## JingleBells 3 years ago In Video lecture 05 round about 37' where Pr. Jerison draws the f(x)=tan(x) and f(x)=arctan(x) the two curve crosses where if I plug in real data I can't get the same results, did he lowered the curve to just to illustrate the symmetry along the y=x line or do you have to do some kind of stretch to get the same picture?

1. keshavsarda

both are inverse function.........so they are symmetric to line y=x

2. JingleBells

Thanks, that bit I get, what I'm not sure about is that does the graph of tanx and arctanx overlap or not.

3. keshavsarda

yes they meet at y=x=1

4. JingleBells

So do you know what I could do to get the same graph as in the video where the two curves cross each other?

5. keshavsarda

no, u cant draw that graps....because we are not accurate. But u can roughly sketch it

6. JingleBells

I have quick graph on my iPhone and the two graphs never cross that's why I'm asking :-)

7. keshavsarda

|dw:1340852698875:dw|

8. JingleBells

Thanks but if you plotted the two graphs with real values of tan(x) and arctan(x) they don't cross at all.

9. kutulu

if you watch the recitation video for Session 15, it addresses your question. The profession drew the graph of arctan wrong. Within the single domain of -pi/2 -> pi/2, the only place that tan and arctan cross is at x=0. Arctan should be *below* y=x everywhere else while tan should be *above* y=x everywhere else.

10. kutulu

erm. make that The professor* drew the graph...

11. JingleBells

Thanks a million! I'm using iTunes U to watch the videos and I will, from now, watch the recitations on the web! :-)