What is the inverse of f(x)=1+2x+3x^2 ?

The inverse of f(x)=1+2x+3x^2 is (-1+sqrt(3x-2))/3 ,-(sqrt(3x-2)+1)/3

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inverse of f(x)=1+2x+3x^2

inverse

f (x) = 1 + 2 x + 3 x^{2}

\frac{- 1 + 3 x - 2}{3}, - \frac{3 x - 2 + 1}{3}

Solution steps

y = 1 + 2 x + 3 x^{2}

Replace

x

with

y

x = 1 + 2 y + 3 y^{2}

Solve for

y, x = 1 + 2 y + 3 y^{2}

\frac{- 1 + 3 x - 2}{3}, - \frac{3 x - 2 + 1}{3}

What is the inverse of f(x)=1+2x+3x^2 ?
The inverse of f(x)=1+2x+3x^2 is (-1+sqrt(3x-2))/3 ,-(sqrt(3x-2)+1)/3