H άσκηση της ημέρας -...

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    lisari.blogspot.gr

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    1

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  • ___________________________________________________________________________ 2015 http://lisari.blogspot.gr

    2

    8

    (1-11-2015)

    8/11/2015

    f 0,1 ,

    :

    2x x 1 f(x) 2x 1, x 0,1

    f 0,1

    : x 3x , f

    C

    . .

    . f(x) 0 ,

    1 2

    , .2 3

    .

    1 1

    10

    2

    f(x)dx f(x)dx

    .

    2

    x 0

    x ln f(x) 1lim

    f(1 x) 1

    . , 0,1 , , :

    2f() f() 2 .

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    3

    1 ( )

    . x 0 : 1 f(0) 1 f(0) 1. :

    x 0

    2

    x 0 x 0 x 0

    f(x) 1 f(x) 1x x f(x) 1 2x x 1 2 lim x 1 lim lim 2

    x x

    1 f (0) 2.

    x 1 :1 f(1) 1 f(1) 1. :

    x 1

    2

    x 1 x 1 x 1

    f(x) 1 f(x) 1x x 2 f(x) 1 2x 2 2 x 2 lim 2 lim lim x 2

    x 1 x 1

    2 f (1) 3.

    : x fC ,

    0,1 , f () 1 .

    f 0,1 ,

    f 0,1 , :

    1 f (0) f () 1 f (0) 1.

    : 3x fC ,

    0,1 , f () 3 .

    f 0,1 ,

    f 0,1 ,

    :3 f () f (1) 3 f (1) 3.

    fC 0, 1 1,1

    , f(0) 0 1 f(1) 3 2.

    . f 0,1 ,

    f (x) f (0) 1 f . 0,1 , f

    f 0,1 f(0),f(1) 1,1 .

    0 1,1 , 0,1 , f() 0 .

    .

    f .. 0,

    ,1 , 1 0, 2 ,1 , :

    1

    f() f(0) 1f ( )

    2

    f(1) f() 1f ( )

    1 1

    .

    f .

    1 2 1 2 f ( ) f ( )

    1 1 1

    1 . 1 2

    f .

    2 2 1 f ( ) f (1)

    1 1 2

    3 1 .1 3 3

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    4

    . 1

    x 0,2

    :

    1 1

    12 2f .

    0 0 1

    10 x f(x) f() 0 f(x)dx 0 f(x)dx f(x)dx 0

    2

    1

    1 12

    10 1

    2

    f(x)dx f(x)dx f(x)dx.

    . 2 x 0

    x 0 x 0

    x 0 x 0

    lim f(x) 1 ln f(x) 1x ln f(x) 1 f(x) 1 ln f(x) 1lim lim

    f(x) 1 f(1 x) 1 f(x) 1 f(1 x) 1f(1 x) 1lim lim

    x x x x

    00,

    1 ( 3)

    :

    f(x) 1

    DLHx 0 0 0 0

    2

    1

    ln lim f(x) 1 ln f(x) 1 lim ln lim lim 0

    1 1

    .

    x 0

    f(x) 1lim f (0) 1.

    x

    1 x u

    x 0 u 1

    f(1 x) 1 f(u) 1lim lim f (1) 3.

    x 1 u

    . g h :

    2g(x) f(x) x x,x 0,1

    h(x) f(x) 2x,x 0,1

    : g(0) f(0) 1 g(1) f(1) 2 3 . g

    0,1 , Bolzano 0,1 ,

    g() 0 2f() 0. (1)

    : h(0) f(0) 1 h(1) f(1) 2 3 . h

    0,1 , Bolzano 0,1 , h() 0 f() 2 0. (2)

    (1) (2) : 2f() f() 2 .

    . : 2 2f() f() 2 0 0

    1 , , 0,1 . .

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    5

    2 ( )

    A. = + (1, (1)) (1) = 1

    (1) = 1 + . ,

    [0,1] . 1 0 ( 1)

    (0) ()

    = 3 + (2, (2)) (2) = 3

    (2) = 32 + . ,

    [0,1] . 2 1 ( 2)

    (1) ()

    2 + 1 () 2 1 , (1) [0,1]

    = 0 (1) 1 (0) 1 (0) = 1

    = 1 (1) 1 (1) 1 (1) = 1

    0 (0) = 0+

    ()+1

    1 (1) = 1

    ()1

    1

    2 + 1 () 2 1 2 + () + 1 2 + 1 ()+1

    2

    ,

    0+

    ( + 1) 0+

    () + 1

    0+

    2 1 (0) 2 ()

    :

    { 1 (0)

    1 (0) (0) = 1(0) = ( 1)

    11 1 = 0

    : (1) = 1 + 1=0 (0) = 0 + =

    2 + 1 () 2 1 2 + 2 () 1 2 2

    2 + 2

    1() + 1

    12 2

    12 2

    1() + 1

    12 + 2

    1

    ,

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    6

    1

    (2 2

    1)

    1

    () 1

    1 1

    2 + 2

    1 2 (1) 3 ()

    : { 3 (1)

    3 (1) (1) = 3(1) = ( 2)

    11 2 = 3

    : (2) = 32 + 2=3 (1) = 3 + =

    . , (

    ) = . () = () 2 + 1 [0,1]

    () 2 1 () 2 + 1 0 () (1

    2) .

    1

    2 Fermat (

    1

    2) = 0 (

    1

    2) 2 = 0 (

    1

    2) =

    2

    ( 1

    2 , (

    1

    2)) : = 2 1

    : () 2 1

    : { () 2 1

    () 2 1

    () = 2 1

    (

    ) .

    =1

    2 (1) (

    1

    2) 0 (

    1

    2) < 0

    =2

    3 (1)

    1

    9 (

    2

    3) (

    2

    3) > 0

    {

    [1

    2,2

    3]

    (1

    2) (

    2

    3) < 0

    Bolzano () = 0

    (1

    2,2

    3)

    :

    0 1 (0) () (1) 1 () 3 () > 0

    () > 0 [0,1] 1-1

    () = 0 [0,1]

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    , () = 0 (1

    2,2

    3) .

    . () 2 1 ()1

    20

    (2 1) =1

    41

    2< 0

    1

    20

    ()1

    20

    < 0

    ()

    12

    0

    < 0 ()

    1

    0

    + ()

    12

    1

    < 0 ()

    1

    0

    < ()

    1

    12

    .

    0+

    (1)1

    =

    1

    ()1

    1= (1) = 3

    2 + 1 () 2 1 2 + () + 1 2

    (2 + ) (() + 1) (2) (2 + ) (() + 1)

    (2)

    {

    0+

    ((2 + )) = = 0

    0+

    ((2)) = = 0

    .. 0+

    ((() + 1)) = 0

    0+

    2(()+1)

    (1)1= 0+

    (()+1)(1)1

    =0

    3= 0

    . () = () + 2 + () 2

    (1

    2,2

    3) () = 0 , =

    () = () + 2 + () 2 () = () + 2 + 2

    (1) = (1) + 2 2 = 3 2 > 0

    () = () + 2 + 2 = 2 = ( 1) < 0

    { [, 1]

    ()(1) < 0 Bolzano (, 1) () = 0

    () + 2 + () 2 = 0 () + 2 + = () + 2

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    3 ( )

    A.

    + () ()

    = 0 (1) (0) = 1

    = 1 (1) (1) = 1

    > 0 (1) 2 + () + 1 + 1 ()(0)

    , f

    lim0( + 1) lim

    0

    ()(0)

    ()

    < 1 (1) + 2 ()(1)

    1 , f

    lim1( + 2) lim

    1

    ()(1)

    1 ()

    [0,1] [0,1] 1-1

    - (1, (1)) = + ,

    (1) = 1

    1 0 (1)

    (0) 1 (0) () () =

    =

    (0,-1) 1 = 0 + =

    - (2, (2)) = 3 + ,

    (2) = 3

    2 1 (2)

    (1) 3 (1) () () =

    =

    (1,1) 1 = 3 + =

    B.

    [ 1

    2 ,2

    3 ]

    (1) = 1

    2 (

    1

    2) 0 ,

    (1

    2) = 0 [ 0,

    1

    2 ] [

    1

    2 ,1 ] f

    , 1 (0,1

    2) , 2

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    (1

    2, 1) , 1 < 2 :

    (1) =(1

    2)(0)

    1

    20

    = 2, (2) =(1

    2)(1)

    1

    21

    = 2 ,

    , (

    ) <

    (1) = 2

    3