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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].1) Find the critical points of f in (a, b).2) Evaluate f at all critical points and at the endpoints.3) 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.
xxxf
2
1)(
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
xxxf
2
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.
xxxf
2
1)(
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
2222
2/
)(
21
)(
)12(1)0)(()(
xx
x
xx
xxxxf
xxxf
2
1)(
Theorem 4.4.4
• Suppose that f is continuous and has exactly one relative extremum on an interval I, say at x0.
a) If f has a relative minimum at x0, then f(x0) is the absolute minimum of f on I.
b) 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 .),0( )3( 23
)( xxexf
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.
),0( )3( 23
)( xxexf
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
),0( )3( 23
)( xxexf
2323 3)3(2/ )2(3)63()( xxxx exxexxxf
Homework
• Section 4.4• Pages 272-273• 1-27 odd
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