The absolute minimum value of the function f( | vedclass

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(A) Let $g(x) = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.For $x \in [-1, 2]$,the range of $g(x)$ is $[\frac{3}{4}, 3]$.Since $g(x) \ge \frac{3

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$\frac{3}{4}$

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(A) Let $g(x) = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.For $x \in [-1, 2]$,the range of $g(x)$ is $[\frac{3}{4}, 3]$.Since $g(x) \ge \frac{3}{4}$,we have $|g(x)| = g(x)$.Thus,$f(x) = g(x) + [g(x)]$.To minimize $f(x)$,we consider the values of $g(x)$ in its range $[\frac{3}{4}, 3]$.If $\frac{3}{4} \le g(x) If $1 \le g(x) If $2 \le g(x) \le 3$,then $[g(x)] = 2$,so $f(x) = g(x) + 2$. The minimum value is $2 + 2 = 4$.Comparing these,the absolute minimum value is $\frac{3}{4}$.

The absolute minimum value of the function $f(x) = |x^2 - x + 1| + [x^2 - x + 1]$,where $[t]$ denotes the greatest integer function,in the interval $[-1, 2]$,is:

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