How do you find the inverse of #f(x)=(2x-3)/(x+1)#?

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Answer 1 · 2015-12-06T15:52:27

#f^-1(x)=-(x+3)/(x-2)#

Write as: #y=(2x-3)/(x+1)#

Switch #x# and #y# and then solve for #y#.

#x=(2y-3)/(y+1)#

#x(y+1)=2y-3#

#xy+x=2y-3#

#xy-2y=-x-3#

#y(x-2)=-(x+3)#

#y=-(x+3)/(x-2)#

The #y# can be replaced by #f^-1(x)#, which simply denotes an inverse function:

#f^-1(x)=-(x+3)/(x-2)#