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

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.

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.

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.

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].

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.

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.

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.

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.

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.

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

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

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

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

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

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

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

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

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.

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.

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

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

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

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.

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.

Finite Open Interval Max Min Neither No Min No max

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

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

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.

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

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.

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

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.

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

Homework Section 4.4 Pages 272-273 1-27 odd