Find the greatest and the least value of f(x) = x³ - 3x + 1 on the closed interval [-2, 3]…

Step 1. The critical points. $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$, so $x = -1$ and $x = 1$. Both lie in $[-2, 3]$.

Step 2. The values. $f(-2) = -8 + 6 + 1 = -1$. $f(-1) = -1 + 3 + 1 = 3$. $f(1) = 1 - 3 + 1 = -1$. $f(3) = 27 - 9 + 1 = 19$.

Step 3. The comparison. The greatest of $-1$, $3$, $-1$ and $19$ is $19$. The least is $-1$.

Step 4. Where. The value $-1$ occurs at $x = -2$ (an end point) and at $x = 1$ (a local minimum). That is $2$ points.