Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 197362 by sonukgindia last updated on 14/Sep/23

Answered by MM42 last updated on 15/Sep/23

$$\frac{e^x-1}{e^x+1}=u\Rightarrow e^{x}u+u=e^x-1\Rightarrow e^x(u-1)=-1-u$$$$e^x=\frac{1+u}{1-u}\Rightarrow f\left(x\right)={\left(\frac{1+x}{1-x}\right)}^2✓$$

Answered by ibrahimabdullayev last updated on 15/Sep/23

$$\frac{e^x-1}{e^x+1}=a\Rightarrow e^x-1={ae}^x+a\Rightarrow e^x(1-a)=a+1$$$$e^x=\frac{1+a}{1-a}\Rightarrow f\left(a\right)={\left(\frac{1+a}{1-a}\right)}^2\Rightarrow f\left(x\right)={\left(\frac{1+x}{1-x}\right)}^2$$$$IbrahimAbdullayev$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com