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