Mathematics - Application of Derivative Question with Solution | TestHub

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

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

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

Now,

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

For the minimum value of $g\left(x\right)$,

$g\text{'}\left(x\right)=0\Rightarrow 2x-1=0$

$\Rightarrow x=\frac{1}{2}$

$\left(x^2-x+1\right)$ attains its minimum value at $x=\frac{1}{2}$

And $\text{min}\left[x^2-x+1\right]=0 \text{as} x^2-x+1>0$

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

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