The value of c in the Lagrange's mean value theorem for the function f(x) = x3

Engineering

Mathematics

Lagranges Mean Value Theorem

Question

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

$\frac{\sqrt{7} - 2}{3}$

$\frac{2}{3}$

$\frac{4 - \sqrt{5}}{3}$

$\frac{4 - \sqrt{7}}{3}$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

f(x) is a polynomial function

$∴$ it is continuous and differentiable in [0, 1]

Here f(0) = 11, f(1) = 1 – 4 + 8 + 11 = 16

f ' (x) = 3x² – 8x + 8

$∴ f' (c) = \frac{f (1) - f (0)}{1 - 0} = \frac{16 - 11}{1} = 3 c^{2} - 8 c + 8$

⇒ 3c² – 8c + 3 = 0

$C = \frac{8 \pm 2 \sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3}$

$∴ c = \frac{4 - \sqrt{7}}{3} \in (0, 1)$