# Domain of g(x) = arctan(2x)

1. Dec 10, 2007

### Goldenwind

http://www.math.yorku.ca/Who/Faculty/Kochman/M1300/solutions/solF04/SFE.pdf
Question 2.

I agree that the range of arctan is [-pi/2, pi/2], however due to the fact that we have arctan(2x), not just arctan(x), wouldn't that mean that the range of the g(x) function would be [-pi/4, pi/4] instead? 'cause then when your pi/4 goes into the arctan, it gets multiplied by the 2, making it pi/2, which is the maximum allowed into the arctan function.

Am I wrong, or is the solution? If it's me, where is my mistake?

Also, in Question 3, wouldn't the answer be 5pi/6? Don't the arctan and tan cancel each other out, like squaring a root?

2. Dec 10, 2007

### Dick

The upper limit of the range of arctan(x) is pi/2 because limit as x->infinity, arctan(x)->pi/2. Changing x to 2x doesn't change that. For question 3) arctan(tan(x))=x only for x in a certain part of the range of tan. You've already said the domain of arctan is [-pi/2,pi/2]. For x outside that range arctan(tan(x))=x CAN'T be true.

Last edited: Dec 10, 2007