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

x

y

y = f(x)f(c)

c

S

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

x

y

y = f(x)

f(c)

cS

Example: f(x) = x2

S = (-∞, ∞) No absolute maximum Absolute minimum:

f(0) = 0 at c = 0

x

y

0

Example: f(x) = x2

S = [0,1] Absolute maximum

f(1) = 1 at c = 1 Absolute minimum:

f(0) = 0 at c = 0

x

y

0 1

Example: f(x) = x2

S = (0,1] Absolute maximum

f(1) = 1 at c = 1 No absolute

minimum,although function isbounded from below:0 < x2 for allx in (0,1] ! x

y

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)

x

yy = f(x)

c

Local minimum values

A function f has a local minimum 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)

x

yy = f(x)

c

Example: y = sin x

π/2

- π/2

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

-1

Applications

Curve sketching Optimization problems (with constraints),

for example: Finding parameters to minimize

manufacturing 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 and cMIN in [a,b]

Extreme Value Theorem - Examples

x

yy = f(x)

cMAX cMINa 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

x

yy = f(x)

cMIN cMAX= ba

Continuity is important

x

y

0

-11

0 xif ,0

0 if ,1

)(

xx

xf

No absolute maximum or minimumon [-1,1]

Closed interval is important

f(x) = x2, S = (0,1] No absolute

minimum in (0,1]

x

y

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

x

y

y = f(x)

c

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

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)

x

y

Problem: f′ not always exists

f(x) = |x|

It has local (and absolute) minimum at 0

However, f′ (0) does not exists!

x

y

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]

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

22

2

2 )1(

1

1)(

x

x

x

xxf

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