Applications of Derivatives
Absolute Maxima and Minima
Grade 12
Question:
<p>The absolute minimum and maximum values of the function \(\frac{x^2 - x + 1}{x^2 + x + 1}\) are given by</p>
<p>(a) 1 and 3</p>
<p>(b) 1/2 and 3</p>
<p>(c) 1/3 and 3</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Convert the function to a quadratic in x and use the discriminant condition for real solutions to find the range.
<p>Let f(x) = (x² - x + 1)/(x² + x + 1)</p><p>Let y = f(x), then y(x² + x + 1) = x² - x + 1</p><p>yx² + yx + y = x² - x + 1</p><p>(y - 1)x² + (y + 1)x + (y - 1) = 0</p><p>For real x, discriminant ≥ 0:</p><p>(y + 1)² - 4(y - 1)² ≥ 0</p><p>(y + 1)² ≥ 4(y - 1)²</p><p>|y + 1| ≥ 2|y - 1|</p><p>Solving: (y + 1)² ≥ 4(y - 1)²</p><p>y² + 2y + 1 ≥ 4y² - 8y + 4</p><p>-3y² + 10y - 3 ≥ 0</p><p>3y² - 10y + 3 ≤ 0</p><p>(3y - 1)(y - 3) ≤ 0</p><p>Therefore: 1/3 ≤ y ≤ 3</p><p>∴ Answer is (c) 1/3 and 3.</p>
Correct Answer: c