Solve the inequality \( \frac{x^{3}-2 x^{2}+x}{x^{2}-4 x+4} \geq 0 \) by testing values in the interval. A \( \quad[0,1] \cup(2,+\infty) \) B \( \quad[0,1] \cup(1,+\infty) \) C \( \quad[0,1] \cup(1,2) \cup(2,+\infty) \) D \( \quad(-\infty, 1) \cup(1,+\infty) \)

To solve the inequality \( \frac{x^{3}-2 x^{2}+x}{x^{2}-4 x+4} \geq 0 \), we start by factoring both the numerator and the denominator. 1. **Numerator**: \( x^3 - 2x^2 + x = x(x^2 - 2x + 1) = x(x-1)^2 \) 2. **Denominator**: \( x^2 - 4x + 4 = (x-2)^2 \) Now we rewrite the inequality: \[ \frac{x(x-1)^2}{(x-2)^2} \geq 0 \] Next, we determine the critical points by setting the numerator and denominator to zero: - \( x = 0 \) (from the numerator) - \( x = 1 \) (from the numerator) - \( x = 2 \) (from the denominator, but it cannot be included since it would make the fraction undefined) Now, we identify intervals based on these critical points: \((-∞, 0)\), \((0, 1)\), \((1, 2)\), and \((2, +\infty)\). Next, we test each interval to determine where the inequality holds true: - For \( x \in (-\infty, 0) \): Choose \( x = -1 \) \( \frac{-1(2)^2}{(3)^2} = \frac{-4}{9} < 0 \) - For \( x \in (0, 1) \): Choose \( x = 0.5 \) \( \frac{0.5(-0.5)^2}{(1.5)^2} = \frac{0.5(0.25)}{2.25} > 0 \) - For \( x \in (1, 2) \): Choose \( x = 1.5 \) \( \frac{1.5(0.5)^2}{(-0.5)^2} = \frac{1.5(0.25)}{0.25} = 1 > 0 \) - For \( x \in (2, +\infty) \): Choose \( x = 3 \) \( \frac{3(2)^2}{(1)^2} = \frac{12}{1} > 0 \) Now we summarize the test results: - The inequality is non-negative in the intervals \( (0, 1) \), \( (1, 2) \), and \( (2, +\infty) \). Finally, we consider the critical points: - Include \( x = 0 \) (numerator zero, works), - Include \( x = 1 \) (numerator zero, works), - Exclude \( x = 2 \) (undefined). Thus, the solution is: \[ [0, 1] \cup (1, 2) \cup (2, +\infty) \] The correct choice is **C: \( \quad[0,1] \cup(1,2) \cup(2,+\infty) \)**.