section 4.1 maximum and minimum values applications of differentiation
TRANSCRIPT
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.