If f(x) = (x - 1)(x - 2)(x - 3) for x in [0, | vedclass

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(B) Given: $f(x) = (x - 1)(x - 2)(x - 3)$.Expanding the function: $f(x) = x^3 - 6x^2 + 11x - 6$.The derivative is $f'(x) = 3x^2 - 12x + 11$.Since $f(x

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

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(B) Given: $f(x) = (x - 1)(x - 2)(x - 3)$.Expanding the function: $f(x) = x^3 - 6x^2 + 11x - 6$.The derivative is $f'(x) = 3x^2 - 12x + 11$.Since $f(x)$ is a polynomial,it is continuous on $[0, 4]$ and differentiable on $(0, 4)$.According to Lagrange's Mean Value Theorem,there exists $c \in (0, 4)$ such that $f'(c) = \frac{f(4) - f(0)}{4 - 0}$.Calculate $f(4) = (4-1)(4-2)(4-3) = 3 	imes 2 	imes 1 = 6$.Calculate $f(0) = (0-1)(0-2)(0-3) = -6$.Thus,$f'(c) = \frac{6 - (-6)}{4} = \frac{12}{4} = 3$.Setting $3c^2 - 12c + 11 = 3$,we get $3c^2 - 12c + 8 = 0$.Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$c = \frac{12 \pm \sqrt{144 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6}$.$c = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3}$.Both values lie in $(0, 4)$.

If $f(x) = (x - 1)(x - 2)(x - 3)$ for $x \in [0, 4]$,then the value of $c \in (0, 4)$ satisfying Lagrange's mean value theorem is

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