<p>Find the domain and range of
\(f(x) = \dfrac{x^2 - 3x + 2}{x^2 + x - 6}\).</p>
Step-by-Step Solution
Key Concept: Factor both numerator and denominator to identify excluded values in domain (zeros of denominator), then use the inverse function method: set y = f(x) and solve for x to find which y-values are impossible.
<p><strong>Step 1: Find Domain</strong></p><p>Factor numerator and denominator:</p><p>Numerator: x² - 3x + 2 = (x - 1)(x - 2)</p><p>Denominator: x² + x - 6 = (x + 3)(x - 2)</p><p>So f(x) = (x - 1)(x - 2) / [(x + 3)(x - 2)]</p><p>Denominator = 0 when x = -3 or x = 2</p><p><strong>∴ Domain = ℝ \ {-3, 2}</strong></p><p><strong>Step 2: Find Range</strong></p><p>Set y = f(x) and solve for x:</p><p>y = (x² - 3x + 2)/(x² + x - 6)</p><p>y(x² + x - 6) = x² - 3x + 2</p><p>yx² + yx - 6y = x² - 3x + 2</p><p>(y - 1)x² + (y + 3)x + (-6y - 2) = 0</p><p><strong>Step 3: Check Validity Conditions</strong></p><p>For x to be real and x ≠ -3, 2, this quadratic must have real solutions.</p><p>When y = 1: 0·x² + 4x - 8 = 0 → x = 2 (excluded from domain)</p><p>When y ≠ 1: Discriminant Δ = (y + 3)² - 4(y - 1)(-6y - 2) ≥ 0</p><p>Δ = y² + 6y + 9 + 4(y - 1)(6y + 2)</p><p>= y² + 6y + 9 + 24y² + 8y - 24y - 8</p><p>= 25y² - 10y + 1 = (5y - 1)² ≥ 0</p><p>Equality holds when y = 1/5, giving x = -3 (excluded from domain)</p><p><strong>∴ Range = ℝ \ {1/5}</strong>
Correct Answer: Domain = R \ {-3, 2}, Range = R \ {1/5}