Maximum and Minimum Values ( Section 3.1)

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Maximum and Minimum Values ( Section 3.1). Alex Karassev. Absolute maximum values. A function f has an absolute maximum value on a set S at a point c in S if f(c) ≥ f(x) for all x in S. y. y = f(x). f(c). x. S. c. Absolute minimum values.

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### Maximum and Minimum Values(Section 3.1)

Alex Karassev

Absolute maximum values
• A function f has an absolute maximum value on a set S at a point c in S if f(c) ≥ f(x) for all x in S

y

y = f(x)

f(c)

x

S

c

Absolute minimum values
• A function f has an absolute minimum value on a set S at a point c in S if f(c) ≤ f(x) for all x in S

y

y = f(x)

x

f(c)

S

c

Example: f(x) = x2
• S = (-∞, ∞)
• No absolute maximum
• Absolute minimum:f(0) = 0 at c = 0

y

x

0

Example: f(x) = x2
• S = [0,1]
• Absolute maximumf(1) = 1 at c = 1
• Absolute minimum:f(0) = 0 at c = 0

y

x

0

1

Example: f(x) = x2
• S = (0,1]
• Absolute maximumf(1) = 1 at c = 1
• No absoluteminimum,although function isbounded from below:0 < x2 for allx in (0,1] !

y

x

0

1

Local maximum values
• A function f has a local maximum value at a point c if f(c) ≥ f(x) for all x near c (i.e. for all x in some open interval containing c)

y

y = f(x)

x

c

Local minimum values
• A function f has a local minimum value at a point cif f(c) ≤ f(x) for all x near c(i.e. for all x in some open interval containing c)

y

y = f(x)

x

c

Example: y = sin x

f(x) = sin xhas local (and absolute) maximumat all points of the form π/2 + 2πk,and local (and absolute) minimumat all points of the form -π/2 + 2πk,where k is an integer

1

- π/2

π/2

-1

Applications
• Curve sketching
• Optimization problems (with constraints),for example:
• Finding parameters to minimizemanufacturing costs
• Investing to maximize profit (constraint: amount of money to invest is limited)
• Finding route to minimize the distance
• Finding dimensions of containers to maximize volumes (constraint: amount of material to be used is limited)
Extreme Value Theorem

If f is continuous on a closed interval [a,b], then f attainsabsolute maximum value f(cMAX) andabsolute minimum value f(cMIN)at some numbers cMAX andcMIN in [a,b]

Extreme Value Theorem - Examples

y

y

y = f(x)

y = f(x)

x

x

a

a

b

cMIN

cMAX

cMIN

cMAX= b

Both absolute max and absolute min are attained in the open interval (a,b) at the points of local max and min

Absolute maximum is attained at the right end point: cMAX = b

Continuity is important

y

x

-1

1

0

No absolute maximum or minimumon [-1,1]

Closed interval is important
• f(x) = x2, S = (0,1]
• No absoluteminimum in (0,1]

y

x

0

1

How to find max and min values?
• Absolute maximum or minimum values of a function, continuous on a closed interval are attained either at the points which are simultaneously the points of local maximum or minimum, or at the endpoints
• Thus, we need to know how to find points of local maximums and minimums
Fermat\'s Theorem
• If f has a local maximum or minimum at c and f′(c) exists, then f′(c) = 0

y

horizontal tangent line at the point of local max (or min)

y = f(x)

x

c

Converse of Fermat\'s theoremdoes not hold!
• If f ′(c) = 0 it does not mean that c is a point of local maximum or minimum
• Example: f(x) = x3, f ′(0) = 0, but 0 is not a point of local max or min
• Nevertheless, points c wheref ′(c) = 0 are "suspicious" points(for local max or min)

y

x

Problem: f′ not always exists
• f(x) = |x|
• It has local (and absolute) minimum at 0
• However, f′ (0) does not exists!

y

x

Critical numbers
• Two kinds of "suspicious" points(for local max or min):
• f′(c) = 0
• f′(c) does not exists
Critical numbers – definition
• A number c is called a critical number of function f if the following conditions are satisfied:
• c is in the domain of f
• f′(c) = 0 or f′(c) does not exist
Closed Interval Method
• The method to find absolute maximum or minimum of a continuous function, defined on a closed interval [a,b]
• Based on the fact that absolute maximum or minimum
• either is attained at some point inside the open interval (a,b) (then this point is also a point of local maximum or minimum and hence isa critical number)
• or is attained at one of the endpoints
Closed Interval Method
• To find absolute maximum and minimumof a function f, continuous on [a,b]:
• Find critical numbers inside (a,b)
• Find derivative f′ (x)
• Solve equation f′ (x)=0 for x and choose solutions which are inside (a,b)
• Find numbers in (a,b) where f′ (x) d.n.e.
• Suppose that c1, c2, …, ckare all critical numbers in (a,b)
• The largest of f(a), f(c1), f(c2), …, f(ck), f(b) is theabsolute maximum of f on [a,b]
• The smallest of these numbers is theabsolute minimum of f on [a,b]
Example
• Find the absolute maximum and minimum values of f(x) = x/(x2+1) on the interval [0,2]

Find the absolute maximum and minimum values of f(x) = x/(x2+1) on the interval [0,2]

Solution
• Find f′(x):
• Critical numbers: f′(x) = 0 ⇔ 1– x2 = 0
• So x = 1 or x = – 1
• However, only 1 is inside [0,2]
• Now we need to compare f(0), f(1), and f(2):
• f(0) = 0, f(1) = 1/2, f(2)= 2/5
• Therefore 0 is absolute minimum and 1/2 is absolute maximum