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

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

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

inverse

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

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

Solution steps

y = x^{2} + x

Replace

x

with

y

x = y^{2} + y

Solve for

y, x = y^{2} + y

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

f (x) = 3 (x + 1)^{2} - 4

y = 2 x^{2} - 3

2 x + 6

x^{2} + 4 x + 7

\frac{x}{( x - 1 ) ( x + 5 )}

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