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?

Printable View

- Aug 2nd 2010, 01:08 PMSyNtHeSiSAsymptotes 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? - Aug 2nd 2010, 01:18 PMMath 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 $\displaystyle -\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 $\displaystyle |arctan(x)| < \frac{\pi}{2} $, so where are those asymptotes? - Aug 3rd 2010, 04:03 AMSyNtHeSiS
- Aug 3rd 2010, 04:14 AMHallsofIvy
If you are talking about vertical asymptotes, it has no vertical asymptotes.

- Aug 3rd 2010, 07:10 AMSyNtHeSiS
- Aug 3rd 2010, 08:26 AMskeeter
- Aug 4th 2010, 05:50 AMSyNtHeSiS
- Aug 4th 2010, 07:06 AMskeeter
- Aug 4th 2010, 01:43 PMSyNtHeSiS
- Aug 4th 2010, 01:50 PMskeeter
- Mar 26th 2013, 05:38 AMMathCrusaderRe: Asymptotes of y = x - arctanx
a