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}

7 x + 4

g (x) = \frac{x - 4}{2 x + 3}

7 x + x + 50

2 x^{2} + 1

x^{2} - 4 x + 4

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