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Question Number 69974 by Shamim last updated on 29/Sep/19

$$\mathrm{Identify}\mathrm{domain}\mathrm{and}\mathrm{range}\mathrm{of}\mathrm{this}$$$$\mathrm{function}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=\ln \frac{4-\mathrm{x}}{4+\mathrm{x}}.$$

Commented by kaivan.ahmadi last updated on 29/Sep/19

$$\frac{4-x}{4+x}>0\Rightarrow -4<x<4$$$$\Rightarrow D_f=(-4,4)$$$$y=ln\frac{4-x}{4+x}\Rightarrow e^y=\frac{4-x}{4+x}\Rightarrow 4-x=4e^y+{xe}^y\Rightarrow$$$$4-4e^y=x+{xe}^y=x(1+e^y)\Rightarrow x=\frac{4-4e^y}{1+e^y}\Rightarrow$$$$f^{-1}\left(x\right)=\frac{4-4e^x}{1+e^x}$$$$R_f=D_{f^{-1}}=R$$

Commented by Shamim last updated on 30/Sep/19

$$\mathrm{plz}\mathrm{explain}\mathrm{with}\mathrm{details}-$$$$\frac{4-\mathrm{x}}{4+\mathrm{x}}>0\Rightarrow -4<\mathrm{x}<4$$

Commented by kaivan.ahmadi last updated on 30/Sep/19

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