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

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

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inverse of f(x)=x^2+x,x>=-1/2

inverse

f (x) = x^{2} + x, x \geq - \frac{1}{2}

\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}

4 t^{2} + 1

f (x) = \frac{x ^{3} + 2}{x ^{3} - 2}

f (x) = \frac{2}{2 x - 1}

y = \frac{x + 1}{2 x + 1}

f (x) = - 5 x - 30

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