Find the domain of \( \mathrm{f}(\mathrm{x}) \). \[ f(x)=\sqrt{x^{2}-9 x+18} \] Answer Attempt out of 2 Answer Type: Interval \( ^{\text {Fin }} \)

To determine the domain of the function \( f(x) = \sqrt{x^{2} - 9x + 18} \), we need to ensure that the expression inside the square root is non-negative: \[ x^{2} - 9x + 18 \geq 0 \] **Step 1: Find the Roots of the Quadratic** First, solve the equation \( x^{2} - 9x + 18 = 0 \) to find the critical points. \[ \text{Discriminant} (D) = (-9)^2 - 4(1)(18) = 81 - 72 = 9 \] \[ x = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2} \] \[ x = 6 \quad \text{or} \quad x = 3 \] **Step 2: Test Intervals Around the Roots** The critical points divide the real number line into three intervals: 1. \( x < 3 \) 2. \( 3 < x < 6 \) 3. \( x > 6 \) Choose a test point from each interval to determine where the quadratic is non-negative. - **For \( x < 3 \) (e.g., \( x = 0 \)):** \[ (0 - 3)(0 - 6) = (-3)(-6) = 18 \geq 0 \] - **For \( 3 < x < 6 \) (e.g., \( x = 4.5 \)):** \[ (4.5 - 3)(4.5 - 6) = (1.5)(-1.5) = -2.25 < 0 \] - **For \( x > 6 \) (e.g., \( x = 7 \)):** \[ (7 - 3)(7 - 6) = (4)(1) = 4 \geq 0 \] **Step 3: Combine the Intervals** The quadratic is non-negative in the intervals where: \[ x \leq 3 \quad \text{or} \quad x \geq 6 \] **Final Domain:** \[ (-\infty, 3] \cup [6, \infty) \] **Answer:** \((-\infty,\,3\,]\,\cup\, [\,6,\,\infty)\)