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

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

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

inverse

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

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

Solution steps

y = x^{2} + x + 1

Replace

x

with

y

x = y^{2} + y + 1

Solve for

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

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

y = - (x - 7)^{2} + 4

\frac{5 x}{x - 7}

f (x) = ln (- x^{2} + 5 x + 6)

3 x^{4} - 24 x^{3} + 48 x^{2}

(x - 1)^{4} \cdot (x - 3)^{2}

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