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(D) Given function is $f(x) = (x-1)(x-2) = x^2 - 3x + 2$ on the interval $[0, 1/2]$.According to Lagrange's Mean Value Theorem,there exists at least o

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$\frac{1}{4}$

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(D) Given function is $f(x) = (x-1)(x-2) = x^2 - 3x + 2$ on the interval $[0, 1/2]$.According to Lagrange's Mean Value Theorem,there exists at least one $c \in (0, 1/2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Here,$a = 0$ and $b = 1/2$.Calculate $f(a) = f(0) = (0-1)(0-2) = 2$.Calculate $f(b) = f(1/2) = (1/2 - 1)(1/2 - 2) = (-1/2)(-3/2) = 3/4$.Calculate $f'(x) = 2x - 3$,so $f'(c) = 2c - 3$.Substitute these into the formula: $2c - 3 = \frac{3/4 - 2}{1/2 - 0}$.$2c - 3 = \frac{-5/4}{1/2} = -5/2$.$2c = 3 - 5/2 = 1/2$.$c = 1/4$.Since $1/4 \in (0, 1/2)$,the value of $c$ is $1/4$.

For the function $f(x)=(x-1)(x-2)$ defined on $\left[0, \frac{1}{2}\right]$,the value of $c$ satisfying Lagrange's mean value theorem is

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