Transcript maxmin

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)
S
c
x
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)
f(c)
S
c
x
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 maximum
f(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 maximum
f(1) = 1 at c = 1
No absolute
minimum,
although function is
bounded from below:
0 < x2 for all
x 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 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
Example: y = sin x
f(x) = sin x
has local (and absolute) maximum
at all points of the form π/2 + 2πk,
and local (and absolute) minimum
at 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 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 attains
absolute maximum value f(cMAX) and
absolute minimum value f(cMIN)
at some numbers cMAX and cMIN in [a,b]
Extreme Value Theorem - Examples
y
y
y = f(x)
x
a c
MAX
cMIN b
Both absolute max and
absolute min are attained in
the open interval (a,b) at the
points of local max and min
y = f(x)
x
a c
MIN
cMAX= b
Absolute maximum is
attained at the right end
point: cMAX = b
Continuity is important
y
 1
 x , if x  0

f ( x)  
0, if x  0


x
-1
No absolute maximum or minimum
on [-1,1]
0
1
Closed interval is important


f(x) = x2, S = (0,1]
No absolute
minimum 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
y
horizontal tangent line
at the point of local max (or min)
y = f(x)
x
c

If f has a local maximum or minimum at c
and f′(c) exists, then f′(c) = 0
Converse of Fermat's theorem
does 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 where
f ′(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 is
a critical number)

or is attained at one of the endpoints
Closed Interval Method

To find absolute maximum and minimum
of 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, …, ck
are all critical numbers in (a,b)

The largest of f(a), f(c1), f(c2), …, f(ck), f(b) is the
absolute maximum of f on [a,b]

The smallest of these numbers is the
absolute 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

x 
1 x2

f ( x)   2   2
2
x

1
(
x

1
)



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