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

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

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

inverse

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

\frac{- 1 + 4 x - 3}{2}, \frac{- 1 - 4 x - 3}{2}

Solution steps

y = 1 + x + x^{2}

Replace

x

with

y

x = 1 + y + y^{2}

Solve for

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

\frac{- 1 + 4 x - 3}{2}, \frac{- 1 - 4 x - 3}{2}

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