Question Number 69974 by Shamim last updated on 29/Sep/19
$$\mathrm{Identify}\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\mathrm{this}\: \\ $$$$\mathrm{function}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{ln}\frac{\mathrm{4}−\mathrm{x}}{\mathrm{4}+\mathrm{x}}. \\ $$
Commented by kaivan.ahmadi last updated on 29/Sep/19
$$\frac{\mathrm{4}−{x}}{\mathrm{4}+{x}}>\mathrm{0}\Rightarrow−\mathrm{4}<{x}<\mathrm{4} \\ $$$$\Rightarrow{D}_{{f}} =\left(−\mathrm{4},\mathrm{4}\right) \\ $$$${y}={ln}\frac{\mathrm{4}−{x}}{\mathrm{4}+{x}}\Rightarrow{e}^{{y}} =\frac{\mathrm{4}−{x}}{\mathrm{4}+{x}}\Rightarrow\mathrm{4}−{x}=\mathrm{4}{e}^{{y}} +{xe}^{{y}} \Rightarrow \\ $$$$\mathrm{4}−\mathrm{4}{e}^{{y}} ={x}+{xe}^{{y}} ={x}\left(\mathrm{1}+{e}^{{y}} \right)\Rightarrow{x}=\frac{\mathrm{4}−\mathrm{4}{e}^{{y}} }{\mathrm{1}+{e}^{{y}} }\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{4}−\mathrm{4}{e}^{{x}} }{\mathrm{1}+{e}^{{x}} } \\ $$$${R}_{{f}} ={D}_{{f}^{−\mathrm{1}} } ={R} \\ $$
Commented by Shamim last updated on 30/Sep/19
$$\mathrm{plz}\:\mathrm{explain}\:\mathrm{with}\:\mathrm{details}- \\ $$$$\frac{\mathrm{4}−\mathrm{x}}{\mathrm{4}+\mathrm{x}}\:>\mathrm{0}\:\Rightarrow\:−\mathrm{4}<\mathrm{x}<\mathrm{4} \\ $$
Commented by kaivan.ahmadi last updated on 30/Sep/19
