Absolute extrema on a closed interval

Problem 3.243 · hard

1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
   Reviewed
2. \[ \frac{d}{d x} \left(- x^{3} + 6 x^{2} - 9 x - 2\right) = \left(9 - 3 x\right) \left(x - 1\right) \]
   Differentiate and factor.✓ Proved
3. Critical numbers inside [-4, 5]: 1, 3.
   Reviewed
4. \[ \left. - x^{3} + 6 x^{2} - 9 x - 2 \right|_{\substack{ x=-4 }} = 194 \]
   f(-4).✓ Proved
5. \[ \left. - x^{3} + 6 x^{2} - 9 x - 2 \right|_{\substack{ x=1 }} = -6 \]
   f(1).✓ Proved
6. \[ \left. - x^{3} + 6 x^{2} - 9 x - 2 \right|_{\substack{ x=3 }} = -2 \]
   f(3).✓ Proved
7. \[ \left. - x^{3} + 6 x^{2} - 9 x - 2 \right|_{\substack{ x=5 }} = -22 \]
   f(5).✓ Proved
8. The largest value is 194 and the smallest is -22.
   Reviewed