Injectivity and range of arctan⁡(1+x1−x)According to the definition, for a function to be injective f(a)=f(b)→a=b for all a,b∈DfUsing this I get:arctan⁡(1+a1−a)=arctan⁡(1+b1−b)⇔(1+a1−a)=(1+b1−b)⇔1+a1−a=1+b1−b⇔2a=2ba=bHence 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 0≤y&lt;π2. But how does one come up with this range?

Question

Injectivity and range of arctan⁡(1+x1−x)According to the definition, for a function to be injective f(a)=f(b)→a=b for all a,b∈DfUsing this I get:arctan⁡(1+a1−a)=arctan⁡(1+b1−b)⇔(1+a1−a)=(1+b1−b)⇔1+a1−a=1+b1−b⇔2a=2ba=bHence 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 0≤y&amp;lt;π2. But how does one come up with this range?

Brianna Sims · Accepted Answer

The domain is −1≤x&amp;lt;1,  limx→1−f(x)=π2, f(−1)=0 and f is a continuous function.Thus, the range is [0,π2)

Brenton Steele · Accepted Answer

We need 1+x1−x≥01+x1−x=0⇒x=−11+x1−x&amp;gt;0⇒−1&amp;lt;x&amp;lt;1WLOG arccos⁡x=2y,x=cos⁡2y where 0&amp;lt;2y≤π using Principal values.arctan⁡1+x1−x=arctan⁡cot⁡y=arctan⁡(tan⁡(π2−y))=π2−arccos⁡x2Now I hope that things should be managed more easily than in the original form.

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

Answered question

Answer & Explanation

New Questions in Precalculus