What is the inverse of f(y)=2y-2y^2 ?

The inverse of f(y)=2y-2y^2 is -(-1+sqrt(-2y+1))/2 ,(sqrt(-2y+1)+1)/2

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inverse of f(y)=2y-2y^2

inverse

f (y) = 2 y - 2 y^{2}

- \frac{- 1 + - 2 y + 1}{2}, \frac{- 2 y + 1 + 1}{2}

Solution steps

y = 2 y - 2 y^{2}

Replace

y

with

y

y = 2 y - 2 y^{2}

Solve for

y, y = 2 y - 2 y^{2}

- \frac{- 1 + - 2 y + 1}{2}, \frac{- 2 y + 1 + 1}{2}

4 - 1 - x

f (x) = \frac{1}{e ^{x + 5}}

- 16 t^{2} + 48 t

f (x) = \frac{1}{4} x + \frac{1}{2}

y = 28.117 x^{0.25}

What is the inverse of f(y)=2y-2y^2 ?
The inverse of f(y)=2y-2y^2 is -(-1+sqrt(-2y+1))/2 ,(sqrt(-2y+1)+1)/2