Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let <span class="inline-math">f(x) = \frac{x^2 + x - 1}{x^2 - x + 1}\</span>, then the largest value of <span class="inline-math">f(x)\</span> for <span class="inline-math">x \in [-1, 3]\</span> is:</p>
<p>(a) <span class="inline-math">\frac{3}{5}\</span></p>
<p>(b) <span class="inline-math">\frac{5}{3}\</span></p>
<p>(c) 1</p>
<p>(d) <span class="inline-math">\frac{4}{3}\</span></p>
Step-by-Step Solution
Key Concept: To find the maximum value of a rational function on a closed interval, we must check critical points (where f'(x) = 0) and endpoints. The denominator x² - x + 1 is always positive (discriminant < 0), ensuring f is continuous on [-1, 3].
<p><strong>Step 1: Find the derivative using the quotient rule.</strong></p><p>Let f(x) = (x² + x - 1)/(x² - x + 1).</p><p>Using quotient rule: f'(x) = [(2x + 1)(x² - x + 1) - (x² + x - 1)(2x - 1)] / (x² - x + 1)²</p><p><strong>Step 2: Expand the numerator of f'(x).</strong></p><p>Numerator = (2x + 1)(x² - x + 1) - (x² + x - 1)(2x - 1)</p><p>= 2x³ - 2x² + 2x + x² - x + 1 - (2x³ - x² + 2x² - x - 2x + 1)</p><p>= 2x³ - x² + x + 1 - (2x³ + x² - 3x + 1)</p><p>= 2x³ - x² + x + 1 - 2x³ - x² + 3x - 1</p><p>= -2x² + 4x = -2x(x - 2)</p><p><strong>Step 3: Find critical points.</strong></p><p>Setting f'(x) = 0: -2x(x - 2) = 0, so x = 0 or x = 2.</p><p>Both critical points lie in [-1, 3].</p><p><strong>Step 4: Evaluate f at critical points and endpoints.</strong></p><p>At x = -1: f(-1) = (1 - 1 - 1)/(1 + 1 + 1) = -1/3</p><p>At x = 0: f(0) = (0 + 0 - 1)/(0 - 0 + 1) = -1/1 = -1</p><p>At x = 2: f(2) = (4 + 2 - 1)/(4 - 2 + 1) = 5/3</p><p>At x = 3: f(3) = (9 + 3 - 1)/(9 - 3 + 1) = 11/7 ≈ 1.57</p><p><strong>Step 5: Compare all values.</strong></p><p>-1/3 ≈ -0.33, -1, 5/3 ≈ 1.67, 11/7 ≈ 1.57</p><p>The largest value is 5/3, which occurs at x = 2.</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B