REVIEWER
Summary of Section 4.1 – 4.4 Definition 3 – Extreme Value Theorem
4.1 Maximum and Minimum Values pp. 276 (REPORTING) If f is continuous on a closed interval [a,b], then f attains an
absolute maximum value f(c) and an absolute minimum value f(d)
Definition 1 at some numbers c and d at interval [a,b].
Let c be a number in the domain D of a function f. Then f(c) In order to use the Extreme Value Theorem we must have
is the an interval that includes its endpoints, often called a closed
absolute maximum value of f on D if f ( c ) ≥ f ( x )for all x in interval, and the function must be continuous on that interval. If we
don’t have a closed interval and/or the function isn’t continuous on
D.
the interval then the function may or may not have absolute
absolute minimum value of f on D if if f ( c ) ≤ f ( x )for all x in extrema.
D.
An absolute minimum or maximum is sometimes called global
maximum and minimum.
The maximum and minimum values of f are called extreme
values of f.
Definition 2
The number f ( c ) is a
local maximum value of f if f ( c ) ≥ f ( x )when x is near c.
local minimum value of f if f ( c ) ≤ f ( x )when x is near c.
This graph is not continuous at x=c, yet it does have both an
absolute maximum (x=b) and an absolute minimum (x=c). Also
note that, in this case one of the absolute extrema occurred at the
point of discontinuity, but it doesn’t need to. The absolute
minimum could just have easily been at the other end point or at
some other point interior to the region. The point here is that this
graph is not continuous and yet does have both absolute extrema.
, Definition 4 – Fermat’s Theorem Find the absolute maximum and minimum values of the
−1
If f has a local maximum or minimum point at x=c and if function f ( x )=x 3−3 x 2 +1 ≤x ≤4
f’(c) exists, then f ' (c )=0 .
2
Since a relative extrema must be a critical point the list of f ' ( x )=3 x 2−6 x=(3 x )(x−2)
all critical points will give us a list of all possible relative extrema.
The absolute maximum and minimum values of will be obtained by
finding the critical values of the function which makes the
derivative equals to zero and considering the endpoints of the
Definition 5 – Critical Number interval. The factors 3 x and x−2gives us the critical points x=0 and
A critical number of a function f is a number c in the domain x=2.
of f such that either f ' ( c )=0∨f ' ( c ) does not exist.
Then, we obtain the intervals (−∞, 0 ) ( 0,2 ) ( 2, ∞ ) .
Example 1
Find the critical numbers of f ( x )=x
3
5
( 4−x ) .
Notice that the function lies between the intervals
( −12 , 4 ).
We then find the value of the f (x) given these four values of x in
The Product Rule gives
the graph of the function.
( )
3 −2
3 f ( 0 )=0 −3 ( 0 ) +1
3 2
f ' ( x )=x 5 (−1 )+ ( 4−x ) x 5
5
f ( 0 )=1
(√ )
3
' 5 3
f ( x )=−x + ( 4−x )
f ( 2 ) =2 −3 ( 2 ) +1
5 3 2
2
5 x
−5 x+3 (4−x )
f ( 2 ) =−3
'
f ( x )= 2
12−8 x
5 x5 f ( )
−1 −1 3
2
=
2
−3
12
2
+1 ( )
f ' ( x )=
( −12 )= 18
2
5x 5 f
3
Therefore, f ' ( x )=0 if 12−8 x=0that is x= . f ( 4 )=4 3−3 ( 42 ) +1
2
And f ' ( x )does not exist if x=0. thus, the critical numbers are
f ( 4 )=17
3 Thus, the absolute minimum of the function is at point f ( 2 ) =−3
and 0.
2 .
Example 2
Summary of Section 4.1 – 4.4 Definition 3 – Extreme Value Theorem
4.1 Maximum and Minimum Values pp. 276 (REPORTING) If f is continuous on a closed interval [a,b], then f attains an
absolute maximum value f(c) and an absolute minimum value f(d)
Definition 1 at some numbers c and d at interval [a,b].
Let c be a number in the domain D of a function f. Then f(c) In order to use the Extreme Value Theorem we must have
is the an interval that includes its endpoints, often called a closed
absolute maximum value of f on D if f ( c ) ≥ f ( x )for all x in interval, and the function must be continuous on that interval. If we
don’t have a closed interval and/or the function isn’t continuous on
D.
the interval then the function may or may not have absolute
absolute minimum value of f on D if if f ( c ) ≤ f ( x )for all x in extrema.
D.
An absolute minimum or maximum is sometimes called global
maximum and minimum.
The maximum and minimum values of f are called extreme
values of f.
Definition 2
The number f ( c ) is a
local maximum value of f if f ( c ) ≥ f ( x )when x is near c.
local minimum value of f if f ( c ) ≤ f ( x )when x is near c.
This graph is not continuous at x=c, yet it does have both an
absolute maximum (x=b) and an absolute minimum (x=c). Also
note that, in this case one of the absolute extrema occurred at the
point of discontinuity, but it doesn’t need to. The absolute
minimum could just have easily been at the other end point or at
some other point interior to the region. The point here is that this
graph is not continuous and yet does have both absolute extrema.
, Definition 4 – Fermat’s Theorem Find the absolute maximum and minimum values of the
−1
If f has a local maximum or minimum point at x=c and if function f ( x )=x 3−3 x 2 +1 ≤x ≤4
f’(c) exists, then f ' (c )=0 .
2
Since a relative extrema must be a critical point the list of f ' ( x )=3 x 2−6 x=(3 x )(x−2)
all critical points will give us a list of all possible relative extrema.
The absolute maximum and minimum values of will be obtained by
finding the critical values of the function which makes the
derivative equals to zero and considering the endpoints of the
Definition 5 – Critical Number interval. The factors 3 x and x−2gives us the critical points x=0 and
A critical number of a function f is a number c in the domain x=2.
of f such that either f ' ( c )=0∨f ' ( c ) does not exist.
Then, we obtain the intervals (−∞, 0 ) ( 0,2 ) ( 2, ∞ ) .
Example 1
Find the critical numbers of f ( x )=x
3
5
( 4−x ) .
Notice that the function lies between the intervals
( −12 , 4 ).
We then find the value of the f (x) given these four values of x in
The Product Rule gives
the graph of the function.
( )
3 −2
3 f ( 0 )=0 −3 ( 0 ) +1
3 2
f ' ( x )=x 5 (−1 )+ ( 4−x ) x 5
5
f ( 0 )=1
(√ )
3
' 5 3
f ( x )=−x + ( 4−x )
f ( 2 ) =2 −3 ( 2 ) +1
5 3 2
2
5 x
−5 x+3 (4−x )
f ( 2 ) =−3
'
f ( x )= 2
12−8 x
5 x5 f ( )
−1 −1 3
2
=
2
−3
12
2
+1 ( )
f ' ( x )=
( −12 )= 18
2
5x 5 f
3
Therefore, f ' ( x )=0 if 12−8 x=0that is x= . f ( 4 )=4 3−3 ( 42 ) +1
2
And f ' ( x )does not exist if x=0. thus, the critical numbers are
f ( 4 )=17
3 Thus, the absolute minimum of the function is at point f ( 2 ) =−3
and 0.
2 .
Example 2
