Absolute extrema on a closed interval

Problem 3.279 · hard

1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
   Reviewed
2. \[ \frac{d}{d x} \left(- x^{3} - 3 x^{2} + 9 x - 4\right) = \left(3 - 3 x\right) \left(x + 3\right) \]
   Differentiate and factor.✓ Proved
3. Critical numbers inside [-4, 5]: -3, 1.
   Reviewed
4. \[ \left. - x^{3} - 3 x^{2} + 9 x - 4 \right|_{\substack{ x=-4 }} = -24 \]
   f(-4).✓ Proved
5. \[ \left. - x^{3} - 3 x^{2} + 9 x - 4 \right|_{\substack{ x=-3 }} = -31 \]
   f(-3).✓ Proved
6. \[ \left. - x^{3} - 3 x^{2} + 9 x - 4 \right|_{\substack{ x=1 }} = 1 \]
   f(1).✓ Proved
7. \[ \left. - x^{3} - 3 x^{2} + 9 x - 4 \right|_{\substack{ x=5 }} = -159 \]
   f(5).✓ Proved
8. The largest value is 1 and the smallest is -159.
   Reviewed