Λύσεις μαθηματικά...

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  • 18 05 16

    (edit) 19 : 00 .

    lisari team

    2016

    4

  • lisari team

    http://lisari.blogspot.gr/2014/10/blog-post_13.html

    4 : 18 05 2015 ( )

    blog

    http://lisari.blogspot.gr

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    lisari tea

    18 05 2016

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  • http://lisari.blogspot.gr

    18 05 2016

    1

    lisari team / 2015 16

    : 18 2016

    :

    &

    : ENNIA (9)

    1. , .262, i

    f (x) 0 0x (,x ) f 0x , f

    0(,x ] .

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

    f (x) 0 0x (x ,) f 0x , f

    0[x ,) . :

    0f (x) f (x ) , 0x [x ,) . (2)

    y

    O

    f (x0)

    f 0

    a x0 x

    , (1) (2), :

    0f (x) f (x ) , x (,) ,

    0f (x ) f (,) .

    A2. , .141

    f g :

    x A f (x) g(x) .

    http://lisari.blogspot.gr/

  • http://lisari.blogspot.gr

    18 05 2016

    2

    f g f g .

    A3. , .246-247

    f :

    [,]

    (,)

    , , (,) , : f () f ()

    f ()

    (,) f f

    f

    , , M(,f ()) fC M

    .

    A4. ()

    ()

    ()

    ()

    ()

    (,f ())

    x

    y

    M(,f ())

    A(,f ())

    http://lisari.blogspot.gr/

  • http://lisari.blogspot.gr

    18 05 2016

    3

    B1. x R :

    22

    2xf x

    x 1

    f :

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

    2. x R :

    2

    32

    2 1 3xf x

    x 1

    :

    f 3

    ,3

    3,

    3

    ,

    3 3,

    3 3

    3 3 3 3

    A ,f ,B , f3 3 3 3

    .

    3. f

    R . :

    2 2

    2 2x x x

    x xlim f x lim lim 1

    x 1 x

    xlim f x 1

    f

    y = 1 .

    http://lisari.blogspot.gr/

  • http://lisari.blogspot.gr

    18 05 2016

    4

    4. f , f x f x x R xx

    . f x 0 x R , x = 0. f xx (0, 0).

    f :

    http://lisari.blogspot.gr/

  • http://lisari.blogspot.gr

    18 05 2016

    5

    1. :

    ln x x 1 x 0 , x 1 .

    x 2xe :

    2 2 2 2x x 2 x x 2ln e e 1 x e 1 e x 1 0

    2x 2e 1 x 0 x 0 .

    2x 2e x 1 0 x 0 .

    2. :

    2 222 x 2 x 2f x e x 1 f x e x 1

    2x 2e x 1 0 x R .

    2x 2f x 0 e x 1 0 x 0

    f R ,0 , 0, .

    :

    x < 0 f x 0 2x 2f x e x 1

    x < 0 f x 0 2x 2f x e x 1

    x > 0 f x 0 2x 2f x e x 1

    x > 0 f x 0 2x 2f x e x 1

    f(0) = 0 :

    2x 2f x e x 1, x R

    2x 2f x e x 1 , x R

    2

    2

    x 2

    x 2

    e x 1 , x 0f x

    e x 1 , x 0

    2

    2

    x 2

    x 2

    e x 1 , x 0f x

    e x 1 , x 0

    3. x R :

    2xf x 2x e 1

    2 22 x xx 0 e 1 e 1 0 f x.

    x R :

    2 2 2 22 x x 2 x xf x 4x e 2e 2 4x e 2 e 1 0

    http://lisari.blogspot.gr/

  • http://lisari.blogspot.gr

    18 05 2016

    6

    f R 0x 0 , f x 0

    x ,0 0, f R f R .

    2 22 x xf x 2 2x e 1 e

    x 0 f 0 0 , :

    2x2 2 2 2 2

    2 2

    e2 x x 2 x x 2 x

    2 x 2 x

    f x 0 2x e 1 e 0 2x e 1 e 2x e 1

    2x 1 e 2x 1 e 1

    x 0 , 22x 1 1 2x 0e e 1 .

    4. x 0 .

    g x f x 3 f x , gD R

    g x f x 3 f x , x R f R f R .

    f

    x 3 x f x 3 f x f x 3 f x 0 g x 0

    0 x f (x) < 0 x .

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

    http://lisari.blogspot.gr/

  • http://lisari.blogspot.gr

    18 05 2016

    8

    3.

    x +

    ( ) f (x) 0

    +

    x + x +

    f

    lim f(x) = +f

    f ( )

    -2 x x 2

    -1 x 1 f (x) f (x) f (x) - 2 x x 2

    -1 x 1 -2 -2lim lim 0

    f(x) f(x)