Question 997307


{{{f(x)= (2x+1)/(x-1)}}}

to find inverse, set {{{f(x)=y}}}, 

{{{y= (2x+1)/(x-1)}}}.... now swap {{{x}}} and {{{y}}}

{{{x= (2y+1)/(y-1)}}}....solve for {{{y}}}

{{{x(y-1)= (2y+1)}}}

{{{xy-x= 2y+1}}}

{{{xy-2y= x+1}}}

{{{(x-2)y= x+1}}}

{{{y= (x+1)/(x-2)}}}

so, your inverse is {{{f^1(x)= (x+1)/(x-2)}}}



the domain and range of both {{{f(x)}}} and {{{f^-1(x)}}}:

for {{{f(x)= (2x+1)/(x-1)}}}

domain:
{ {{{x}}} element {{{R}}} : {{{x<>1}}} }
range:
{ {{{f(x)}}} element {{{R}}} : {{{f(x)<>2}}} }


for {{{f^1(x)= (x+1)/(x-2)}}}

domain:
{ {{{x}}} element {{{R}}} : {{{x<>2}}} }
range:
{ {{{f(x)}}} element {{{R}}} : {{{f(x)<>1}}} }