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Absolute extrema on a closed interval practice problems

The closed-interval method: check the critical numbers and the endpoints. 70 problems with worked solutions; in 70 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 3.46

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + \frac{9 x^{2}}{2} + 6 x - 1 \) on \( \displaystyle [-4, 1] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} + \frac{9 x^{2}}{2} + 6 x - 1\right) = \left(x + 2\right) \left(3 x + 3\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 1]: -2, -1.
    Reviewed
  4. \[ \left. x^{3} + \frac{9 x^{2}}{2} + 6 x - 1 \right|_{\substack{ x=-4 }} = -17 \]
    f(-4).✓ Proved
  5. \[ \left. x^{3} + \frac{9 x^{2}}{2} + 6 x - 1 \right|_{\substack{ x=-2 }} = -3 \]
    f(-2).✓ Proved
  6. \[ \left. x^{3} + \frac{9 x^{2}}{2} + 6 x - 1 \right|_{\substack{ x=-1 }} = - \frac{7}{2} \]
    f(-1).✓ Proved
  7. \[ \left. x^{3} + \frac{9 x^{2}}{2} + 6 x - 1 \right|_{\substack{ x=1 }} = \frac{21}{2} \]
    f(1).✓ Proved
  8. The largest value is 21/2 and the smallest is -17.
    Reviewed
Answer \( \text{max } \frac{21}{2} \text{ at } x=1;\ \text{min } -17 \text{ at } x=-4 \)
The graph of f(x) = x³ + 9x²/2 + 6x - 1 on [-4, 1]. The candidates are the endpoints and the critical points inside; the highest is 21/2 and the lowest -17.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.47

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 3 x \) on \( \displaystyle [-4, 3] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} - 3 x\right) = \left(x + 1\right) \left(3 x - 3\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 3]: -1, 1.
    Reviewed
  4. \[ \left. x^{3} - 3 x \right|_{\substack{ x=-4 }} = -52 \]
    f(-4).✓ Proved
  5. \[ \left. x^{3} - 3 x \right|_{\substack{ x=-1 }} = 2 \]
    f(-1).✓ Proved
  6. \[ \left. x^{3} - 3 x \right|_{\substack{ x=1 }} = -2 \]
    f(1).✓ Proved
  7. \[ \left. x^{3} - 3 x \right|_{\substack{ x=3 }} = 18 \]
    f(3).✓ Proved
  8. The largest value is 18 and the smallest is -52.
    Reviewed
Answer \( \text{max } 18 \text{ at } x=3;\ \text{min } -52 \text{ at } x=-4 \)
The graph of f(x) = x³ - 3x on [-4, 3]. The candidates are the endpoints and the critical points inside; the highest is 18 and the lowest -52.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.48

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{3 x^{2}}{2} - 2 \) on \( \displaystyle [-4, 3] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} - \frac{3 x^{2}}{2} - 2\right) = - 3 x \left(x + 1\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 3]: -1, 0.
    Reviewed
  4. \[ \left. - x^{3} - \frac{3 x^{2}}{2} - 2 \right|_{\substack{ x=-4 }} = 38 \]
    f(-4).✓ Proved
  5. \[ \left. - x^{3} - \frac{3 x^{2}}{2} - 2 \right|_{\substack{ x=-1 }} = - \frac{5}{2} \]
    f(-1).✓ Proved
  6. \[ \left. - x^{3} - \frac{3 x^{2}}{2} - 2 \right|_{\substack{ x=0 }} = -2 \]
    f(0).✓ Proved
  7. \[ \left. - x^{3} - \frac{3 x^{2}}{2} - 2 \right|_{\substack{ x=3 }} = - \frac{85}{2} \]
    f(3).✓ Proved
  8. The largest value is 38 and the smallest is -85/2.
    Reviewed
Answer \( \text{max } 38 \text{ at } x=-4;\ \text{min } - \frac{85}{2} \text{ at } x=3 \)
The graph of f(x) = -x³ - 3x²/2 - 2 on [-4, 3]. The candidates are the endpoints and the critical points inside; the highest is 38 and the lowest -85/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.49

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{15 x^{2}}{2} - 18 x - 1 \) on \( \displaystyle [-4, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} + \frac{15 x^{2}}{2} - 18 x - 1\right) = \left(9 - 3 x\right) \left(x - 2\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 5]: 2, 3.
    Reviewed
  4. \[ \left. - x^{3} + \frac{15 x^{2}}{2} - 18 x - 1 \right|_{\substack{ x=-4 }} = 255 \]
    f(-4).✓ Proved
  5. \[ \left. - x^{3} + \frac{15 x^{2}}{2} - 18 x - 1 \right|_{\substack{ x=2 }} = -15 \]
    f(2).✓ Proved
  6. \[ \left. - x^{3} + \frac{15 x^{2}}{2} - 18 x - 1 \right|_{\substack{ x=3 }} = - \frac{29}{2} \]
    f(3).✓ Proved
  7. \[ \left. - x^{3} + \frac{15 x^{2}}{2} - 18 x - 1 \right|_{\substack{ x=5 }} = - \frac{57}{2} \]
    f(5).✓ Proved
  8. The largest value is 255 and the smallest is -57/2.
    Reviewed
Answer \( \text{max } 255 \text{ at } x=-4;\ \text{min } - \frac{57}{2} \text{ at } x=5 \)
The graph of f(x) = -x³ + 15x²/2 - 18x - 1 on [-4, 5]. The candidates are the endpoints and the critical points inside; the highest is 255 and the lowest -57/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.50

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 3 x^{2} + 9 x - 4 \) on \( \displaystyle [-4, 4] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} + 3 x^{2} + 9 x - 4\right) = \left(9 - 3 x\right) \left(x + 1\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 4]: -1, 3.
    Reviewed
  4. \[ \left. - x^{3} + 3 x^{2} + 9 x - 4 \right|_{\substack{ x=-4 }} = 72 \]
    f(-4).✓ Proved
  5. \[ \left. - x^{3} + 3 x^{2} + 9 x - 4 \right|_{\substack{ x=-1 }} = -9 \]
    f(-1).✓ Proved
  6. \[ \left. - x^{3} + 3 x^{2} + 9 x - 4 \right|_{\substack{ x=3 }} = 23 \]
    f(3).✓ Proved
  7. \[ \left. - x^{3} + 3 x^{2} + 9 x - 4 \right|_{\substack{ x=4 }} = 16 \]
    f(4).✓ Proved
  8. The largest value is 72 and the smallest is -9.
    Reviewed
Answer \( \text{max } 72 \text{ at } x=-4;\ \text{min } -9 \text{ at } x=-1 \)
The graph of f(x) = -x³ + 3x² + 9x - 4 on [-4, 4]. The candidates are the endpoints and the critical points inside; the highest is 72 and the lowest -9.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.52

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{9 x^{2}}{2} - 6 x - 1 \) on \( \displaystyle [-3, 3] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} - \frac{9 x^{2}}{2} - 6 x - 1\right) = \left(- 3 x - 3\right) \left(x + 2\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-3, 3]: -2, -1.
    Reviewed
  4. \[ \left. - x^{3} - \frac{9 x^{2}}{2} - 6 x - 1 \right|_{\substack{ x=-3 }} = \frac{7}{2} \]
    f(-3).✓ Proved
  5. \[ \left. - x^{3} - \frac{9 x^{2}}{2} - 6 x - 1 \right|_{\substack{ x=-2 }} = 1 \]
    f(-2).✓ Proved
  6. \[ \left. - x^{3} - \frac{9 x^{2}}{2} - 6 x - 1 \right|_{\substack{ x=-1 }} = \frac{3}{2} \]
    f(-1).✓ Proved
  7. \[ \left. - x^{3} - \frac{9 x^{2}}{2} - 6 x - 1 \right|_{\substack{ x=3 }} = - \frac{173}{2} \]
    f(3).✓ Proved
  8. The largest value is 7/2 and the smallest is -173/2.
    Reviewed
Answer \( \text{max } \frac{7}{2} \text{ at } x=-3;\ \text{min } - \frac{173}{2} \text{ at } x=3 \)
The graph of f(x) = -x³ - 9x²/2 - 6x - 1 on [-3, 3]. The candidates are the endpoints and the critical points inside; the highest is 7/2 and the lowest -173/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.54

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 12 x - 4 \) on \( \displaystyle [-4, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} + 12 x - 4\right) = \left(6 - 3 x\right) \left(x + 2\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 5]: -2, 2.
    Reviewed
  4. \[ \left. - x^{3} + 12 x - 4 \right|_{\substack{ x=-4 }} = 12 \]
    f(-4).✓ Proved
  5. \[ \left. - x^{3} + 12 x - 4 \right|_{\substack{ x=-2 }} = -20 \]
    f(-2).✓ Proved
  6. \[ \left. - x^{3} + 12 x - 4 \right|_{\substack{ x=2 }} = 12 \]
    f(2).✓ Proved
  7. \[ \left. - x^{3} + 12 x - 4 \right|_{\substack{ x=5 }} = -69 \]
    f(5).✓ Proved
  8. The largest value is 12 and the smallest is -69.
    Reviewed
Answer \( \text{max } 12 \text{ at } x=-4;\ \text{min } -69 \text{ at } x=5 \)
The graph of f(x) = -x³ + 12x - 4 on [-4, 5]. The candidates are the endpoints and the critical points inside; the highest is 12 and the lowest -69.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.59

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{9 x^{2}}{2} - 6 x + 3 \) on \( \displaystyle [-4, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} + \frac{9 x^{2}}{2} - 6 x + 3\right) = \left(6 - 3 x\right) \left(x - 1\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 5]: 1, 2.
    Reviewed
  4. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 6 x + 3 \right|_{\substack{ x=-4 }} = 163 \]
    f(-4).✓ Proved
  5. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 6 x + 3 \right|_{\substack{ x=1 }} = \frac{1}{2} \]
    f(1).✓ Proved
  6. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 6 x + 3 \right|_{\substack{ x=2 }} = 1 \]
    f(2).✓ Proved
  7. \[ \left. - x^{3} + \frac{9 x^{2}}{2} - 6 x + 3 \right|_{\substack{ x=5 }} = - \frac{79}{2} \]
    f(5).✓ Proved
  8. The largest value is 163 and the smallest is -79/2.
    Reviewed
Answer \( \text{max } 163 \text{ at } x=-4;\ \text{min } - \frac{79}{2} \text{ at } x=5 \)
The graph of f(x) = -x³ + 9x²/2 - 6x + 3 on [-4, 5]. The candidates are the endpoints and the critical points inside; the highest is 163 and the lowest -79/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.60

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + \frac{3 x^{2}}{2} - 18 x - 3 \) on \( \displaystyle [-4, 4] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} + \frac{3 x^{2}}{2} - 18 x - 3\right) = \left(x + 3\right) \left(3 x - 6\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 4]: -3, 2.
    Reviewed
  4. \[ \left. x^{3} + \frac{3 x^{2}}{2} - 18 x - 3 \right|_{\substack{ x=-4 }} = 29 \]
    f(-4).✓ Proved
  5. \[ \left. x^{3} + \frac{3 x^{2}}{2} - 18 x - 3 \right|_{\substack{ x=-3 }} = \frac{75}{2} \]
    f(-3).✓ Proved
  6. \[ \left. x^{3} + \frac{3 x^{2}}{2} - 18 x - 3 \right|_{\substack{ x=2 }} = -25 \]
    f(2).✓ Proved
  7. \[ \left. x^{3} + \frac{3 x^{2}}{2} - 18 x - 3 \right|_{\substack{ x=4 }} = 13 \]
    f(4).✓ Proved
  8. The largest value is 75/2 and the smallest is -25.
    Reviewed
Answer \( \text{max } \frac{75}{2} \text{ at } x=-3;\ \text{min } -25 \text{ at } x=2 \)
The graph of f(x) = x³ + 3x²/2 - 18x - 3 on [-4, 4]. The candidates are the endpoints and the critical points inside; the highest is 75/2 and the lowest -25.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.137

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 6 x^{2} - 9 x - 1 \) on \( \displaystyle [-4, 2] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} - 6 x^{2} - 9 x - 1\right) = \left(- 3 x - 3\right) \left(x + 3\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 2]: -3, -1.
    Reviewed
  4. \[ \left. - x^{3} - 6 x^{2} - 9 x - 1 \right|_{\substack{ x=-4 }} = 3 \]
    f(-4).✓ Proved
  5. \[ \left. - x^{3} - 6 x^{2} - 9 x - 1 \right|_{\substack{ x=-3 }} = -1 \]
    f(-3).✓ Proved
  6. \[ \left. - x^{3} - 6 x^{2} - 9 x - 1 \right|_{\substack{ x=-1 }} = 3 \]
    f(-1).✓ Proved
  7. \[ \left. - x^{3} - 6 x^{2} - 9 x - 1 \right|_{\substack{ x=2 }} = -51 \]
    f(2).✓ Proved
  8. The largest value is 3 and the smallest is -51.
    Reviewed
Answer \( \text{max } 3 \text{ at } x=-4;\ \text{min } -51 \text{ at } x=2 \)
The graph of f(x) = -x³ - 6x² - 9x - 1 on [-4, 2]. The candidates are the endpoints and the critical points inside; the highest is 3 and the lowest -51.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.138

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} + 6 x + 3 \) on \( \displaystyle [-3, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} + \frac{3 x^{2}}{2} + 6 x + 3\right) = \left(6 - 3 x\right) \left(x + 1\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-3, 5]: -1, 2.
    Reviewed
  4. \[ \left. - x^{3} + \frac{3 x^{2}}{2} + 6 x + 3 \right|_{\substack{ x=-3 }} = \frac{51}{2} \]
    f(-3).✓ Proved
  5. \[ \left. - x^{3} + \frac{3 x^{2}}{2} + 6 x + 3 \right|_{\substack{ x=-1 }} = - \frac{1}{2} \]
    f(-1).✓ Proved
  6. \[ \left. - x^{3} + \frac{3 x^{2}}{2} + 6 x + 3 \right|_{\substack{ x=2 }} = 13 \]
    f(2).✓ Proved
  7. \[ \left. - x^{3} + \frac{3 x^{2}}{2} + 6 x + 3 \right|_{\substack{ x=5 }} = - \frac{109}{2} \]
    f(5).✓ Proved
  8. The largest value is 51/2 and the smallest is -109/2.
    Reviewed
Answer \( \text{max } \frac{51}{2} \text{ at } x=-3;\ \text{min } - \frac{109}{2} \text{ at } x=5 \)
The graph of f(x) = -x³ + 3x²/2 + 6x + 3 on [-3, 5]. The candidates are the endpoints and the critical points inside; the highest is 51/2 and the lowest -109/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.200

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 18 x \) on \( \displaystyle [-3, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} - \frac{15 x^{2}}{2} + 18 x\right) = \left(x - 2\right) \left(3 x - 9\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-3, 5]: 2, 3.
    Reviewed
  4. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x \right|_{\substack{ x=-3 }} = - \frac{297}{2} \]
    f(-3).✓ Proved
  5. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x \right|_{\substack{ x=2 }} = 14 \]
    f(2).✓ Proved
  6. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x \right|_{\substack{ x=3 }} = \frac{27}{2} \]
    f(3).✓ Proved
  7. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x \right|_{\substack{ x=5 }} = \frac{55}{2} \]
    f(5).✓ Proved
  8. The largest value is 55/2 and the smallest is -297/2.
    Reviewed
Answer \( \text{max } \frac{55}{2} \text{ at } x=5;\ \text{min } - \frac{297}{2} \text{ at } x=-3 \)
The graph of f(x) = x³ - 15x²/2 + 18x on [-3, 5]. The candidates are the endpoints and the critical points inside; the highest is 55/2 and the lowest -297/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.206

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} - 1 \) on \( \displaystyle [-3, 4] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{3} - 3 x^{2} - 1\right) = - 3 x \left(x + 2\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-3, 4]: -2, 0.
    Reviewed
  4. \[ \left. - x^{3} - 3 x^{2} - 1 \right|_{\substack{ x=-3 }} = -1 \]
    f(-3).✓ Proved
  5. \[ \left. - x^{3} - 3 x^{2} - 1 \right|_{\substack{ x=-2 }} = -5 \]
    f(-2).✓ Proved
  6. \[ \left. - x^{3} - 3 x^{2} - 1 \right|_{\substack{ x=0 }} = -1 \]
    f(0).✓ Proved
  7. \[ \left. - x^{3} - 3 x^{2} - 1 \right|_{\substack{ x=4 }} = -113 \]
    f(4).✓ Proved
  8. The largest value is -1 and the smallest is -113.
    Reviewed
Answer \( \text{max } -1 \text{ at } x=-3;\ \text{min } -113 \text{ at } x=4 \)
The graph of f(x) = -x³ - 3x² - 1 on [-3, 4]. The candidates are the endpoints and the critical points inside; the highest is -1 and the lowest -113.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.240

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{9 x^{2}}{2} + 6 x + 1 \) on \( \displaystyle [0, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} - \frac{9 x^{2}}{2} + 6 x + 1\right) = \left(x - 1\right) \left(3 x - 6\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [0, 5]: 1, 2.
    Reviewed
  4. \[ \left. x^{3} - \frac{9 x^{2}}{2} + 6 x + 1 \right|_{\substack{ x=0 }} = 1 \]
    f(0).✓ Proved
  5. \[ \left. x^{3} - \frac{9 x^{2}}{2} + 6 x + 1 \right|_{\substack{ x=1 }} = \frac{7}{2} \]
    f(1).✓ Proved
  6. \[ \left. x^{3} - \frac{9 x^{2}}{2} + 6 x + 1 \right|_{\substack{ x=2 }} = 3 \]
    f(2).✓ Proved
  7. \[ \left. x^{3} - \frac{9 x^{2}}{2} + 6 x + 1 \right|_{\substack{ x=5 }} = \frac{87}{2} \]
    f(5).✓ Proved
  8. The largest value is 87/2 and the smallest is 1.
    Reviewed
Answer \( \text{max } \frac{87}{2} \text{ at } x=5;\ \text{min } 1 \text{ at } x=0 \)
The graph of f(x) = x³ - 9x²/2 + 6x + 1 on [0, 5]. The candidates are the endpoints and the critical points inside; the highest is 87/2 and the lowest 1.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.241

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{9 x^{2}}{2} - 1 \) on \( \displaystyle [-4, 4] \).
  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
Answer \( \text{max } 135 \text{ at } x=-4;\ \text{min } -1 \text{ at } x=0 \)
The graph of f(x) = -x³ + 9x²/2 - 1 on [-4, 4]. The candidates are the endpoints and the critical points inside; the highest is 135 and the lowest -1.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.242

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 18 x - 4 \) on \( \displaystyle [-3, 5] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} - \frac{15 x^{2}}{2} + 18 x - 4\right) = \left(x - 2\right) \left(3 x - 9\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-3, 5]: 2, 3.
    Reviewed
  4. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 4 \right|_{\substack{ x=-3 }} = - \frac{305}{2} \]
    f(-3).✓ Proved
  5. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 4 \right|_{\substack{ x=2 }} = 10 \]
    f(2).✓ Proved
  6. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 4 \right|_{\substack{ x=3 }} = \frac{19}{2} \]
    f(3).✓ Proved
  7. \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 4 \right|_{\substack{ x=5 }} = \frac{47}{2} \]
    f(5).✓ Proved
  8. The largest value is 47/2 and the smallest is -305/2.
    Reviewed
Answer \( \text{max } \frac{47}{2} \text{ at } x=5;\ \text{min } - \frac{305}{2} \text{ at } x=-3 \)
The graph of f(x) = x³ - 15x²/2 + 18x - 4 on [-3, 5]. The candidates are the endpoints and the critical points inside; the highest is 47/2 and the lowest -305/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.243

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 6 x^{2} - 9 x - 2 \) on \( \displaystyle [-4, 5] \).
  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
Answer \( \text{max } 194 \text{ at } x=-4;\ \text{min } -22 \text{ at } x=5 \)
The graph of f(x) = -x³ + 6x² - 9x - 2 on [-4, 5]. The candidates are the endpoints and the critical points inside; the highest is 194 and the lowest -22.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.246

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{15 x^{2}}{2} - 18 x \) on \( \displaystyle [-4, 3] \).
  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
Answer \( \text{max } 16 \text{ at } x=-4;\ \text{min } - \frac{297}{2} \text{ at } x=3 \)
The graph of f(x) = -x³ - 15x²/2 - 18x on [-4, 3]. The candidates are the endpoints and the critical points inside; the highest is 16 and the lowest -297/2.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.279

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} + 9 x - 4 \) on \( \displaystyle [-4, 5] \).
  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
Answer \( \text{max } 1 \text{ at } x=1;\ \text{min } -159 \text{ at } x=5 \)
The graph of f(x) = -x³ - 3x² + 9x - 4 on [-4, 5]. The candidates are the endpoints and the critical points inside; the highest is 1 and the lowest -159.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.281

hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 3 x - 2 \) on \( \displaystyle [-4, 4] \).
  1. A continuous function on a closed interval has its extreme values at critical points or endpoints.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} - 3 x - 2\right) = \left(x + 1\right) \left(3 x - 3\right) \]
    Differentiate and factor.✓ Proved
  3. Critical numbers inside [-4, 4]: -1, 1.
    Reviewed
  4. \[ \left. x^{3} - 3 x - 2 \right|_{\substack{ x=-4 }} = -54 \]
    f(-4).✓ Proved
  5. \[ \left. x^{3} - 3 x - 2 \right|_{\substack{ x=-1 }} = 0 \]
    f(-1).✓ Proved
  6. \[ \left. x^{3} - 3 x - 2 \right|_{\substack{ x=1 }} = -4 \]
    f(1).✓ Proved
  7. \[ \left. x^{3} - 3 x - 2 \right|_{\substack{ x=4 }} = 50 \]
    f(4).✓ Proved
  8. The largest value is 50 and the smallest is -54.
    Reviewed
Answer \( \text{max } 50 \text{ at } x=4;\ \text{min } -54 \text{ at } x=-4 \)
The graph of f(x) = x³ - 3x - 2 on [-4, 4]. The candidates are the endpoints and the critical points inside; the highest is 50 and the lowest -54.

✓ Nihil obstat Lines: 5 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More absolute extrema on a closed interval practice

Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 3 x + 4 \) on \( \displaystyle [-1, 1] \).
Problem 3.143medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{3 x^{2}}{2} + 18 x + 3 \) on \( \displaystyle [-3, 2] \).
Problem 3.146medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} - 2 \) on \( \displaystyle [0, 1] \).
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Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} + 18 x - 2 \) on \( \displaystyle [-2, 3] \).
Problem 3.204medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} + 9 x - 4 \) on \( \displaystyle [-3, 1] \).
Problem 3.280medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + 3 x^{2} - 9 x - 3 \) on \( \displaystyle [-3, 1] \).
Problem 3.323medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} + 9 x + 1 \) on \( \displaystyle [-3, 1] \).
Problem 3.362medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} + 6 x - 4 \) on \( \displaystyle [-1, 2] \).
Problem 3.365medium Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 12 x + 2 \) on \( \displaystyle [-4, 3] \).
Problem 3.51hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{9 x^{2}}{2} + 1 \) on \( \displaystyle [-2, 4] \).
Problem 3.53hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 3 x + 3 \) on \( \displaystyle [-1, 2] \).
Problem 3.55hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{9 x^{2}}{2} - 6 x \) on \( \displaystyle [-2, 3] \).
Problem 3.56hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 3 x^{2} + 9 x - 3 \) on \( \displaystyle [-2, 4] \).
Problem 3.57hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 18 x - 2 \) on \( \displaystyle [-1, 3] \).
Problem 3.58hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 3 x^{2} + 3 \) on \( \displaystyle [-1, 2] \).
Problem 3.136hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{3 x^{2}}{2} - 18 x - 3 \) on \( \displaystyle [-2, 4] \).
Problem 3.139hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} - 2 \) on \( \displaystyle [-2, 3] \).
Problem 3.140hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 12 x + 1 \) on \( \displaystyle [-2, 4] \).
Problem 3.141hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 18 x - 2 \) on \( \displaystyle [-2, 3] \).
Problem 3.142hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} - 1 \) on \( \displaystyle [-2, 1] \).
Problem 3.144hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} + 9 x - 2 \) on \( \displaystyle [-4, 4] \).
Problem 3.145hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + 12 x - 3 \) on \( \displaystyle [-4, 3] \).
Problem 3.147hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{3 x^{2}}{2} - 4 \) on \( \displaystyle [-3, 4] \).
Problem 3.148hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} + 2 \) on \( \displaystyle [-1, 1] \).
Problem 3.149hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 18 x - 4 \) on \( \displaystyle [-4, 3] \).
Problem 3.150hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + \frac{9 x^{2}}{2} + 6 x + 1 \) on \( \displaystyle [-2, 1] \).
Problem 3.199hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{9 x^{2}}{2} - 6 x + 3 \) on \( \displaystyle [-2, 2] \).
Problem 3.201hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} + 18 x + 3 \) on \( \displaystyle [-3, 3] \).
Problem 3.202hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 6 x^{2} + 9 x + 3 \) on \( \displaystyle [-1, 3] \).
Problem 3.205hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{15 x^{2}}{2} - 18 x - 2 \) on \( \displaystyle [-4, 2] \).
Problem 3.239hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{15 x^{2}}{2} - 18 x + 2 \) on \( \displaystyle [-4, 4] \).
Problem 3.244hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + 6 x^{2} + 9 x \) on \( \displaystyle [-3, 5] \).
Problem 3.245hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - 3 x^{2} + 9 x - 4 \) on \( \displaystyle [-3, 2] \).
Problem 3.282hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{9 x^{2}}{2} + 4 \) on \( \displaystyle [-2, 5] \).
Problem 3.283hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{3 x^{2}}{2} - 18 x + 3 \) on \( \displaystyle [-2, 4] \).
Problem 3.284hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{3 x^{2}}{2} + 18 x \) on \( \displaystyle [-3, 3] \).
Problem 3.285hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{3 x^{2}}{2} - 6 x + 1 \) on \( \displaystyle [-2, 2] \).
Problem 3.286hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + 3 x^{2} - 9 x \) on \( \displaystyle [-4, 3] \).
Problem 3.319hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{9 x^{2}}{2} + 4 \) on \( \displaystyle [-4, 4] \).
Problem 3.320hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + \frac{3 x^{2}}{2} - 6 x + 3 \) on \( \displaystyle [-2, 3] \).
Problem 3.321hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 12 x + 4 \) on \( \displaystyle [-2, 5] \).
Problem 3.322hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + \frac{3 x^{2}}{2} - 6 x \) on \( \displaystyle [-3, 2] \).
Problem 3.324hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 6 x^{2} + 9 x + 4 \) on \( \displaystyle [-4, 3] \).
Problem 3.325hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} + 3 x^{2} - 2 \) on \( \displaystyle [-2, 4] \).
Problem 3.326hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} - 4 \) on \( \displaystyle [-2, 3] \).
Problem 3.359hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{9 x^{2}}{2} + 4 \) on \( \displaystyle [-1, 3] \).
Problem 3.360hard Video✓ Nihil obstat
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 3 x^{2} - 1 \) on \( \displaystyle [0, 3] \).
Problem 3.361hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} - \frac{3 x^{2}}{2} + 18 x - 2 \) on \( \displaystyle [-3, 3] \).
Problem 3.363hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{3 x^{2}}{2} + 1 \) on \( \displaystyle [-3, 1] \).
Problem 3.364hard Video✓ Every equation proved
Find the absolute maximum and minimum values of \( \displaystyle f(x) = - x^{3} + \frac{9 x^{2}}{2} - 1 \) on \( \displaystyle [-1, 4] \).
Problem 3.366hard Video✓ Nihil obstat