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f : D R → R

R

R

R 2

G(f ) = (x, y) R 2 ; x D , y = f (x) .

(x, y) f x D

x y f f

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0

h : R \ {0} → R

h(x) = x

|x|

− 5 − 4 . 5 − 4 − 3 . 5 − 3 − 2 . 5 − 2 − 1 . 5 − 1 − 0 .5 0 .5 1 1 . 5 2 2 . 5 3 3 . 5 4 4 . 5 5

− 2

− 1

0

1

2

x

y

x = 0

h

f (x) = 1 se x > 0

−1 se x < 0

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h

(0, −1) (0, 1)

f

p : R →R p(x) = 2 x2 −3x + 1

p

−3 3

x p(x)

−3 28

−2 15

−1 60 11 02 33 10

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q 0 0, 12

q

q 0

12 , 1 q

[0, 1]

x q (x)

0, 1 0, 0720, 2 0, 0960, 3 0, 0840, 4 0, 0480, 5 00, 6 −0, 0480, 7 −0, 0840, 8

−0, 096

0, 9 −0, 072

q q [0, 1]

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o

o

g : R →R g(x) = x4 −x3 −2x2

g(x) = x4

−x3

−2x2

= x2

(x2

−x −2).

g x1 = −1 x2 = 0 x3 = 2

g(x) < 0

−1 < x < 0 0 < x < 2

g(x) > 0 x < −1 x > 2

g

∆ x = 0, 5

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x g(x)

−1, 5 3, 9375

−1 0

−0, 5 −0, 31250 00, 5 −0, 56251 −21, 5 −2, 81252 02, 5 10, 9375

g

r : R \ {1} → R r(x) = x −2x −1

r x = 2 r(x) > 0 x < 1 x > 2 r(x) < 0

1 < x < 2 x

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x r (x)

−5 7/ 6

−4 6/ 5

−3 5/ 4

−2 4/ 3

−1 3/ 20 22 03 1/ 24 2/ 35 3/ 4

x

r (x)

1 x

−2

−1

x − 2

x − 1 = x

x = 1

r

x = 1

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∆ x = 0 , 5

∆ x = 0 , 5

g

] −1, 0[

]0, 2[ g(−1) = g(0) = g(2) = 0 g(x) < 0 ] − 1, 0[

]0, 2[

g

x = 1 x

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o

x2 > 1

x > 1

f (x) = 0 f

f (x) > 0 f (x) 0

f

x4 −x3 −2x2 < 0 ]−1, 0[ ]0, 2[ x4 −x3 −2x2 0 [−1, 2] x4 −x3 −2x2 > 0 ]− ∞, −1[ ]2, + ∞[ x4 −x3 −2x2 0 ]−∞, −1] [2, + ∞[

x3 −4x2 + 3 x 0

x R f : R → R f (x) = x3 − 4x2 + 3x

f f (x) = x (x −1) (x −3)

f x1 = −1 x2 = 0 x3 = 3 f (x) < 0 0 < x < 1

x > 3 f (x) > 0 x < 0 1 < x < 3

x3 −4x2 + 3 x 0 [0, 1] [3, + ∞[

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2x −1x −2

> 3 x R

o 2x −1 > 3x −6 x < 5

x = 2

x −2

• x −2 > 0 x > 2

2x −1x

−2

> 3 2x −1 > 3x −6 x < 5.

2 < x < 5

• x −2 < 0 x < 2

2x −1x −2

> 3 2x −1 < 3x −6 x > 5.

]2, 5[

g : R \{2} →R g(x) = 2x −1

x −2

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y = √ x

[0, + ∞[

[−1, + ∞[

[1, + ∞[ N f 1 : [0, + ∞[→ R f 2 : [1, + ∞[→ R f 3 : N → R

R

1

x

S (x) = x12 −x .

x R

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0 < x < 12

S : 0, 1

2 → Rx → x 1

2 −x

x = 14

14

0, 25 n

p : N

→ R

n → 0, 25 n

p

0, 25 n

n p N

12

π

p N

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a,b,c,d

f (x) = c sen (d x + b) + a

y = sin( x) y = sin( x) + 1 y = sin x − π

4

y = sin( x) y = 2sin( x)

y = sin x

2

y = sin( x) y = sin( x) + 1 y = sin x − π

4

y = sin( x) y = 2sin( x) y = sin x

2

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• a b

• c d

a

•

•

b

bx

b

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•

•

f, f 1 : R → R

f (x) = sen( x) f 1 (x) = sen x − π

4

x x − π

4 f 1 (x)

x − π

4 x f 1 (x)0 π

4 0π

23 π

4 1π 5 π

4 03 π

27 π

4 −12 π 9 π

4 0

c

• 1

• 0 1

•

−1

•

−1 0

d

d

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d

• 1

• 0 1

•

−1

•

−1

0

f, f 2 : R → R

f (x) = sen( x) f 2 (x) = sen x

2

x

12 x f 2 (x)

12 x x y0 0 0π

2 π 1π 2π 0

3 π

2 3π −12 π 4π 0

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f : D R →R

f x1 , x2 Dx1 < x 2 f (x1 ) < f (x2 )

f x1 , x2 Dx1 < x 2 f (x1 ) f (x2 )

f x1 , x 2 Dx1 < x 2 f (x1 ) > f (x2 )

f x1 , x2 Dx1 < x 2

f (x1 ) f (x2 )

f : D R →R

f M R f (x) M

x D

f M R f (x) M

x D

x0 D f f (x0 ) f (x)

x D

x0 D f f (x0 ) f (x)

x D

x0 D f r > 0

f (x0 ) f (x) x D ∩ ]x0 −r, x 0 + r [

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x0 D f r > 0

f (x0 ) f (x) x D ∩ ]x0 −r, x 0 + r [

h :]−1, 6] →R

h(x) = 3x −x2 se x 2

|x −4|+ 1 se x > 2

h

•

32 , 9

4

• (2, 2) (4, 1)

• (6, 3)

•• −1, 3

2 [ 4, 6 ]

•

32 , 2 ]2, 4 ]

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f : R → R

a, b R f (x) = ax + b x R

f : R → R f (x) = x x R

f : R →R f (x) = x + b

f (x) = ax

f (x) = b

f : R → R a b

f : x → ax + b x

b a

f : x → ax + b

P 1 = ( x1 , ax 1 + b)P 2 = ( x2 , ax 2 + b) P 3 = ( x3 , ax 3 + b)

x1 x2 x3 x1 < x 2 < x 3

d(P 1 , P 2 ) = (x2 −x1 )2 + a2 (x2 −x1 )2

= ( x2 −x1 )√ 1 + a2 ;d(P 2 , P 3 ) = ( x3 −x2 )√ 1 + a2 ;d(P 1 , P 3 ) = ( x3 −x1 )√ 1 + a2 .

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d(P 1 , P 3 ) = d(P 1 , P 2 ) + d(P 2 , P 3 ).

f f (x1 ) f (x2 ) x1 = x2 f

b

f : x → ax + b OY a a

OX a

a > 0 f

a < 0

OY

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f : D R → R

y = f (x) D

R

y = x3 + x2 + x y = x4 −5x2 + 4

y = x |x| y = |x2 −1| y = x + |x| y =

xx2 −1

x R

(x2 −1)2 1 x3 −2 x2 −x + 2 > 0

2x + 1x + 1

< 3.

h : R

→ R h(x) =

|x2

−1

|

h h1 (x) = h(x + 1) −2 h2 (x) = 3 h(2 x)

h3 (x) = 12 h(3 x −1) −2

R

q (x) = p(a x + b) + c a b c a b c

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N

◦ N ◦ C

N

f : R → R

f (x1 ) f (x2 ) x1 = x2

r

n

X Y X

Y

a1 , a 2 , . . . , a n f (1), f (2), . . . , f (n)

a i

f OX

x = i − 12

x = i + 12

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S = a1 + a2 + · · ·+ an

f OX

x = 12

x = n + 12

S = a 1 + a n

2 n

f : R

→R f (x) =

1

x2

+ 1 f : [−1, 2[→R f (x) = |x| f :]−1, 1] →R f (x) =

x + 1 se x < 0x −1 se x 0

f : [0, 4] →R f (x) = 3x se x < 1

x2 −6x + 8 se x 1

f : [0, 5] →R f (x) = 3x se x < 5

x2

−6x + 8 se x 4

g : [0, 5] →R

g(x) = 4x −x2 , se x < 3

x −2, se x 3

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g(x) = −1 g(x) = 0 g(x) = 3

g(x) = 4 g(x) < 3 g(x) 3

f : R → R g : R → R

f f

f f

f

f f

f x0 f

f 2

x0 f 2

f f g f ◦ g

f g f ·g f A R B R f

A B R

f : D R → R f f (x) =f (−x) x D f f (x) =

−f (−x) x D

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f g

f + g f ·g

f g

f + g f ·g

f g

f + g f ·g

h

]2, 4 ] [ 4, 6 ]

f : [−2, 3] →R

h : D R →R

f D

h

h(x) = |f (x)| h(x) = f (

|x

|)

h(x) = ( f (x)) 2

h(x) = 1f (x)

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D

N

C H

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y = a x 2 + b x + c

y = a x 2 + b x + c

= a x2 + ba

x+ + c

= a x2

+ ba x +

b2

4a2 − b2

4a + c

= a x + b2a

2

+ 4ac −b2

4a .

y = a (x −x0 )2 + y0

x0 = − b2a

y0 = 4ac −b2

4a = −

∆4a

y = a x 2 x0 y0

a

a

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xv = − b2a

yv = −∆4a

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f f (x) = ax + b

a f

f

x f (x)

a = 0 f (x) = ax + b

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4a