What is the inverse of (2x^2+x-1)/9 ?

The inverse of (2x^2+x-1)/9 is (-1+3sqrt(8x+1))/4 ,(-1-3sqrt(8x+1))/4

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inverse of (2x^2+x-1)/9

inverse

\frac{2 x ^{2} + x - 1}{9}

\frac{- 1 + 3 8 x + 1}{4}, \frac{- 1 - 3 8 x + 1}{4}

Solution steps

y = \frac{2 x ^{2} + x - 1}{9}

Replace

x

with

y

x = \frac{2 y ^{2} + y - 1}{9}

Solve for

y, x = \frac{2 y ^{2} + y - 1}{9}

\frac{- 1 + 3 8 x + 1}{4}, \frac{- 1 - 3 8 x + 1}{4}

f (x) = x - 3 + 2

f^{8}

f (x) = x^{2} - 3 x

y = 46 x - \frac{5}{6}

tan (2 x)

What is the inverse of (2x^2+x-1)/9 ?
The inverse of (2x^2+x-1)/9 is (-1+3sqrt(8x+1))/4 ,(-1-3sqrt(8x+1))/4