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Review 3.1-3.3
- Increasing or Decreasing
- Relative Extrema
- Absolute Extrema
- Concavity
Increasing/Decreasing/Constant
y
8
7
6
5
4
3
2
1
x
-3
-2
-1
1
-1
2
3
4
5
6
Increasing/Decreasing/Constant
y
8
7
6
5
4
3
2
1
x
-3
-2
-1
1
-1
2
3
4
5
6
Increasing/Decreasing/Constant
If f  x   0 for each value of x in an interval a, b ,
then f is increasing on a, b .
If f x   0 for each value of x in an interval a, b ,
then f is decreasing on a, b .
If f  x   0 for each value of x in an interval a, b ,
then f is constant on a, b .
Generic Example
y
8
f ( x)  0
7
6
5
4
3
f ( x)  DNE
2
1
x
-3
-2
-1
1
2
3
4
5
6
-1
The corresponding values of x are called
Critical Points of f
Critical Points of f
A critical number of a function f is a number c in
the domain of f such that
a. f (c)  0
b. f (c ) does not exist
Example
Find all the critical numbers of
x3 - 2
f (x) =
x
2x 3 + 2
f ¢(x) =
2
x
When set = 0
Excluded values
x = -1
x=0
Determine where the function is
increasing or decreasing:
Plug into the derivate around each critical #
-----
+++
-1
+++
0
Increasing: (-1, ∞)
Decreasing: (-∞, -1)
Example
Find all the critical numbers of
f (x) = 1- x
f ¢(x) =
2
-x
1- x 2
When set = 0
Excluded values
x=0
x = ±1
Determine where the function is
increasing or decreasing:
Plug into the derivate around each critical #
Not in Domain
-----
+++
-1
0
Increasing: (-1, 0)
Decreasing: (0, 1)
Not in Domain
1
Example
Find all the critical numbers of
f ( x)  3 x 3  3 x .
x2 1
f ( x) 
3
x
3
 3x
When set = 0

2
x  1
Excluded values x  0,  3
Determine where the function is
increasing or decreasing:
Plug into the derivate around each critical #
+++ +++ ----- ----- 3
-1
0
+++ +++
1
Increasing: (- ∞, -1) U (1, ∞)
Decreasing: (-1, 1)
3
Graph of f ( x)  x  3 x .
3
3
y
Local max. f (1)  3 2
2
1
x
-2
-1
1
2
3
-1
-2
-3
Local min. f (1)   3 2
If the price of a certain item is p(x) and the total
cost to produce x units is C(x), at what production
levels is profit increasing and decreasing?
p(x) =115- 2x
Now find P’(x)
C(x) = x 3 - 26x 2 + 7x + 30
P'(x) = -3x 2 + 48x +108
P(x) = R(x) -C(x)
0 = -3(x -18)(x + 2)
P(x) = x × p(x)-C(x)
P(x) = x (115 - 2x ) - ( x 3 - 26x 2 + 7x + 30)
P(x) = -x 3 + 24x 2 +108x - 30
Now test around 18, -2
Increasing: (0, 18)
Decreasing: (18, ∞)
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)
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)
Relative
Minima
The First Derivative Test
Determine the sign of the derivative of f to
the left and right of the critical point.
left
right




No change
conclusion
f (c) is a relative maximum
f (c) is a relative minimum
No relative extremum
The First Derivative Test
Find all the relative extrema of f ( x)  x3  6 x 2  1.
f ( x)  3x 2  12 x  0
Excluded Values: None
3x( x  4)  0
Relative max.
f (0) = 1
f
f
x  0, 4
+
0
0
Relative min.
f (4) = -31
-
0
4
+
The First Derivative Test
y
5
(0,f(0)=1)
-2
-1
1
2
x
3
4
5
6
7
-5
-10
-15
-20
-25
-30
(4,f(4)=-31)
-35
8
9
10
The First Derivative Test
Find all the relative extrema of f (x) = 3x 5 - 5x 3
f ¢(x) =15x 4 -15x 2 = 0
15x (x -1)(x +1) = 0
2
Excluded Values: None
x = 0,1, -1
+++ ----- -----
-1
0
Rel. Min. (1, -2)
Rel. Max. (-1, 2)
+++
1
Example from before:
f ( x) 
f ( x)  x  3 x .
3
3
3
x
x  1
Relative max.
f ( 1)  3 2
x2 1
3
 3x

2
Relative min.
f (1)   3 2
Exclude Values:
x  0,  3
f +
f
ND + 0
 3
-1
-
ND
0
-
0
+ ND +
1
3
Graph of f ( x)  x  3 x .
3
3
y
2
Rel. max.
f ( 1)  3 2
Rel. min.
f (1)   3 2
1
x
-2
-1
1
2
3
-1
-2
-3
Absolute Extrema
Let f be a function defined on a domain D
Absolute
Maximum
Absolute
Minimum
Absolute Extrema
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.
The number f (c) is called the absolute maximum
value of f in D
Absolute
Maximum
f (c )
c
Absolute Extrema
A 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 )
Absolute
Minimum
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].
Example
 1 
Find the absolute extrema of f ( x)  x  3 x on   ,3 .
 2 
2
f ( x)  3x  6 x  3x( x  2)
3
2
Critical values of f inside the interval (-1/2,3) are x = 0, 2
Evaluate
f (0)  0
Absolute Max.
f (2)  4
Absolute Min.
7
 1
f    
8
 2
f  3  0
Absolute Max.
Example
 1 
Find the absolute extrema of f ( x)  x  3 x on   ,3 .
 2 
3
2
Critical values of f inside the interval (-1/2,3) are x = 0, 2
Absolute Max.
-2
-1
1
2
3
4
5
6
Absolute Min.
-5
Example
 1 
Find the absolute extrema of f ( x)  x  3x on   ,1 .
 2 
2
f ( x)  3x  6 x  3x( x  2)
3
2
Critical values of f inside the interval (-1/2,1) is x = 0 only
Evaluate
f (0)  0
Absolute Max.
7
 1
f    
8
 2
f 1  2
Absolute Min.
Example
 1 
Find the absolute extrema of f ( x)  x  3x on   ,1 .
 2 
2
f ( x)  3x  6 x  3x( x  2)
3
2
Critical values of f inside the interval (-1/2,1) is x = 0 only
Absolute Max.
-2
-1
1
2
3
4
5
6
Absolute Min.
-5
Concavity
Let f be a differentiable function on (a, b).
1. f is concave upward on (a, b) if f ' is increasing on
aa(a, b). That is f ''(x) > 0 for each value of x in (a, b).
2. f is concave downward on (a, b) if f ' is decreasing
on (a, b). That is f ''(x) < 0 for each value of x in (a, b).
concave upward
concave downward
Inflection Point
A point on the graph of f at which f is continuous
and concavity changes is called an inflection point.
To search for inflection points, find any point, c in the
domain where f ''(x) = 0 or f ''(x) is undefined.
If f '' changes sign from the left to the right of c, then
(c,f (c)) is an inflection point of f.
Example: Inflection Points
Find all inflection points of
f ( x)  x  6 x  1.
2

f ( x)  3x  12 x
3
2
f ( x)  6 x  12
Possible inflection points are solutions of
a) f ( x)  0
b) f ( x)  DNE
6 x  12  0
no solutions
x2
Inflection
point at x
2
f 
f
-
0
2
+
y
5
(0,f(0)=1)
-2
-1
1
2
x
3
4
5
6
7
-5
-10
-15
-20
-25
-30
(4,f(4)=-31)
-35
8
9
10