Question

Qual a função inversa (e seu domínio) de $y=x^2-x, \:\:\:x\geq \frac{1}{2}$

Answer 1

$\sf{y=x^2-x}\\\sf{x^2-x+\dfrac{1}{4}-\dfrac{1}{4}=y}\\\sf{4x^2-4x+1-1=4y}\\\sf{4x^2-4x+1=4y+1}\\\sf{(2x-1)^2=4y+1}\\\sf{2x-1=\pm\sqrt{4y+1}}\\\sf{2x=1\pm\sqrt{4y+1}}\\\sf{x=\dfrac{1+\sqrt{4y+1}}{2}}\\\sf{y^{-1}=\dfrac{1+\sqrt{4x+1}}{2}}$

$\sf{D=\{x\in\mathbb{R}/x\geq~-\dfrac{1}{4}\}}$