, how does it approach y=x? I thought the lines it approaches were y= x+/- π/2?
August 4th 2010, 07:06 AM
skeeter

Quote:

Originally Posted by SyNtHeSiS

Sorry but I am a bit confused. I understand the first line, but for the second line, how does it approach y=x? I thought the lines it approaches were y= x+/- π/2?

I'll agree with that ...

as $x \to \infty$ , the function approaches $y = x - \frac{\pi}{2}$

as $x \to -\infty$ , the function approaches $y = x + \frac{\pi}{2}$
August 4th 2010, 01:43 PM
SyNtHeSiS

Quote:

Originally Posted by skeeter

I'll agree with that ...

as $x \to \infty$ , the function approaches $y = x - \frac{\pi}{2}$

as $x \to -\infty$ , the function approaches $y = x + \frac{\pi}{2}$

One more question, if you use a limit with x tending to +/- infinity, why isnt it y = ∞ - π/2 and y = ∞ + π/2?
August 4th 2010, 01:50 PM
skeeter

Quote:

Originally Posted by SyNtHeSiS

One more question, if you use a limit with x tending to +/- infinity, why isnt it y = ∞ - π/2 and y = ∞ + π/2?

there is a difference between the limit of a function and the limit of a function value. $\infty \pm \, any \, constant$ is still $\infty$
March 26th 2013, 05:38 AM
MathCrusader

Re: Asymptotes of y = x - arctanx

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