Absolute extrema on a closed interval

Problem 3.246 · hard

1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
   Reviewed
2. \[ \frac{d}{d x} \left(- x^{3} - \frac{15 x^{2}}{2} - 18 x\right) = \left(- 3 x - 6\right) \left(x + 3\right) \]
   Differentiate and factor.✓ Proved
3. Critical numbers inside [-4, 3]: -3, -2.
   Reviewed
4. \[ \left. - x^{3} - \frac{15 x^{2}}{2} - 18 x \right|_{\substack{ x=-4 }} = 16 \]
   f(-4).✓ Proved
5. \[ \left. - x^{3} - \frac{15 x^{2}}{2} - 18 x \right|_{\substack{ x=-3 }} = \frac{27}{2} \]
   f(-3).✓ Proved
6. \[ \left. - x^{3} - \frac{15 x^{2}}{2} - 18 x \right|_{\substack{ x=-2 }} = 14 \]
   f(-2).✓ Proved
7. \[ \left. - x^{3} - \frac{15 x^{2}}{2} - 18 x \right|_{\substack{ x=3 }} = - \frac{297}{2} \]
   f(3).✓ Proved
8. The largest value is 16 and the smallest is -297/2.
   Reviewed