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x
    (Function)

                 F(x)
:A       B                   f
A      B
       f                                          xA
         y,z  B
            (x,y)  f      (x,z)  f          y=z
                 A = {1,2,3,4}       B = {a,b,c}

    f1 = {(1,a),(2,b),(3,a)}   f2 = {(1,a),(2,b),(4,c),(1,b)}

       1                              1
                      a                             a
       2                              2
                      b                             b
       3                              3             c
                      c
       4                              4
Df = { x | y     (x,y)  f
                  }
                  Rf = { y | x     (x,y)  f
                  }
 f                       (x,y)  f
             y
                        x
              y                        f
     x     (1,2)  f  f(1) = 2
                       f(x)
y = f(x)
                                  f(x) =
(x,y1)  f    (x,y2)  f

              y1 = y2




(x,y1)  f    (x,y2)
 f
   y1  y2
f={(x,y)RR|y = x2+1}

(x,y1)f      (x,y2)f
      y1  y2
 x2+1  x2+1
   x2  x 2

        y1 = y2
f={(x,y)RR|x2+y2 = 1}

 f
     (0,1)f    ( 02+12= 1)
     (0,-1)f    (02+(-1)2 = 1)
      1  -1
Rf  B

     f                        A     B (f is function from A to
B)
                                   f:AB

                 Rf = B               f                     A
         B
                                          (f is function from A
onto B)

                     A = {1,2,3} f BA { a,b,c}
                                      : = B
             f1= {(1,a),(2,a),(3,b)}      f1 : A  B
             f2 = {(1,a),(2,b),(3,c)}     f2 : A  B
             f3 = {(1,b),(2,c),(3,a)}     f3 : A  B
f:AB

                         yB , x  A   f(x)
         =y
B = Rf
                 y                       B
         y = f(x)

                                   x
                     x A
                      x A
f:QQ            f(x) = 3x+4

                    yQ          y=
  f(x)
               y = 3x+4
         x = (y-4)/3  Q


            f:NN           f(x) = 3x+4
              yN      y = f(x)
         y = 3x+4
         X = (y-4)/3  N        y=5       x=
1/3
f:RR       f(x) = x2

            yR       y = f(x)
                 y = x2
             x = y R          y<0
      f



              yB      xA         f(x) y
                           y      f(x) y
  xA
                   y = -1     xA
x2  -1
:f:AB               f
    (one-to-one)
                               x1,x2  A          y
B
       (x1,y)  f   (x2,y)  f         x1 = x 2
                    1-1

                          f:A  B


       x1  x2  f(x1)  f(x2)

          f(x1) = f(x2)  x1 = x2
Direct method                   Contradiction method


              Contrapositive method
    f(x1) =
  f(x2)
                                     f(x1) = f(x2)   x1
                                x2
                      x1 x2
  x1 = x 2
                                x1 x2

                f(x1)  f(x2)
f(x) = 2x+1
x

                 f(x1) = f(x2)
             2x1+1 = 2x2+1
                2x1 = 2x2
                  x1 = x 2
x1 x2
       f(x1) = f(x2)   x1      f(x1) = f(x2)
x2                           2x1+1 = 2x2+1
     2x1+1 = 2x2+1             2x1 = 2x2
        2x1 = 2x2                x1 = x 2
         x1 = x 2             f(x1)  f(x2)
x1,x2  Df        x1 
                                 x2
                                              f(x1) = f(x2)
           f(x) = x2

         x1 = 2        x2 = -2
f(x1) = f(2) = 4
f(x2) = f(-2)= 4                                         f(x1)
                                  = f(x2)
                                                  (x1)2 = (x2)2
                                                    x1 =  x2
:                 f:AB
        f-1                              f-1
                          f
               f-1 : f(A)  A
               A = {1,2,3} B = {a,b,c}
               f = {(1,b),(2,a),(3,c)}
              f-1 = {(a,2),(b,1),c,3)}

        f                           f-1
                 a              a
1                                              1
                 b              b
2                                              2
                 c              c
3                                              3
A = {1,2,3} B = {a,b,c}
          f = {(1,b),(2,b),(3,c)}
        f-1 = {(b,1),(b,2),c,3)}

    f                               f-1
         a                   a
1                            b            1
         b                                2
2        c                   c
3                                         3

                              f-1

                                    f
:                    f:AB
      f                                  f


                   f                             f

                       f
          (             f                            )
           x1,x2  A         yB
           (x1,y),(x2,y)  f     (       f(x1)=f(x2)=
y )
          (y,x1),(y,x2)  f-1
                 f -1                (       f
                    )
f                    f

      f
  y B         x1,x2 A
(y,x1),(y,x2)  f-1
 (x1,y),(x2,y)  f
      x1 = x 2
  f-1
1-1                   1-1
        :       f:AB                   f :BA
                                         -1



                 1-1
                  f:AB                       f   f-1:
B A
                 f-1
                  f-1(y1) = f-1(y2)
          x          x = f-1(y1) = f-1(y2)
         (y1,x) f-1          (y2,x)f-1
       (x,y1)f           (x,y2)f
                 y1 = y 2           (f
               )
           f-1
f-1

(              Rf-1 = A)

        xA
      yB        f(x) = y
(x,y)  f
(y,x)  f-1
 x  Rf-1
         A  f-1
     Rf-1  A
 Rf-1 = A
f-1
:       A       B
               A               B
                       B
                                   AB

       :       f:AB                 g:BC
                                      f  g
gof
                                A      C
                           (gof)(x) = g(f(x))        xA
      gof = {(x,z) |           yB       (x,y)f   (y,z)
g}
:   f1-1A  B
             :               1-1 :
                              g      BC    1-1 f
                                             go     :
A C

                       1-1
             1-1                 1-1
              f:AB                  g:B
  C
                     gof(x1) = gof(x2)
              g(f(x1)) = g(f(x2))
              f(x1) = f(x2)
                 x 1 = x2
:       f:AB           g:BC          gof : A
C


                   f:AB          g:BC
                    zC (            C
 Rgof )
              Rg = C
               z  Rg
           y B       (y,z)  g
               Rf = B
               yRf
          xA        (x,y)  f
            (x,z)  gof
               z  Rgof           C  Rgof
                   R o C
:   f:AB         kN
                 f          k       1 (k-1)
               y B   x                        k
        f(x) = y


0            a              0                 a
1            b              1                 b
2            c              2                 c
3            d              3                 d

            2-                            3-
    1                           1
:        f:AB               k-1
         n(B)
                          A                  n(A) 
         k.n(B)
               f:AB                 k-1      n(B)
               x                      k
              yB
                                 A
k.n(B)
                A                    n(A)  k.n(B)
                    x1
                     .
                                             y1
                    xk                       y2
                                              .
                                              .
                                             ym
(Pigeon-Hole Principle )
                  n
m       n>m
                      1
2                             1
                              2
                              3
                              4


                                     J.P.G.L.
    Dirichlet
x (floor of x)
x
                       x

          4= 4
          4.0000001= 4
          4.9999999= 4

                   x(ceiling of x)
x
                       x
        4 = 4
        4.000000001 = 5
        4.999999999 = 5
(Generalized Pigeon-Hole
    Principle)
             :   f:AB      n(A) = m        n(B) = m
                                                   k
                                                   
                    yB                      xA 
                                                     k 

                 f:AB       n(A) = m          n(B) = k m 
                            yB                         k  1
                                                       xA
                                                           
                                               m 
                                                k  1
                                                
                                           A                    k(
                      m            m
)                      k  1
                                   k
                                              m 
                   m  n(        ) < k((       k   1m
                                               
                                            +1) -1) =
                                              m 
                                              k 
                                               
                  y B                     xA
 89 
       12 
          
               =

= 7.4167
=8
1.

                              n2    ; n 0
     1.1 f : Z  Z   f(n) =
                              -n2 ; n < 0
                               n+1 ; n
     1.2 f : Z  Z   f(n) =
                               n3   ; n
                              x+1   ;xQ
     1.3 f : R  R   f(x) =
                              2x   ;xQ
                              3x+2 ; x  Q
     1.4 f : R  R   f(x) =
                               3
x 1
     1.5 f : R  R   f(x) =                    ;x0
                              x
                                  x
     1.6 f : R  R   f(x) =   2
                              x   1
                           3x - 1
     1.7 f : R  R   f(x) =                    ;x0
                             x
                           x 1
     1.8 f : R  R   f(x) =x  1               ;x1

2.            f:RR           f(x) = 2x
                                           3
                                      2 x  2x
                 g:RR                g(x)=
                                        2

                   f=g                 x   1
3.          f:RR          f

       3.1 f(x) = 6x-9
      3.2 f(x) = 3x2-3x+1
      3.3 f(x) = sin x
      3.4 f(x) = 2x3-4
      3.5 f(x) = 3x-2x
                   2
                 x     1
     3.6 f(x) =
4.           f      g           Z+   Z+   ff,
gg , fg, gf
   4.1 f(n) = 2n+1
            g(n) = 3n-1
   4.2       f(n) = n2
            g(n) = 2n
1.                 44                  4

2.           35                                 7

     (       ABCD           F)
3.                                     35,000
                        4
                                 999

             135

4.       6

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บทที่ 4 ฟังก์ชัน

  • 1. x (Function) F(x)
  • 2. :A B f A B f xA y,z  B (x,y)  f (x,z)  f y=z A = {1,2,3,4} B = {a,b,c} f1 = {(1,a),(2,b),(3,a)} f2 = {(1,a),(2,b),(4,c),(1,b)} 1 1 a a 2 2 b b 3 3 c c 4 4
  • 3. Df = { x | y (x,y)  f } Rf = { y | x (x,y)  f } f (x,y)  f y x y f x (1,2)  f  f(1) = 2 f(x) y = f(x) f(x) =
  • 4. (x,y1)  f (x,y2)  f y1 = y2 (x,y1)  f (x,y2) f y1  y2
  • 5. f={(x,y)RR|y = x2+1} (x,y1)f (x,y2)f y1  y2 x2+1  x2+1 x2  x 2 y1 = y2
  • 6. f={(x,y)RR|x2+y2 = 1} f (0,1)f ( 02+12= 1) (0,-1)f (02+(-1)2 = 1) 1  -1
  • 7. Rf  B f A B (f is function from A to B) f:AB Rf = B f A B (f is function from A onto B) A = {1,2,3} f BA { a,b,c} : = B f1= {(1,a),(2,a),(3,b)} f1 : A  B f2 = {(1,a),(2,b),(3,c)} f2 : A  B f3 = {(1,b),(2,c),(3,a)} f3 : A  B
  • 8. f:AB yB , x  A f(x) =y B = Rf y B y = f(x) x x A x A
  • 9. f:QQ f(x) = 3x+4 yQ y= f(x) y = 3x+4 x = (y-4)/3  Q f:NN f(x) = 3x+4 yN y = f(x) y = 3x+4 X = (y-4)/3  N y=5 x= 1/3
  • 10. f:RR f(x) = x2 yR y = f(x) y = x2 x = y R y<0 f yB xA f(x) y y f(x) y xA y = -1 xA x2  -1
  • 11. :f:AB f (one-to-one) x1,x2  A y B (x1,y)  f (x2,y)  f x1 = x 2 1-1 f:A  B x1  x2  f(x1)  f(x2) f(x1) = f(x2)  x1 = x2
  • 12. Direct method Contradiction method Contrapositive method f(x1) = f(x2) f(x1) = f(x2) x1 x2 x1 x2 x1 = x 2 x1 x2 f(x1)  f(x2)
  • 13. f(x) = 2x+1 x f(x1) = f(x2) 2x1+1 = 2x2+1 2x1 = 2x2 x1 = x 2
  • 14. x1 x2 f(x1) = f(x2) x1 f(x1) = f(x2) x2 2x1+1 = 2x2+1 2x1+1 = 2x2+1 2x1 = 2x2 2x1 = 2x2 x1 = x 2 x1 = x 2 f(x1)  f(x2)
  • 15. x1,x2  Df x1  x2 f(x1) = f(x2) f(x) = x2 x1 = 2 x2 = -2 f(x1) = f(2) = 4 f(x2) = f(-2)= 4 f(x1) = f(x2) (x1)2 = (x2)2 x1 =  x2
  • 16. : f:AB f-1 f-1 f f-1 : f(A)  A A = {1,2,3} B = {a,b,c} f = {(1,b),(2,a),(3,c)} f-1 = {(a,2),(b,1),c,3)} f f-1 a a 1 1 b b 2 2 c c 3 3
  • 17. A = {1,2,3} B = {a,b,c} f = {(1,b),(2,b),(3,c)} f-1 = {(b,1),(b,2),c,3)} f f-1 a a 1 b 1 b 2 2 c c 3 3 f-1 f
  • 18. : f:AB f f f f f ( f ) x1,x2  A yB (x1,y),(x2,y)  f ( f(x1)=f(x2)= y ) (y,x1),(y,x2)  f-1 f -1 ( f )
  • 19. f f f y B x1,x2 A (y,x1),(y,x2)  f-1 (x1,y),(x2,y)  f x1 = x 2 f-1
  • 20. 1-1 1-1 : f:AB f :BA -1 1-1 f:AB f f-1: B A f-1 f-1(y1) = f-1(y2) x x = f-1(y1) = f-1(y2) (y1,x) f-1 (y2,x)f-1 (x,y1)f (x,y2)f y1 = y 2 (f ) f-1
  • 21. f-1 ( Rf-1 = A) xA yB f(x) = y (x,y)  f (y,x)  f-1 x  Rf-1 A  f-1 Rf-1  A Rf-1 = A f-1
  • 22. : A B A B B AB : f:AB g:BC f g gof A C (gof)(x) = g(f(x)) xA gof = {(x,z) | yB (x,y)f (y,z) g}
  • 23. : f1-1A  B : 1-1 : g BC 1-1 f go : A C 1-1 1-1 1-1 f:AB g:B C gof(x1) = gof(x2) g(f(x1)) = g(f(x2)) f(x1) = f(x2) x 1 = x2
  • 24. : f:AB g:BC gof : A C f:AB g:BC zC ( C Rgof ) Rg = C z  Rg y B (y,z)  g Rf = B yRf xA (x,y)  f (x,z)  gof z  Rgof C  Rgof R o C
  • 25. : f:AB kN f k 1 (k-1) y B x k f(x) = y 0 a 0 a 1 b 1 b 2 c 2 c 3 d 3 d 2- 3- 1 1
  • 26. : f:AB k-1 n(B) A n(A)  k.n(B) f:AB k-1 n(B) x k yB A k.n(B) A n(A)  k.n(B) x1 . y1 xk y2 . . ym
  • 27. (Pigeon-Hole Principle ) n m n>m 1 2 1 2 3 4 J.P.G.L. Dirichlet
  • 28. x (floor of x) x x 4= 4 4.0000001= 4 4.9999999= 4 x(ceiling of x) x x 4 = 4 4.000000001 = 5 4.999999999 = 5
  • 29. (Generalized Pigeon-Hole Principle) : f:AB n(A) = m n(B) = m k   yB xA  k  f:AB n(A) = m n(B) = k m  yB  k  1 xA   m   k  1   A k( m  m )  k  1   k m  m  n( ) < k((  k   1m   +1) -1) = m  k    y B xA
  • 30.  89   12    = = 7.4167 =8
  • 31. 1. n2 ; n 0 1.1 f : Z  Z f(n) = -n2 ; n < 0 n+1 ; n 1.2 f : Z  Z f(n) = n3 ; n x+1 ;xQ 1.3 f : R  R f(x) = 2x ;xQ 3x+2 ; x  Q 1.4 f : R  R f(x) = 3
  • 32. x 1 1.5 f : R  R f(x) = ;x0 x x 1.6 f : R  R f(x) = 2 x 1 3x - 1 1.7 f : R  R f(x) = ;x0 x x 1 1.8 f : R  R f(x) =x  1 ;x1 2. f:RR f(x) = 2x 3 2 x  2x g:RR g(x)= 2 f=g x 1
  • 33. 3. f:RR f 3.1 f(x) = 6x-9 3.2 f(x) = 3x2-3x+1 3.3 f(x) = sin x 3.4 f(x) = 2x3-4 3.5 f(x) = 3x-2x 2 x 1 3.6 f(x) = 4. f g Z+ Z+ ff, gg , fg, gf 4.1 f(n) = 2n+1 g(n) = 3n-1 4.2 f(n) = n2 g(n) = 2n
  • 34. 1. 44 4 2. 35 7 ( ABCD F) 3. 35,000 4 999 135 4. 6