Definition: Absolute Extreme Values
Let f be a function with domain D
  is the
(a) absolute maximum value on D if and only if
     for all x in D .
(b) absolute minimum value on D if and only if
     for all x in D .
Absolute (or global ) maximum and minimum values are also called absolute extrema . Sometimes the word “absolute” or “global” is simply omitted…
The Extreme Value Theorem (EVT)
If f is continuous on a closed interval [ a , b ], then has both a maximum value and a minimum value on the interval.
f

For our first example, let’s complete the following table:
Function Domain D Absolute Extrema on D y
 x
2
 
No absolute max.
Absolute min. of 0 at x = 0 y
 x
2 Abs. max. of 4 at x = 2
Abs. min at (0, 0) y
 x
2

0, 2

Abs. max. of 4 at x = 2
No absolute min.
y
 x
2
No abs. extrema
How about a graph for each of these???

Definition: Local (Relative) Extreme Values
Let c
Then
  is a f
(a) local maximum value at c if and only if
     for all x in some open interval containing c .
(b) local minimum value at c if and only if for all x in some open interval containing c .
    
Note: all absolute extrema are also local extrema!!!

Local Max.
y
  
Abs. Max.
Local
Min.
Abs.
Min.
Local Min.
Let’s classify extreme values in the function shown above x

Guided Practice – Identify Any Extreme Values y
7
3
–7 –4 –2
5 10 x
Min of – 6 at x = – 2
–6
Local Min of 0 at x = – 7
Max of 7 at x = – 4 and x = 5
Local Min of 3 at x = 10

The “Do Now”
Writing: How can the derivative help in identifying extreme values of functions (maxima and minima)?
Theorem: Local Extreme Values
If a function has a local maximum value or a local minimum value at an interior point c of its domain, and if f
 exists at c, then f
   
0

Definition: Critical Point
A point in the interior of the domain of a function f f
  f

 To find extrema analytically , we only need to investigate the critical points and the endpoints of the function!!!

Guided Practice
Find the extreme values of
First, find the first derivative: f
   x
2 3 on the interval [ –2, 3].
   
2
3 x
 1
3

3 3
2 x
Next, identify any critical points:
No zeros, but undefined at x = 0
(i.e., the function has a critical point at x = 0)

Guided Practice
Find the extreme values of
   x
2 3 on the interval [ –2, 3].
Finally, evaluate and analyze the original function at this critical point, and at the endpoints:
Crit. Pt: f
  
0  Abs. Min.
Endpts: f
    2
3

3 4

1.587
f
    2
3

3 9

2.080
 Local Max.
 Abs. Max.
Can we support these answers graphically ???

Guided Practice
Find the extreme values of y
 x
2 
2 x

4
 Find answer both analytically and graphically!
Domain is all real numbers…need to check critical points… y
 
2 x

2 The derivative is never undefined.
y
 
The derivative is zero for x = 1.
 
2 1

4

3
As x moves away from 1 in either direction, f increases.
Absolute Minimum of 3 at x = 1

Guided Practice
Find the extreme values of the given function.
( )
 e
 x x 1
    e
 x  
1 e x g
  e
   e
The derivative has no zeros.
We only need to check endpoints.
g
   e
   
1 e
Maximum value is e at x = –1
Minimum value is 1/e at x = 1