Absolute Max/Min

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


• 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


• 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


• f /(x) = 6x2 -30x + 36 = 6(x – 2)(x – 3) c.p. @ x = 2, 3

Example 1


• 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


• 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


• f/(x) = 8x1/3 – x-2/3 = x-2/3(8x – 1) c.p. @ x = 0, 1/8

Example 2


• 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


• 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.



• If the polynomial is an odd degree, there will be no absolute max or min.



• 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


• 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


• 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.


• 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.


• 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


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


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


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

Absolute Max/Min

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