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Identify-domain-and-range-of-this-function-that-f-x-ln-4-x-4-x-




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

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