The absolute minimum value, of the function f(x)=|x^2-x+1|+[x^2-x+1], where [t] denotes th…

Question

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 :

Accepted Answer

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

Let $g(x)=x^{2}-x+1$

$$
=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}
$$

$$
\because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text { and }\left[\mathrm{x}^{2}-\mathrm{x}+1\right]
$$

Both have minimum value at $\mathrm{x}=1 / 2$

$\Rightarrow$ Minimum $\mathrm{f}(\mathrm{x})=\frac{3}{4}+0$

$=\frac{3}{4}$

Solutions · 1