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LESSON 14 LOCAL MAXIMUM AND MINIMUM VALUES
Definition Let a function f be defined on an interval ( a , b ) and let x 1 and x 2 be
any numbers in the interval. Then
(1)
f is increasing on the interval if f ( x 1 )  f ( x 2 ) whenever x 1  x 2 ,
(2)
f is decreasing on the interval if f ( x 1 )  f ( x 2 ) whenever x 1  x 2 ,
(3)
f is constant on the interval if f ( x 1 )  f ( x 2 ) for every x 1 and x 2 in the
interval.
Part (1) of the definition is saying that the function f is increasing on the interval
( a , b ) if the functional values, which are given by the y-coordinates of the points
on the graph of y  f ( x ) are increasing as you move left to right across the
interval on the x-axis.
Part (2) of the definition is saying that the function f is decreasing on the interval
( a , b ) if the functional values, which are given by the y-coordinates of the points
on the graph of y  f ( x ) are decreasing as you move left to right across the
interval on the x-axis.
Example Determine the interval(s) on which the function f is increasing and
decreasing.
f is increasing on ( a , b )  ( c , d )
f is decreasing on (   , a )  ( b , c )  ( d ,  )
Copyrighted by James D. Anderson, The University of Toledo
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Example Determine the interval(s) on which the function
f ( x )   2 x 2  12 x  19 is increasing and decreasing.
2
Then y   2 x  12 x  19 . We can find the vertex of the
2
parabola by completing the square on y   2 x  12 x  19 . Thus,
Set y  f ( x ) .
y   2 x 2  12 x  19  y  19 __   2 ( x 2  6 x  __ ) 
y  19  18   2 ( x 2  6 x  9 )  y  1   2 ( x  3 ) 2
Thus, the vertex of the parabola is V ( 3 ,  1 ) and the parabola opens down since
a   2  0.
NOTE: You could have also found the x-coordinate of the vertex of the parabola
b
12

  (  3 )  3 . The y-coordinate of the vertex of the
by x  
2a
4
parabola is given by y ( 3 )   18  36  19   1 .
2
The graph of y   2 x  12 x  19 :
Answer:
Increasing: (   , 3 )
Decreasing: ( 3 ,  )
Theorem If a function f is continuous on the closed interval [ a , b ] and
differentiable on the open interval ( a , b ) , then
Copyrighted by James D. Anderson, The University of Toledo
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1.
if f ( x )  0 for all x in the interval ( a , b ) , then the function f is
increasing on the interval ( a , b ) ,
2.
if f ( x )  0 for all x in the interval ( a , b ) , then the function f is
decreasing on the interval ( a , b ) .
Proof Will be provided at a later date.
Examples Find the interval(s) on which the following functions are increasing and
decreasing.
1.
f ( x )   2 x 2  12 x  19
f ( x )   4 x  12  4 ( 3  x )
NOTE: The domain of the function f is the set of all real numbers.
Sign of f ( x ) :

+

3
Answer:
Increasing: (   , 3 )
Decreasing: ( 3 ,  )
2.
y
3x
9  x2
NOTE: The domain of the function y is the set of all real numbers such that
x   3 and x  3 .
We differentiated this function in Lesson 9 obtaining that
y 
3( 9  x2 )
3( 9  x2 )
3( 9  x 2 )
=
=
[ ( 3  x ) ( 3  x ) ]2
( 9  x2 )2
( 3  x )2 ( 3  x )2
Copyrighted by James D. Anderson, The University of Toledo
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Sign of y  :
+
+
+


3
3
Answer:
Increasing: (   ,  3 )  (  3 , 3 )  ( 3 ,  )
Decreasing: Nowhere
3.
y
6
4t  3t  22
2
NOTE: The domain of the function y is the set of all real numbers such that
11
t
and t  2 .
4
We differentiated this function in Lesson 9 obtaining that
 6 ( 8t  3 )
 6 ( 8t  3 )
 6 ( 8t  3 )
dy

2 =
2
2 =
dt ( 4t  3t  22 )
[ ( t  2 ) ( 4t  11) ]
( t  2 ) 2 ( 4t  11) 2
Sign of
dy
:
dt
+

+


Answer:
11
4


Increasing:    , 



3
8

3
8
4.

2
11 
3
 11
,  
  
4 
8
 4


Decreasing:   , 2   ( 2 ,  )
g( w) 

3w  22
4w 2  3w  22
Copyrighted by James D. Anderson, The University of Toledo
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NOTE: The domain of the function g is the set of all real numbers such that
11
w
and w  2 .
4
We differentiated this function in Lesson 9 obtaining that
g ( w ) 
4w ( 44  3w )
4w ( 44  3w )
=
( 4w 2  3w  22 ) 2
[ ( w  2 ) ( 4w  11) ]2

Sign of g ( w ) :

+


Answer:
11
4


Increasing: ( 0 , 2 )   2 ,


Decreasing:    , 
5.

0

+

2

44
3
44 

3 
11 
 11

 44

, 0  
, 
  
4 
 4

 3

7 x 3  13x
h( x )  4
x  8 x 2  16
NOTE: The domain of the function h is the set of all real numbers such that
x   2 and x  2 .
We differentiated this function in Lesson 9 obtaining that
 7 x 6  17 x 4  232 x 2  208
h( x ) 
( x 4  8 x 2  16 ) 2
We used the Factor Theorem and synthetic division to factor the numerator of
this fraction in Lesson 9 obtaining that
Copyrighted by James D. Anderson, The University of Toledo
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(1  x ) (1  x ) ( 7 x 2  52 )
h( x ) 
( x 2  4 )3

Sign of h ( x ) :

+


Answer:

1
1
 2

+

2
Increasing: (  2 ,  1)  (1, 2 )
Decreasing: (   ,  2 )  (  1, 1)  ( 2 ,  )
6.
y
5 x  17
9  x2
NOTE: The domain of the function y is the set of all real numbers such that
x   3 and x  3 .
We differentiated this function in Lesson 9 obtaining that
y 
( x  5 ) ( 5x  9 )
( 9  x2 )2
Sign of y  
( x  5 ) ( 5x  9 )
:
[ ( 3  x ) ( 3  x) ] 2
+
Answer:

+

+




3
9
5
3
5


Increasing: (   ,  3 )    3 ,
9
  ( 5,  )
5
9

, 3   ( 3, 5 )
5

Decreasing: 
Copyrighted by James D. Anderson, The University of Toledo
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7.
h( z )  10 3 ( z 2  3z  18 ) 2
NOTE: The domain of the function h is the set of all real numbers.
We differentiated this function in Lesson 10 obtaining that
h( z ) 
20 ( 2 z  3 )
20 ( 2 z  3 )
=
3 ( z 2  3z  18 )1 / 3
3[ ( z  6 ) ( z  3 ) ]1 / 3

Sign of h( z ) :

6


Increasing:   6 , 
Answer:

+


3
3

2
3
  ( 3,  )
2


Decreasing: (   ,  6 )   
8.
+
3

, 3
2

2 x 2  11
f ( x) 
4
x 2  16
NOTE: The domain of the function f is given by (   ,  4 )  ( 4 ,  )
2
since we need that x  16  0 .
2 x 2  11
f ( x) 
( x 2  16 )1 / 4
1
4 x ( x 2  16 )1 / 4  ( 2 x 2  11) ( x 2  16 )  3 / 4 2 x
4
f ( x ) 
=
2
1/ 2
( x  16 )
Copyrighted by James D. Anderson, The University of Toledo
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1
x ( x 2  16 )  3 / 4 [ 8 ( x 2  16 )  ( 2 x 2  11) ]
2
=
( x 2  16 )1 / 2
x ( 8 x 2  128  2 x 2  11)
3x ( 2 x 2  39 )
x ( 6 x 2  117 )
=
=
2 ( x 2  16 ) 5 / 4
2 ( x 2  16 ) 5 / 4
2 ( x 2  16 ) 5 / 4
39
However, 0 is not in the domain of the
2
39
39
  4 and
 4 , then the numbers
2
2
f ( x )  0  x  0 , x  
function

f .
Since 
39
are in the domain of the function f.
2
f ( x ) is undefined when x   4
Sign of f ( x ) :


+


39
2

Answer:
Increasing:  




4
4
39
2


39
,  4   
2



Decreasing:    , 

9.
+
39
2

39
,  
2



   4,




39
2




s( t )  2 t 3  13t 2  20 t  18
NOTE: The domain of the function s is the set of all real numbers.
s( t )  6 t 2  26 t  20 = 2 ( 3 t 2  13 t  10 ) = 2 ( t  5 ) ( 3 t  2 )
Copyrighted by James D. Anderson, The University of Toledo
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Sign of s ( t ) :

+

Answer:

2
3
5




10.

Increasing:    , 
Decreasing:  
+
2
  ( 5,  )
3
2

, 5
3

y  x 4  8x 3  12
NOTE: The domain of the function y is the set of all real numbers.
dy
 4 x 3  24 x 2 = 4 x 2 ( x  6 )
dx
Sign of
dy
:
dx

+

6
Answer:
+

0
Increasing: (  6 ,  )
Decreasing: (   ,  6 )
11.
g ( x )  5x 
4
x2
NOTE: The domain of the function g is the set of all real numbers such that
x  0.
g ( x )  5x  4 x  2
Copyrighted by James D. Anderson, The University of Toledo
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g ( x )  5  8 x
3
= x
3
5x 3  8
(5 x  8 ) =
x3
3
3
3
NOTE: g ( x )  0  5 x  8  0  x  
Sign of g ( x ) :

+

3
Answer:
+

0
2

8
2
 x
3 5
5
5

2 
 ( 0,  )
Increasing:    ,  3

5 



2



,
0
Decreasing:  3

5


12.
f ( x )  4 x (15  x)
NOTE: The domain of the function f is the interval [ 0 ,  ) .
f ( x )  x1 / 4 (15  x) = 15 x1 / 4  x 5 / 4
f ( x ) 
5  3/ 4
5( 3  x )
15  3 / 4 5 1 / 4
x
 x
x
(
3

x
)
=
=
4
4
4
4 x 3/ 4
f ( x )  0  x  3 and f ( x ) is undefined when x  0 .

Sign of f ( x ) :

0
+

3
Copyrighted by James D. Anderson, The University of Toledo
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Answer:
Increasing: ( 3 ,  )
Decreasing: ( 0 , 3 )
Definition Let c be a number in the domain of the function f. Then
1.
f ( c ) is a local maximum of f if there exists an open interval ( a , b )
containing c such that f ( x )  f ( c ) for all x in the interval ( a , b ) .
2
f ( c ) is a local minimum of f if there exists an open interval ( a , b )
containing c such that f ( x )  f ( c ) for all x in the interval ( a , b ) .
On the graph of y  f ( x ) , f ( c ) is the y-coordinate of the point with an xcoordinate of c. Part (1) of the definition is saying that f ( c ) is a local maximum
of the function f if it is the largest y-coordinate of all the y-coordinates of points
on the graph of y  f ( x ) in a neighborhood of x  c .
On the graph of y  f ( x ) , f ( c ) is the y-coordinate of the point with an xcoordinate of c. Part (1) of the definition is saying that f ( c ) is a local minimum
of the function f if it is the smallest y-coordinate of all the y-coordinates of points
on the graph of y  f ( x ) in a neighborhood of x  c .
Theorem If a function f has a local maximum or a local minimum at a number c,
then either f ( c ) = 0 or f ( c ) is undefined.
The converse of this theorem is not true. The converse of this theorem would read:
If f ( c ) = 0 or f ( c ) is undefined, then the function f has a local maximum or a
local minimum at a number c. Unfortunately, most calculus students use this
statement.
The theorem above tells you where to look for local maximums and local
minimums. It doesn’t guarantee that you will find one.
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The theorem works like the following. If I tell you at 4:00 today, I will either be in
my office, in the math conference room, UH 1000, or in the hallway between two
of these three places, then you would not go to the library looking for me. And if
you go to the math conference room, you will not find me there if I am in my
office.
Definition A number c in the domain of a function f is called a critical number
of f if either f ( c ) = 0 or f ( c ) is undefined.
NOTE: Critical numbers are possible places where local maximums or local
minimums occur.
NOTE: The critical numbers of a function f are a subset of the number(s) on the
number line used in the three-step method for finding the sign of f  .
Theorem (The First Derivative Test for Local Maximums and Minimums)
Suppose that c is a critical number of a function f and ( a , b ) is an open interval
containing c. If f is continuous on the closed interval [ a , b ] and differentiable on
the open interval ( a , b ) , except possibly at c. Then
1.
if f ( x )  0 for all x in the interval ( a , c ) and f ( x )  0 for all x in the
interval ( c , b ) , then f ( c ) is a local maximum of the function f,
2.
if f ( x )  0 for all x in the interval ( a , c ) and f ( x )  0 for all x in the
interval ( c , b ) , then f ( c ) is a local minimum of the function f,
3.
if f ( x )  0 or f ( x )  0 for all x in the interval ( a , b ) , except possibly
at c, then f ( c ) is neither a local maximum nor a local minimum.
Proof Will be provided at a later date.
In Part (1) of the theorem, the interval ( a , c ) is on the left-hand side of the critical
number c and the interval ( c , b ) is on the right-hand side of the critical number c.
Thus, the statement f ( x )  0 for all x in the interval ( a , c ) is saying that the
function f is increasing on the left-hand side of the critical number c, and the
Copyrighted by James D. Anderson, The University of Toledo
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statement f ( x )  0 for all x in the interval ( c , b ) is saying that the function f
is decreasing on the right-hand side of the critical number c. If you would visualize
this, you would see the local maximum being formed.
In Part (2) of the theorem, the statement f ( x )  0 for all x in the interval ( a , c )
is saying that the function f is decreasing on the left-hand side of the critical
number c, and the statement f ( x )  0 for all x in the interval ( c , b ) is saying
that the function f is increasing on the right-hand side of the critical number c. If
you would visualize this, you would see the local minimum being formed.
Examples Find the local maximum(s) and local minimum(s) of the functions in
the previous examples using the First Derivative Test.
1.
f ( x )   2 x 2  12 x  19
f ( x )   4 x  12  4 ( 3  x )
NOTE: The domain of the function f is the set of all real numbers.
Inc
+
Sign of f ( x ) :
Dec


3
Critical Number(s): 3
NOTE: A local maximum occurs when x  3 and the local maximum is
f ( 3) .
f ( 3 )   18  36  19   1
Answer:
Local Maximum(s):  1
Local Minimum(s): None
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2.
y
3x
9  x2
NOTE: The domain of the function y is the set of all real numbers such that
x   3 and x  3 .
We differentiated this function in Lesson 9 obtaining that
3( 9  x2 )
3( 9  x2 )
3( 9  x 2 )
y 
=
=
[ ( 3  x ) ( 3  x ) ]2
( 9  x2 )2
( 3  x )2 ( 3  x )2
Sign of y  :
+
+

3
+

3
NOTE:  3 and 3 are not in the domain of the function y
Critical Number(s): None
Answer:
Local Maximum(s): None
Local Minimum(s): None
3.
y
6
4t 2  3t  22
NOTE: The domain of the function y is the set of all real numbers such that
11
t
and t  2 .
4
We differentiated this function in Lesson 9 obtaining that
 6 ( 8t  3 )
 6 ( 8t  3 )
 6 ( 8t  3 )
dy

dt ( 4t 2  3t  22 ) 2 = [ ( t  2 ) ( 4t  11) ] 2 = ( t  2 ) 2 ( 4t  11) 2
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dy
Sign of
:
dt
Inc
Inc
Dec
Dec
+
+




NOTE: 

11
4


2
3
8
11
and 2 are not in the domain of the function y
4
Critical Number(s): 
3
8
NOTE: A local maximum occurs when t  
3
and the local maximum is
8
 3
y  .
 8
6
6
 3
y
  =
=
,
then
( t  2 ) ( 4t  11)
4t 2  3t  22
 8
6
6
6
96

.
361 =
 3
 3
 =  19   19  =
361

   2     11 



16
 8
 2

 8  2 
Since y 
Answer:
Local Maximum(s): 
96
361
Local Minimum(s): None
4.
g( w) 
3w  22
4w  3w  22
2
NOTE: The domain of the function g is the set of all real numbers such that
11
w
and w  2 .
4
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We differentiated this function in Lesson 9 obtaining that
g ( w ) 
4w ( 44  3w )
4w ( 44  3w )
=
( 4w 2  3w  22 ) 2
[ ( w  2 ) ( 4w  11) ]2
Sign of g ( w ) :
Dec




NOTE: 
Inc
+
11
4

0
Inc
+

2
Dec


44
3
11
and 2 are not in the domain of the function g
4
Critical Number(s): 0 and
44
3
NOTE: A local minimum occurs when w  0 and the local minimum is
44
g ( 0 ) . A local maximum occurs when w 
and the local maximum is
3
 44 
g
.
 3 
g( 0 ) 
 22
1
 22
22
 44 
g

2
=  2  22
 3 
 44 
 44 
4
4
  3
  22
3
3




 3
2
=

  44  22

1
1
22
=
= 352
9 =
  4  22 
 =  88 
4  22 2 

4

1
22 4 


  22
  1
9
9
9 
 9 
  9 

22

4 

22
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
1
9
361 = 361 or
9
Since g ( w ) 
3w  22
3w  22
 44 
 =
=
, then g 
4w  3w  22 ( w  2 ) ( 4w  11)
 3 
2
44  22
22
11  9
22  9
=
=
=
=
38
209
44
6
176
33






19 ( 209 )
38 ( 209 )
 






3  3
3 
 3  3 
 3
1 9
9

.
19 (19 ) 361
Answer:
Local Maximum(s):
9
361
Local Minimum(s): 1
5.
7 x 3  13x
h( x )  4
x  8 x 2  16
NOTE: The domain of the function h is the set of all real numbers such that
x   2 and x  2 .
We differentiated this function in Lesson 9 obtaining that
 7 x 6  17 x 4  232 x 2  208
h( x ) 
( x 4  8 x 2  16 ) 2
We used the Factor Theorem and synthetic division to factor the numerator of
this fraction in Lesson 9 obtaining that
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
(1  x ) (1  x ) ( 7 x 2  52 )
h( x ) 
( x 2  4 )3
Sign of h ( x ) :
Inc
+


Dec
Inc
+



1
1
 2


2
NOTE:  2 and 2 are not in the domain of the function h
Critical Number(s):  1 , 1
NOTE: A local maximum occurs when x   1 and the local maximum is
h(  1 ) . A local minimum occurs when x  1 and the local minimum is
h(1 ) .
h(  1) 
Answer:
 7  13
6 2
 
1  8  16 9 3
Local Maximum(s):
h(1 ) 
7  13
6
2


1  8  16
9
3
2
3
Local Minimum(s): 
2
3
NOTE: If you didn’t find the domain of the function h to know that the
number  2 is not in the domain of the function, then by the First Derivative
Test, you would have concluded that a local minimum is occurring at
x   2 . You would have recognized the domain problem when you would
have tried to find the local minimum by calculating h(  2 ) . You would have
division by zero. In the same way, if you didn’t find the domain of the
function h to know that the number 2 is not in the domain of the function,
then by the First Derivative Test, you would have concluded that a local
maximum is occurring at x  2 . You would have recognized the domain
problem when you would have tried to find the local maximum by
calculating h( 2 ) . You would also have division by zero.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
6.
y
5 x  17
9  x2
NOTE: The domain of the function y is the set of all real numbers such that
x   3 and x  3 .
We
y 
differentiated
( x  5 ) ( 5x  9 )
( 9  x2 )2
Sign of y  
this
function
in
Lesson
9
obtaining
that
( x  5 ) ( 5x  9 )
:
[ ( 3  x ) ( 3  x) ] 2
Inc
+
+
Dec
Dec

Inc
+





3
9
5
3
5
NOTE:  3 and 3 are not in the domain of the function y
9
,5
5
Critical Number(s):
NOTE: A local maximum occurs when x 
9
and the local maximum is
5
9
y   . A local minimum occurs when x  5 and the local minimum is
5
y( 5 ) .
8
25
 9  9  17
 8 ( 25 )
 8 ( 25 )
 8 ( 25 )
y  

=
=
=
=
=
81 25
 5  9  81
9 ( 25 )  81 9 ( 25  9)
9 (16)
9
25
25
 1 ( 25 )
25

= 9 ( 2) =
18
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
y( 5 ) 
25  17
8
1


9  25
 16
2
Answer:
Local Maximum(s): 
25
18
Local Minimum(s): 
1
2
NOTE: A graph of this function was given in Lesson 9.
7.
h( z )  10 3 ( z 2  3z  18 ) 2
NOTE: The domain of the function h is the set of all real numbers.
We differentiated this function in Lesson 10 obtaining that
h( z ) 
20 ( 2 z  3 )
20 ( 2 z  3 )
=
3 ( z 2  3z  18 )1 / 3
3[ ( z  6 ) ( z  3 ) ]1 / 3
Sign of h( z ) :
Dec
Inc
+


6
Critical Number(s):  6 , 
Dec
Inc
+


3

2

3
3
,3
2
NOTE: A local minimum occurs when z   6 and the local minimum is
3
h(  6 ) . A local maximum occurs when z   and the local minimum is
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
 3
h 
 2
h( 3 ) .

 . A local minimum occurs when x  3 and the local minimum is

h(  6 )  10 3 ( 36  18  18 ) 2 = 10 3 0 = 0
 3
9 9

h     10 3    18 
 2
4 2

10
3
 81 


 4 
2
10 3 4  38
3
= 10
10 3 36
=
43
3
 81 
 
4 
3
2
= 10 3
2
= 10
=
2
=
4  ( 34 ) 2
812
3
= 10
=
43
42
5  32
4  32
4
3
 9 18 72 

 

4
4 
4
3
49
2
45 3 36
=
2
h( 3 )  10 3 ( 9  9  18 ) 2 = 10 3 0 = 0
45 3 36
Local Maximum(s):
2
Answer:
Local Minimum(s): 0
8.
2 x 2  11
f ( x) 
4
x 2  16
NOTE: The domain of the function f is given by (   ,  4 )  ( 4 ,  )
2
since we need that x  16  0 .
We differentiate this function in Example 8 above obtaining that
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
3x ( 2 x 2  39 )
f ( x ) 
2 ( x 2  16 ) 5 / 4
Sign of f ( x ) :
Dec
Inc
+

Dec


39
2

Inc
+



4
4
39
2
NOTE: The domain of the function f is given by (   ,  4 )  ( 4 ,  )
39
,
2
Critical Number(s): 
39
2
39
and the local minimum
2
NOTE: A local minimum occurs when x  

f
is  


 . A local minimum occurs when x 


 39 
.
minimum is f 

2


39
2

f  

39
2
28
4
7
4
2
4
4
=
 39 
2
  11
39 
2


 
=

2 
39
4
 16
2


  f




28
4
4
7
2
=
28 4 2
4
7

4
73
4
73
=
39  11
4
39
32

2
2
28 4 2  7 3
4
39
and the local
2
74
686
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
=
28
=
4
7
2
28 4 2  343
7
=
=
Answer:
Local Maximum(s): None
Local Minimum(s): 4 4 686
9.
s( t )  2 t 3  13t 2  20 t  18
NOTE: The domain of the function s is the set of all real numbers.
s( t )  6 t 2  26 t  20 = 2 ( 3 t 2  13 t  10 ) = 2 ( t  5 ) ( 3 t  2 )
Inc
+
Sign of s ( t ) :
Dec


Critical Number(s): 
Inc
+


2
3
5
2
,5
3
NOTE: A local maximum occurs when t  
2
and the local maximum is
3
 2
s    . A local minimum occurs when t  5 and the local minimum is
 3
s( 5 ) .
16
52
40
16
156
360
486
674
 2
s    


 18 = 



=
27
9
3
27
27
27
27
27
 3
s( 5 )  250  325  100  18   157
Answer:
Local Maximum(s):
674
27
Local Minimum(s):  157
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
10.
y  x 4  8x 3  12
NOTE: The domain of the function y is the set of all real numbers.
dy
 4 x 3  24 x 2 = 4 x 2 ( x  6 )
dx
Sign of
dy
:
dx
Dec
Inc
Inc

+
+

6

0
Critical Number(s):  6 , 0
NOTE: A local minimum occurs when x   6 and the local minimum is
y (  6 ) . Neither a local maximum nor a local minimum occurs at x  0
since the function y is increasing on the left-hand side of x  0 and it is
still increasing on the right-hand side of x  0 .
Using a calculator: y (  6 )  1296  1728  12   444 or
y(  6 )  6 4  8  63  12 = 6 ( 6 3  8  6 2  2 ) = 6[ 6 2 ( 6  8 )  2 ] =
6 [ 36 (  2 )  2 ] = 6 (  72  2 ) = 6 (  74 ) =  444
Answer:
Local Maximum(s): None
Local Minimum(s):  444
11.
g ( x )  5x 
4
x2
NOTE: The domain of the function g is the set of all real numbers such that
x  0.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
g ( x )  5  8 x
3
= x
3
5x 3  8
(5 x  8 ) =
x3
3
Inc
+
Sign of g ( x ) :
Dec



0
2

3
+
5
NOTE: 0 is not in the domain of the function g
Critical Number(s): 
2
3
5
NOTE: A local maximum occurs when x  
2
3
5
and the local maximum

2 

g

is  3
.
5 

 8
5    4


5x  4
4
 2    5
g
g
(
x
)

5
x


Since
, then  3 5 
=
4
x2
x2


3 25
3
8  4
= 
4
3
25
Answer:
12
4
3
= 
12
3
25
4
=  3 3 25
25
Local Maximum(s):  3 3 25
Local Minimum(s): None
12.
f ( x )  4 x (15  x)
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
NOTE: The domain of the function f is the interval [ 0 ,  ) .
f ( x )  x1 / 4 (15  x) = 15 x1 / 4  x 5 / 4
f ( x ) 
5  3/ 4
5( 3  x )
15  3 / 4 5 1 / 4
x
 x
x
(3  x) =
=
4
4
4
4 x 3/ 4
Dec
Inc
+

Sign of f ( x ) :

0

3
Critical Number(s): 3
NOTE: A local minimum occurs when x  3 and the local minimum
is f ( 3 ) .
Since f ( x )  4 x (15  x) , then f ( 3 )  12 4 3
Answer:
Local Maximum(s): None
Local Minimum(s): 12 4 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850