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

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Answer 1 · 2018-01-26T16:52:15

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

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

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

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

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

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