Absolute extrema on a closed interval

Problem 3.241 · hard

1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
   Reviewed
2. \[ \frac{d}{d x} \left(- x^{3} + \frac{9 x^{2}}{2} - 1\right) = - 3 x \left(x - 3\right) \]
   Differentiate and factor.✓ Proved
3. Critical numbers inside [-4, 4]: 0, 3.
   Reviewed
4. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 1 \right|_{\substack{ x=-4 }} = 135 \]
   f(-4).✓ Proved
5. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 1 \right|_{\substack{ x=0 }} = -1 \]
   f(0).✓ Proved
6. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 1 \right|_{\substack{ x=3 }} = \frac{25}{2} \]
   f(3).✓ Proved
7. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 1 \right|_{\substack{ x=4 }} = 7 \]
   f(4).✓ Proved
8. The largest value is 135 and the smallest is -1.
   Reviewed