Answer
$y=\frac{1}{2}+\sqrt {x+\frac{1}{4}}$
Work Step by Step
$y=x^2-x, x\geq\frac{1}{2}$
$=(x^2-x+\frac{1}{4})-\frac{1}{4}$
$=(x-\frac{1}{2})^2-\frac{1}{4}$
$(x-\frac{1}{2})^2=y+\frac{1}{4}$
$x-\frac{1}{2}=\pm \sqrt {y+\frac{1}{4}}$
$\because x\geq\frac{1}{2}, x-\frac{1}{2}\geq0$
$\therefore x-\frac{1}{2}=\sqrt {y+\frac{1}{4}}$
$x=\frac{1}{2}+\sqrt {y+\frac{1}{4}}$
$y\rightleftharpoons x$
$f^{-1}(x)=y=\frac{1}{2}+\sqrt {x+\frac{1}{4}}$
