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:
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:
ƒ(x) = x3 - 4x2 + 8x + 11, when x \( \in \) [0, 1] is:
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Lagrange's MVT: \(f'(c)=\dfrac{f(1)-f(0)}{1-0}\). Solve the resulting quadratic and keep the root lying in (0, 1).
Updated On: Oct 31, 2026
- \( {2 \over 3} \)
- \( {{\sqrt 7 - 2} \over 3} \)
- \( {{4 - \sqrt 5 } \over 3} \)
- \( {{4 - \sqrt 7 } \over 3} \)
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The Correct Option is D
Solution and Explanation
Step 1: \(f(0)=11\), \(f(1)=1-4+8+11=16\), so \(\dfrac{f(1)-f(0)}{1-0}=5\).
Step 2: \(f'(x)=3x^2-8x+8\). Set \(f'(c)=5\): \(3c^2-8c+3=0\).
Step 3: \(c=\dfrac{8\pm\sqrt{64-36}}{6}=\dfrac{8\pm2\sqrt7}{6}=\dfrac{4\pm\sqrt7}{3}\).
Step 4: \(\dfrac{4+\sqrt7}{3}\approx2.22\) lies outside (0, 1); \(\dfrac{4-\sqrt7}{3}\approx0.45\) lies inside.
Answer: (D) \(\dfrac{4-\sqrt7}{3}\)
Step 2: \(f'(x)=3x^2-8x+8\). Set \(f'(c)=5\): \(3c^2-8c+3=0\).
Step 3: \(c=\dfrac{8\pm\sqrt{64-36}}{6}=\dfrac{8\pm2\sqrt7}{6}=\dfrac{4\pm\sqrt7}{3}\).
Step 4: \(\dfrac{4+\sqrt7}{3}\approx2.22\) lies outside (0, 1); \(\dfrac{4-\sqrt7}{3}\approx0.45\) lies inside.
Answer: (D) \(\dfrac{4-\sqrt7}{3}\)
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