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Absolute Max/Min

Absolute Max/Min. Objective: To find the absolute max/min of a function over an interval. Definition 4.4.1.

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Absolute Max/Min

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  1. Absolute Max/Min Objective: To find the absolute max/min of a function over an interval.

  2. Definition 4.4.1 • Let I be an interval in the domain of a function f. We say that f has an absolute maximum at a point x0 in I if f(x) < f(x0) for all x in I, and we say that f has an absolute minimum at x0 if f(x0) < f(x) for all x in I.

  3. Definition 4.4.1 • Let I be an interval in the domain of a function f. We say that f has an absolute maximum at a point x0 in I if f(x) < f(x0) for all x in I, and we say that f has an absolute minimum at x0 if f(x0) < f(x) for all x in I. • If f has an absolute maximum at the point x0 on an interval I, then f(x0) is the largest value of f on I. If f has an absolute minimum at the point x0 on an interval I, then f(x0) is the smallest value of f on I.

  4. Definition 4.4.1 • Let I be an interval in the domain of a function f. We say that f has an absolute maximum at a point x0 in I if f(x) < f(x0) for all x in I, and we say that f has an absolute minimum at x0 if f(x0) < f(x) for all x in I. • If f has an absolute maximum at the point x0 on an interval I, then f(x0) is the largest value of f on I. If f has an absolute minimum at the point x0 on an interval I, then f(x0) is the smallest value of f on I. • Always be aware of what they are asking for. Where the extrema occur is the x coordinate and the max/min value is the y coordinate.

  5. Extreme Value Theorem • Theorem 4.4.2 (Extreme Value Theorem) • If a function f is continuous on a finite closed interval [a, b] then f has both an absolute maximum and an absolute minimum on [a, b].

  6. Extreme Value Theorem • Theorem 4.4.2 (Extreme Value Theorem) • If a function f is continuous on a finite closed interval [a, b] then f has both an absolute maximum and an absolute minimum on [a, b]. • This is an example of what mathematicians call an existence theorem. Such theorems state conditions under which certain objects exist.

  7. Max/Min • If f is continuous on a finite closed interval [a, b], then the absolute extrema of f occur either at the endpoints of the interval or inside the open interval (a, b). If they fall inside, we will use this theorem to find them.

  8. Max/Min • If f is continuous on a finite closed interval [a, b], then the absolute extrema of f occur either at the endpoints of the interval or inside the open interval (a, b). If they fall inside, we will use this theorem to find them. • Theorem 4.4.3 • If f has an absolute extremum on an open interval (a, b), then it must occur at a critical point of f.

  9. Finding a Max/Min on a Closed Interval • A Procedure for finding the absolute extrema of a continuous function f on a finite closed interval [a, b]. • Find the critical points of f in (a, b). • Evaluate f at all critical points and at the endpoints. • The largest value in step 2 is the maximum value and the smallest is the minimum value.

  10. Example 1 • Find the absolute maximum and minimum values of the function f(x) = 2x3-15x2+36x on the interval [1, 5] and determine where these values occur.

  11. Example 1 • Find the absolute maximum and minimum values of the function f(x) = 2x3-15x2+36x on the interval [1, 5] and determine where these values occur. • f /(x) = 6x2 -30x + 36 = 6(x – 2)(x – 3) c.p. @ x = 2, 3

  12. Example 1 • Find the absolute maximum and minimum values of the function f(x) = 2x3-15x2+36x on the interval [1, 5] and determine where these values occur. • f /(x) = 6x2 -30x + 36 = 6(x – 2)(x – 3) c.p. @ x = 2, 3 • f(1) = 23 • f(2) = 28 • f(3) = 27 • f(5) = 55

  13. Example 1 • Find the absolute maximum and minimum values of the function f(x) = 2x3-15x2+36x on the interval [1, 5] and determine where these values occur. • f /(x) = 6x2 -30x + 36 = 6(x – 2)(x – 3) c.p. @ x = 2, 3 • f(1) = 23 Min value of 23 @ x = 1 • f(2) = 28 • f(3) = 27 • f(5) = 55 Max value of 55 @ x = 5

  14. Example 2 • Find the absolute extrema of f(x) = 6x4/3 – 3x1/3 on the interval [-1, 1] and determine where these values occur.

  15. Example 2 • Find the absolute extrema of f(x) = 6x4/3 – 3x1/3 on the interval [-1, 1] and determine where these values occur. • f/(x) = 8x1/3 – x-2/3 = x-2/3(8x – 1) c.p. @ x = 0, 1/8

  16. Example 2 • Find the absolute extrema of f(x) = 6x4/3 – 3x1/3 on the interval [-1, 1] and determine where these values occur. • f/(x) = 8x1/3 – x-2/3 = x-2/3(8x – 1) c.p. @ x = 0, 1/8 • f(-1) = 9 • f(0) = 0 • f(1/8) = -9/8 • f(1) = 3

  17. Example 2 • Find the absolute extrema of f(x) = 6x4/3 – 3x1/3 on the interval [-1, 1] and determine where these values occur. • f/(x) = 8x1/3 – x-2/3 = x-2/3(8x – 1) c.p. @ x = 0, 1/8 • f(-1) = 9 Maximum value of 9 @ x = -1 • f(0) = 0 • f(1/8) = -9/8 Minimum value of -9/8 @ x = 1/8 • f(1) = 3

  18. Absolute Extrema on Infinite Intervals • When looking at an infinite interval, we can make some generalizations.

  19. Absolute Extrema on Infinite Intervals • When looking at an infinite interval, we can make some generalizations. • If the polynomial is an odd degree, there will be no absolute max or min.

  20. Absolute Extrema on Infinite Intervals • When looking at an infinite interval, we can make some generalizations. • If the polynomial is an odd degree, there will be no absolute max or min. • If the polynomial is an even degree and is positive, there will be a min but no max. • If the polynomial is an even degree and is negative, there will be a max but no min. • This max/min will occur at a critical point.

  21. Example 4 • Determine whether p(x) = 3x4 + 4x3 has any absolute extrema by calculus and by looking at the graph.

  22. Example 4 • Determine whether p(x) = 3x4 + 4x3 has any absolute extrema by calculus and by looking at the graph. • We know since this is a positive 4th degree, we will have a min and no max. • p /(x) = 12x3 + 12x2 = 12x2(x + 1) c.p. @ x = 0, -1

  23. Example 4 • Determine whether p(x) = 3x4 + 4x3 has any absolute extrema by calculus and by looking at the graph. • We know since this is a positive 4th degree, we will have a min and no max. • p /(x) = 12x3 + 12x2 = 12x2(x + 1) c.p. @ x = 0, -1 • f(0) = 0 • f(-1) = -1 Minimum value of -1 @ x = -1

  24. Finite Open Interval • Lets go back to Theorem 4.4.3 • If f has an absolute extremum on an open interval (a, b) it must occur at a critical point of f.

  25. Finite Open Interval • Lets go back to Theorem 4.4.3 • If f has an absolute extremum on an open interval (a, b) it must occur at a critical point of f. • Notice the word if. On an open interval, there may be a max, a min, both, or neither. We will follow the same procedure we did on a closed interval. If the max or min occurs at one of the endpoints, that means there is no max or min.

  26. Finite Open Interval • Max Min Neither • No Min No max

  27. Example 5 • Determine whether the function has any absolute extrema on the interval (0, 1). If so find them.

  28. Example 5 • Determine whether the function has any absolute extrema on the interval (0, 1). If so find them. • This is a little different than what we have looked at so far since the endpoints are asymptotes of the function. We need to first look at the behavior around each asymptote. ____|___-___|___ 0 1

  29. Example 5 • Determine whether the function has any absolute extrema on the interval (0, 1). If so find them. ____|___-___|___ 0 1 • This tells us that the function approaches negative infinity at both asymptotes, so there will be a max, but no min. If they approached + infinity, there would be no max or min.

  30. Example 5 • Determine whether the function has any absolute extrema on the interval (0, 1). If so find them. • We need to find the critical points. • The only critical point on (0, 1) is ½, so this must be the max. f(1/2) = -4 so the maximum value is -4 @ x = 1/2

  31. Theorem 4.4.4 • Suppose that f is continuous and has exactly one relative extremum on an interval I, say at x0. • If f has a relative minimum at x0, then f(x0) is the absolute minimum of f on I. • If f has a relative maximum at x0, then f(x0) is the absolute maximum of f on I.

  32. Example 6 • Find the absolute extrema, if any, of the function on the interval .

  33. Example 6 • Find the absolute extrema, if any, of the function on the interval . • Since the limit of this function as x approaches infinity is positive infinity, there will be no max. We need to look for a min.

  34. Example 6 • Find the absolute extrema, if any, of the function on the interval . • Since the limit of this function as x approaches infinity is positive infinity, there will be no max. We need to look for a min. • The critical numbers are x = 0 and x = 2. f(0) = 1 f(2) = .0183 Min of .0183 @ x = 2

  35. Homework • Section 4.4 • Pages 272-273 • 1-27 odd

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