A value of c according to the Lagrange's mean | vedclass

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(A) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$.According to Lagrange's Mean Value Theorem,there exists at least one $c \in (0, 4)$ such tha

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$2+\frac{2}{\sqrt{3}}$

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(A) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$.According to Lagrange's Mean Value Theorem,there exists at least one $c \in (0, 4)$ such that $f'(c) = \frac{f(4) - f(0)}{4 - 0}$.First,calculate $f(4) = (4-1)(4-2)(4-3) = 3 	imes 2 	imes 1 = 6$.Next,calculate $f(0) = (0-1)(0-2)(0-3) = -1 	imes -2 	imes -3 = -6$.Then,$f'(c) = \frac{6 - (-6)}{4} = \frac{12}{4} = 3$.Now,find $f'(x) = 3x^2 - 12x + 11$.Set $f'(c) = 3c^2 - 12c + 11 = 3$.$3c^2 - 12c + 8 = 0$.Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $c = \frac{12 \pm \sqrt{144 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6} = 2 \pm \frac{4\sqrt{3}}{6} = 2 \pm \frac{2}{\sqrt{3}}$.Since $c \in (0, 4)$,both values $2 + \frac{2}{\sqrt{3}}$ and $2 - \frac{2}{\sqrt{3}}$ are valid. Option $A$ matches this result.

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

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