If f(x)=(x-1)(x-2)(x-3) for x in[0,4],then th | vedclass

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(B) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. Since $f(x)$ is a polynomial,it is continuous on $[0, 4]$ and differentiable on $(0, 4)$. C

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

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(B) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. Since $f(x)$ is a polynomial,it is continuous on $[0, 4]$ and differentiable on $(0, 4)$. Calculate $f(4)$ and $f(0)$: $f(4) = 4^3 - 6(4)^2 + 11(4) - 6 = 64 - 96 + 44 - 6 = 6$. $f(0) = 0^3 - 6(0)^2 + 11(0) - 6 = -6$. By Lagrange's Mean Value Theorem,there exists $c \in (0, 4)$ such that $f'(c) = \frac{f(4) - f(0)}{4 - 0}$. $f'(x) = 3x^2 - 12x + 11$. So,$3c^2 - 12c + 11 = \frac{6 - (-6)}{4} = \frac{12}{4} = 3$. $3c^2 - 12c + 8 = 0$. Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $c = \frac{12 \pm \sqrt{144 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3}$.

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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