Question

The value of c in the Lagrange's mean value theorem for the function
ƒ(x) = x³ - 4x² + 8x + 11, when x \( \in \) [0, 1] is:

• A. \( {2 \over 3} \)
• B. \( {{\sqrt 7 - 2} \over 3} \)
• C. \( {{4 - \sqrt 5 } \over 3} \)
• D. \( {{4 - \sqrt 7 } \over 3} \)

Hint:

Lagrange's MVT: \(f'(c)=\dfrac{f(1)-f(0)}{1-0}\). Solve the resulting quadratic and keep the root lying in (0, 1).

Answer 1

The Correct Option is D

Step 1: MVT requires \(f'(c)=f(1)-f(0)\) with \(c\in(0,1)\). Compute \(f(1)-f(0)=(1-4+8)=5\) (the constant 11 cancels).
Step 2: Test the options in \(f'(c)=3c^2-8c+8\). For \(c=\tfrac{4-\sqrt7}{3}\): \(c^2=\dfrac{23-8\sqrt7}{9}\), so \(3c^2=\dfrac{23-8\sqrt7}{3}\) and \(8c=\dfrac{32-8\sqrt7}{3}\).
Step 3: \(f'(c)=\dfrac{23-8\sqrt7-32+8\sqrt7}{3}+8=\dfrac{-9}{3}+8=5\). Correct.
Step 4: The other options do not give 5 (e.g. \(c=2/3\) gives \(f'=4/3-16/3+8=4\)). Also \(\tfrac{\sqrt7-2}{3}\approx0.22\) gives about 6.4, and \(\tfrac{4-\sqrt5}{3}\approx0.59\) gives about 4.3. Option (D).