If L.M.V.T. is true for f(x) = x(x-1)(x-2) on | vedclass

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(B) Given $f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 2x$.According to $L.M.V.T.$,there exists at least one $c \in (0, 1/2)$ such that $f'(c) = \frac{f(1/2) -

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$1 - \frac{\sqrt{3}}{6}$

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(B) Given $f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 2x$.According to $L.M.V.T.$,there exists at least one $c \in (0, 1/2)$ such that $f'(c) = \frac{f(1/2) - f(0)}{1/2 - 0}$.First,calculate $f'(x) = 3x^2 - 6x + 2$.Next,calculate $f(1/2) = \frac{1}{2}(\frac{1}{2} - 1)(\frac{1}{2} - 2) = \frac{1}{2} 	imes (-\frac{1}{2}) 	imes (-\frac{3}{2}) = \frac{3}{8}$.Also,$f(0) = 0$.So,$f'(c) = \frac{3/8 - 0}{1/2} = \frac{3}{8} 	imes 2 = \frac{3}{4}$.Now,set $3c^2 - 6c + 2 = \frac{3}{4}$.$12c^2 - 24c + 8 = 3 \implies 12c^2 - 24c + 5 = 0$.Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$c = \frac{24 \pm \sqrt{576 - 240}}{24} = \frac{24 \pm \sqrt{336}}{24} = \frac{24 \pm 4\sqrt{21}}{24} = 1 \pm \frac{\sqrt{21}}{6}$.Since $c \in (0, 1/2)$,we check the values: $1 + \frac{\sqrt{21}}{6} > 1$ and $1 - \frac{\sqrt{21}}{6} \approx 1 - 0.76 = 0.24$,which is in $(0, 1/2)$.Thus,$c = 1 - \frac{\sqrt{21}}{6}$.

If $L.M.V.T.$ is true for $f(x) = x(x-1)(x-2)$ on the interval $x \in [0, 1/2]$,then find the value of $C$.

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