The absolute minimum value, of the function f(x) = | x ^ 2 - x + 1|...

$f\left(x\right) = \left[{x}^{2} - x + 1\right] + \left|{x}^{2} - x + 1\right|x \in \left[- 1, 2\right]$

$∵ \left|{x}^{2} - x + 1\right| \to \left({x}^{2} - x + 1\right) > 0$

$∴ f\left(x\right) = \left({x}^{2} - x + 1\right) + \left[{x}^{2} - x + 1\right]$

Now,

Consider $g ( x ) = \left({x}^{2} - x + 1\right)$

For the minimum value of $g ( x )$,

$g ' ( x ) = 0 ⇒ 2x - 1 = 0$

$⇒ x = \dfrac{1}{2}$

\left({x}^{2} - x + 1\right)$attains its minimum value at$x = \dfrac{1}{2}$</p><p>And$\text{min}$\left[{x}^{2} - x + 1\right] = 0\text{as}{x}^{2} - x + 1 > 0$

$⇒ f\left(x\right)$ attains its minimum at $x = \dfrac{1}{2}$

So, $f\left(\dfrac{1}{2}\right) = \dfrac{3}{4} + 0 = \dfrac{3}{4}$