Applications of Derivatives
Lagrange's Mean Value Theorem
Grade 12
Question:
<p>The value of <i>c</i> in the Lagrange's mean value theorem for the function \(f(x) = x^3 - 4x^2 + 8x + 11\), when \(x \in [0, 1]\) is</p>
<p>(a) \(\frac{7 - 2\sqrt{3}}{3}\)</p>
<p>(b) \(\frac{2}{3}\)</p>
<p>(c) \(\frac{4 - \sqrt{5}}{3}\)</p>
<p>(d) \(\frac{4 - \sqrt{7}}{3}\)</p>
Step-by-Step Solution
Key Concept: Apply Lagrange's mean value theorem by equating the derivative at c to the average rate of change, then solve the resulting quadratic equation.
<p><strong>Given:</strong> $f(x) = x^3 - 4x^2 + 8x + 11$ on $[0, 1]$</p><p><strong>Step 1:</strong> The function is continuous on [0, 1] and differentiable on (0, 1).</p><p><strong>Step 2:</strong> By Lagrange's mean value theorem, for $c \in (0, 1)$:</p><p>$$f'(c) = \frac{f(1) - f(0)}{1 - 0}$$</p><p><strong>Step 3:</strong> Calculate $f'(x) = 3x^2 - 8x + 8$</p><p><strong>Step 4:</strong> Calculate $f(1) = 1 - 4 + 8 + 11 = 16$ and $f(0) = 11$</p><p>$$3c^2 - 8c + 8 = \frac{16 - 11}{1} = 5$$</p><p><strong>Step 5:</strong> Solve $3c^2 - 8c + 3 = 0$</p><p>$$c = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{4 \pm \sqrt{7}}{3}$$</p><p>Since $c \in (0, 1)$, we have $c = \frac{4 - \sqrt{7}}{3}$</p><p>∴ Answer is (d).</p>
Correct Answer: D