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Section 4.1
Maximum and
Minimum Values
Applications of Differentiation
Applications of Differentiation
Maxima and Minima
Applications of Maxima and Minima
Absolute Extrema
Absolute Minimum
Let f be a function defined on a domain D
Absolute Maximum
The number f (c) is called the absolute maximum value of f in D
A function f has an absolute (global) maximum at x = c if f (x) f (c) for all x in the domain D of f.
Absolute Maximum
Absolute Extrema
c
( )f c
Absolute Minimum
Absolute ExtremaA function f has an absolute (global) minimum at x = c if f (c) f (x) for all x in the domain D of f.
The number f (c) is called the absolute minimum value of f in D
c
( )f c
Generic Example
-3 -2 -1 1 2 3 4 5 6
-1
1
2
3
4
5
6
7
8
x
y
Generic Example
-3 -2 -1 1 2 3 4 5 6
-1
1
2
3
4
5
6
7
8
x
y
Generic Example
-3 -2 -1 1 2 3 4 5 6
-1
1
2
3
4
5
6
7
8
x
y
Relative Extrema
A function f has a relative (local) maximum at x c if there exists an open interval (r, s) containing c such
that f (x) f (c) for all r x s.
Relative Maxima
Relative Extrema
A function f has a relative (local) minimum at x c if there exists an open interval (r, s) containing c such
that f (c) f (x) for all r x s.
Relative Minima
Fermat’s TheoremIf a function f has a local maximum or minimum at c, and if exists, then ( )f c
0
0
0
( ) ( )( ) lim
( ) ( )lim
( ) ( )lim
h
h
h
f c h f cf c
hf c h f c
hf c h f c
h
( ) 0f c
Proof: Assume f has a maximum
( ) ( )f c h f c
( ) ( )0 if 0
f c h f ch
h
0
( ) ( )( ) lim 0 if 0
h
f c h f cf c h
h
( ) ( )f c h f c
( ) ( )0 if 0
f c h f ch
h
0
( ) ( )( ) lim 0 if 0
h
f c h f cf c h
h
( ) 0 and ( ) 0f c f c
Then ( ) 0f c
The Absolute Value of x.( )f x x (0)f DNE
1 if 0( )
1 if 0
xf x
x
Generic Example
-3 -2 -1 1 2 3 4 5 6
-1
1
2
3
4
5
6
7
8
x
y
( ) 0f x
( )f x DNE
The corresponding values of x are called Critical Points of f
Critical Points of f
a. ( ) 0f c
A critical number of a function f is a number c in the domain of f such that
b. ( ) does not existf c(stationary point)
(singular point)
Candidates for Relative Extrema
1.Stationary points: any x such that x is in the domain of f and f ' (x) 0.
2.Singular points: any x such that x is in the domain of f and f ' (x) undefined
3. Remark: notice that not every critical number correspond to a local maximum or local minimum. We use “local extrema” to refer to either a max or a min.
Fermat’s Theorem
If a function f has a local maximum or minimum at c, then c is a critical number of f
Notice that the theorem does not say that at every critical number the function has a local maximum or local minimum
Generic Example
-3 -2 -1 1 2 3 4 5 6
-1
1
2
3
4
5
6
7
8
x
y
( )
not a local extrema
f x DNE
Two critical points of f that donot correspond to local extrema
( ) 0
not a local extrema
f x
Example
3 3( ) 3 .f x x x
2
233
1( )
3
xf x
x x
Find all the critical numbers of
0, 3x Stationary points: 1x Singular points:
Graph of 3 3( ) 3 . f x x x
-2 -1 1 2 3
-3
-2
-1
1
2
x
y
Local max. 3( 1) 2f
Local min. 3(1) 2f
Extreme Value TheoremIf a function f is continuous on a closed interval [a, b], then f attains an absolute maximum and absolute minimum on [a, b]. Each extremum occurs at a critical number or at an endpoint.
a b a ba b
Attains max. and min.
Attains min. but not max.
No min. and no max.
Open Interval Not continuous
Finding absolute extrema on [a , b]
1. Find all critical numbers for f (x) in (a , b).
2. Evaluate f (x) for all critical numbers in (a , b).
3. Evaluate f (x) for the endpoints a and b of the interval [a , b].
4. The largest value found in steps 2 and 3 is the absolute maximum for f on the interval [a , b], and the smallest value found is the absolute minimum for f on [a , b].
ExampleFind the absolute extrema of 3 2 1
( ) 3 on ,3 .2
f x x x
2( ) 3 6 3 ( 2)f x x x x x
Critical values of f inside the interval (-1/2,3) are x = 0, 2
(0) 0
(2) 4
1 7
2 8
3 0
f
f
f
f
Absolute Max.
Absolute Min.Evaluate
Absolute Max.
ExampleFind the absolute extrema of 3 2 1
( ) 3 on ,3 .2
f x x x
Critical values of f inside the interval (-1/2,3) are x = 0, 2
Absolute Min.
Absolute Max.
-2 -1 1 2 3 4 5 6
-5
ExampleFind the absolute extrema of 3 2 1
( ) 3 on ,1 .2
f x x x
2( ) 3 6 3 ( 2)f x x x x x
Critical values of f inside the interval (-1/2,1) is x = 0 only
(0) 0
1 7
2 8
1 2
f
f
f
Absolute Min.
Absolute Max.
Evaluate
-2 -1 1 2 3 4 5 6
-5
ExampleFind the absolute extrema of 3 2 1
( ) 3 on ,1 .2
f x x x
2( ) 3 6 3 ( 2)f x x x x x
Critical values of f inside the interval (-1/2,1) is x = 0 only
Absolute Min.
Absolute Max.