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2.7
Study of properties of functions
1.Monotonicity of functions
2.Exterme values of functions
3.Global maxima and minima
4.Convexity of functions
5.Construct the graph of a function
2.7.1 Monotonicity of functions
Th2.7.1Let f ( x) be continuous on I , and differentiable in I , then
(1)The necessary and sufficient condition for the function f to be
monotone increasing (decrea sin g ) on I is f   0( 0).
(2) If f   0( 0) in I , then f is strictly monotone increasin g
(decreasin g ) on I .
Example Discuss the monotonicity of the function
Solution:
2

f ( x)  6 x  18 x  12  6( x  1)( x  2)
x  1, x  2
let f ( x)  0 , we have
x
( , 1)
1
(1 , 2)
f (x)
f (x)

0

2
2
0
( 2 ,  )

1
y
so
is strictly monotone increasing in
  ,1 , and  2,   
And the monotone decreasing interval is
(1 , 2).
2
1
o
1 2
x
Note:
The partition points may be the points where the function f
is not differentiable.
y y  3 x2
For instance,
o
x

Exampl e 2. 7. 2 Prove t hat tanx > x if x  (0, ).
2
Example 2.7.3 Prove that
1 x
e 
i f 0  x  1.
1 x
2x
2.7.2 Extreme values of functions
Lemma 2.5.1 (Fermat lemma) if f:U  x0  
is
differentiable at x0 , and x0 is a local maximum
(local minimum)point,then f ( x0 )  0.
So an extreme point of a differentiable function must be a
stationary point, but the converse is not necessarily true.
say, f ( x )  x i s not di f f er ent i abl e at x  0,
but x  0 i s t he mi ni mal poi nt of f .
Note：
f ( x0 )  0

the possible extreme points x 0 
 f ( x0 ) doesn ' t exi
Th2.7.2
1)i f f ' ( x )  0f or x  ( x0   , x0 )，and f ' ( x )  0 for x  ( x0 , x0   )，
t hen f ( x )has a maxi mum at x0；
2)I f f ' ( x )  0 for x  ( x0   , x0 )，and f ' ( x )  0 when x  ( x0 , x0   )，
t hen f ( x )has a mi ni mum at x0；
0
3)I f f ( x )has t he same si gh inU ( x0， )，
'
t hen x0 i s not an ext r eme poi nt of f .
Example Find the extreme values of
Solution
 13
2
1) f ( x )  ( x  1)  ( x  4)  3 ( x  1) 
2
3
5( x 1)
3 3 x 1
2) let f ( x)  0 , we get a stationary point x1  1;
x2  1 is non  differentiable point of f .
3) Make a table as follows:
x ( ,  1) 1

f (x) 
0
f (x)
( 1 , 1)

1
0
3 3 4
(1,   )

Th2.7.3
and
then
then
has a maximum at
x0 
has a minimum at. x0

Example Find the extreme value of
Solution 1)
f ( x)  6 x ( x 2  1) 2 ,
f ( x)  6 ( x 2  1)(5 x 2  1)
2)
Let f ( x)  0 , we get x1  1, x2  0 , x3  1
3) Judge
f (0)  6  0, so x2  0 is a minimal point and f (0)  0.
y
f (1)  0, but byTh 2.7.2, x  1 are
not extreme point s.
1
1
x
Th2.7.4
f
and
then:1) If n is even
( n)
( x0 )  0 ,
is an extreme point
is a minimum point ;
is a maximum point .
2) when n is odd,


is not an extreme point .
f ( n ) ( x0 )
( x  x0 ) n
f ( x)  f ( x0 )  f ( x0 )( x  x0 )   
n!
 o(( x  x0 ) n )
y
In above example
f ( x )  6 x( x 2  1)2 ,
f ( x )  6( x 2  1)(5 x 2  1)
f ( x)  24 x (5 x 2  3) , f (1)  0
so
1
are not extreme points .
Note:
Th2.7.3 and Th2.7.4 are both sufficient.
1
x
3.Global maxima and minima
Continuous function f(x) on a closed interval must have a
global maximum and a global minimum.
How to find the global maximum and minimum:
(2) The global maximum
M  max
Global minimum
f (a) , f (b)
Example Find the global maximum and minimum of
Solution
si nce f ( 3) 20,
f (4)  6
f (1)  0
3 1
f( )
2 4
f (2)  0
1
Example2.7.6.Let p  1, x  [0,1], prove： p1  x  (1  x )  1
2
p
p
Note :
(1) If f ( x) is monotone increasing on the interval  a , b  ,
then the global maximum and minimum must be abttained
at the end po int .
(2)if continuous function on  a , b  has only one extreme
po int, then this point must be the global point .
Example2.7.7 A cylindrical container with volumeV0 and without
cov er is to be made of a sheet of iron.How should we design it if
we wish to use the least amount of material ?
4.Convexity of functions
If the graph of a function is convex down, then the function
is called a convex function, while a function whose graph is
convex up, is called a concave function.
The property of being convex up or convex down for the graph
of a function is called the convexity of the function.
Definition2.7.2 Let f : I  , if x1 , x2  I
the inequality
x1  x2
1
f(
)  ( f ( x1 )  f ( x2 ))
2
2
holds, then f is called a convex function on I ;
if x1 , x2  I and x1  x2 , we have
x1  x2
1
f(
)  ( f ( x1 )  f ( x2 ))
2
2
then f is called a strictly convex function on I .
If the inequality sign is reversed , then f is called a concave
function or strictly concave function on I respectively.
Th2.7.5 Assume that f is twice differentiable in the int erval I ,
then
(1) f is a strictly convex (convex) function if f ( x )  0

( 0), x  I ;
(2) f is a strictly concave (concave) function if
f ( x)  0 ( 0), x  I

The transition point on the cruve between the concave arc
and convex arc is called an inf lection po int of f ( x).
Example Find the convex and concave interval of
Sou: 1)
3
2

y  12 x  12 x ,
 36 x( x  23 )
2)
(0,1) ( 2 , 11 )
3 27
2,
x

0
,
x

we
get
let y  0
1
2 3
3)
x ( , 0)

y
y convex
(0 , 23 )
0

0
1 concave
( 23 ,  )

0
2
3
2
3
11
27
convex
So f is convex on( , 0)and ( 2 ,  );and is concave on
3
)
( 0 , 1 ) and ( 23 , 11
27 are both inflection points.
(0, 23 )
Exampl e 2. 7. 10 St udy t he convexi t y of t he f ol l owi ng
f unct i ons
（1）f ( x )  x ,( x  0,   1),
(2) f ( x )  ln x( x  0)
Example2.7.11
Prove that  a  b   16(a  b ), a  0, b  0, a  b.
5
5
5
Th2.7.6 If f is continuous and strcitly convex in an
interval I , then f has at most one global minimum
point, and if there exists a unique local minimum
point in I , then it must be global minimum po int in I .
Method of proving inequality：
1.By mean value theorem
2.By the monotone theorem
3.By the global maximum
4.By the convexity of the function
5.Construct the graph of a function
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Oblique asymptote
If
( P79 T1)
( k x  b)
( k x  b)
Then y  k x  b .
f ( x) b
k  lim [
 ]
x x
x
1
lim [ f ( x)  kx  b ]  0 
x  x
f ( x)
k  lim
x x
(or x  )
f ( x)
b
lim [
k  ]0
x x
x
b  lim [ f ( x)  k x]
x
(or x  )
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Solution:
x3
 y
,
( x  3)(x  1)
so
and
lim y   ,
x3
(or x  1)
y  x2
x  3 andx  1
f ( x)
x2
k  lim
 lim 2
x x
x  x  2 x  3
3
1
 2 x 2  3x
b  lim [ f ( x) x]  lim 2
x
x  x  2 x  3
 y  x  2 is oblique asymptote. .
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Example Paint the graph of
Solution : 1) domain
2) y  x 2  2 x , y  2 x  2 ,
l et y  0,
l et y  0,
3)
4)
x ( , 0)
y

y

y
x 1 3
y 23 2
1
0
0
2
(max)
(0 ,1)


1

4
3
2
3
(inflec
tion）
(min)
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2
2 ( 2 ,  )
0


(1, 2)

0
1
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3