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#f# is a function on real where #f(x)=2x^2-x# How do you find #f^-1(x)# ?

↳Redirected from "How to prove this ? #2|k^2# then #2|k# for some #k\inZZ#"

1 Answer

Shwetank Mauria

Jan 26, 2018

#f^(-1)(x)=(1+-sqrt(1+8x))/4#

Explanation:

We have #y=f(x)=2x^2-x#

or #2x^2-x-y=0#

or #x=(1+-sqrt(1+8y))/4#

Hence #f^(-1)(x)=(1+-sqrt(1+8x))/4#

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