In the Mean Value Theorem,f'(c) = frac{f(b) - | vedclass

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(C) Given $f(x) = x(x - 1)(x - 2) = x(x^2 - 3x + 2) = x^3 - 3x^2 + 2x$. First,calculate $f(a)$ and $f(b)$ where $a = 0$ and $b = \frac{1}{2}$: $f(0) =

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$1 - \frac{\sqrt{21}}{6}$

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(C) Given $f(x) = x(x - 1)(x - 2) = x(x^2 - 3x + 2) = x^3 - 3x^2 + 2x$. First,calculate $f(a)$ and $f(b)$ where $a = 0$ and $b = \frac{1}{2}$: $f(0) = 0(0 - 1)(0 - 2) = 0$. $f(\frac{1}{2}) = \frac{1}{2}(\frac{1}{2} - 1)(\frac{1}{2} - 2) = \frac{1}{2}(-\frac{1}{2})(-\frac{3}{2}) = \frac{3}{8}$. Now,calculate the slope $\frac{f(b) - f(a)}{b - a} = \frac{3/8 - 0}{1/2 - 0} = \frac{3/8}{1/2} = \frac{3}{4}$. Find the derivative $f'(x) = 3x^2 - 6x + 2$. Set $f'(c) = \frac{3}{4}$: $3c^2 - 6c + 2 = \frac{3}{4} \implies 12c^2 - 24c + 8 = 3 \implies 12c^2 - 24c + 5 = 0$. Using the quadratic formula $c = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: $c = \frac{24 \pm \sqrt{(-24)^2 - 4(12)(5)}}{2(12)} = \frac{24 \pm \sqrt{576 - 240}}{24} = \frac{24 \pm \sqrt{336}}{24} = \frac{24 \pm 4\sqrt{21}}{24} = 1 \pm \frac{\sqrt{21}}{6}$. Since $c \in (0, 1/2)$,we choose $c = 1 - \frac{\sqrt{21}}{6}$.

In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

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