If the Mean Value Theorem holds for the funct | vedclass

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(B) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. First,calculate the values at the endpoints of the interval $[0, 4]$: $f(4) = (4-1)(4-2)(4-

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

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(B) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. First,calculate the values at the endpoints of the interval $[0, 4]$: $f(4) = (4-1)(4-2)(4-3) = 3 	imes 2 	imes 1 = 6$. $f(0) = (0-1)(0-2)(0-3) = -1 	imes -2 	imes -3 = -6$. According to the Lagrange's Mean Value Theorem $(LMVT)$,there exists at least one $c \in (0, 4)$ such that $f'(c) = \frac{f(4) - f(0)}{4 - 0}$. $f'(c) = \frac{6 - (-6)}{4} = \frac{12}{4} = 3$. Now,find the derivative $f'(x)$: $f'(x) = 3x^2 - 12x + 11$. Set $f'(c) = 3$: $3c^2 - 12c + 11 = 3$. $3c^2 - 12c + 8 = 0$. Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $c = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6}$. $c = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3} = 2 \pm \frac{2}{\sqrt{3}}$.

If the Mean Value Theorem holds for the function $f(x)=(x-1)(x-2)(x-3)$ on the interval $x \in [0, 4]$,then the values of $c$ as per the theorem are:

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