Προσομοιωτικό...

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  • c 20

    16

    .

    - .

    .

    16 2016

    1 - 16 2016

  • - .

    1

    1. x0 .

    2. .

    3. - .

    () f - A f (x) 6= 0 x A, f A.

    () f (x) = ln(x) g(x) = ex (f g

    )(x) = x.

    () limx10

    (f (x) g(x)) = f (10) g(10).() f x0 Df , .

    () f [, ] - (, ) f () = 0, f () f () 0

    3. - ;

    6 + 7 + 12 = 25

    3 - 16 2016

  • - .

    3

    1. f : R R :

    x, y R, |f (x) f (y)| |(x) (y)|

    () f 2.

    () f R.() f

    2 -

    f (

    2

    ).

    3 + 5 + 7 = 15

    2. f [, ], . x (, ) :

    f (x) f ()x

    f () f ()

    =1

    2(x)f (), (, )

    10

    - 16 2016 4

  • - .

    4

    f : R R f (0) = > 0 :

    x, y R (xy + 1 6= 0) f (x)f (y) = f( x + y

    1 + xy

    )(1)

    , 1 8, f .

    1. f (1) = f (1) = 0 f (0) = 1.2. f (x) = 0 1

    1.3. x (1, 1) f f (x) >

    0.4. x1 R 6= 1, 0, 1,

    () t |t| 1 & |y| > 1

    ) x + y1 + xy

    < 1(2)

    5 - 16 2016

  • - . (|x| < 1 & |y| > 1

    )(|x| > 1 & |y| < 1

    ) x + y1 + xy

    > 1(3)

    8. g R :

    g(x) =

    {x [1, 1] g(x) = f (x)x / [1, 1] g(x) = f (x)

    g(x) .

    9. x 6= 1, f (x) =

    1 + x1 x/2

    .

    10. : f (x), - 9, f (1) = 0, .

    11. , .

    (1) (2) = 3.5, (3) = 3.5, (4) (5) =5 (6) (7) (8) = 6 (9) (10) (11) = 7

    - 16 2016 6

  • c 20

    16

    .

    - .

    1 :

    1. : ( . 217 ) x 6= x0 :

    f(x) f(x0) =f(x) f(x0)

    x x0 (x x0)

    :

    limxx0

    [f(x) f(x0)] = limxx0

    (f(x) f(x0)

    x x0 (x x0)

    )= lim

    xx0

    f(x) f(x0)x x0

    limxx0

    (x x0)= f (x0) 0 = 0

    , limxx0

    f(x) = f(x0), f x0.

    2. ( . 194) f , [, ]. :

    () f [, ] () f() 6= f(), f() f() , x0 (, ) : f(x0) = .

    3. () () () () ()

    7 - 16 2016

  • - .

    2 :

    1. v . - 2v.

    AB

    A= () A = 4

    (), t1 =

    A

    v=

    4

    v()

    B

    AB= () B = 4(), t2 =

    B

    2v+B

    2v=

    7

    2v+

    4()2v

    2. t2 > t1, :

    7 + 4()2v

    >4

    v() 7 + 4()

    2 4

    ()> 0

    f() =7 + 4()

    2 4

    ()> 0

    f(x) =7 + 4()

    2 4

    ()=

    7() + 4() 82()

    , .

    3. .

    f () = 21 2()2()

    f (x) = 0 x =

    6. 0 f(x) . >

    6

    f () < 0 f(x) . x = 6

    . ,

    f(

    6

    ) 0, 03 > 0

    f(0) = 0, 5 < 0

    lim

    2f() =

    f() = 0 1 2. (1, 2) f() > 0.: 1 2. ( : 1 = 0, 39 rad 2 = 0, 64 rad).

    - 16 2016 8

  • - .

    3 :

    1. () x R, |f(x) f(x+ 2)| |(x) (x+ 2)| = 0

    , x R, f(x) = f(x+ 2), f 2.() f x0 R

    :

    0 |f(x) f(x0)| |(x) (x0)|

    :0 lim

    xx0|f(x) f(x0)| lim

    xx0|(x) (x0)|

    : limxx0

    |f(x) f(x0)| = 0

    limxx0

    f(x) = f(x0)

    () :

    x, y R,f(x) f(

    2

    ) (x) (2

    )

    x 2

    x R {2

    }:

    f(x) f(

    2

    )x

    2

    (x)

    (2

    )x

    2

    limx

    2

    (x)

    (2

    )x

    2

    = (

    2

    )= 0

    ,

    limx

    2

    f(x) f

    (2

    )x

    2

    0

    = 0 : f

    (2

    )= 0.

    2. x (, )

    =

    f(x) f()x

    f() f()

    x (4)

    f(x) :

    f(x) = f() +f() f()

    (x ) + (x )(x ) (5)

    9 - 16 2016

  • - .

    :

    h(t) = f(t)

    (f() +

    f() f()

    (t ) + (t )(t )

    )

    h(x) [, x]. h() = 0 h(x) = 0. Rolle 1 (, x)

    h(1) = 0 (6)

    [x, ]. Rolle 2 (x, )

    h(2) = 0 (7)

    h(x) , [1, 2] [, ], h(1) = h

    (2) = 0. Rolle (1, 2)

    h() = 0 (8)

    , h(x) = f (t) 2 h() = f () 2 8= 0, , = 12f ().

    : 4 :

    1

    2f () =

    f(x) f()x

    f() f()

    x

    - 16 2016 10

  • - .

    4 :

    1. () (1): x = x y = 0. : f(x)f(0) = f(x) f(x))f(0) 1) = 0. f(x) = 0 f (0) = 0 = , . :f(0) = 1.

    () (1): x = x y = 1.

    f(x)f(1) = f(x+ 1

    1 + x

    ) f(x)f(1) = f(1) f(1)(f(x) 1) = 0

    f(x) = 1 f (x) = 0 . , f(1) = 0.() (1): x = x y = 1.

    f(x)f(1) = f(x 1

    1 x

    ) f(x)f(1) = f(1) f(1)(f(x) 1) = 0

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

    2. / 1, 1 f() = 0. (1) x = y = , f()f() = f(0) = 1. f() = 0, f()f() = 0, . , 1 1.

    3. f [1, 1]! (1) : x = y. , x R : f(x)2 = f

    ( 2x1 + x2

    ). ,

    f( 2x

    1 + x2

    )> 0. x =

    2 (, ).

    2x

    1 + x2= (). , f(()) > 0, (, ).

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

    4. x1 6= 1, 0, 1.

    ()

    x1 + y

    1 + x1y= x1 + t

    x1 + t = (x1 + t)(1 + x1y)y 6= 1x1

    y(1 (x+ 1 + t)x1) = x1 + t x1 = ty 6= 1x1

    (9)

    x1 + t =1

    x1 t = 1 x

    2

    x16= 0 t = 0 .

    .

    , y = t1 x21 x1t

    .

    11 - 16 2016

  • - .

    :t

    1 x21 x1t6= 1

    x1:

    t

    1 x21 x1t6= 1

    x1 tx1 + 1 x1(x1 + t)

    x1(1 x1(x1 + t))6= 0

    . . .

    x21 1

    x1(1 x1(x1 + t))6= 0

    , t =t

    1 x21 x1t.

    () : limt0

    f(x1 + t) f(t)t

    .

    f(x1 + t) f(t)t

    9=

    f(x1+t1+x1t

    ) f(x1)

    t(1)=

    f(x1)f(t) f(x1)t

    =f(x1)

    (f(t) 1

    )t

    = f(x1) f(t) 1

    t tt

    ,tt

    = = 11 x21 x1t

    , t 0 11 x21

    .

    , limt0

    f(t) 1t

    = f (0) = , t 0 t 0.

    : limt0

    f(x1 + t) f(t)t

    =f(x1)

    1 x21. : f -

    x1 f (x1) =f(x1)

    1 x21 x1 6= 1, 0, 1.

    5. f (x) =f(x)

    1 x2 f(x) > 0 x 6= 1, 0, 1, :

    f (x)

    f(x)=

    1 x2=

    2

    ((1 + x)

    1 + x (1 x)

    1 x

    )

    =

    2 ln

    (1 + x

    1 x

    )

    , ln(f(x)

    )=

    2 ln

    (1 + x

    1 x

    )+ c. , f(0) = 1, , c = 0 x

    (1, 1), f(x) =(1 + x

    1 x

    )/2.

    6. f(x) = 0 x 6=1, 1. 1 1. ,

    - 16 2016 12

  • - .

    (,1) (1,+). x = x (1), f(x)f(x) = f(0) = 1. f(x) f(x) . : f (,1) (1,+).

    7.1 x+ y

    1 + xy= = (1 x)(1 y)

    1 + xy(10)

    1 +x+ y

    1 + xy= = (1 + x)(1 + y)

    1 + xy(11)

    11 (10) x = x y = y. (10) (11), :

    () |x| < 1 & |y| < 1 1 x > 0 1 y > 0 1 + xy > 0. :

    1 x+ y1 + xy

    > 0 & 1 +x+ y

    1 + xy< 0

    x+ y1 + xy

    < 1()

    |x| > 1 & |y| > 1

    1x

    + 1y

    1 + 1xy

    < 1 x+ y1 + xy < 1()

    |x| < 1 & |y| > 1 1 >

    x+1y

    1 + x 1y

    = 1 + xyx+ y x+ y1 + xy > 1() |x| > 1 & |y| < 1 .

    8. [1, 1] f(x) = g(x) g(0) = > 0, (1) g(x) xy 6= 1.

    () |x| < 1 & |y| < 1

    x+ y1 + xy < 1. , f(x) = g(x).

    () |x| > 1 & |y| > 1

    x+ y1 + xy < 1. ,

    g(x)g(y) = (f(x))(f(y)) = f(x)f(y) = f

    (x+ y

    1 + xy

    )= g

    (x+ y

    1 + xy

    )

    () |x| < 1 & |y| > 1

    x+ y1 + xy > 1. ,

    g(x)g(y) = f(x)(f(y)) = f(x)f(y) = f

    (x+ y

    1 + xy

    )

    =

    [ g

    (x+ y

    1 + xy

    )]= g

    (x+ y

    1 + xy

    )

    13 - 16 2016

  • - .

    () |x| > 1 & |y| < 1 .: g(x) .

    9. x 6= 1 f (1, 1)

    f(x) =

    (1 + x

    1 x

    )/2=

    1 + x1 x/2

    1 + x

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

    x A = (,1) (1,+) f(x) > 0, 6,

    : f(x) =

    1 + x1 x/2

    .

    , :f (x)

    f(x)=

    1 x2 A :

    ln(f(x)

    )=

    2 ln

    1 + x1 x

    f(x) > 0 1 + x

    1 x< 0 x A. ,

    ln(f(x)

    )= ln

    1 + x1 x/2

    + ln c f(x)2 = c

    1 + x1 x

    (1), f 2(x) = f

    (2x

    1 + x2

    ).

    limx+

    f 2(x) = limx+

    f

    (2x

    1 + x2

    )(12)

    X =2x

    1 + x2 x + X 0. , f(x)

    0 f(0) = 1 limx+

    f(x) = 1. , 12 :

    1 = c 1 c = 1. : x 6= 1, f(x) =

    1 + x1 x/2

    .

    10. f(x) =

    1 + x1 x/2

    x 6= 1 f(1) = 0

    .

    f

    (x+ y

    1 + xy

    )=

    x+y1+xy

    + 1x+y1+xy

    1

    /2

    =

    (1 + x)(1 + y)(1 x)(1 y)/2

    = f(x)f(y)

    - 16 2016 14

  • - .

    x = 1 y = 1 f(x)f(y) = f

    (x+ y

    1 + xy

    )= 0. :

    ln f(x) =

    2 ln 1 + x

    1 x

    (1, 1), f (x) = f(x)1 x2

    f (0) = .

    : f(x), 9, f(1) = 0, .

    11. 2 :

    () f(x) =

    1 + x1 x/2

    , x 6= 1, f(1) = 0.

    () g(x) ={x [1, 1] g(x) = f(x)x / [1, 1] g(x) = f(x)

    15 - 16 2016