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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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absolute maximum values
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
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 x 2
Example: f(x) = x2
  • S = (-∞, ∞)
  • No absolute maximum
  • Absolute minimum:f(0) = 0 at c = 0

y

x

0

example f x x 21
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 x 22
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
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
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
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
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
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
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
Continuity is important

y

x

-1

1

0

No absolute maximum or minimumon [-1,1]

closed interval is important
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
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
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 theorem does not hold
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
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
Critical numbers
  • Two kinds of "suspicious" points(for local max or min):
    • f′(c) = 0
    • f′(c) does not exists
critical numbers definition
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
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 method1
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
Example
  • Find the absolute maximum and minimum values of f(x) = x/(x2+1) on the interval [0,2]
solution

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
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