91Ó°ÊÓ

Set the denominator of the fraction equal to zero, and solve for 'x'. \n\[2x - 1 = 0\]Adding 1 to both sides \n\[2x = 1\]Dividing both sides by 2 \n\[x = 0.5\]Thus, 'x' cannot be equal to 0.5, since it would make the denominator equal to zero.

Now, the expression inside the square root must be greater than or equal to zero. Hence, we set what's inside the square root greater than or equal to zero and solve for 'x'. \[\frac{x}{2x-1}-1 \geq 0\]This simplifies into: \[2x^2 - 3x \geq 1\]To find the roots, put it into the quadratic form, \(0 = ax²+bx+c\): \[2x^2 - 3x -1 = 0\]Using the quadratic formula, \(x=\frac{-b±\sqrt{b²-4ac}}{2a}\), we find: \[x_1 = 1, \; x_2 = -\frac{1}{2}\]We have two regions to consider, \(x \leq -0.5\), \(-0.5 < x\leq 0.5\), and \(x > 1\). By choosing test points in each region, we can determine which portions satisfy the inequality. The test points chosen are \(x = -1, x = 0, x = 2\), representing each region respectively. When these values are inserted in the inequality:When \(x = -1\), \(\frac{-1}{2* -1 -1}-1 > 0\), it is valid.When \(x = 0\), \(\frac{0}{2*0 -1}-1 > 0\), this is not valid.When \(x = 2\), \(\frac{2}{2*2-1}-1 > 0\), it is valid.

Combining the restrictions from Step 1 and Step 2, we find that the domain of the function is \(x \leq -0.5\) and \(x > 1\). Note, 0.5 is excluded from the domain due to Step 1.