Finding critical numbers of a piecewise function

For $x<1$, $f(x)=x^2$ so $f'(x)=2x$, which equals zero at $x=0$; this lies in the domain $x<1$, so $x=0$ is a critical number. For $x>1$, $f(x)=-x+2$ so $f'(x)=-1$, which is never zero, so this piece contributes no critical numbers.

At $x=1$, check differentiability: the left-hand derivative is $2(1)=2$ and the right-hand derivative is $-1$. Since these don't match, $f'(1)$ is undefined, so $x=1$ is also a critical number. Combining both pieces, the critical numbers in $(-2,3)$ are $x=0$ and $x=1$.

Evaluate $f$ at each candidate point: $f(-2) = (-2)^2 = 4$, $f(0) = 0^2 = 0$, $f(1) = 1^2 = 1$, and $f(3) = -(3)+2 = -1$. These are the only points where an absolute extremum can occur, per the closed interval method.

Comparing the values $4, 0, 1, -1$, the largest is $4$ at $x=-2$ and the smallest is $-1$ at $x=3$. So the absolute maximum value is $4$ (occurring at $x=-2$) and the absolute minimum value is $-1$ (occurring at $x=3$).