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

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Answer 1 · 2018-01-26T16:52:15

$f^{- 1} (x) = \frac{1 \pm \sqrt{1 + 8 x}}{4}$ $f^(-1)(x)=(1+-sqrt(1+8x))/4$

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

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

or $x = \frac{1 \pm \sqrt{1 + 8 y}}{4}$ $x=(1+-sqrt(1+8y))/4$

Hence $f^{- 1} (x) = \frac{1 \pm \sqrt{1 + 8 x}}{4}$ $f^(-1)(x)=(1+-sqrt(1+8x))/4$

fff is a function on real where f(x)=2x^2-xf(x)=2x2−xf(x)=2x^2-x How do you find f^-1(x)f−1(x)f^-1(x) ?