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Absolute Maxima and Minima: Finding Extrema on Closed Intervals

Learn how to find the absolute maximum and minimum of a function on a closed interval using the Extreme Value Theorem and the Second Derivative Test.

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Absolute Maxima and Minima: Finding Extrema on Closed Intervals

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  1. Chapter 4GraphingandOptimization Section 5 Absolute Maxima and Minima

  2. Absolute Maxima and Minima f(c) is a local maximum if f(x) <f(c) for x near c and a local minimum if f(x) >f(c) for x near c. In this section, we seek the largest and smallest values of f(x) throughout the domain of f.

  3. Definition Definition Absolute Maxima and Minima If f(c) >f(x) for all x in the domain of f, then f(c) is called the absolute maximum of f. • If f(c) <f(x) for all x in the domain of f, then f(c) is called the absolute minimum of f. • An absolute maximum or absolute minimum is called an absolute extremum.

  4. Visual Examples-Absolute Extremum f(x) has no absolute maximum or minimum. x = –2 gives a local maximum at f(–2). This corresponds to the point (–2, 16/3). x = 2 gives a local minimum at f(2). This corresponds to the point (2, –16/3).

  5. Visual Examples-Absolute Extremum f(x) has a local maximum at x = 0. The local maximum at x = 0 is also an absolute maximum. f(x) does not have a local or absolute minimum.

  6. Visual Examples-Absolute Extremum f(x) has no absolute maximum. x = 0 gives a local minimum at f(0), corresponds to the point (0, 0), and is an absolute minimum.

  7. Theorem 1 Extreme Value Theorem Theorem 1 Extreme Value Theorem A function f that is continuous on a closed interval [a, b] has both an absolute maximum and an absolute minimum on that interval.

  8. Extreme Value Theorem Examples Find absolute extrema for f(x) = x3 – 21x2 + 135x – 170 on the closed interval [2, 12]. Solution The graph of f(x) is shown. The absolute maximum for f(x) occurs at the right endpoint of [2, 12]. The absolute minimum for f(x) occurs at the left endpoint of [2, 12].

  9. Extreme Value Theorem Examples Find absolute extrema for f(x) = x3 – 21x2 + 135x – 170 on the closed interval [4, 10]. Solution The graph of f(x) is shown. The absolute maximum for f(x) occurs at x = 5 on [4, 10]. The absolute minimum for f(x) occurs at x = 9 on the interval [4, 10].

  10. Extreme Value Theorem Examples Find absolute extrema for f(x) = x3 – 21x2 + 135x – 170 on the closed interval [4, 8]. Solution The graph of f(x) is shown. The absolute maximum for f(x) occurs at x = 5 on [4, 8]. The absolute minimum for f(x) occurs at the right hand endpoint x = 8 on the interval [4, 8].

  11. Extreme Value Theorem Examples Find absolute extrema for f(x) = x3 – 21x2 + 135x – 170 on the closed interval [3, 11]. Solution The graph of f(x) is shown. Absolute maxima for f(x) occur at both x = 5 and x = 11 on [3, 11]. Absolute minima for f(x) occur at both x = 3 and x = 9 on the interval [3, 11].

  12. Theorem 2 Locating Absolute Extrema Theorem 2 Locating Absolute Extrema Absolute extrema (if they exist) must occur at critical numbers or at endpoints.

  13. Procedure for Finding Absolute Extrema on a Closed Interval Step 1 Check to make certain the f is continuous over [a, b]. Step 2 Find the critical numbers in the interval (a, b). Step 3 Evaluate f at the endpoints a and b and at the critical numbers found in step 2. Step 4 The absolute maximum of f on [a, b] is the largest value found in step 3. Step 5 The absolute minimum of f on [a, b] is the smallest value found in step 3.

  14. Example 1 Finding Absolute Extrema Find the absolute maximum and absolute minimum of f(x) = x3 – 6x2 on [–7, 7]. Solution Step 1 The polynomial function f is continuous. Step 2 f ´(x) = 3x2 – 12x = 3x(x – 4). Critical values are x = 0 and x = 4. Step 3 Evaluate f at the endpoints and critical values. f(-1) = –7, f(0) = 0 f(4) = –32 f(7) = 49 Smallest Largest

  15. Example 1 Finding Absolute Extrema continued Solution Step 4 The absolute maximum of f on the interval is the largest of the values found in step 3. The absolute maximum of f is 49 when x = 7. Step 5 The absolute minimum of f on the interval is the smallest of the values found in step 3. The absolute minimum of f is –32 when x = 4.

  16. Second Derivative and Extrema Example f ´(c) = 0 and f ″(x) > 0 implies f(c) is a local minimum.

  17. Second Derivative and Extrema Example f ´(c) = 0 and f ″(x) < 0 implies f(c) is a local maximum.

  18. Second Derivative Test for Local Extrema Let c be a critical number of f(x) such that f ´(x) = 0. If the second derivative f ″(x) > 0, then f(c) is a local minimum. If f ″(x) < 0, then f(c) is a local maximum.

  19. Theorem 3 Second Derivative Test Theorem 3 Second-Derivative Test for Absolute Extrema on an Interval Let f be continuous on an interval I from a to b with only one critical number c in (a, b). If f ´(c) = 0 and f ″(c) > 0, then f(c) is the absolute minimum of f on I. • If f ´(c) = 0 and f ″(c) < 0, then f(c) is the absolute maximum of f on I.

  20. Example 2 Find Absolute Extrema on an Open Interval Critical numbers are x = –2 and x = 2. The only critical number in the interval (0, ∞) is x = 2. Since f ″(2) = 1 > 0, f(2) = 4 is the absolute minimum of f on (0, ∞). f(x) has no maximum on (0, ∞).

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