Extrema of |x² − 3x + 2| on [0, 3]

Without the absolute value.
- On $[0,1]\cup[2,3]$, $(x-1)(x-2)\ge0$ and $f(x)=x^2-3x+2$ (decreasing on $[0,1]$ from $2$ to $0$; increasing on $[2,3]$ from $0$ to $2$).
- On $[1,2]$, $f(x)=-(x^2-3x+2)$, which rises from $0$ at $x=1$ to its maximum $\frac14$ at $x=\frac32$, then falls back to $0$ at $x=2$.

Extrema.
- Minimum: $f=0$, attained at $x=1$ and $x=2$.
- Maximum: $2$, attained at the endpoints $x=0$ and $x=3$ (the interior local maximum $\frac14$ at $\frac32$ is much lower).

Non-differentiable points. At $x=1$ and $x=2$, where the expression inside the absolute value changes sign. At $x=1$ the left derivative is $-1$ and the right derivative is $+1$; the same holds at $x=2$ (left $-1$, right $+1$). Both are corners.

Remark. The minima occur *exactly at the points where $f$ is not differentiable*. This is why the theorem "an extremum of a differentiable function has zero derivative" requires differentiability at the extremum.