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

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Answer 1 · 2015-12-03T01:07:50

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

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

Swap #x# and #y#.

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

Solve for #y#.

#xy=2y+1#

#xy-2y=1#

#y(x-2)=1#

#y=1/(x-2)#

You can rewrite #y# as #f^-1(x)#, which denotes an inverse function.

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