What is the inverse of y=x^2-x,x>= 1/2 ?

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

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inverse of y=x^2-x,x>= 1/2

inverse

y = 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) = ln (x^{2} - 1)

y = 4 x + 1

y = - 2 - x

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

f (x) = 2 y + 3

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