Injectivity and range of \arctan (\sqrt{\frac{1+x}{1-x}}) According to the definition, for

Sydney Stanley

Sydney Stanley

Answered question

2022-04-21

Injectivity and range of arctan(1+x1x)
According to the definition, for a function to be injective f(a)=f(b)a=b for all a,bDf
Using this I get:
arctan(1+a1a)=arctan(1+b1b)
(1+a1a)=(1+b1b)
1+a1a=1+b1b
2a=2b
a=b
Hence the function is injective. Now this could be total bs but I also am not able to find the proper range for this one. WolframAlpha says 0y<π2. But how does one come up with this range?

Answer & Explanation

Brianna Sims

Brianna Sims

Beginner2022-04-22Added 19 answers

The domain is 1x<1,  limx1f(x)=π2, f(1)=0 and f is a continuous function.
Thus, the range is [0,π2)
Brenton Steele

Brenton Steele

Beginner2022-04-23Added 18 answers

We need 1+x1x0
1+x1x=0x=1
1+x1x>01<x<1
WLOG arccosx=2y,x=cos2y where 0<2yπ using Principal values.
arctan1+x1x=arctancoty=arctan(tan(π2y))=π2arccosx2
Now I hope that things should be managed more easily than in the original form.

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