How do you find the inverse of #f(x)=(x+1)/(x-2)# and graph both f and #f^-1#?

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Answer 1 · 2017-01-22T09:09:43

The answer is #f^-1(x)=(2x+1)/(x-1)#

Let #y=(x+1)/(x-2)#

#y(x-2)=x+1#

#yx-2y=x+1#

#yx-x=2y+1#

#x(y-1)=2y+1#

#y=(2y+1)/(y-1)#

Therefore,

#f^-1(x)=(2x+1)/(x-1)#

Verification,

#f(f^-1(x))=f((2x+1)/(x-1))=((2x+1)/(x-1)+1)/((2x+1)/(x-1)-2)#

#=(2x+1+x-1)/(2x+1-2x+2)#

#=3x/3=x#

graph{(y-(x+1)/(x-2))(y-x)=0 [-10, 10, -5, 5]}

graph{(y-(2x+1)/(x-1))(y-x)=0 [-10, 10, -5, 5]}

The graphs are symmetric wrt #y=x#