Animated Solution for Mathematics - Functions: 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 :

The function provided is $f(x)=|x^2-x+1|+[x^2-x+1]$. To simplify our environment, let us define the quadratic component as:

Our function can now be expressed as $f(x)=|g(x)|+[g(x)]$.

To understand the behavior of $g(x)$, we calculate the discriminant $D=b^2-4ac$. For our quadratic, $a=1$, $b=-1$, and $c=1$:

A negative discriminant indicates that the parabola $g(x)$ never intersects the $x$-axis. Since the leading coefficient $a=1$ is positive, the parabola opens upwards and remains strictly positive for all real $x$.

Because $g(x)>0$ for all $x$, the absolute value $|g(x)|$ is simply $g(x)$. Consequently, the function simplifies to:

Next, we find the vertex of the parabola $g(x)$. The $x$-coordinate is given by:

Substituting this back into $g(x)$, we find the minimum value of the quadratic part:

We now examine the behavior of $f(x)=g(x)+[g(x)]$ near the vertex. Since the minimum value of $g(x)$ is $\frac{3}{4}$, and $\frac{3}{4}<1$, the greatest integer function $[g(x)]$ evaluates to $0$ in the neighborhood of the vertex.

In this region, the function simplifies to:

If we move to regions where $g(x)\ge 1$, the value of $[g(x)]$ becomes at least $1$, resulting in $f(x)=g(x)+1$, which is strictly greater than $\frac{3}{4}$. Therefore, the absolute minimum value of the function is: