1. **Stating the problem:** We analyze the function $$f(x) = \frac{x}{2x^{2} - 3x + 1} + \sqrt{\frac{x-1}{1-2x}}$$ which consists of a rational function and a square root of a rational expression. 2. **Simplify the denominator of the rational function:** The denominator is $$2x^2 - 3x + 1$$. Factor it: $$2x^{2} - 3x + 1 = (2x -1)(x -1)$$. Therefore, $$\frac{x}{2x^{2} - 3x + 1} = \frac{x}{(2x -1)(x -1)}$$. 3. **Analyze the domain restrictions from the rational function:** The denominator cannot be zero: \begin{align*} (2x -1)(x -1) &\neq 0 \\ x &\neq \frac{1}{2}, \ x \neq 1 \end{align*} 4. **Analyze the square root term:** $$\sqrt{\frac{x-1}{1-2x}}$$ requires the expression inside the root to be \(\geq 0\) and denominator \(\neq 0\). We must have: $$\frac{x-1}{1-2x} \geq 0$$ and $$1-2x \neq 0 \Rightarrow x \neq \frac{1}{2}$$. 5. **Solve the inequality:** Sign analysis on \(\frac{x-1}{1-2x}\) : - Numerator zero at \(x=1\). - Denominator zero at \(x=\frac{1}{2}\). Intervals: - For \(x < \frac{1}{2}\): denominator is positive (since \(1 - 2x > 0\)) - For \(x > \frac{1}{2}\): denominator negative Check sign: - When \(x < \frac{1}{2}\), numerator \(< 1\) and if \(x < 1\), numerator negative. In \(( -\infty, \frac{1}{2} )\), numerator negative, denominator positive, so fraction negative \(<0\). - In \(( \frac{1}{2}, 1 )\): numerator \(< 1\) positive (since \(x > 1/2\) but less than 1 means numerator negative), denominator negative, fraction \(\frac{negative}{negative} = positive\). - At \(x=1\), numerator zero, fraction zero. - For \(x > 1\), numerator positive, denominator negative, fraction negative. So the inequality \(\geq 0\) holds in \([\frac{1}{2}, 1]\) except at \(x= \frac{1}{2}\) where denominator zero. 6. **Domain combined:** From rational and root domain: - Denominator of rational cannot be zero: \(x \neq \frac{1}{2}, 1\) - Root expression nonnegative: \(x \in (\frac{1}{2},1]\) Removing \(x=1\) from denominator, domain for root part compatible: \((\frac{1}{2}, 1)\) Therefore, the function's domain is: $$\boxed{\left(\frac{1}{2}, 1\right)}$$ 7. **Summary:** - The function is defined only on the open interval \((\frac{1}{2}, 1)\). - The function combines a rational expression and a radical with a rational expression inside. - Vertical asymptotes at \(x=\frac{1}{2}, 1\) due to zero denominators. This detailed analysis aids in understanding the function behavior and domain restrictions.