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}

f (x) = \frac{x - 2}{7} + 8

y = 9 - x^{2}

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

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

f (x) = - 3 x^{4} - 8 x^{3} - 6 x^{2} + 1

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