Answer to Question #134287 in Algebra for Mike

Question

Please help !
if f(x)=1-x/1+x
then
f(-x)=? f(x+1)=? f(x)+1=? f(1/x)=? 1/f(x)=?

Answer 1

Given : $\mathbf{f(x)=\dfrac{1-x}{1+x}}$

1. $for\;f(-x)\; ,replace\;x\;by\;(-x)\;in\;f(x),\;then$

$\mathbf{f(-x)=\dfrac{1-(-x)}{1+(-x)}=\dfrac{1+x}{1-x}}$

2.$\;for\;f(x+1)\;,replace\;x\;by\;(x+1)\;in\;f(x),\;then$

$\mathbf{f(x+1)=\dfrac{1-(x+1)}{1+(x+1)}=\dfrac{1-x-1}{1+x+1}=\dfrac{-x}{2+x}}$

3.$\;for\;f(x)+1,\;we\;have-$

$\mathbf{f(x)+1=\dfrac{1-x}{1+x}+1=\dfrac{1-x+(1+x)}{1+x}=\dfrac{2}{1+x}}$

4.$\;$ $for\;f(1/x)\; ,replace\;x\;by\;(1/x)\;in\;f(x),\;then$

$\mathbf{f(1/x)=\dfrac{1-\dfrac{1}{x}}{1+\dfrac{1}{x}}=\dfrac{\dfrac{x-1}{x}}{\dfrac{x+1}{x}}=\dfrac{x-1}{x+1}}$

5.$\;\;for\;1/f(x),\;we\;have-$

$\mathbf{\dfrac{1}{f(x)}=\dfrac{1}{\dfrac{1-x}{1+x}}=\dfrac{1+x}{1-x}}$