Математика са збирком задатака за 4. разред гимназије - 24077 by Zavod za udžbenike

MATEMATIKA SA ZBIRKOM ZADATAKA

za IV razred gimnazije

y = f (x)

F (x,y )=0

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

N0 N ∪{0}

Q R R+

R+ 0 R+ ∪{0}

a,b ∈ R

(a,b)= {x | a<x<b}, [a,b)= {x | a x<b},

(−∞,a)= {x | x<a},

(a,b]= {x | a<x b}, [a,b]= {x | a x b}, (a, +∞)= {x | x>a}, (−∞, +∞)= R

a ∈ R ε ∈ R+ (a ε,a + ε) ε a

a,b ∈ R a = b (a,b) (2, 3)=3

a = b (a,b)= a(= b) (4, 4)=4

n a1 ,a2 ,...,an (a1 ,a2 ,...,an ) (a1 ,a2 ,...,an ) 1 k n ak

⊂ R

(an ) ε

(an )

→∞ an = a.

ε (an ) (an ) M> 0 (an ) M (an ) +∞ n→∞ an =+∞. P< 0 (an )

(an )

→∞ an = −∞. +∞

a (an ) ε> 0 n0 |an a| <ε n>n0

(an ) +∞ M> 0 n0

an >M n>n0

(an ) −∞ P< 0 n0

an n0 a (an ) ε> 0 n0 ∈ N n>n0 |an a| <ε

n→∞ an = n→∞ bn = a

an xn bn n ∈ N (xn ) n→∞ xn = a

n→∞ an = a n→∞ bn = b an bn n ∈ N a b

n→∞ an = a n→∞ bn = b

n→∞ can = ca (c ∈ R)

n→∞(an + bn )= a + b n→∞(an bn )= a b

n→∞(an bn )= ab n→∞ an bn = a b , (b =0)

(an ) an <an+1 n =1, 2,... (an ) an an+1 n =1, 2,... ) an+1 <an n =1, 2,... an+1 an n =1, 2,...

A B A B f x A y B x ∈ A y ∈ B y f (x) y = f (x) A B f f A B f : A → B f : A → B f (A) A

f (A)= {y | y ∈ f (A),x ∈ A}. f (A) ⊂ B f (A)= B f A B f 1 1 f : A → B x1 x2 ∈ A x1 = x2 ⇒ f (x1 ) = f (x2 ) x1 ,x2 ∈ A ,

f (x1 )= f (x2 ) ⇒ x1 = x2 x1 ,x2 ∈ A , f 1 1 1 1

A B

(x,y ) x ∈ A y ∈ B A B A × B

A × B = {(x,y ) | x ∈ A,y ∈ B }

f A B A × B x ∈ A f

(x,y ) x (x,y ) ∈ f y = f (x) A × B A B A = B A

A = {1, 2} B = {a,b}

A × B = {(1,a), (1,b), (2,a), (2,b)}

A × B 24 =16 A B

f1 = {(1,a), (2,a)},f2 = {(1,a), (2,b)},f3 = {(1,b), (2,a)},f4 = {(1,b), (2,b)}. f2 f3 1 1

A = {1, 2, 3} B = {a,b} A B

A × B = {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)};

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

f3 = {(1,a), (2,b), (3,a)},f4 = {(1,b), (2,a), (3,a)},

f5 = {(1,a), (2,b), (3,b)},f6 = {(1,b), (2,a), (3,b)},

f7 = {(1,b), (2,b), (3,a)},f8 = {(1,b), (2,b), (3,b)}.

f2 f3 f4 f5 f6 f7 1 1

A = {1, 2} B = {a,b,c}

A × B = {(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)} A B

f1 = {(1,a), (2,a)},f2 = {(1,a), (2,b)},f3 = {(1,a), (2,c)},

f4 = {(1,b), (2,a)},f5 = {(1,b), (2,b)},f6 = {(1,b), (2,c)},

f7 = {(1,c), (2,a)},f8 = {(1,c), (2,b)},f9 = {(1,c), (2,c)}.

f2 f3 f4 f6 f7 f8 1 1

A = {1, 2, 3} B = {a,b,c} i A B

(x,y )

{x,y } = {y,x} {x,x} = {x}

(a,b)=(c,d) a = c b = d

ii A × B

iii A B

iv A × B

v A B

vi A B 1 1

vii 1 1 A B A B A n B m

i m n A B

ii n>m 1 1 A B

iii n<m A B

iv 1 1 n = m n!

f R R f = {(x, 3x +5) | x ∈ R}

f x ∈ R 3x +5 ∈ R 1 1 R R R × R

f = {(x,x 2 ) | x ∈ R} R R 1 1 f R+ 0 R+ 0

f = {(x,x 2 ) | x ∈ R+ 0 } 1 1

i f : A → B Åx1 x2 x3 ... y1 y2 y3 ...ã , x1 x2 x3 ,... A y1 y2 y3 ,...

A = {1, 2} B = {a,b} A

f1 = 12 aa ,f2 = 12 ab ,f3 = 12 ba ,f4 = 12 bb A = {1, 2, 3} 3!=6 A 123 123 , 123 132 , 123 213 , 123 231 , 123 312 , 123 321 .

ii f N f

123 ...

y1 y2 y3 ...ã

(y1 ,y2 ,y3 ,... ) 1 2 3 ... N (y1 ,y2 ,y3 ,... )

(yn ) (yn )=(y1 ,y2 ,y3 ,... )

f : N → R f (n)= 1 n

123 1 1 2 1 3 å , 1, 1 2 , 1 3 ,... 1 n . iii f f (x) f f : A → B f (x) x

f f x f (x) ∈ B

f (x) x x f (x) f (x) f f (x)

f = {(x,f (x)) | x ∈ A}.

x → f (x) x ∈ A f (x) ∈ B

f = {(x,x 2 ) | x ∈ R} (1) x x 2 1 f

f (x)= x 2 x ∈ R x → x 2 x ∈ R

f = {(x, 2x ) | x ∈ R} R R+ f (x)=2x

a> 0 a =1

a : R → R

a = {(x, a x) | x ∈ R}. a x → a x x ∈ R+

f (x)= a x A B f R A = N B = R

f (i) A ⊂ R (ii) B ⊂ R (iii) x ∈ A f (x) ∈ B

f1 : A1 → B1 f2 : A2 → B2

i A1 = A2 B1 = B2

ii f1 (x)= f2 (x) x ∈ A1 (= A2 ) f1 :[0, +∞) → [0, +∞) f2 : R → [0, +∞) x x 2

1 f2

f2 f2 ( 1)= f2 (1)=1 x = 1,x> 0 0,x =0 1,x< 0 f : R → [0, +∞) g : R → [0, +∞) f (x)= x x; g (x)= |x| f (x)= x,x 0 x,x< 0 , f (x)= g (x) x ∈ R f = g

(x)= x 2

f (x)= a x f (x)= x 2 f (x)= a x f (x)= √1 x A = {x | x 1} B =[0, +∞) f : A → B A B R f ∗ A0 A f ∗ (x)= f (x) x ∈ A0 A0 f ∗ A f A0 f ∗ f A0 f (x)= 3 x 2 g (x)=2 3 x f R \{0} g R+ f = g x> 0 x ∈ R+ 3 x 2 =2 3 x x ∈ R+ f (x)= g (x) g f R

x,y )

)

(x,y ) P (x,y )

(x,y )

=(x,y ) (1, 2) ( 2, 3) (0, 0) (x,y )

= a x = a y = b y x =0 x y =0 (x,y ) x> 0 y y x 0 y 0

II III IV I = {(x,y ) | x> 0,y> 0},II = {(x,y ) | x< 0,y> 0}, III = {(x,y ) | x< 0,y< 0},IV = {(x,y ) | x> 0,y< 0}.

Gf f (x,y )

y = f (x)

(x,y ) ∈ Gf ⇔ y = f (x) y = f (x)

Gf f (x)= x 2

y = x 2 f (x)= x x x [x] f (x)=[x] f

y = f (x)

GJ J (x,y )=0 x y (x,y ) ∈ GJ ⇔ J (x,y )=0, y x = a b1 b2 a b1 b2 1 1 y = a y 2 = x x = a> 0 (x,y ) 1 x 2 +y 2 =1 x 2 + y 2 =1 x = a 1 <a< 1

f (x)= x 2 f (x)= x 3 ,...

f (x)=2x f (x)= ex ,...

f (x)= 2 x f (x)= 10 x,...

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

f (x)= x 2 + xex ,f (x)= x √1+ x,f (x)= x √1 x2 f g

F1 (x)= f (x)+ g (x),F2 (x)= f (x) g (x), F3 (x)= f (x) · g (x),F4 (x)= f (x) g (x) f g

f + g f g fg f g

f + g f g fg x f g

Df +g = Df g = Dfg = Df ∩ Dg .

g (x)=0

f g x ∈ Dg

Df /g = Df ∩ Dg \{x | g (x)=0}

Df ∗g ⊂ Df Df ∗g ⊂ Dg ∗ f f (x)= x 2 g (x)= √x

x → x 2 + √x,x → x 2 √x,x → x 2 √x,x → x 2 √x

Df = R Dg =[0, +∞)

Df +g = Df g = Dfg = R ∩ [0, +∞)=[0, +∞)

g (x)=0 √x =0 x =0

Df /g = R ∩ [0, +∞) \{0} =(0, +∞)

f (x)= x + √x g (x)= √x

Df = Dg = R+ 0 F f g

f (x)+ g (x)= x + √x √x = x

F (x)= x

F (x)= x R

f + g Df ∩ Dg = R+ 0 ∩ R+ 0 = R+ 0 f g

F (x)= x (x 0)

F (x)= x;

f (x) g (x) x → f (x)+ g (x) f + g Df Dg f g

f (x)= x 1 x g (x)= √x

f (x)= √x g (x)= 1 x

f (x)=2x g (x)= 1 x +1

f (x)= x g (x)= x 2 1

f (x)= x 2 1 g (x)= 1 x

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

f (x)= 2 x g (x)= x 2 1

f (x)= x g (x)= 2 x

A B C R f : B → C g : A → B F : A → C

F (x)= f (g (x)) f g F f g f ◦ g (f ◦ g )(x)= f (g (x)). f g Df Dg f ◦ g x ∈

Dg g (x) ∈ Df f g

f (x)= √x g (x)= x (f ◦ g )(x)= f (g (x))= √ x.

Dg = R Df =[0, +∞) f ◦ g x ∈ R x ∈ [0, +∞) x 0 Df ◦g = {x | x ∈ [2kπ, (2k +1)π ],k ∈ Z}

f (x)= 2 x g (x)=2x x 2 6 (f ◦ g )(x)= f (g (x))= 2 (2x x 2 6)

Dg = R Df =(0, +∞) f ◦ g x ∈ R

2x x 2 6 > 0 x ∈ R 2x x 2 6= (x 1)2 +5 < 0,

Df ◦g = ∅

f (x)= x 2 g (x)= √x 1

f (g (x))= f (√x 1)=(√x 1)2 = x 1, (1) f ◦ g Df = R Dg = {x | x 1} Df ◦g x ∈ Dg √x 1 ∈ R Df ◦g = Dg = {x | x 1} 1 R (f ◦ g )(x)= x 1(x 1) f (g (x))

x → f (g (x)) f ◦ g f g Df +g ⊂ Df

Df +g ⊂ Dg f g Df Dg Df ◦g f ◦ g x ∈ Dg g (x) ∈ Df Df ◦g ⊂ Dg Df ◦g Df f (x)= 1 x ,g (x)= x 2 +1,

Df ⊂ Df ◦g

Df ◦g ⊂ Df

f (x)= 2 x,g (x)= x 2 ,

f ◦ g

f (x)= x 2 g (x)= x f (x)= x g (x)= x 2

f (x)= √x g (x)= x f (x)= x g (x)= √x

f (x)= x 2 4 g (x)= √x f (x)= √x g (x)= x 2 4

f (x)= x 2 4 g (x)= 3 x f (x)= 3 x g (x)= x 2 4 f

f (x)= x3 +7x 8 f (x)= x √x f (x)= 5x 2 10 x2 3x +5

f (x)= x 2 9 x2 +1 f (x)= x 2 4 x2 1 f (x)= 3x +10 2 (8 2x x2 )

f (x)= 2x 3 √5x 7 f (x)= 3x 2 4x +7 2x2 + x +6 f (x)= 4 (x 2 +5)

f (x)= 2 (x 2 4x +4) f (x)= (2x 2 x 10) x 4 f (x)= √13 2x 3 (5x 3)

f (x)= x 2 x 6 2 (5 x) f (x)=2x 5 x f (x)=2 21/x

f (x)= 3√x+3 4 (x2 8) f (x)= 2√x2 1 2 |x2 1| f (x)= 3 x

f (x)=2 1 x f (x)= x 1 x f (x)= 3 x

f (x)= 3 | x|

f (x)= x 2 3x +1

i f (t) ii f 1 t iii f (x)+ f (t +1)

iv f (2t) v f (x 2 +1) vi f (t2 )+1

f (x)= x 1 3x f (x +1) f 1 x f x + 1 x

f (x 3)= 2x 5 x +1 f (x)

f 1+ 1 x = x 2 2 x +1 f (x)

f (x)= 2x +1 x 2 f (f (x)) f (f (f (x)))

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

f (x)= x 3 g (x)=2x f (g (x)) g (f (x))

a> 0 f (x)= a x + a x f (x + y )+ f (x y )= f (x)f (y )

f (x)+ f (y )= f (z ) z

i f (x)= 1 x ii f (x)= e 1+ x 1 x

f Ä 1 x ä

i f (x)= x + 1 x ii f (x)= x 2 + x + 1 x + 1 x2 iii f (x)= g (x)+ g Ä 1 x ä f

f (x)= 1, 2 x 0 x, 0 <x< 1 1, 1 x 4

a ∈ [ 2, 4] a

i f (a)=0 ii f (a)= 1 2 iii f (a)=1

f Ä 1 x ä x → f Ä 1 x ä

f (x)= f Ä 1 x ä A B R f 1 1 A B f y ∈ B x ∈ A f (x)= y f x f 1 1

g : B → A y ∈ B g (y )= x, x A y = f (x) g f f 1 f A B B f 1 : B → A f A f (A) ⊂ B f 1 : f (A) → A f 1 f (A) A

f (x) ∈ f (A) x ∈ A x ∈ A f 1 Äf (x)ä = x. (1) f 1 f f f 1

y ∈ f (A) f Äf 1 (y )ä = y. (2) f : A → B f 1 f 1 1 f 1 2

f : A → B f 1 1 f (x) = x f 1 2 f (x) = x x ∈ A,

f 1 1 (y )= f 1 2 (y ) y ∈ B,

f 1 1 = f 1 2 1 2

y = f (x) ⇔ x = f 1 (y ). (3) 1 y = f (x) f 1 (y )= x 2

x = f 1 (y ) f (x)= y f 1 f f x f (x) y = f (x) x 3 x = f 1 (y ) f 1 f f (x)=2x 1 1 1 R R y =2x 1 x x = 1 2 y + 1 2 f 1 f

f 1 (x)= 1 2 x + 1 2 f f 1

y = x (a,b) f b = f (a)

a = f 1 (b) (b,a) f 1 (a,b) (b,a) y = x f f 1 f f 1 1

f : R → R+

f 1 : R+ → R f 1 (x)= e x

f f (x)= 1 x

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

f (x)= e x

A = R \{0} A

y = 1 x x = 1 y

f (x)= x 2 R R+ 0 1 1 2 2 g A0 =(−∞, 1) ∪ [0, 1] 1 1 A0 R+ 0 g 1 : R+ 0 → A0

g 1 (x)= ß√x, 0 x 1 √x, 1 <x . g g 1

f (x)=3x +2

(x)= x +3 2x

(x)= 4x +2 x 3 f (x)= x 2

(x)= x 2 x 0

(x)= 2 (x 1) f (x)= 3 x 2 x

f (x)= −| x|

f (x)= | x|

f (x)= x 2 +3x 28

f (x)=3x 2 x 3

f (x)= x 2 +1

f (x)=2 x

f (x)=5x 2 +13x 6

f (x)=5+ x

f (x)= 2x +4 x 3 f (x)= x 2 2x2 +1 0 f (x) > 0 f f (x) < 0 f f (x) 0 f (x) 0

: A → R x ∈

(x) f (x) < 0 f (x)=0 f (x) > 0

f (x) < 0 3 <x< 1 2 A = R,A+ =(−∞, 3) ∪ 1 2 , +∞ ,A = 3, 1 2 ,A0 = 3, 1 2 . a 0 <a< 1 A = R A+ =(0, 1) A0 = {1} A = (1, +∞) A = R A+ = {x | 2kπ<x< (2k +1)π,k ∈ Z} A0 = {x | x = kπ,k ∈ Z} A = {x | (2k 1)π<x< 2kπ,k ∈ Z} A A+ A A0

f (x)=5x 2 + x 4

f (x)= x(x 1)(x 2)

f (x)= x 2 4 x2 1

f (x)= 4+4x 3x 2 x2 4x 5

f (x)= 3x +1 x 3

f (x)= x(x +2) x 1

f (x)= x 2 +2x 15 2x2 3x 2

f (x)= 3x 5 √5x 7

f (x)= 3 x √2x +4 f (x)= 2 (x +3)

f : A → R x ∈ A

f (x)= f ( x).

f : A → R x ∈ A f (x)= f ( x). (1) A x ∈ A x ∈ A y 0 ∈ A 1 f (0)=0

f (x)= ax 2 + b a,b ∈ R R x ∈ R a( x)2 + b = ax 2 + b a b x → ax 3 + bx R x ∈ R a( x)3 + b( x)= ax 3 bx = (ax 3 + bx)

f : R → R x →|f (x)|

f (x)= |x|

f (x)= 2 x

f (x)= x 4 x 6

f (x)=2x x

f (x)=2 x (1+2x)2

f (x)= 16x 1 4x

f (x2 )

f (x)=2x 5 + x 3 4x

f (x)= | x|

f (x)= x + x

f (x)= x 4 + 2 x x

f (x)= a (x + 1+ x2 )

f (x)= x a 1+ x + x 2 1 x + x2

f :(a,b) → R x1 <x2 x1 x2 (a,b) f (x1 ) <f (x2 ) f (a,b) x1 <x2 f (x1 ) f (x2 ) f (a,b) x1 <x2 f (x1 ) >f (x2 ) f (x1 )

(xn ) xn <xn+1 n ∈ N f (x)= 2x 1 5 x x1 <x2 f (x1 ) f (x2 )= 2x1 1 5 x1 2x2 1 5 x2 = 9(x1 x2 ) (5 x1 )(5 x2 ) x1 x2 ∈ (5, +∞) x1 > 5 x2 > 5 5 x1 > 0 5 x2 > 0 (5 x1 )(5 x2 ) > 0 x1 x2 < 0 f (x1 ) f (x2 ) < 0 f (x1 ) <f (x2 ) f x ∈ (5, +∞) x1 x2 ∈ (−∞, 5) x1 < 5 x2 < 5 5 x1 > 0 5 x2 > 0 (5 x1 )(5 x2 ) > 0 f f (x)= 2x 1 5 x x< 5

5 <x< +∞ x =5 < f

f (x)=4x 2 7x +3

f (x1 ) f (x2 )=4x 2 1 7x1 +3 4x 2 2 +7x2 3 =(4x1 +4x2 7)(x1 x2 ).

x1 <x2 x1 ,x2 ∈ Ä 7 8 , +∞ä x1 > 7 8 x2 > 7 8

4x1 +4x2 7 > 4 7 8 +4 7 8 7=0 4x1 +4x2 7 > 0 x1 x2 < 0 f (x1 ) f (x2 ) < 0 f (x1 ) <f (x2 ) f Ä 7 8 , +∞ä x1 <x2 x1 ,x2 ∈ Ä−∞, 7 8 ä x1 < 7 8 x2 < 7 8

4x1 +4x2 7 < 4 7 8 +4 7 8 7=0 4x1 +4x2 7 < 0 x1 x2 < 0 f (x1 ) f (x2 ) > 0 f (x1 ) >f (x2 ) Ä−∞, 7 8 ä f x = 1,x> 0 0,x =0 1,x< 0 R x1 <x2 x1 = x2 x1 ,x2 ∈ (−∞, 0) x1 ,x2 ∈ (0, +∞); x1 < x2 x1 ∈ (−∞, 0) x2 ∈ [0, +∞), x1 ∈ (−∞, 0] x2 ∈ (0, +∞) x1 x2 ∈ R x1 <x2 x1 x2 .

⊂ R B ⊂ R f 1 B A B f f f x1 x2 ∈ A x1 = x2 x1 <x2 f f (x1 ) <f (x2 ) f (x1 ) = f (x2 ) f 1 1 f f 1 : B → A f 1 B x1 x2 ∈ A f A B y1 y2 ∈ B y1 = f (x1 ) y2 = f (x2 ) f A x1 <x2 ⇒ f (x1 ) <f (x2 ), x1 x2 ⇒ f (x1 ) f (x2 ), f (x1 ) >f (x2 ) ⇒ x1 >x2 (1) y1 = f (x1 ) y2 = f (x2 ) x1 = f 1 (y1 ) x2 = f 1 (y2 ) 1 y1 >y2 ⇒ f 1 (y1 ) >f 1 (y2 ), f 1 B

f (x)=2x 3 f (x)=4 3x f (x)= x 2 +1

f (x)=2x 2 8x +5 f (x)= x 4 2x +1 f (x)=2x

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

f (x)= 1/2 x

:[a,b] → R x1 x2 ∈ [a,b]

a,b] x1 <x2

,f (x

1 A2 A A

(x1 )+ f (x2 )

1 =(x1 , 0) M2 =(x2 , 0) A1 =(x1 ,f (x1 ))

f f (x)= ax 2 + bx + c a,b,c ∈ R

a b c

i f R

ii f R f I

f (x)= x 3 I = R+

f (x)= √x I = R+

f (x)= e x I = R

f (x)= x + e x I = R f : A → R p

f (x + p)= f (x) x ∈ A. (1) f 1

f (x)= f (x ± p)= f (x ± 2p)= ··· f

(pn )

f (x)= f (x + pn ), (2) x ∈ A n ∈ N A x ∈ A x + kp ∈ A k ∈ Z

f : A → R 2 f f p fp [0,p) A [0,p) ⊂ A [0,p) p [0,p)

[mp, (m +1)p) m ∈ Z m =0

f f (x)= x [x] p =1 [x +1]=[x]+1 x ∈ R

f (x +1)= x +1 [x +1]= x +1 ([x]+1)= x [x]= f (x).

f : R → [0, 1] f (x)= x 2 x ∈

[ 1, 1] f (x +2)= f (x) f 2

f (x)= ß 1, 2k x< 2k +1 1, 2k +1 x< 2k +2 (k ∈ Z)

f (x +2)= ß 1, 2k x +2 < 2k +1 1, 2k +1 x +2 < 2k +2 (k ∈ Z) = ß 1, 2k 2 x< 2k 1 1, 2k 1 x< 2k (k ∈ Z) = ß 1, 2m x< 2m +1 1, 2m +1 x< 2m +2 (m ∈ Z) = f (x) f

f (x)= 2x f (x)= Ä x 2 π 3 ä f (x)=3 2 3 x +1

f (x)=1 x f (x)= 2 x f (x)=2 3 2 x

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

f (x)= 2 x + 2 x f (x)= 4 x + 4 x f (x)= 2 ( x)

ax a> 0 x x

(i) a n an

a 1 = a,an+1 = an · a (n ∈ N) a =0

a 0 =1,a n = 1

an (n ∈ N),

an n a =0

(ii) a n

f (x)= xn x ∈ R xn ∈ R n

f (x)= xn x =0 R+ 0

1 1 R R+ 0 x ∈ (−∞, 0) x ∈ (0, +∞) n

f (x)= xn x> 0 x< 0 R 1 1 R R R

(x,xn )

f (x)= xn

f (x)= xn [0, +∞) x 0 n R+ 0 R+ 0

1 1 R+ 0 R+ 0 R+ 0

f (x)= x n x 0

R+ 0 R+ 0 x 0 x n 0

f (x)= x n a b bn = a

f (x)= x n R+ 0 R+ 0

(iii) f (x)= xn 1 1 R+ 0 R+ 0 a n

b bn = a b n √a n a

bn = a ⇔ b = n √a (a 0,b 0,n ∈ N). n a a 0 n √0=0 a> 0 ⇒ n √a> 0 a< 0 n n

a< 0 n

n √a = n √ a n √a a< 0 n √ a a> 0 (iv ) a x m n x = m/n ax = a m n = n √am m/n = p/q am/n = ap/q

(v ) a> 0 x ax

f (x)= ax Q R ax x a =2

= Mx . x> 2π k

x k ( 7) x x 1 x 1 x Mx k x → x f (x)= x R [ 1, 1] x x (x, x)

f (x)= x

f (x)= xn n ∈ N

f (x)= ax a> 0

f (x)= x R [ 1, 1] X ∈ [ 1, 1] x ∈ R x = X

f (x)= x 1 1 R [ 1, 1] x +2kπ k ∈ Z Mx (= Mx+2kπ ) k x = (x +2kπ ) f f (x)= a a

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

f + g,f g,f · g, f g ,f ◦ g,f 1 R R f

i f (x)= x + x ii f (x)= x 2 x iii f (x)=( e x) 3

iv f (x)=6+ 5 e x v f (x)= x 4 5 x

vi f (x)= x + √x + 2x f

i f (x)= x ii f (x)=[x] x

i f (x)= xn n ∈ N f (x)= an xn + an 1 xn 1 + ··· + a1 x + a0 , n ∈ N a0 a1 ... an R

f (x)= am xm + am 1 xm 1 + ··· + a1 x + a0 bn xn + bn 1 xn 1 + + b1 x + b0 , (1) m n a0 a1 ... am b0 b1 ... bn 1 x

bn xn + bn 1 xn 1 + + b1 x + b0 =0 iii n f (x)= xn x 0 R+ 0 R+ 0 f 1 : R+ 0 → R+ 0 f 1 (x)= n √x. (2) x 0 f (f 1 (x))= f ( n √x)=( n √x)n = x, f 1 (f (x))= f 1 (xn )= n √xn = x, 2 n ( n √x)n =

√xn = x, x 0 x< 0 n =2 x< 0 (√x)2 √x2 x ∈ R x< 0 √x2 = x ( 2)2 = ( 2)=2 f (x)=(√x)2 R+ 0 f (x)= √x2 R R+ 0 f (x)= n √x n f (x)= n √xn n

iv n f (x)= xn R

R R f 1 : R → R

f 1 (x)= n √x. (3) x 0 iii x< 0

f (f 1 (x))= f Ä n √xä = f ( n √ x) n √x = n √ x x< 0 =( 1)n ( n √ x)n

= ( n √ x)n ( 1)n = 1 n

= ( x)= x. x< 0

f 1 (f (x))= f 1 Äxn ä = n √xn = n √ xn x< 0 ⇒ xn < 0 = n »( x)n xn =( x)n n = ( x)= x. 3 n x ∈ R n √x n = n √xn = x

f (x)= n √x n f (x)= n √xn f (x)= x

v f (x)= a

f (x)= x f (x)= 3 √x 2 2+ √x + x2 ,f (x)= 1+ x3 x √x , f (x)= √x +2 3 √x 2+ 5 √x2 x + x2 8 √x3 ,

f (x)= √x f (x)= x 2 R+ 0

vi f (x)= ex R R+ R f 1 R+ R R+ f 1 (x)= x x ∈ R+ ex = x (x ∈ R),e x = x (x ∈ R+ ).

a> 0 a =1 f (x)= ax ax = ex a R R+ f 1 (x)= a x R+ R a ax = x (x ∈ R),a a x = x (x ∈ R+ )

a> 1 f (x)= ax R f 1 (x)= a x R+ 0 <a< 1

f (x)= ax R f 1 (x)= a x R+ a x x a a = e e x x x x

vii a x> 0 xa xa = ea x .

f (x)= ( x) [ 1, 1]

f (x)= ( x) R 2π

f (x)= x x f (x)= x R [ 1, 1] 1 1

[0,π ] [0,π ] 1 1

[ 1, 1] f 1 : [ 1, 1] → [0,π ]

( x)= x (x ∈ [0,π ]) ( x)= x (x ∈ [ 1, 1]);

y = x x = y (x ∈ [ 1, 1],y ∈ [0,π ])

( x) x ∈ [ 1, 1] ( x) x ∈ R

( x)= ß(k +1)π x,k x kπ,k ; kπ x (k +1)π.

f (x)= ( x) [ 1, 1]

f (x)= ( x) R 2π

(x)= ( x)

f (x)=2x 2 x f1 f2 f3 f4

f1 (x)= xf (x) f2 (x)= f (2x) f3 (x)= f (x 1) f4 (x)=3f (2x +1) f

i f (x)= ® x 2 ,x< 0 3x,x 0 ii f (x)= ⎧

4 x,x< 1 5, 1 x 0 x 2 +5,x> 0

iii f (x)= 1 2 x iv f (x)= |x|

v f (x)= 1 2 ( x) vi f (x)= x + | x| f

i f (x)= Ä x 2 1ä ii f (x)= (3x 6)

iii f (x)= (2x 5) iv f (x)= √4x 3

v f (x)= ( x) vi f (x)= (x 2 +2)

i ( (3 √2)) ii 0+ 1 √2 + √3 2 + 1

iii 2 1 5 5 12 iv 6 ( 1) 12 √3 2 +5 1 √2

i x + x = π 2 ii x = ® 1 x2 , 0 x 1 π 1 x2 , 1 x< 0

ii +∞ A M ∈ R x ∈ A

x ∈ (M, +∞) x>M A iii ii i ii iii A ⊆ R x0 A

x0 +∞ −∞ f : A → R y0 y0 +∞ −∞ x0 (xn ) xn ∈ A n→∞ xn = x0 n→+∞ f (xn )= y0 x→x0 f (x)= y0 .

y0 x0 (x1 ,x2 ,...,xn ,... ) x0 +∞ −∞ f (x1 ),f (x2 ),...,f (xn ),... y0 +∞ −∞ (xn ) x0

(f (xn )) y0 A ⊆ R x0 ∈ R A f : A → R y0 ∈ R x0

ε> 0 δ> 0 x ∈ A \{x0 } |x x0 | <δ (2) |f (x) y0 | <ε. x ∈

(x0 δ,x0 + δ ) x ∈ A \{x0 }

f (x) ∈ (y0 ε,y0 + ε) f ε> 0 δ> 0 ε

x0 ∈ R y0 ∈ R

x0 <x<x0 + δ f x→x0 +0 f (x)= y0

x0 δ<x<x0 , f x→x0 0 f (x)= y0 .

→x0 f (x)= y0 x→x0 +0 f (x)= y0 ∧ x→x0 0 f (x)= y0 . x0 =0

x→x0 f (x)=+∞, x→−∞ f (x)= y0 , x→+∞ f (x)=+∞, x0 y0 Df f i x→x0 f (x)=+∞ M> 0 δ> 0 x ∈ Df |x x0 | <δ f (x) >M ii x→x0 f (x)= −∞ M< 0 δ> 0 x ∈ Df |x x0 | <δ f (x) <M iii x→+∞ f (x)= y0 ε> 0 M> 0 x ∈ Df x>M |f (x) y0 | <ε iv x→−∞ f (x)= y0 ε> 0 M< 0 x ∈ Df x<M |f (x) y0 | <ε v x→+∞ f (x)=+∞ L> 0 M> 0 x ∈ Df x>M f (x) >L vi x→+∞ f (x)= −∞ L< 0 M> 0

x ∈ Df x>M f (x) <L

vii x→−∞ f (x)=+∞ L> 0 M< 0

x ∈ Df x<M f (x) >L

(

)=4

0 <a< 1

→−∞

x =+∞ 0 <a< 1 x→+∞ x x =0 i f x → x0 x0

x→x0 f1 (x)= x→x0 f2 (x)= y0 x0 f1 (x) f (x) f2 (x) x→x0 f (x)= y0

x→x0 f1 (x)= y0 x→x0 f2 (x)= Y0 x0 f1 (x) f2 (x) y0 Y0

x→x0 f1 (x)= y0 x→x0 f2 (x)= Y0 x→x0 cf1 (x)= cy0 c x→x0 (f1 (x)+ f2 (x))= y0 + Y0

x→x0 (f1 (x) f2 (x))= y0 Y0

x→x0 (f1 (x)f2 (x))= y0 Y0 x→x0 f1 (x) f2 (x) = y0 Y0 Y0 =0

(11)

(12)

(13)

(14)

(15)

(16)

y = ax + b a =0 f

f f (x)= e 1 x x =0 x→0+ f (x)=+∞ x→0 f (x)=0 0 f (x)= x 2 +1 x2 1 y =1 x =1 x = 1 f f (x)= x 2 √x2 +1 y = 1 y =1 x→+∞ f (x)=1 x→−∞ f (x)= 1 f (x)= (e x +1) (e x +1)= e x (1+ e x )= x + (1+ e x ) x→+∞ (1+ e x )=0; f f (x)= x + α(x) x→+∞ α(x)=0 y = x x → x2 +1 y = x y = x x→+∞ √x2 +1 x =1 x→+∞ Ä x2 +1 xä =0, x→−∞ √x2 +1 x = 1 x→−∞ Ä x2 +1+ xä =0. f (x)= x 4 (1+ x)3 x = 1 y = x 3 x→±∞ f (x) x = x→±∞ x 3 (1+ x)3 =1 x→±∞ (f (x) x)= x→±∞ x 4 x(1+ x)3 (1+ x)3 = 3.

f (x)= 3 2x2 x3 3 2x2 x3 = x 3 …1 2 x = x 1+ 2 x 1/3 . t→0 (1+ t)a 1 t = a (1+ t)a 1 t = a + α(t), t→0 α(t)=0, (1+ t)a =1+ at + tα(t), t→0 α(t)=0.

2 x 1/3 =1+ 1 3

α 2 x =0, 3 2x2 x3 = x + 2 3 +2α 2 x , x→±∞ 2α 2 x =0,

x , x

y = x + 2 3 f f (x)= (x +1)3 (x 1)2 x =1 y = x +5

(x) x = x

(x +1)3 x(x 1)2 =1,

(f (x) x)= x→±∞ (x +1)3 x(x 1)2 (x 1)2 =5. (x +1)3 (x 1)2 (x 3 +3x 2 +3x +1):(x 2 2x +1)= x +5 x 3 2x 2 + x 5x 2 +2x +1 5x 2 10x +5 12x 4 (x +1)3 (x 1)2 = x +5+ α(x), x

α(x)= x→±∞ 12x 4 x2 2x +1 =0, f y = x +5

f (x)= 2x x 1 f (x)= 3x 2 4x f (x)= x 5 2 x f (x)= x 2 +4 x2 4 f (x)= 4x 2 2 x2 9 f (x)= 2x 2 4 x2

f (x)= x 2 +3x 4 x 1 f (x)= 2x 2 x 1 x f (x)= 2x x2 25

f (x)= 3x +2 2x2 +5x 3 f (x)= 4x2 3x f (x)= x2 4

f (x)= x2 4x +3 f (x)= e 1 x f (x)= 2 x

f (x)= x x

f : A → R A ⊆ R x0 ∈ A

→x0 f (x) f (x0 ) x→x0 f (x)= f (x0 ) x0 ∈ A ⊂ R f : A → R f x0 x→x0 f (x)= f (x0 ). (1) f x0 f x0 f : A → R x ∈ A A 1 x0 f

x0 xã , f 1 x x0 f (x) f (x0 ) x0 ∈ A f A

)= 1

f (x)= x x x =0 f (0) g g (x)= x x ,x =0 2,x =0 x =0 g (0) x→0 g (x) g (0)= 2 = x→0 g (x)=1 h h(x)= x x ,x =0 1,x =0 x =0 h(0)= x→0 h(x) f g h 0

f (a)

a =1 f (x)= ®x 2 ,x< 1 2 x,x> 1

a =0 f (x)= ®2x ,x> 0 2x ,x< 0

a =3 f (x)= ®x 2 5,x< 3 x +1,x> 3

a =2 f (x)= (x 2) x 2 i f x0 x0 f (x0 ) f x0

x→x0 f (x)= b f (x0 )= b Df Df ∪{x0 }

f (x0 )= b

x0 f x0

f (x)= x x 0 f (0) x→0 f (x)=1 f (0)=1 f (x)= x x ,x =0 1,x =0 x =0 ii f x0 x→x0 f (x)= f (x0 ) (1) 1 x→x0 +0 f (x)= f (x0 ), f x0 x→x0 0 f (x)= f (x0 ), f x0 f (x)= ß3,x 1 2,x< 1 x→1 0 f (x)=2 x→1+0 f (x)=3 x→1 f (x) f 1 f (1)=3 x→1+0 f (x)= f (1) f 1 x→1 0 f (x) = f (1) f 1 f [a,b] f a b f a b f x→a+0 f (x)= f (a) f b x→b 0 f (x)= b f [a,b] (a,b) a b f [a,b] f (x)= (2 x)(x 3) [2, 3] (2, 3) f (2)=0 x→2+0 f (x)=0 f 2 f (3)= x→3 0 f (x)=0

3 f [2, 3]

f (x0 +0) x→x0 +0 f (x) f (x0 0) x→x0 0 f (x) f (x0 +0) f (x0 0) f (x0 +0) =

f (x0 0) f x0 f x0

|f (x0 +0) f (x0 0)| f (x)= x 0 2

f (x0 +0) f (x0 0) f x0 f (x)= 1 x 1 f (1+0) x→1+0 f (x)=+∞,f (1 0) x→1 0 f (x)= −∞. 1

f (x)= e1/x 0 f (0 0)=0 f (0+0)=+∞ f x ∈ R f (x)= 2x 1 x 1 x ,x =0 0,x =0 0 f (0 0) f (0+0) f x = a a x =0 0 x =0 f x0 x0 f x0 f f (x0 ) x0 f (x)= x x (2) x =0 f (x)= ß1,x> 0 2,x< 0 (3) x =0 f (x)= 1 x (4)

x =0 x =0 0 x =0 x0 x0

2 3 4 D = R \{0} 0

D D = R

i f g x0 f + g

f g fg x0 f g x0 g (x0 ) =0 iv f g x0 x→x0 f (x)= f (x0 )

x→x0 g (x)= g (x0 ) x→x0 (f (x) ∗ g (x))= x→x0 f (x) ∗ x→x0 g (x)= f (x0 ) ∗ g (x0 ), ∗ + · :

ii f : B → R g : A → B A B ⊆ R F : A → R F (x)= f (g (x)) x0 ∈ A x→x0 F (x)= F (x0 ), x→x0 f (g (x))= f (g (x0 )). f g x0 ∈ A y0 ∈ B x→x0 g (x)= g (x0 ), y →y0 f (y )= f (y0 ). g (x)= y g (x0 )= y0 x→x0 y = y0 x→x0 f (g (x))= y →y0 f (y )= f (y0 )= f (g (x0 ))

iii f [a,b] [c,d] f 1 [c,d] [a,b]

[c,d]

f [a,b] f 1

f g x0

f + g i f g x0 ii x0 f + g

i f + g x0 f

f (x0 ) x→x0 f (x) f (x0 ) = x→x0 f (x) g x0 f + g x→x0 f (x) x→x0 g (x) x→x0 (f (x)+ g (x))

ii f + g x0

f g f (x)= g (x)= x x x =0 x → 2 x x

f g f (x)= ß2,x< 1 3,x 1 ,g (x)= ß3,x< 1 2,x 1 x =1 h h(x)=5 x ∈ R x =1

f x = a g f ◦ g g ◦ f f ◦ g g ◦ f x = a

f (x)= x x =0 g (x)= x

f (g (x))= g (f (x))= x f ◦ g g ◦ f x =0 f (x)= ß1,x 0 1,x< 0 ,g (x)= x 2 f x =0 g

f (g (x))= g (f (x))=1 f ◦ g g ◦ f x =0 f (x)= x 2 +1 g (x)= 1 x f x =0 g

f (g (x))= 1+ x 2 x2 ,g (f (x))= 1 1+ x2 f ◦ g g ◦ f x =0

f g x = a fg x = a

f g x = a

fg x = a

f [ a,a] a> 0 x →|f (x)|

i f x0 ∈ [ a,a] a> 0

ii x →|f (x)| [ a,a] f

i f x =1 x =2 x =3 ii x →|f (x)| R

f (x)= c c f (x0 )= c x→x0 f (x)= c x0 ∈

(x)= x f (x0 )= x0

a> 0).

f (x)= 1+ x 1+ x2

f (x)= x 2 2x +1 x3 3x +2

f (x)= 2 x 1 + x x 5

f (x)= e 1 x

f (x)= 1+ x 1+ x3

f (x)= x + 1 x

f (x)= 8x 2x2 2x 1

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

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

f (x)= x (x 1)(x +2)

f (x)= x e x

f (x) > 0 x→a f (x) > 0

f (x)= … 1 x

f (x)= ( x)

x + 1 x x = e.

x→+∞ x +2 2x +1 x→±∞ (2x +15)14 (3x 1)23 (4x +5)7 (9x +2)30 x→1 x 4 1 x 1 x→1 x 2 4x +3 x2 3x +2 x→2 x 3 2x 2 +5x 10 x2 8x +12 x→1 1 1 x 3 1 x3 x→0 (1+ x)5 (1+5x) x2 + x5 x→0 (1+ x)6 1 x x→+∞ √x2 +9 x √4x2 +3 2x x→−∞ √x2 +9 x √4x2 +3 2x x→2 √x2 +21 5 x 2 x→−2 √x +18 2√x +6 x2 4 x→3 √3x +7 √2x +10 √4x +13 √x +22 x→1 √x +8 √8x +1 √5 x √7x 3 x→7 3 √x 6 1 x 7 x→0 3 √x +1 3 √3x +1 6x x→a 3 √x 3 √a x a x→+∞ Ä x3 + x2 +1 x3 x2 +1ä

x→0 x x x→0 3x 4x 2x x→0 5x 3x x→0 ax bx b =0 x→ π 2 1 x 2 x x→0 x x 3 x x→ π 2 1 x x x→ π 6 1 4 2 x 3x x→0 x 5x 2x + 3x x→0 x x x + x

x→ π 2 x 2 x 2 x x→0 x x x3

(∀ε> 0)(∃δ> 0)(¬|x x0 | <δ ∨|f (x) f (x0 )| <ε) (p ⇒ q ) ⇔ (¬p ∨ q )

(∀ε> 0)((∃δ> 0)(¬|x x0 | <δ ) ∨|f (x) f (x0 )| <ε)(∃x)(A ∨ B ) ⇔ (∃x)A ∨ B x B

(∀ε> 0)(¬(∀δ> 0)(|x x0 | <δ ) ∨|f (x) f (x0 )| <ε)(∃x)¬A ⇔¬(∀x)A

(∀ε> 0)(¬|x x0 | =0 ∨|f (x) f (x0 )| <ε)(∀δ> 0)|a| <δ ⇔ a =0

(∀ε> 0)(x = x0 ∨|f (x) f (x0 )| <ε) ¬|x x0 | =0 ⇔ x = x0

x = x0 ∨ (∀ε> 0)(|f (x) f (x0 )| <ε)(∀x)(A ∨ B ) ⇔ A ∨ (∀x)B x A

x = x0 ∨|f (x) f (x0 )| =0(∀ε> 0)|a| <δ ⇔ a =0

x = x0 ∨ f (x)= f (x0 ) |a| =0 ⇔ a =0

x = x0 ⇒ f (x)= f (x0 )(¬p ∨ q ) ⇔ (p ⇒ q ) 3 x0 3

i f : A → R A ⊆ R a ∈ A f (a) > 0

δ> 0 f (x) > 0 x ∈ (a δ,a + δ ) f (a) < 0 δ> 0 f (x) < 0 x ∈ (a δ,a + δ )

ii f :[a,b] → R f (a) < 0 f (b) > 0

c ∈ (a,b) f (c)=0 f (a) > 0 f (b) < 0 iii f :[a,b] → R [a,b] iv f :[a,b] → R [a,b] xm xM ∈ [a,b] f (xm ) f (x) f (xM ) x ∈ [a,b]

i f (a) > 0 f (a) x δ a f y = f (x) x x x

ii f (a) < 0 A =(a,f (a)) x f (b) > 0 B =(b,f (b)) x A B x (c, 0) c ∈ (a,b) f (c)=0

f (a) > 0 A x f (b) < 0 B x

c ∈ (a,b) f (c)=0 iii iv [a,b]

i iv f f (x)=[x] 1 2

f (1)= 1 2 > 0 1 f (x) > 0 x f 1 2 = 1 2 < 0,f 3 2 = 1 2 > 0,

c ∈ Ä 1 2 , 3 2 ä f (c)=0 iii iv f [a,b] (a,b) iii iv

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

(0, 1) xm xM ∈ (0, 1) f (xm ) f (x) f (xM ) x ∈ (0, 1) [0, 1] xm =0 xM =1 ii

f (a)= A f (b)= B A = B f [a,b] C A B c ∈ (a,b) f (c)= C AB

g g (x)= f (x) C [a,b] g (a)=

f (a) C = A C< 0 g (b)= f (b) C = B C> 0 ii

c ∈ (a,b) g (c)=0 g (c)= f (c) C

c ∈ (a,b) f (c) C =0 f (c)= C A<B C ∈ (A,B ) c ∈ (a,b) f (c)= C f [a,b] [A,B ] iv f [a,b]

[A,B ] f 1 [A,B ] [a,b] f 1 f x1 = x2 f (x1 ) = f (x2 ) |f (x2 ) f (x1 )| > 0. h ∈ (0,b a) 0 <hh |f (x1 ) f (x2 )| m. f 1 |x1 x2 | h ⇒|f (x1 ) f (x2 )| m |f (x1 ) f (x2 )| <m ⇒|x1 x2 | <h, y1 = f (x1 ) y2 = f (x2 ) |y1 y2 | <m ⇒|f 1 (y1 ) f 1 (y2 )| <h. ε> 0 δ> 0 y |y1 y | <δ |f 1 (y1 ) f 1 (y )| <ε, f 1 y1 ∈ [A,B ] (i) (iv ) i iv i iv i ε = 1 2 f (a) > 0 ε> 0 δ> 0 |f (x) f (a)| < 1 2 f (a) (1) x |x a| <δ 1 f (x) >f (a) 1 2 f (a)= 1 2 f (a) > 0,

x |x a| <δ x ∈ (a δ,a + δ ) (ii) X = {x | x ∈ [a,b] f (x) < 0} a ∈ X b c = X f (c)=0

f (c) =0 f (c) < 0 f (c) > 0

f (c) < 0 (i) δ c f (x) < 0 x ∈ (c δ,c + δ ) c X c = X

f (c) > 0 (i) δ c

f (x) > 0 x ∈ (c δ,c] c X f (c)=0 (iii) f a δ> 0

[a,λ] f (a) δ<f (x) <f (a)+ δ [a,λ] X1 x ∈ [a,b] f [a,x] X =[a,b] \ X1 X a σ = X σ ∈ [a,b] f σ δ> 0 (σ ε,σ + ε) ε> 0 f (σ ) δ<f (x) <f (σ )+ δ. f (σ ε,σ + ε) σ ε / ∈ X X σ ε ∈ X1 f [a,σ ε] f [a,σ ε] (σ ε,σ + ε) [a,σ + ε) [a,c] a c σ + ε [a,σ + ε] ⊆ X1 [a,σ + ε] ∩ X = ∅ σ X

X1 =[a,b] [a,b] (iv ) M = {f (x) | x ∈ [a,b]} x ∈ [a,b] f (x) M. xM ∈ [a,b] f (xM )= M δ> 0 x ∈ [a,b] M f (x) <δ, 1 M f (x) > 1 δ . x → 1 M f (x) (iii) x → M f (x) x → 1 M f (x) M f (x)=0 xM ∈ [a,b]

M f (xM )=0 f (xM )= M xm ∈ [a,b] f (xm )= m = {f (x) | x ∈ [a,b]} f (x) x ∈ [a,b]

xm xM ∈ [a,b] x ∈ [a,b] f (xm ) f (x) f (xM ),

(T )

(x0 ,y0 ) (P ) y = x 2 y = ax + b (1) a b (x0 ,y0 ) (T ) (P ) y0 = x 2 0 y0 = ax0 + b, x 2 0 = ax0 + b x 2 0 ax0 b =0 (2) 1 2

a =2x0 3 b = 1 4 (2x0 )2 = x 2 0 . a b 1 (x0 ,y0 ) y =2x0 x x 2 0 . y = x 3 (P ) y = x 2 x0 x0 + h A Ah (P ) x 2 0 (x0 + h)2 A =(x0 ,x 2 0 ) Ah =(x0 + h, (x0 + h)2 ) A Ah y x 2 0 = (x0 + h)2 x 2 0 h (x x0 ), y x 2 0 = 2x0 h + h2 h (x x0 ) y x 2 0 =(2x0 + h)(x x0 ). (4) Ah (P )

h → 0 Ah →

4 h → 0

y x 2 0 =2x0 (x x0 ),

y =2x0 x x 2 0 .

y = f (x) A

AAh h → 0 Ah Ah = A A =(x0 ,f (x0 )),Ah (x0 + h,f (x0 + h)), A Ah

y f (x0 )= f (x0 + h) f (x0 ) h (x x0 ) y = f (x0 + h) f (x0 ) h (x x0 )+ f (x0 ). (5) 5 h → 0 y = Çh→0

f (x0 + h) f (x0 ) h å (x x0 )+ f (x0 )

y = k (x x0 )+ f (x0 ), k = h→0

f (x0 + h) f (x0 ) h (6) 6

f (x0 + h) f (x0 ) h (1)

y = f (x) x0 h→0

f (x0 )= h→0

f x0 f (x0 )

f (x0 + h) f (x0 ) h . f x0 f (a,b) x ∈ (a,b) f

f (x)= h→0

f (x + h) f (x) h (x ∈ (a,b))

f (a,b) x0 1 f h→0+ f (x0 + h) f (x0 ) h , h→0 f (x0 + h) f (x0 ) h . f x0

f+ (x0 ) f (x0 ) x0

f+ (x0 )= f (x0 )= f (x0 ) f x0 f f x0 h→0

f (x0 + h) f (x0 ) h = f (x0 ).

f (x0 + h) f (x0 )= f (x0 + h) f (x0 ) h h, h→0 (f (x0 + h) f (x0 ))= h→0 Å f (x0 + h) f (x0 ) h hã = h→0 f (x0 + h) f (x0 ) h h→0 h = f (x0 ) h→0 h =0, h→0 (f (x0 + h) f (x0 ))=0, (2)

f (x)=5x 9 f (x)=3x 2 +4x 8 f (x)= 2 x f (x)= x +1 x f (x)= 2x +1 x 5 f (x)= √4x 3

(ii) 1 f f (x)= x x ∈ R

h→0 f (x + h) f (x) h = h→0 x + h x h = h→0 h h =1.

(iii) x → ex

f (x)= ex x ∈ R

h→0 f (x + h) f (x) h = h→0 ex+h ex h = h→0 ex eh 1 h = ex h→0 eh 1 h = ex ,

h→0 eh 1 h =1

(iv ) f (x)= x x ∈ R h→0 f (x + h) f (x) h = h→0 (x + h) x h = h→0 2 h 2 2x + h 2 h = h→0 h 2 h 2 2x + h 2 . (1) h→0 h 2 h 2 =1 h→0 2x + h 2 = x ) 1 ( x) = x. f g x

(f (x) ± g (x)) = f (x) ± g (x)

(cf (x)) = cf (x)

(f (x)g (x)) = f (x)g (x)+ f (x)g (x) Å f (x) g (x) ã = f (x)g (x) f (x)g (x) g (x)2 (g (x) =0)

(i) (ii)

(f (x) ± g (x)) = h→0

(f (x + h) ± g (x + h)) (f (x) ± g (x)) h = h→0

(f (x + h) f (x)) ± (g (x + h) g (x)) h = h→0

f (x + h) f (x) h ± h→0 g (x + h) g (x) h

= f (x) ± g (x),

(cf (x)) = h→0

(cf (x + h) cf (x)) h = h→0 c(f (x + h) f (x)) h = c h→0 f (x + h) f (x) h = cf (x)(c ). (iii)

(f (x)g (x)) = h→0 f (x + h)g (x + h) f (x)g (x) h (1)

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

(f (x)g (x)) = h→0 f (x + h)g (x + h) f (x + h)g (x)+ f (x + h)g (x) f (x)g (x) h = h→0 g (x) f (x + h) f (x) h + h→0 f (x + h) g (x + h) g (x) h

= g (x) h→0 f (x + h) f (x) h + h→0 f (x + h) Åh→0 g (x + h) g (x) h ã

= f (x)g (x)+ f (x)g (x). (iv )

Å f (x)

g (x) ã = h→0

f (x + h) g (x + h) f (x) g (x) h = h→0

f (x + h)g (x) g (x + h)f (x)

hg (x + h)g (x) = h→0

f (x + h)g (x) f (x)g (x)+ f (x)g (x) g (x + h)f (x)

hg (x + h)g (x)

f (x + h) f (x) h g (x) f (x) g (x + h) g (x) h

= h→0

g (x + h)g (x)

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

g (x)2 . (iii) h→0 f (x + h)= f (x) (iv ) h→0

g (x + h)= g (x)

f g x

f g

f + g f g fg f /g g (x) =0

x f u = g (x) F (x)= f (g (x)) F (x)= f (g (x))g (x). (2) F (x)= h→0 F (x + h) F (x) h = h→0 f (g (x + h)) f (g (x)) h = h→0 f (g (x + h))

(g (x))

(x + h) g (x) h→0

(x + h)

(x)

(x + h) g (x) h .

= g (x)

→0 f (g (x + h)) f (g (x)) g (x + h) g (x) (3) g (x)= u g (x + h)= u + k. (4) g h → 0 4 k → 0 3 k →0 f (u + k ) f (u) u + k u = k →0 f (u + k ) f (u) k = f (u). h→0 f (g (x + h)) f (g (x)) g (x + h) g (x) = f (g (x)), F (x)= f (g (x))g (x), 2 f 1 f f f 1 (x) f (f 1 (x)) =0 f 1 x (f 1 (x)) = 1 f (f 1 (x)) (5) (f 1 (x)) = h→0 f 1 (x + h) f 1 (x) h u = f 1 (x),u + k = f 1 (x + h), x = f (u),x + h = f (u + k ). (6)

f 6 h → 0 k → 0 6

h→0 f 1 (x + h) f 1 (x) h = h→0 f 1 (x + h) f 1 (x) x + h x = k →0 u + k u f (u + k ) f (u) = h→0 k f (u + k ) f (u) = k →0 1 f (u + k ) f (u) k = 1 f (u) = 1 f (f 1 (x)) , 5 n ∈ N (x n ) = nx n 1 . (7) n =1 n (x n+1 ) =(x n x) =(x n ) x + x n x = nx n 1 x + x n =(n +1)x n , 7 n ∈ N 7 n ∈ Z (x n ) = nx n 1 (x =0 n ∈ N) 7 (x n ) = 1 xn = 1 x n (x n ) 1 x2n = nx n 1 x2n = nx n 1 7 n ∈ Z x =0 n ∈ N n √x = 1 n x 1 n n (x> 0) (8) x 0 x → n √x x → x n n √x = 1 n n √x n 1 = 1 n x 1 n n (x> 0). 8 x 1 n = 1 n x 1 n n (x> 0). p ∈ Z q ∈ N x → xp/q f (x)= xp g (x)= x1/q F F (x)= f (g (x))

F (x)= xp/q

F (x)= f (g (x))g (x), (9)

2 x +2x + e x 3 2 x +3x e x x + x x + x (x 2 2x)(x 4 +7) (x 2 3x +1)(x 3 3x +1) (x 5 x 2 +2)(x 4 3x 2 +4) (x +2)(√x +2)

√x(x 5 3x 2 +5) (√x +3x)(5x 2 3 √x2 xe x x 2 e x x n e x x a a x x 2 x e x x 2x

5 2x (1+ x 2 ) x x x2 1 x 3 x 1 x 2 +1 x2 +4 x 2 1 x3 +4 x 4 x 2 +1 x4 + x2 +1 x +1 √x √x x2 +4

1+ x√x 1 x√x 3x 4 √x + x 2 3x x +5 2x 2 +3x x 4 5x 3 √x x x x x + x x x e x e x ex + e x (7x 5 3x 2 )20 (2x 3)3 (x +5)2 (2x 3)2 (x +5)2 3 (2+ x2 )4 2x x2 +1

x 3 2x2

x 2 + x +1 x2 x +1

3x 2 +2x 1 3x2 2x 1 x 1 x2 x 2 +1 x2 1 x 2 + x +1 x2 x +1 1 x 2 +2 x 3 x 2 (x + a2 + x2 ) … 2x 2 +1 2x

2x 2 +1 2x f (x) f (x) e ax 2 bx 2x + 2x 2 x 2 x 2 x + 2 x x 2 2x x 2 4 x 4 x x + 1 3 3 x 2 x 3 x +3 x x +3x x x 1 3 2 x 3 x 3 x +3x √x √ 3x 2 (3x +1) x 4 2x 2 (5x +2) 2 (ax + b) 2 x 2 x 2 x x2 1+ x 1 x 1+ x 1 x (1 3 x) x + 1 x x + 1 x 1 2 x + π 4

1 x2 x x 1 2 x 2 1 x 1+ x 1 x x √1 x2 x x (x 2 )x (2x )2 (2x )x x x x

(a,b)

f f f (x)=(f (x)) f (3) (x)=(f (x)) ,f (4) (x)=(f (3) (x)) ,... f (n) (x)=(f (n 1) (x)) , (n =2, 3,... ). n f (x)= x m m ∈ N f (x)= mx m 1 ,f (x)= m(m 1)x m 2 ,f (3) (x)= m(m 1)(m 2)x m 3 , m ≥ 3 m ≥ n f (n) (x)= m(m 1) (m n +1)x m n , (1) n>m f (n) (x)=0 1 f (x)= x x> 0 f (x)= 1 x ,f (x)= 1 x = 1 x2 ,f (3) (x)= 1 x2 = 2 x3 f (4) (x)= 2 · 3 x4 ,f (5) (x)= 2 · 3 · 4 x5 (2) f (n) (x)=( 1)n 1 (n 1)! xn (n ∈ N) (3) 3 n =1, 2, 3, 4, 5 2 3 3 n =1 n ∈ N 3 f (n+1) (x)=( 1)n 1 (n 1)!( n)x n 1 =( 1)n n! xn+1 3 n ∈ N x> 0

f (x)= x

f (x)= x = Äx + π 2 ä ,f (x)= x = (x + π ), f (3) (x)= x = x + 3π 2 ,f (4) (x)= x = (x +2π )

f (n) (x)= Äx + nπ 2 ä , (n ∈ N) (4) n =1 4 4 n ∈ N f (n+1) (x)= Äx + nπ 2 ä = Äx + nπ 2 + π 2 ä = Åx + (n +1)π 2 ã , 4 n ∈ N

f (x) f (x) f (3) (x)

f (x)=5x 3 3x 2 + x 1 f (x)= x 6 6x 4 +1 f (x)= x√x +1

f (x)= x x f (x)= x f (x)= e x 2 f f (x)=(x 2 1)6 (x 3 +5)10 (i)42 f (42) (x) (ii)50 f (50) (x)

f (x)= Aex + Be x A B

f (x)= f (x)

f (x)= A x + B x A B

f (x)+ f (x)=0 n f f (x) x m m ∈ N (x a)m m ∈ Z

x ax a ∈ R)

n e x

(x)e x y = f (x) (x0 ,f (x0 ))

y = f (x) y = Çh→0

f (x0 + h) f (x0 ) h å (x x0 )+ f (x0 ). f x0 y = f (x)

(x0 ,f (x0 ))

y = f (x0 )(x x0 )+ f (x0 ). (1)

(x0 ,f (x0 ))

y = f (x) 1 y = 1 f (x0 ) (x x0 )+ f (x0 )

F (x,y )=0 G(x,y )=0

(x0 ,y0 ) α F (x,y )=0 G(x,y )=0

(1, 2)

y =2+ x x 2

f (x)=2+ x x 2 f (1)=2 f (x)=1 2x f (1)= 1 y = 1(x 1)+2, y =1(x 1)+2, y = x +3, y = x +1. y = x 2 6y =7 x 3 (1, 1) y = f (x) y = g (x)

f (x)= x 2 g (x)= 7 6 1 6 x 3 .

f (x)=2x g (x)= 1 2 x 2 . (1, 1) y =2(x 1)+1, y = 1 2 (x 1)+1, f (1)=2 g (1)= 1 2 y =2x 1, y = 1 2 x + 3 2 .

2 · Ä 1 2 ä = 1

y =2x 2 4x +5 (3, 11) y = 6x x2 1 (2, 4)

y = 2 x Ä π 3 , 3 4 ä y = 5x 2 1+ x2 (2, 4)

√x + √y =5 (9, 4)

y = x 3 12x +4 x y (a 2 + x 2 )= a 2 x x

y 2 =8x x 2 =4y 12 xy =6 x 2 y =12 P y = x 3 x T N P x |OT | =2|TN | O y = x n n ∈ N

F (x,y )=0

F (x,y )=0 (1) (x,y ) x y 1 1 1 1 y y = f (x), f 1 y 1 y A( 4, 2) x 2 + y 2 =20 (2)

x2/3 + y 2/3 = a2/3 (a> 0) (12) = (x0 ,y0 ) 12 12 2 3 x 1/3 + 2 3 y 1/3 y =0, y = y 1/3 x1/3 , (x0 ,y0 ) y = y 1/3 0 x1/3 0 (x x0 )+ y0 .

. 13 x =0 y = a2/3 y 1/3 0 13 x A =(a2/3 x1/3 0 , 0) y B =(0,a2/3 y 1/3 0 ) |AB |2 =(a2/3 x1/3 0 ) 2 +(a2/3 y 1/3 0 ) 2 =

2 |AB | (x0 ,y0 ) y 3 = x 2 (8, 4) x 2 + y 2 3x +4y 31=0 ( 2, 3) 9x 2 4y 2 =108 (4, 3) x 4 + y 4 =17 ( 2, 1) 2x 2 xy +3y 2 =18 (3, 1)

x 2 +25y 2 =225 x 2 y 2 =8

0 t 2

f :(a,b) → (c,d) f 1 :(c,d) → (a,b) x = f (t) ⇔ t = f 1 (x)(t ∈ (a,b),x ∈ (c,d)), (8)

y = g (f 1 (x))(x ∈ (c,d)). (9)

F x F (x)= g (f 1 (x))

= F (x)

F (x)=(g (f 1 (x))) = g (f 1 (x))(f 1 (x)) = g (f 1 (x)) f (f 1 (x)) , F (x)= g (t)

)

x =2t t2 ,y =3t t3 (0, 0) (1, 2) x 2 a2 + y 2 b2 =1 (a t,b t) M y 2 =4ax (at2 , 2at) ϕ M x t = ϕ y = kx + n (1) 1 y 2 =2px p =2kn 1 F (x,y )=0. (2) 1 2 x y y y = kx + n,F (x,y )=0,y = k, (F (x,y )) =0 x y y x y y 1 2 1 y 2 =2px y = kx + n,y 2 =2px,y = k, 2yy =2p. y ky = p y = p k y p k = kx + n, p 2 k 2 =2px, p k n = kx, p 2k 2 = x, x p k n = p 2k , p kn k = p 2k , p =2kn 1 y = x 3 + ax 2 a y = kx + n,y = x 3 + ax 2 ,y = k,y =3x 2 +2ax, x y y y kx + n = x 3 + ax 2 y k = 3x 2 +2ax x x 3 + ax 2 kx n =0, 3x 2 +2ax k =0

k +2a 2 )x 9n + ak =0, x = ak 9n 6k +2a2

ak 9n)2 +2a(

)(6

3 + ax 2 kx n =0

+2a 2 )

(6k +2a 2 )2

(x + h) f (x) h . (1)

(x)= h→0

f (x + h) f (x) h = f (x)+ λ (h→0 λ =0) f (x + h) f (x)= f (x)h + λh, h→0 λ =0 (2) Δf (x) f Δf (x)= f (x + h) f (x) f (x)= x Δx = h 2 Δf (x)= f (x)Δx + λΔx. f f (x) f (x)= f (x)Δx. f (x)= x x =Δx f (x)= f (x) x, f (x) x = f (x). (3) f f x 3 x f f

= f (x)

y x = f (x)

y = f (z ) z = g (x) f (g (x)) = f (g (x))g (x), y x = y z · z x . y x = 1 x y .

(f (x) ± g (x))= f (x) ± g (x); (f (x)g (x))= f (x) g (x)+ g (x) f (x);

f (x) g (x) å = g (x) f (x) f (x) g (x) g (x)2 . f (g (x))= f (g (x)) g (x)= f (g (x))g (x) x, x 2 f (x) f (x) 2 f (x)= ( f (x)). 3 f (x)= ( 2 f (x)),..., n f (x)= ( n 1 f (x)). x x

2 f (x)= ( f (x))= (f (x) x)= f (x) x = f (x) x x = f (x) x 2 ,

x 2 ( x)2

3 f (x)= ( 2 f (x)) (f (x) x 2 )= Äf (x) x 2 ä = f (3) (x) x 3 , x 3 =( x)3

n f (x)= f (n) (x) xn , n f (x) xn = f (n) (x), xn =( x)n (x 2 ) 1 x (xe x ) ( x) x x (x 4 x 2 +2) (x2 ) x (x 2 x) 2 x2 ( x)

3 (x 5 ) (x +1) (x 2 x) 3 x2 (x 5 ) 3 x3 (x 5 )

n (x n e x ) n ( x) n ( ax)

h(=Δx) Δf (x) f (x) Δf (x)= f (x + h) f (x); f (x)= f (x)Δx. Δf (x) f (x)

f (x) f (x)

x Δx Δf (x) f (x) f (x)

x 2 3 f (x)=2x 3 3x +1 Δf (x)=(6x 2 3)h +6xh2 +2h3 (h =Δx), f (x)=(6x 2 3)h. x =1

f (1)=3h +6h2 +2h3 , f (1)=3h. h

i h =10 Δf (1)=30+600+2000=2630, f (1)=30, Δf (1) f (1)=2600

ii h =1 Δf (1)=3+6+2=11 f (1)=3 Δf (1) f (1)=8

iii h = 1 10 Δf (1)= 3 10 + 6 100 + 2 1000 = 362 1000 , f (1)= 3 10 , Δf (1) f (1)= 62 1000 =0, 062.

iv h = 1 100 Δf (1)= 3 100 + 6 10000 + 2 1000000 = 30602 1000000 , f (1)= 3 100 , Δf (1) f (1)= 602 1000000 =0, 000602 h(=Δx) Δf (1) f (1) Δf (1) f (1) Δf (x)= f (x)Δx = f (x)h (1) Δf (x) ≈ f (x)h Δf (x) f (x) 1 f (x + h) f (x) ≈ f (x)h, f (x + h) ≈ f (x)+ f (x)h, (2)

(5, 123)2 2 f (x)= x 2 x =5 h =0, 123 (5, 123)2 ≈ 52 +10 · 0, 123( f (x)=2x) (5, 123)2 ≈ 25+1, 23=26, 23

i √5 ≈ 2, 25(a =2,x =1) ii √120 ≈ 10, 9545(a =11,x = 1)

f (a)= f (b)

(a,f (a)) (b,f (b))

(a,f (a)) (b,f (b)) c a<c<b c y = f (x) x c x f (c)=0

i f [a,b]

ii f (a,b) iii

f (a)= f (b)

c ∈ (a,b)

f (c)=0

[a,b] a b i ii i x→a+0

f (x)= f (a) x→b 0 f (x)= f (b)

ii f (a,b)

f [a,b] (a,b) x0 ∈

(a,b) f (x0 ) > 0 δ> 0

f (x) <f (x0 ) x ∈ (x0 δ,x0 ); f (x) >f (x0 ) x ∈ (x0 ,x0 + δ )

f (x0 + h) f (x0 ) h = f (x0 ), ε> 0 δ> 0 |h| <δ

h→0

f (x0 + h) f (x0 ) h f (x0 ) <ε

f (x0 ) ε< f (x0 + h) f (x0 ) h <f (x0 )+ ε. (1)

f (x0 ) > 0 ε f (x0 ) ε> 0 1

f (x0 + h) f (x0 ) h > 0,

f (x0 + h) f (x0 ) h

f (a)= f (b)=0

f (x)=0 x ∈ (a,b) f (x)=0 x ∈ (a,b)

f (x) f (x) x ∈ (a,b) f M c ∈ (a,b)

f (c)= M f (x) M x ∈ [a,b] f (x) x ∈ (a,b) c ∈ (a,b) f (c)= m m f

f (c) > 0 x c

f (x) >M M

f (c) < 0 f (c)=0

f (a)= f (b)= k =0 F F (x)= f (x) k

F (a)= F (b)=0 c ∈ (a,b) F (c)= f (c)=0

f f (x)= x 3 4x +1 f [ 2, 2] f (2)= f ( 2) c ∈ ( 2, 2)

f (c)=0 f (x)=3x 2 4 f (c)=0

3c 2 4=0 c 2 √3 2 √3 ( 2, 2) c1 c2 ∈ (a,b) f (c1 )= f (c2 )=0 f

f (x)= x 2 3 f ( 1)= f (1)=1

c ∈ ( 1, 1)

f (c)=0 f (x)= 2 3 3 √x

f (x)=0 f x =0 ∈ ( 1, 1) 6x 5 5x 4 +4x 3 3x 2 +2x 1=0 (2) (0, 1) f f (x)= x 6 x 5 + x 4 x 3 + x 2 x. [0, 1] f (0)= f (1)=0 c ∈ (0, 1) f (c)=0

f (x)=6x 5 5x 4 +4x 3 3x 2 +2x 1 c 2 f (x)= x 3 3x +4 f ( 2)= f (1)

c ∈ ( 2, 1) f (c)=0 P (0, 1) i P (x)=3x 2 6x +2 ii P (x)=4x 3 +3x 2 +2x 3 iii P (x)=8x 3 3x 2 +4x 3 iv P (x)=5x 4 8x 3 9x 2 +2x +3 P P f g f g

[a,b] f (x)g (x) f (x)g (x) =0 x ∈ [a,b] f g f g x → a x + b x a b =0

(kπ, (k +1)π ) k ∈ Z

f (x)= af (x) g (x)= ag (x) a ∈ R g (x) =0 C f (x)= Cg (x)

(1)

f i [a,b]

ii (a,b)

c ∈ (a,b)

f (b) f (a) b a = f (c).

f (b) f (a) b a (b x) (2) F i [a,b]

F (x)= f (b) f (x)

ii (a,b)

iii F (a)= F (b)(=0)

F (c)=0 2

F (c)=0

c ∈ (a,b)

F (x)= f (x)+ f (b) f (a) b a ,

f (c)+ f (b) f (a) b a =0, 1 f [a,b] (a,b)

y = f (x)

f (b) f (a) b a , A =(a,f (a)) B =(b,f (b)) 1 y = f (x) c t AB 1

f (b)= f (a)+ f (c)(b a) c ∈ (a,b)

f (b)= f (1)=2

f (x)= x 3 +1 a = 2 b =1 f (a)= f ( 2)= 7

f (x)=3x 2

f (b) f (a) b a = 2 ( 7) 1 ( 2) = 9 3 =3

c ∈ ( 2, 1) f (c)=3 3x 2 =3 x =1 x = 1 1 1/ ∈ ( 2, 1)

f ( 1)=3 1 ∈ ( 2, 1) f f (x)= |x| a = 1 b =2

f [ 1, 2]

f (b) f (a)

b a = 1 3 .

f (x)= 1 x< 0 f (x)= 1 x> 0 f (x) = 1 3 x ∈ ( 1, 2) c ∈ ( 1, 2) f (b) f (a) b a = f (c) f 0 ( 1, 2) 0 <a 0 1 b (b a) < 1 c (b a) < 1 a (b a). (5) 4 b a = 1 c (b a), (6) 5 6 3 x F (x)=0, (7) F (a,b) x ∗ 7 x = f (x). (8) f f (x)= x + λF (x) λ =0 f 7 8

x ∗ 8 (a,b) f (a,b) (a,b) |f (x)| q< 1 x ∈ (a,b) (9) x0 ∈ (a,b) (xn ) x1 = f (x0 ),x2 = f (x1 ),x3 = f (x2 ),..., xn+1 = f (xn )(n =0, 1, 2,... ) (10)

). (12) i x 3 4x +1=0 (0, 1) x ∗ F (x)= x 3 4x +1 x = 1 4 (x 3 +1) 8 f (x)= 1 4 (x 3 +1) f (x)= 3 4 x 2 x ∈ (0, 1) |f (x)| 3 4 9 x0 =0, 5 x1 = 1 4 (x 3 0 +1)=0, 28125,x2 = 1 4 (x 3 1 +1)=0, 2555618, x3 = 1 4 (x 3 2 +1)=0, 2541728,x4 = 1 4 (x 3 3 +1)=0, 2541051, x5 = 1 4 (x 3 4 +1)=0, 2541019. x5 12 |x5 x ∗ | < 3 4 5 =0, 2373047. F (0, 2541) > 0 F (0, 2542) < 0 x5 ii 5x =4 x Ä0, π 2 ä y =5x y =4 x x = 4 5 x 8 f (x)= 4 5 x |f (x)| = 4 5 | x| 4 5

9 (xn )

x0 ∈ Ä0, π 2 ä ; xn+1 = 4 5 xn (n =0, 1, 2,... ) x ∗ 5x =4 x x0 =1 x1 = 4 5 1=0, 4322418, x2 = 4 5 x1 =0, 7264231,...,x15 = 4 5 x14 =0, 6411282. x15 |x15 x ∗ | < 4 5 15 π 2 =0, 0552675

c ∈ (a,b)

f (b) f (a) b a = f (c)

f (x)= x 2 +1 a =1 b =2 f (x)= x 3 + x 2 2x a =0 b =1

f (x)= x 10 + x 4 +1 a = 1 b =1 f (x)= 1 x a =1 b =4

f (x)= x + 1 x a = 1 2 b =2 f (x)= x 3 a = 5 b =3

f (x)= e x a =0 b =1 f (x)= x a =0 b = π 2

f (x)= x a =1 b = e f (x)= x(x 1)(x 2) a =0 b = 1 2

| b a| |b a| pap 1 (b a) bp ap pbp 1 (b a) 0 <a 1 b a 1+ b2 < b a 1+ ab < b a 1+ a2 (a<b) x = 1 (x +1)2 x = 3 √5 x 4 3x = x x 2 = x x 2 = e x +2 2x =4x f f (a,b) f [a,b] c ∈ (a,b)

f (b)= f (a)+(b a)f (a)+ 1 2 (b a)2 f (c)

F (x)= f (b) f (x) (b x)f (x) b x b a 2 (f (b) f (a) (b a)f (a))

i [a,b] ii (a,b)

iii F (a)= F (b)(=0) c ∈ (a,b) F (c)=0 1

F (x)= f (x)+ f (x) (b x)f (x)+2 b x (b a)2 (f (b) f (a) (b a)f (a)) =(x b) Åf (x) 2 (b a)2 (f (b) f (a) (b a)f (a))ã ,

F (c)=0 f (c)= 2 (b a)2 (f (b) f (a) (b a)f (a))=0, 1 2 (b a)2 f (c)=

)= f (a)+(b a)f (a)+ 1 2 (b a)2 f (c) (2) n (a,b) f (n 1) [a,b]

c ∈ (a,b) f (b)= f (a)+(b a)f (a)+ 1 2! (b a)2 f (a)+ ··· + 1 (n 1)! (b a)n 1 f (n 1) (a)+ 1 n! (b a)n f (n) (c) n =1 n =2 2 f (x + h)= f (x)+ f (x)h + 1 2 f (x + θh)h2 , (3)

0 <θ< 1 a = x b = x + h x<c<x + h c = x + θh

0 <θ< 1 a b c 2 3 f (x + h) ≈ f (x)+ f (x)h, (4) 3 4 3 f (x + θh)h2 f f |f (x)| M h→0 1 2 f (x + θh)h2 =0, h 3

1 2 f (x + θh)h2 f (x + h) f (x)+ f (x)h x x h h 3 f (x)= x f (x)= x f (x)= x 3 (x + h)= x + h x 1 2 h2 (x + θh) x =0 h = h 1 2 h2 θh. R R = | h h| = 1 2 h2 θh 1 2 h2 . |h| < 10 R< 50 h ≈ h |h| < 1 100 R< 1 20000 h ≈ h h √1+ h ≈ 1+ 1 2 h h 1 80000 3 f (x)= √1+ x x =0 √1+ h =1+ 1 2 h 1 8 1 (1+ θh)3/2 h2 , R R = √1+ h 1+ 1 2 h = 1 8 1 (1+ θh)3/2 h2 = 1 8 1 |1+ θh|3/2 h2 0 <θ< 1 h> 0 |1+ θh| =1+ θh> 1 (1+ θh)3/2 > 1 1 |1+ θh|3/2 < 1 R = 1 8 1 |1+ θh|3/2 h2 < 1 8 h2 0 <h< 1 100 h2 < 1 10000 R< 1 80000 x ≈ x |x| < 10 1 (1+ x) ≈ x x ∈ [0, 10 3 ] (1+ e x ) ≈ 2+ x 2 |x| < 10 2

f (a)= g (a)=0, x→a f (x)= x→a g (x)=0 (1) x→a f (x) g (x) 1 x a 0/0 f g a g (a) =0 1

→a f (x) g (x) = f (a) g (a) (2) 1 f (x) g (x) = f (x) f (a) g (x) g (a) , x = a f (x) g (x) = f (x) f (a) x a g (x) g (a) x a x→a f (x) g (x) = x→a f (x) f (a) x a g (x) g (a) x a = f (a) g (a) . f g a 2

→a f (x) g (x) =

→a f (x) g (x) . f g n a f (a)= g (a)= f (a)= g (a)= ··· = f (n 1) (a)= g (n 1) (a)=0,g (n) (a) =0, x→a f (x) g (x) = x→a f (x) g (x) = ··· = x→a f (n) (x) g (n) (x) . n

)+3 (2+ x) 2 x3

→+∞ e x x x→+∞ e x xk (k ∈ R+ ) x→+∞ x xk (k ∈ R+ )

→1 x 1 1 x x→1(1 x) πx 2 (a,b) f x ∈ (a,b) f (x)=0 f x1 x2 (a,b) c x1 x2 f (x1 ) f (x2 ) x1 x2 = f (c). (1) f (c)=0 1 f (x1 )= f (x2 ) f (a,b) f (a,b) f g x ∈ (a,b) f (x)= g (x) (a,b)

f (x)= g (x)+ C (C ).

)

x1 x2 (a,b) x1 <x2 c f (x2 ) f (x1 ) x2 x1 = f (c)(x1 <c<x2 )

c ∈ (a,b) f (c) > 0

f (x2 ) f (x1 ) x2 x1 x1 <x2

f (x1 ) <f (x2 ) f (a,b) f (a,b)

i f (x) 0 (a,b) f ii f (x) < 0 (a,b) f

iii f (x) 0 (a,b) f f f (x)

2x 2 x 3 x 2 +5x 6 x 3 1 (x 5)3 x 3 +2x 2 5x +6 x 3 2x 2 4x +8 x 3 +2x 2 + x +2 x 4 x 2 +1 1 4 x 4 + 2 3 x 3 4x 2 1 2x x 2 x +3 5 x 2x +5 x +2

3x +4 (x2 +1)2 3x 2 +5x +25 x +2 x 2 +3x +1 x2 + x 1 x 2 e x 1 x x√3 x 2 x +2 x +1 x x x → ax + b cx + d i R ii R

f :[a,b] → R f x0 ∈ (a,b) δ> 0

f (x) f (x0 ) x ∈ (x0 δ,x0 + δ ) x0 ∈ (a,b) δ> 0

f (x) f (x0 ) x ∈ (x0 δ,x0 + δ ). f

x0 (a,b) f

x1 x3 x5 x2 x4 x3 f (x) f (x3 ) x f f x0 f

f (x0 )=0

f (x0 ) =0

f (x0 ) > 0 δ> 0

f (x) <f (x0 ) x ∈ (x0 δ,x0 ) f (x) >f (x0 ) x ∈ (x0 ,x0 + δ ), f x0

f (x0 )=0

f (x0 ) < 0

f (x0 )=0 f f (x)= x 3 f (x)=3x 2

f (0)=0 0 f (x) >f (0)

x> 0 f (x) <f (0) x< 0 f x0

f (x0 )=0 f (x0 ) > 0 f x0

f (x0 ) < 0 f x0

f (x0 + h)= f (x0 )+ hf (x0 )+ 1 2! h2 f (c), (1)

c x0 x0 + h

f (x0 + h) f (x0 )= 1 2! h2 f (c) (2)

f (x0 )=0 1

f (x0 ) > 0 f x0 (x0 δ,x0 + δ ) f (x) > 0 h x0 + h ∈

(x0 δ,x0 + δ ) |h| <δ c ∈ (x0 δ,x0 + δ ) f (c) > 0 2

f (x0 + h) f (x0 ) > 0(|h| <δ ), f (x) >f (x0 )

x ∈ (x0 δ,x0 + δ ) f x0 f (x0 ) < 0 f x0 x0 f (x0 )=0

i f (x) < 0 x<x0 f (x) > 0 x>x0 f x0 f

ii f (x) > 0 x<x0 f (x) < 0 x>x0 f x0 f

iii f (x) x0 f

f (x)=(x 1)2 1

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

(x)= e x

f (x) > 0 R f f (x)= 3 √x2 x =0 x =0 f (0) f x0

x → x 2 x =0 (x 2 ) =2 > 0 0 x =0 [0, 2π ]

3π 2 f (x)= x f Ä π 2 ä = f 3π 2 =0,f Ä

2 ä < 0,f 3π 2 > 0. f (x)=3x 4 8x 3 18x 2 +24 f (x)=12x 3 24x 2 36x =12x(x 2 2x 3)=12x(x +1)(x 3), f 0 1 3 f (x)=36x 2 48x 36 f (0)= 36 f ( 1)= 48 f (3)=144 f 0 1 3 f f (x)= x 2 e x 2 f (x)=2xe x 2 2x 3 e x 2 =2xe x 2 (1 x 2 ), 0 1 1 x< 1 0 <x< 1 f (x) f x =0 f x = 1 x =1 iv [a,b] f (a,b) x1 ,x2 ,...,xp ∈ (a,b) f f [a,b] a b x1 ,x2 ,...,xp x1 ,x2 ,...,xq f [a,b] a b x1 ,x2 ,...,xq f f (x) x 3 3x +1 10+3x x 2 2x 3 + 3 2x 1 x (1+ x)2 √x 1+ x x + 2 x f

f (x)= ax + b f (x)= ax + b cx + d f (x)= x 2 4 x2 +2x 3 x 2 +3x +1 x2 + x 1 f 1 2 4 f (0)=8 f x → x 3 + ax 2 + bx + c 0 x = 2 x = 2 3 a b c

f [a,b]

(x) 0 x ∈ [a,b] f x0 f f (x0 )=0 f (x) x0 f (x0 a)f (x0 + a) < 0 a> 0 f f (x)= x k x> 0 k ∈ R k> 1 k< 0 0 <k< 1 R+ (x k ) = k (k 1)x k 2 x> 0 k (k 1) k> 1 k< 0 0 k 1 x → e x (e x ) > 0 x R+ ( x) = 1 x2 < 0.

(0, 2π ) ( x) = x ( x) < 0 x ∈ (0,π ), ( x) > 0 x ∈ (π, 2π ), x ∈ (0,π ) x ∈ (π, 2π ) x = π ( x) = 0 x = π

f (x)= x + x

f (x)= x 3 √x2 1

f (x)= 1+ x 1 x 4

f (x)= x x

f (x)= 1 4 2x 1 9 3x f (x)= 1 1+ x2

f (x)= 1 12 x 4 + 1 6 x 3 x 2 +3x 5 (1+ x)(1+ y ) 1+ 1 2 (x + y )(x> 1,y> 1)

2 x + y 2 x + y Ä|x| π 2 , |y | π 2 ä 2 x+y 2 2x 1 +2y 1 (x ∈ R,y ∈ R

y = f (x)

x x = 1 x =1

x→±1+0 f (x)=+∞,

x→±1 0 f (x)= −∞,

x =1 x = 1 f x> 0 f

x→±∞ f (x) x = x→±∞ x 2 +1 x2 1 =1,

x→±∞(f (x) x)= x→±∞ 2x x2 1 =0, y = x

x =0 f 1 <x< 0 1 <x

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

f (x) < 0

f (x)= (3x 2 +1)(x 2 1) (x 3 + x)(2x) (x2 1)2 = x 4 4x 2 1 (x 1)2 , f (x)=0 x 4 4x 2 1=0 t2 4t 1=0 2

2+ √5 f x = 2+ √5 f f x = 2+ √5 f (x)= (4x 3 8x)(x 2 1)2 (x 4 4x 2 1)4x(x 2 1) (x2 1)4 = 4x(x 2 +3) (x2 1)3 , f (x) > 0 x> 1 f (x) < 0 0 <x< 1 f (0)=0 x> 1 0 <x< 1 1 <x< 0 x< 1 x =0 y =0 f f f (x)= x 2 2 x,

5 x 9 x2 x 2 +1 x2 4 x 2 1 x2 25 x 2 5x +4 x2 9 x 2 2x 3 2x x2 2x x2 1

x + 2x x2 1 √3x +1 x +1 √x +3 2x 1 2 x x x 1 x e x xe 1 x (1 x 2 )e x 1 x 1 x x x x x x 2 2 x |x 3 2x 2 |− x

f g (a,b) x0 ∈ (a,b)

y = f (x) y = g (x) A x0

f (x0 )= g (x0 ) y = f (x) A

y = f (x0 )(x x0 )+ f (x0 ), (1) y = g (x)

y = g (x0 )(x x0 )+ g (x0 ) (2) 1 2

f (x0 )= g (x0 ),f (x0 )= g (x0 ) (3) y = f (x) y = g (x) x0 3 n f g

f (x0 )= g (x0 ),f (x0 )= g (x0 ),...,f (n) (x0 )= g (n) (x0 ), y = f (x) y = g (x) x0 n

f (n+1) (x0 ) = g (n+1) (x0 ) y = f (x)

y = g (x) x0 n

y = f (x) y = g (x) x0 n y = g (x) y = f (x) x0

f g f (x)= x 2 g (x)=0 f (0)= g (0),f (0)= g (0),f (0) = g (0), y = x 2 y =0 x0 x x =0

(0)= g (0),f (0)= g (0),f (0)= g (0),f (3) (0)= g (3) (0),f (4) (0) = g (4) (0),

P (x)= a1 +2a2 (x x0 )+3a3 (x x0 )2 + ··· + nan (x x0 )n 1 ,

P (x)=2a2 +3 2a3 (x x0 )+ + n(n 1)an (x x0 )n 2 ,

P (n) (x)= n!an ,

P (x0 )= a0 ,P (x0 )= a1 ,P (x0 )=2!a2 ,...,P (n) (x0 )= n!an . (5)

f (x0 )= P (x0 ),f (x0 )= P (x0 ),...,f (n) (x0 )= P (n) (x0 ), 5 a0 = f (x0 ),a1 = f (x0 ),a2 = 1 2! f (x0 ),...,an = 1 n! f (n) (x0 )

P (x)= f (x0 )+ f (x0 )(x x0 )+ 1 2! f (x0 )(x x0 )2 + + 1 n! f (n) (x0 )(x x0 )n .

f (x)= x x0 =1 n =2

f (x)= e x x0 =0 y =1+ x + x 2 2! + ··· + x n n! y = e x n x0 y = f (x)

i f (x)=3x 4 6x 3 +1 ii f (x)= 2x 1+ x2

iii f (x)= x iv f (x)= a 2 x + b 2 x a b ∈ R

f x0 f (x0 ) =0

Åx x0 + f (x0 )(1+ f (x0 )2 ) f (x0 ) ã2 + Åy f (x0 ) 1+ f (x0 )2 f (x0 ) ã2 = (1+ f (x0 )2 )3 f (x0 )2 (6)

y = f (x) (x0 ,f (x0 ))

6 y = f (x) (x0 ,f (x0 ))

(3, 4) xy =12

f (x + h) ≈ f (x)+ f (x)h, (1)

f (x + h)= f (x)+ f (x)h + 1 2 f (x + θh)h2 (0 <θ< 1), (2) x ∈ (a,b) f h x + h ∈ (a,b) 1 2 f (x + θh)h2 (0 <θ< 1), f (x + h) f (x)+ f (x)h x0 ∈ (a,b) 1 2 x = x0 x = x0 x0 + h = t 2

f (t)= f (x0 )+ f (x0 )(t x0 )+ 1 2 f (x0 + θ (t x0 ))(t x0 )2 , t = x0 + h (a,b) t x

f (x)= f (x0 )+ f (x0 )(x x0 )+ 1 2 f (x0 + θ (x x0 ))(x x0 )2 , (3) 0 <θ< 1 f (x) ≈ f (x0 )+ f (x0 )(x x0 ) (4) 3 4 x0 (a,b) f (x0 ) f (x0 ) 4 f (a,b) l l (x)= Ax + B,

A = f (x0 ) B = f (x0 ) f (x0 )x0 y = f (x) y = l (x)

y = f (x0 )+ f (x0 )(x x0 ),

y = f (x) (x0 ,f (x0 )) 3 4

|R| = 1 2 |f (x0 + θ (x x0 ))||x x0 |2 (0 <θ< 1). f [a,b] M |f (x)| M x ∈ [a,b] |R| 1 2 M |x x0 |2 , x x0 |R| y = l (x) y = f (x) x0 [a,b] x0 y = l (x) y = f (x) x1 ∈ [a,b] f [a,b] (a,b)

f (b)= f (a)+ f (c)(b a)(a<c<b) f (a,b) f [a,b]

f (b)= f (a)+ f (a)(b a)+ 1 2 f (c)(b a)2 (a<c<b) f (n+1) (a,b) f (n) [a,b]

f (b)= f (a)+ f (a)(b a)+ 1 2 f (a)(b a)2 + + 1 n! f (n) (a)(b a)n + 1 (n +1)! f (n+1) (c)(b a)n+1 , a<c<b x0 x ∈ [a,b] [x0 ,x] [x,x0 ]

f (x)= f (x0 )+ f (x0 )(x x0 )+ f (x0 ) 2! (x x0 )2 + ···

+ f (n) (x0 ) n! (x x0 )n + f (n+1) (c) (n +1)! (x x0 )n+1 ,

c x0 x f

f (x) ≈ f (x0 )+ f (x0 )(x x0 )+ f (x0 ) 2! (x x0 )2 + ··· + f (n) (x0 ) n! (x x0 )n (1)

Rn = f (n+1) (c) (n +1)! (x x0 )n+1

c x0 x c = x0 + θ (x x0 ) 0 <θ< 1 Rn = f (n+1) (x0 + θ (x x0 )) (n +1)! (x x0 )n+1 (2) 1 f n Tn

Tn (x)= f (x0 )+ f (x0 )(x x0 )+ f (x0 ) 2! (x x0 )2 + ··· + f (n) (x0 ) n! (x x0 )n f x0 Rn (n +1) f (n+1) [a,b] |f (n+1) (x)| M x ∈ [a,b] 2 |Rn | = f (n+1) (x0 + θ (x x0 )) (n +1)! (x x0 )n+1 M (n +1)! |x x0 |n+1 , x [a,b] n Tn f f (x)= e x f (n) (x)= e x x ∈ R n ∈ N x0 =0 e x =1+ x + x 2 2! + + x n n! + e θx (n +1)! x n+1 . (3) e x ≈ 1+ x + x 2 2! + + x n n! Rn |Rn | = e θx (n +1)! x n+1 e θx (n +1)! |x|n+1 n→∞ e θx |x|n+1 (n +1)! =0, n y = f (x) y = Tn (x) n x0

3 e 3 x =1 e =2+ 1 2! + + 1 n! + 1 (n +1)! e θ (0 <θ< 1) (4) e p q e = p q 4 n n>q n> 2 n!e = n! p q ∈ Z n>q q n! 4 n!e =2n!+ n! 2! + ··· + n! n! + e θ n +1 . (5) θ< 1 e< 3 e θ < 3 n> 2 n +1 > 3 1 n +1 < 1 3 1 n +1 e θ < 1 3 3=1 5 n!e 2n!+ n! 2! + ··· + n! n! 1 n +1 e θ 0 1 e f (x)= x x0 =0 f (n) (x)= Äx + n π 2 ä n =1, 2,... f (n) (0)= nπ 2 , f (2k +1) (0)=( 1)k ,f (2k ) (0)=0(k =1, 2,... ) n ∈ N x ∈ R f (x)= f (0)+ f (0)x + f (0) 2! x 2 + + f (n) (0) n! x n + f (n+1) (θx) (n +1)! x n+1 0 <θ< 1 n n =2k +1 k =0, 1,... x = x x 3 3! + x 5 5! + ( 1)k x 2k +1 (2k +1)! + ( 1)k +1 θx (2k +2)! x 2k +2 , (6) f (n+1) (θx)= f (2k +2) (θx)= (θx +(k +1)π )=( 1)k +1 θx 2k +1 R2k +1 |R2k +1 | = ( 1)k +1 θx (2k +2)! x 2k +2 |x|2k +2 (2k +2)! . n→∞ |x|n n! =0 R2k +1 k x x ∈ î 1 2 , 1 2 ó |x| 1 2 k =0 6 x = x + R1 , |R1 | |x|2 2! 1 22 2! = 1 8 =0, 125.

î 1 4 , 1 4 ó f f (x)= (1+ x)

0, 0005 f (x)= (1+ x) f (x)= 1 1+ x ,f (x)= 1 (1+ x)2 ,..., f (n) (x)= ( 1)n 1 (n 1)! (1+ x)n (n =1, 2,... ), f (n) (0)=( 1)n 1 (n 1)!, (1+ x)= x x 2 2 + x 3 3 +( 1)n 1 x n n + Rn , Rn = ( 1)n x n+1 (n +1)(1+ θx)n+1 (0 <θ< 1) 1 4 <x< 1 4 1 4 θ<θx< 1 4 θ 1 4 θx 1 4 , 3 4 1+ θx 5 4 , 4 5 1 1+ θx

|Rn | = |x|n+1 (n +1)|1+ θx|n+1 4n+1 (n +1)4n+1 3n+1 = 1 (n +1)3n+1 n |Rn | < 0, 0005 1 (n +1)3n+1 < 0, 0005 r (n)= 1 (n +1)3n+1 r (1)= 1 18 =0, 05555 ...,r (2)= 1 81 =0, 01234 ..., r (3)= 1 4 34 =0, 003086 ...,r (4)= 1 5 35 =0, 000823 ..., r (5)= 1 6 · 36 =0, 0002286 (1+ x) ≈ x x 2 2 + x 3 3 x 4 4 + x 5 5 1 4 x 1 4 , 0, 0005 a> 0 [ a,a] x ≈ 1 x 2 2 (8) 0, 0001

)

(1)

(x0 )=

∈ (x0 ,x0 + h)

(x0 + h)

<c<x0 + h c

θh

(x0 + h)

(x0 )

(x)= f (x),

(x)= f (x) x

2 F2 (x)= x 2 +3

F1 (x)= F2 (x)= f (x)

(x)=2x

x)= F (x)+ C,

(x)=Φ (x) (2)

x)= F (x)+ C (C ).

f (x) x = F (x)+ C (C )

2x x = x 2 + C x 2 x = 1 3 x 3 + C x 2 x x 2

(x)=

x → ex2 x → x x

F (x) x = F (x)+ C,

f (x) x = f (x) x,

F (x)= F (x)+ C,

f (x) x = f (x).

Kf (x) x = K f (x) x (K =0 ); (1)

(f (x) ± g (x)) x = f (x) x ± g (x) x. (2) K f (x) x = K f (x) x = Kf (x) x, 1

f (x) x ± g (x) x = f (x) x ± g (x) x

= f (x) x ± g (x) x =(f (x) ± g (x)) x, 2

F F (x)= f (x)

f (x) x

(2, 5) (0, +∞) ( 4, 1) (−∞, 0)

x =0 x ∈ ( 2, 3) x =0 x ( 2, 0) ∪ (0, 3)

4 8 9 10 11 12 (3x 2 x +1) x = x 3 1 2 x 2 + x + C x +1 √x x = (x1/2 + x 1/2 ) x = 2 3 x3/2 +2x1/2 + C 3 √x +1 x x = (x 2/3 + x 1 ) x =3x1/3 + |x| + C (2x +3x ) x = 2x 2 + 3x 3 + C (x +1) x = ( x 1+ 1 x) x = ( 1) x x +( 1) x x = 1 x + 1 x + C = (x +1)+ C 6+2x + x 2 x4 x 1+3x +5x 2 √x x 1+ x 2 +2x 4 2x2 x 1+ x 2 +2x 4 3 √x x (1 3x)2 x (1+ x 2 )3 x (ax + b)2 x4 x

3+4x 2 x ã2 x ax 2 + bx + c xn x (x n + a n )2 x x m + x n xp x ax 2 + b √x x (2 x +3 x) x (x +1) x (x + a) x e x +1 2 x e x 1 2 x (e x +5) x 1+ xe x x x 1+ x 2 x x2 x f (x) x t x 3x x. (1) x x 1 t =3x. (2)

(g (x))

(

)

= F (g (x))+ C. f (t) t = F (t)+ C

(g (x))g (x) x = F (g (x))+ C (F (g (x))) = f (g (x))g (x)(f = F )

(g (x)) g (x).

t = x x = ± 1 2 x x, x = ± 1

1 t2 t

± t3 √1 t2 t, 9 3 x = 2 x x =(1 2 x) x, x = x x,

3 x x = (1 2 x) x = ( 2 x 1) x, t = x (t2 1) t = 1 3 t3 t + C, 3 x x = 1 3 3 x x + C.

f (x) x

f (x)=(5+3x)55 f (x)= √1 2x

f (x)=(a + bx)n n = 1 b =0

f (x)= 1 2+3x2 f (x)= x 2+3x2 f (x)= 1 a2 + b2 x2

f (x)= x(4+7x 2 )9 f (x)= x(a + bx2 )n n = 1 b =0

f (x)= 3 x x f (x)= 3 x x f (x)= 6 x x

f (x)= 5 x x f (x)= n x x f (x)= n x x

f (x)= 1 x2 1 x f (x)= 2 x x f (x)= n x x

f (x)= x 1+ x2

f (x)= ( x)n √1 x2 f (x)= x 3 (1+3x 4 )18

f (x)= 1 √x2 + a t = x + x2 + a

f (x)= e x (a + bex )n

f (x)= √x √x f (x)= x 2 x

f (x)= 1 (1+ x) 2 x f (x)= x 1+ 2 x f (x)= e√x √x(1+ e√x )

u v (u(x)v (x)) = u (x)v (x)+ u(x)v (x), (1) (u(x)v (x))= v (x) u(x)+ u(x) v (x). (2) 1 2

)

xe x x 3 u(x)= x v (x)= e x u (x)=1 v (x)= e x xe x x = xe x e x 1 x = xe x e x + C.

u(x)= x v (x)= e x x

u(x)= x v (x)= e x xe x x = xe x e x x = xe x e x + C.

(x)= x g (x)= e x f (x)=1

g (x) x = e x 5 xe x x = xe x e x 1 x = xe x e x + C. 3 4 5

u v xe x x u(x)= e x v (x)= x u (x)= e x v (x)= 1 2 x 2 xe x x = 1 2 x 2 e x 1 2 x 2 e x x, xe x x x 2 e x x

= e x x x

= e x x e x x e x x x,

= e x x e x x J, J = 1 2 e x ( x x)+ C. f (x) x

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

f (x)= e 3x 2x f (x)= e x 2 x f (x)= e ax ax

f (x)= √x 2 x f (x)= 1 x2 2 x f (x)= x 2 e x

f (x)= x 2 x f (x)= 1 x2 x f (x)= √x

f (x)= x n x f (x)= e ax bx

y = f (x)

y = f (x) y =0 x = a x = b

P (x)

I [a,b] ⊂

a<x<b P (x) x

P (x) x P :[a,b] → [0, +∞) P (a)=0 h> 0 f [a,b]

x,x + h] h [x,x + h] ⊂ [a,b]

m(x,h) M (x,h) f h→0 M (x,h)= h→0 m(x,h)= f (x)

P (x + h) P (x) hm(x,h) P (x + h) P (x) hM (x,h), (1) hM (x,h)

(x,h)

(x + h) P (x) h M (x,h) h → 0

(x)= f (x). (2)

F (x)= f (x) x ∈ I 2

P (x)= F (x)+ C, (3)

P (a)=0 x = a 3 C = F (a) P (x)= F (x) F (a). (4) 4 Π y = f (x) x = a x = b y =0

P (b)= F (b) F (a), (5) F (x)= f (x) x ∈ I 5 x x = a x = b y = f (x) f [a,b] A =(a, 0) B =(b, 0) C =(b,f (b)) D =(a,f (a)) E =(b,f (a)) F =(a,f (b))

ABED

ABCF

Π(ABED ) < Π(ABCD ) < Π(ABCF ). (1) A1 =(a1 , 0) A B

G =(a1 ,f (a1 )) E1 =(b,f (a1 )) F1 =(a,f (a1 )) G1 =(a1 ,f (b)) G2 =(a1 ,f (a))

Π(AA1 G2 D ) < Π(AA1 GD ) < Π(AA1 GF1 ),

Π(A1 BE1 G) < Π(A1 BCG) < Π(A1 BCG1 ),

Π(AA1 G2 D )+Π(A1 BE1 G) < Π(ABCD ) < Π(AA1 GF1 )+Π(A1 BCG1 ) (2)

Π(AA1 GD )+Π(A1 BCG)=Π(ABCD ) Π(ABCD ) 2 1

Π(AA1 G2 D )+Π(A1 BE1 G) > Π(ABED ), Π(AA1 GF1 )+Π(A1 BCG1 ) < Π(ABCF ). A2 = A1 A B

Π(ABCD ) f [a,b] P y = f (x) x = a x = b x [a,b] x0 ,x1 ,...,xk 1 ,xk ,xk +1 ...,xn x0 = a xn = b x0 <x1 < <xn Pk P x = xk 1 x = xk P = n k =1 Pk mk Mk f [xk 1 xk ] πk Ak Bk Ck Dk Πk Ak Bk Ek Fk πk <Pk < Πk (k =1, 2,...,n).

πk = mk (xk xk 1 ), Πk = Mk (xk xk 1 ),

mk (xk xk 1 ) <Pk <Mk (xk xk 1 )(k =1, 2,...,n) n k =1 mk (xk xk 1 ) <P< n k =1 Mk (xk xk 1 ),

k =1

(an ) an = n k =1 mk (xk xk 1 )

(An )

n = n k =1 Mk (xk xk 1 )

n→∞ n k =1 mk (xk xk 1 )= n→∞ n k =1 Mk (xk xk 1 ),

a,b] n [x0 ,x1 ] [x1 ,x2 ] ..., [xn 1 ,xn ] (P ) a = x0 ,x1 ,...,xn = b (x0 <x1 < ··· <xn ) (T )

1 ,x

]

k = xk xk 1 (k =1, 2,...,n) d = (d1 ,d2 ,...,dn ) ε> 0 δ> 0 P

x ∈ [a,b] m f (ξk ) M (k =1, 2,...,n) m(xk xk 1 )

(ξk )(xk xk 1 )

∈ [a,b] x a

a,x] a x b

a,b]

(t) t F :

a,b] → R x

(t) t. (1)

a,b] F 1 F (x)= f (x) x ∈ [a,b] h> 0

(x,x + h)

(x)= f (x)

F (x)= f (x) [a,b] a = x0 ,x1 ,...,xn = b (x0 <x1 < ··· <xn )

) F (b) F (a)= F (xn ) F (x0 ) = F (xn ) F (xn 1 )+ F (xn 1 ) F (xn 2 )+ F (xn 2 )+ + F (x1 ) F (x0 ) = n k =1 (F (xk ) F (xk 1 )).

(a,b)

(xk 1 ,xk ) ξk ∈ (xk 1 ,xk ) F (xk ) F (xk 1 ) xk xk 1 = F (ξk ), F (x)= f (x) F (xk ) F (xk 1 )= f (ξk )(xk xk 1 ).

(b) F (a)= n k =1

(ξk )(xk xk 1 ) (2)

[xk 1 ,xk ] d

→0 (F (b) F (a))= d→0 n k =1

(b) F (a)= d→0 n k =1

(ξk )(xk xk 1 ),

(ξk )(xk xk 1 ) (3) f [a,b] b a f (x) x 3

)(xk xk 1 ) (4) 3 4 1

(x) b a

(x) x = F (x)

= F (b) F (a)

(5)

(9)

[a,b]

y = f (x) x = a x = b x P = b a f (x) x. (1)

[a,b]

P = F (b) F (a), 5

a = x0 <x1 < ··· <xn = b.

mk Mk f [xk 1 xk ] f ξk ,ξk ∈ [xk 1 ,xk ]

f (ξk )= mk ,f (ξk )= Mk (k =1, 2,...,n).

k =1 mk (xk xk 1 ) P

k =1 Mk (xk xk 1 ),

f (ξk )(xk xk 1 ). (2)

k =1

f (ξk )(xk xk 1 ) P n k =1

[a,b]

= 1 k n (xk xk 1 )

d→0 n k =1

f (ξk )(xk xk 1 )= d→0 n k =1 f (ξk )(xk xk 1 )= b a f (x) x, 2

y = f (x) x = a x = b x f

y = f (x) x b a f (x) x y = f (x)

y = |f (x)| y = f (x) x P = b a |f (x)| x.

x = a(t t),y = a(1 t),t ∈ [0, 2π ]

f g [a,b] f (x) g (x)

x ∈ [a,b] y = f (x) y = g (x) x = a x = b

P = b a (f (x) g (x)) x.

f (x) g (x) x ∈ [a,b]

y = f (x) y = g (x) x = a x = b P = b a |f (x) g (x)| x.

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

c f (x) g (x) x ∈ [a,c] g (x) f (x)

x ∈ [c,b] x = a

c a (f (x) g (x)) x,

a<c1 <c2 < ··· <cn <b f (a,c1 ) (c1 ,c2 ) (cn ,b)

c1 ,c2 ,...,cn y = f (x) x

, 2,...,n

)

3 1=4+

= f (x) x x = a x = b f (x)=5+ 1 10 x 2 a =2 b =10 f (x)=1+2x +6x 2 a =2 b =3

f (x)= 1 2 (e x + e x ) a =0 b = 2 y = f (x) x f (x)= x 3 3x 2 6x +8

(x)= x 2 (1 x 2 ) f (x)=4 3 √x2

y = x 2 2x 3 y = 3 y =2x 2 y =4x

y 2 =4x x + y =0 y = x 2 y = x +2

y 2 =4x 3y 2x =4 y 2 =16x y =3x y 2 =2x x 2 =16y ax = y 2 ay = x 2 a> 0

y = x y = x + 2 x 0 x π x 2 =3y x 2 + y 2 +6x =0

x 2 + y 2 2x +2y 1=0 x 2 a2 + y 2 b2 =1 a,b> 0

y 2 = x 2 (4 x 2 ) 2x 2 + xy +3y 2 =1 y 2 =2x x 2 + y 2 =8 P =(2, 4) y 3 =4x 4 OP a ∈ (0, 5) x = a y = x 2 x 0 x x =5 y = x 5 4x 4 +4x 3 x ( 9, 0) y 2 =4x y = 2 1 x 1 2 y =9 x 2 y y = 3x 4 +1 x3 , x =1 x =2 x =2t t2 y =2t2 t3 x2/3 + y 2/3 = a2/3 (a> 0) f y = f (x) x x = a x = b

f (x)= x a = 1 b =2 f (x)=[x] a = 1 2 b = 5 2

f (x)= ⎧ ⎨

x,x ∈ [0, 1) x 2 +1,x ∈ [1, 2) 3,x ∈ [2, 5] a = 1 2 b = 9 2

[a,b]

S y = f (x) A = (a,f (a)) B =(b,f (b)) S = b a »1+(f (x))2 x. (1)

[a,b] a = x0 x1 xk 1 xk xn = b x0 <x1 < <xn M0 = A Mk =(xk ,f (xk )) k =1, 2,...,n 1 Mn = B S M0 M1

P =2π 2π 0 a(1 t) a2 (1 t)2 + a2 2 t t =2πa 2 2π 0 a(1 t)√2 2 t t =4πa 2 2π 0 (1 t) t 2 t =8πa 2 2π 0 1 2 t 2 t 2 t

16πa 2

2 1+ 1 3 1+ 1 3 = 64 3 πa 2 x x 2 a2 + y 2 b2 =1 y = x 0 x π 2y = e x e x 0 x 3 y 2 = x(2 x) y 2 = x 2 (1 x 2 ) x y 2 =16x 5 x 12 y = x 3 0 x 1 2 x 2 a2 + y 2 b2 =1 9y 2 = x(x 3)2 0 x 3 2y = e x + e x 1 x 1 y 2 =4ax x 2 =4ay a> 0 x y = x 2 + 9 x2 x x =1 x =3 x O y 2 =4ax a> 0 P =(X,Y ) OP x 2 3 aX 2 3y 2 = x(x 1)2 x x y = x 2 +1 y =2 y 2 =12x 6 x x x2/3 + y 2/3 = a2/3 x

x → y (x) (a,b) y (x)=0(x ∈ (a,b)) (1) y (x)= C (x ∈ (a,b)) (2) C y (x)= f (x)(x ∈ (a,b)) (3) f (a,b)

(x)= f (x) x. (4) 1 3 y 2 4 1 3

)

(x) y 1 3

y (x),y (x),... y,y ,... y,y ,... x y (x),y (x),... 5 y = y,x 2 y + xy + y =0,yy =(y )2 . y =(5x +8)e x

y 2y + y =0 y y (x)= (5x +8)e x y (x) 2y (x)+ y (x)=0

= Cex C

= y. (6) 6 e x

e x ye x =0 (7) (ye x ) (ye x ) = y e x ye x , 7 (ye x ) =0 (8) 8 8 ye x = C (C ), y = Cex (9) 6 9 9 y 9 6 9 6 y = f (x)g (y ), (1) f g x → y (x) 1 1 g (y ) =0 y x = f (x)g (y ), y g (y ) = f (x) x, (2)

(a)= b, x = a y = b, (9) a b C 9 (a,b) (1+ x)y +1+ y =0, (10) y (2)=8. (11)

)(1+ y )= A, (12)

(= ±e C )

(2, 8)

x)(1+ y )=27

y +1) x y Ä π 4 ä = 1 2

y Cx) =0 y Cx = D D 1

y = f (x), (3)

=3 y = 3 2 x 2 + Cx + D

= f (x) x x + Cx + D,

f (x) x x =0, y R y (0)=1 y (0)=2 y (x)=6x +2 y =6x +2 y = x 3 + x 2 + Cx + D C D y =6x +2 y (x)= x 3 + x 2 + Cx + D y (x)=3x 2 +2x + C. x =0 y (0)=1 y (0)=2 y (0)= D =1,y (0)= C =2. y (x)= x 3 + x 2 +2x +1

y + k 2 y =0, (4) k k =0 4 1 4

1 k y + ky =0

1 k y kx + ky kx =0

1 k y kx y kx + y kx + ky kx =0

Å 1 k y kxã +(y kx) =0

Å 1 k y kx + y kxã =0

1 k y kx + y kx = C C

y kx + ky kx = Ck y kx + ky kx 2 kx Ck 2 kx =0

Å y kx ã (C kx) =0

Å y kx C kxã =0 y kx C kx = D D y = C kx + D kx. (5) 4 5 C D 5 y = Ck kx Dk kx y = Ck 2 kx Dk 2 kx = k 2 (C kx + D kx) = k 2 y, 4 5 4 y +9y =0 (6) y (0)=2,y Ä π 2 ä = 1 6 y (x)= C 3x + D 3x, y (0)= D =2 y Ä π 2 ä = C = 1 C =1 D =2 y (x)= 3x +2 3x 6 y (0)=2 y (π )= 1

F (x)= f (x)

f (a,b) x ∈ (a,b)

f (x)=0 C f (x)= C x ∈ (a,b) f (a,b)

[a,b] (a,b)

(−∞,a) (a, +∞)

a<b c<d D f

D =(a,b) ∪ (c,d) f (x)=0 x ∈ D f

f (x)=0 x ∈ (a,b) ∪ (c,d) f (x)=0 x ∈ (a,b)

f (x)=0 x ∈ (c,d)

(a,b) (c,d) f (a,b) (c,d) C1 C2

f (x)= C1 x ∈ (a,b)

f (x)= C2 x ∈ (c,d) f

f (x)= ßC1 ,a<x<b C2 ,c<x<d

f (x)=0 x ∈ (a,b) ∪ (c,d)

f (x)= g (x) x ∈ (a,b) ∪ (c,d),

(f (x) g (x)) =0 x ∈ (a,b) ∪ (c,d) f (x) g (x)= C1 x ∈ (a,b)

f (x) g (x)= C2 x ∈ (c,d),

f (x)= ßg (x)+ C1 ,a<x<b g (x)+ C2 ,c<x<d

C1 C2 1 x = 1 x2 (x =0), f g

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

(−∞, 0) ∪ (0, +∞)

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

f (x)= ßg (x)+ C1 ,x< 0

g (x)+ C2 ,x> 0 1 x2 x = ⎧ ⎨ ⎩ 1 x + C1 ,x< 0 1 x + C2 ,x> 0

C1 C2

a<b c<d

Df f (a,b) ∪ (c,d)

Dg g (a,b) Dh h (c,d)

f (x)= g (x) x ∈ (a,b)

f (x)= h (x) x ∈ (c,d).

f (x)= g (x)+ C1 x ∈ (a,b) f (x)= h(x)+ C2 x ∈ (c,d),

f (x)= ßg (x)+ C1 ,a<x<b h(x)+ C2 ,c<x<d

C1 C2 f (x)= 1 x x,g (x)= ( x),h(x)= x.

Df =(−∞, 0) ∪ (0, +∞) Dg =(−∞, 0) Dh =(0, +∞) f (x)= g (x) x ∈ (−∞, 0) f (x)= h (x) x ∈ (0, +∞) f (x)= g (x)+ C1 x ∈ (−∞, 0) f (x)+ h(x)+ C2 x ∈ (0, +∞) 1 x x = ß ( x)+ C1 ,x< 0 x + C2 ,x> 0

C1 C2 iv x → 2x 3 x2 3x +2 (−∞, 1) ∪ (1, 2) ∪ (2, +∞) 2x 3 x2 3x +2 = 1 x 1 + 1 x 2 , 2x 3 x2 3x +2 x = 1 x 1 x + 1 x 2 x = ß (1 x)+ C1 ,x< 1 (x 1)+ C2 ,x> 1 + ß (2 x)+ C3 ,x< 2 (x 2)+ C4 ,x> 2 = ⎧ ⎨ ⎩ (1 x)+ (2 x)+ C1 + C3 ,x< 1 (x 1)+ (2 x)+ C2 + C3 , 1 <x< 2 (x 1)+ (x 2)+ C2 + C4 ,x> 2

⎧ ⎨ ⎩ (x 2 3x +2)+ K1 ,x< 1 (3x 2 x 2 )+ K2 , 1 <x< 2 (x 2 3x +2)+ K3 ,x> 2 K1 (= C1 + C3 ) K2 =(C2 + C3 ) K3 (= C2 + C4 )

= F (b) F (a), (F = f ). (5)

a,b] 4

1 (2+ (x 1)) x y =2+ (x 1) y =0 x = 1 x =2 f 2 [0, 1] F (1) F (0) 1 F f 3 x y = f (x) y =0 x =0 x =1 1

1, 2, 3}

}

);

S = {A,B,C

A B C

(A,B,C ), (A,C,B ), (B,A,C ), (B,C,A), (C,A,B ), (C,B,A

(A,B ), (A,C ), (B,A), (B,C ), (C,A), (C,B ); (1)

1 (A,B ) (B,A)

(A,C ) (C,A) (B,C ) (C,B ) A B C

S = {a1 ,a2 ,...,an } n a1 ,a2 ,...,an

n (a1 ,a2 ,...,an ) a1 a2 ...an

S = {a1 ,a2 } a1 ,a2 a1 a2 a2 a1

S = {a1 ,a2 ,a3 }

a1 a2 a3 ,a1 a3 a2 ,a2 a1 a3 ,a2 a3 a1 ,a3 a1 a2 ,a3 a2 a1 a1 ,a2 ,a3 6

a1 ,a2 ,a3 ,a4 S = {a1 ,a2 ,a3 ,a4 } a1 a2 a3 a4

2 a1 a3

; a4 a1 a2 a3 ,a4 a1 a3 a2 ,a4 a2 a1 a3 ,a4 a2 a3 a1 ,a4 a3 a1 a2 ,a4 a3 a2 a1 a1 a2 a3 a4 24

P (n) a1 ,a2 ,...,an P (2)=2 P (3)=6 P (1)=1 P (n)

a1 ,a2 ,...,an

2=3!=6

,a2 ,...,an

n+1

n+1 a1 ,a2 ,...,an ... n (n +1) n +1 n +1 n P (n +1)= (n +1)P (n)=(n +1)n!=(n +1)! 1 n ∈ N 1 n +1 1

3! 2! = 6 2 =3

P2 (3)= 3! 2! =3

0101 0101010 P (4)=4!=24 P2 (4)=12 P2 (4)=12

Å a1 a2 ...an ai1 ai2 ...ain ã

i1 ,i2 ,...,in {1, 2,...,n} 1

a1 ,a2 ,...,an ai1 ,ai2 ,...,ain 1 ai1 ,ai2 ,...,ain S = {a1 ,a2 ,a3 }

a1 a2 a3

1 a2 a3 ,

a1 a2 a3 ,a1 a3 a2 ,a2

3 ,a2 a3 a1 ,a3

2 ,a3 a2 a1 f = Å abcd bcad ã ,g = Å abcd cadb ã

S = {a,b,c,d}

f ◦ g = Å abcd bcad ã ◦ Å abcd cadb ã = Å abcd adcb ã , f a b g b a f ◦ g a a f b c g c d f ◦ g b d adcb bcad cadb i a,b,c,d ii 5, 6, 7, 9 a,b,c,d i b c ii ca iii ca i ii 0, 2, 4, 6, 8 i ii 4000 30 43215 i ii i ii iii

i 3, 4, 5, 5, 5 ii 3, 3, 4, 4, 4

abcabdacbacdadbadc bacbadbcabcdbdabdc cabcadcbacbdcdacdb dabdacdbadbcdcadcb

3 4 =4 3 2=24

n ∈ N n = 1 n = 2 4(2n +3) (n +4)(n +3) = 2 3 , n 2 5n 6=0 n =6

9V 3 n =5V 3 n n

S = {a1 ,a2 ,...,an } k k n n S S k k S { } k k n n S k S = {a1 ,a2 ,a3 } 2 S

a1 a2 ...ak k k n S = {a1 ,a2 ,...,an } k S k !

C k n k n k !

k !C k n = V k n ,

C k n = 1 k ! V k n = n(n 1) ··· (n k +1) k ! (2) 2 n k n k (n k )! n k = n! k !(n k )! .

S = {a,b,c,d,e,f }

C 3 6 = Ä 6 3 ä = 6! 3!3! =20

3 abc,abd,abe,abf,bcd,bce,bcf,cde,cdf,def. acd,ace,acf, bde,bdf, cef, ade,adf, bef, aef,

20 C 2 5 = Ä 5 2 ä = 5! 2!3! =10 a

C 2 4 = Ä 4 2 ä = 4! 2!2! =6 b C 2 3 = Ä 3 2 ä =3

c C 2 2 = Ä 2 2 ä =1 d 125 4 10

C 3 9 = Ä 9 3 ä =84 C 3 8 = Ä 8 3 ä =56 125 84+56 > 125 C 2 7 = Ä 7 2 ä =21

84+21=105 Ä 6 2 ä =15

105+15=120 Ä 5 2 ä =10

120+10 > 125 10 Ä 4 1 ä =4 S 9 i ii iii S i Ä 9 3 ä = 9 8 7 3 2 1 =84

ii Ä 9 2 ä = 9 8 2 =36 iii

Ä 9 4 ä = 9 8 7 6 4 3 2 1 =126

i Ä n k ä = Ä n n k ä ii Ä n k ä + Ä n k +1 ä = Ä n +1 k +1 ä

Ä n k ä = n! k !(n k )! Ä n n k ä = n! (n k )!(n (n k ))! = n! (n k )!k ! i

Ä n k ä + Ä n k +1 ä = n! k !(n k )! + n! (k +1)!(n k 1)! = n! k !(n k 1)! 1 n k + 1 k +1 = n!(n +1) k !(k +1)(n k 1)!(n k ) = (n +1)! (k +1)!(n k )! = Ä n +1 k +1 ä, ii S = {a1 ,a2 ,...,an } k n S k 2 S = {a1 ,a2 ,a3 }

a1 a1 ,a1 a2 ,a1 a3 ,a2 a1 ,a2 a2 ,a2 a3 ,a3 a1 ,a3 a2 ,a3 a3 . a1 a2 a2 a1 a1 a3 a3 a1 a2 a3 a3 a2 2

a1 a1 ,a1 a2 ,a1 a3 ,a2 a2 ,a2 a3 ,a3 a3 .

a1 a2 ...ak k n

a1 a2 an

a1 a2+1 a3+2 a4+3 ...ak +k 1

a1 ,a2 ,...,an ,...,an+k 1 k n

k n + k 1

C k n k n

C k n = C k n+k 1 ,

C k n = n + k 1 k .

S = {a1 ,a2 ,a3 } 2

a1 a1 ,a1 a2 ,a1 a3 ,a2 a2 ,a2 a3 ,a3 a3 . {a1 ,a2 ,a3 ,a4 } a1 a2 ,a1 a3 ,a1

C 2 3 = C 2 4 = Ä 4 2 ä =6

S = {a,b,c} 3 S aaa,aab,aac,abb,abc,acc,bbb,bbc,bcc,ccc. 3 S =

{a,b,c,d,e} abc,abd,abe,acd,ace,ade,bcd,bce,bde,cde.

C 3 3 = C 3 5 = Ä 5 3 ä =10

S = {0, 1, 2, 3, 4} i 66 4 S ii 1224

i 0 C 3 5 = Ä 7 3 ä =35 1 C 3 4 = Ä 6 3 ä =20 2 C 3 3 = Ä 5 3 ä =10 65 66 3333 ii 0 35 11 C 2 4 = Ä 5 2 ä =10 122 C 1 3 = Ä 3 1 ä =3 1224 48 C 2 n =276 n C 2 n = Ä n +1 2 ä = (n +1)n 2! (n +1)n 2 =276 n 2 + n 552=0 23 24 n =23 2 3 4 a,b,c,d,e,f

15 4

80 13 6 3 4 7 3 2 2 n

i C 3 n + C 2 n =15(n 1)(n ∈ N) ii C 2 n 3 =21 n ∈ N iii C 3 n = 5 4 n(n 3) n ∈ N iv C n 4 n+1 = 7 15 n(n +1)(n 1) n ∈ N n 4

v V 3 n + C n 2 n =14n n ∈ N n 2 vi V 5 n =336 C n 5 n 2 n ∈ N n 5

vii C n 2 n+1 +2C 3 n 1 =7(n 1) n ∈ N n 2 x y

i C y x+1 : C y +1 x : C y 1 x =6:5:2 ii C y +1 x+1 : C y x+1 : C y 1 x+1 =5:5:3

i C 7 n+1 = C n 6 n+1 ii C 9 n + C 8 n = C 9 n+1 iii C 8 n C 8 n+1 + C 7 n =0 iv C k +1 n + C k 1 n +2C k n = C k +1 n+2

v C k n + C k n 1 + ··· + C k n 10 = C k +1 n+1 C k +1 n 10 2 3 4 a,b,c,d 24488 12 4 10 (a + b)n a,b ∈ R n ∈ N (a + b)1 = a + b; (a + b)2 = a 2 +2ab + b2 ; (a + b)3 = a 3 +3a 2 b +3ab2 + b3 . (1)

1 a + b (a + b)4

(2) 2 a + b

(a + b)4 = a 4 +4a 3 b +6a 2 b2 +4ab3 + b4

(a + b)5

(a + b)n n (a + b)(a + b) ··· (a + b) (3) 3

i a a a a = an ii b a an 1 b n(= C 1 n ) n an 1 b iii 1 k n 3 k C k n b n k a an k bk

C k n = n k iv b bn (a + b)n = an + C 1 n an 1 b + ··· + C k n an k bk + ··· + C n 1 n abn 1 + bn , (a + b)n = an + n 1 an 1 b + ··· + n k an k bk + ··· + n n 1 abn 1+ bn . (4) 4 4 4 n =1 (a + b)1 = a 1 + b1 n =2, 3, 4 1 2 4 n ∈ N 4 a + b (a + b)n+1 = an+ n 1 an 1 b + + n k an k bk + + n n 1 abn 1 + bn (a + b) = an+1 + n 1 an b + ··· + n k an k +1 bk + ··· + n n 1 a 2 bn 1 + abn + an b + n 1 an 1 b2 + + n k an k bk +1 + + n n 1 abn + bn+1

= an+1 + 1+ n 1 an b + ··· + n k + n k 1 an+1 k bk + ...

n k b

n 0 ä = n! 0!(n 0)! =1

Ä n n ä = n! n!(n n)! =1

(2a +3b)5 =(2a)5 + Ä 5 1 ä(2a)4 3b + Ä

(2a +3b)

2 ä(2a)3 (3b)2 + Ä 5 2 ä(2a)2 (3b)3 + Ä 5 4 ä(2a)(3b)4 +(3b)5 =32a 5 +240a 4 b +720a 3 b2 +1080a 2 b3 +810ab4 +243b5 . 5 0, 9974 0, 997=1 0, 003 0, 9974 =(1 0, 003)4 =1 Ä 4 1 ä(0, 003)+ Ä 4 2 ä(0, 003)2 Ä 4 3 ä(0, 003)3 +(0, 003)4 =1 4 · 3 · 10 3 +6 · 9 · 10 6 4 · 27 · 10 9 +81 · 10 12 =1 12 10 3 +54 10 6 108 10 9 +81 10 12 . 5 0, 98805 √x + 1 4 √x 9 x Ä 9 k ä(√x)9 k Å 1 4 √x ãk (k =0, 1,..., 9).

∈{

,...,

}

3k =0

r 5 r 1 7 r ∈{1, 2,..., 25} 25 r 5 r =5, 10, 15, 20, 25 r 1 7 r =8, 15, 22

( x + i x)5 ( x + i x)5 = 5 x + Ä 5 1 äi 4 x x Ä 5 2 ä 3 x 2 x Ä 5 3 äi 2 x 3 x + Ä 5 4 ä x 4 x + i 5 x. (5) ( x + i x)5 = 5x + i 5x. (6) 5 6 5x = 5 x 10 3 x 2 x +5 x 4 x, 5x = 5 x 10 3 x 2 x +5 x 4 x. 1+ x + 2x + ··· + nx = (n +1)x 2 x 2 nx 2 , (7)

x + 2x + ··· + nx = (n +1)x 2 x 2 nx 2 , (8) x 2 =0 (1 z )(1+ z + z 2 + + z n )=1 z n+1 , 1+ z + z 2 + ··· + z n = 1 z n+1 1 z (z =1) (9) 9 z = x + i x (=1) 1+( x + i x)+( x + i x)2 + ··· +( x + i x)n = 1 ( x + i x)n+1 1 ( x + i x) , (1+ x + 2x + ··· + nx)+ i( x + 2x + ··· + nx) = 1 ( x + i x)n+1 1 ( x + i x) (10) 1 ( x + i x)n+1 1 ( x + i x) = 1 (n +1)x i (n +1)x 1 x i x = 2 2 (n +1)x 2 2i (n +1)x 2 (n +1)x 2 2 2 x 2 2i x 2 x 2 = (n +1)x 2 x 2 (n +1)x 2 i (n +1)x 2 x 2 i x 2 = (n +1)x 2 x 2 (n +1)x 2 + i (n +1)x 2 x 2 + i x 2 = (n +1)x 2 x 2 Ä nx 2 + i nx 2 ä . 10 7 8 (x + a)6 x + 1 x 5 Ä x 2 + y 3 ä4 (2m 3n)6 (a 2 + b3/2 )6 ( 3 √m 3 √n)7 (√3 3√2)4 ( 3 √x √x)4 (1+ i)5

(1 i√2)6 (1+ i)4 (1 i)3 (1+ i)8 (1 i)8 (1+ i)6 (1 i)6

(√3+ √5)5 (√5 √3)5

i (1+ x +3x 2 )4 ii (1 2x + x 4 )6 (1+0, 04)15 (2√x x√8)10

i 1, 035 ii 1, 15 iii 0, 977 Ä

+1)12 x 6 (√x + √2)15 x 5 a i 1 3 √a2 + 4 √a2 17 ii 3 √a2 1 √a 15 iii a 3 √a

n ∈ N i (1+ x)n

(ax 1/2 + xa 1/2 )n

(x 1 + x 2 )n

z √z + 1 3 √z n z 5 128 5 √x2 1 2 6 √x n 153 x Än + 1 n än 14400 ak k (√3+ √2)n ak +2 : ak +1 : ak =28:8√6: 9 ak (2x + x 2 )n x x (1+ x 3 )30 i (√3+ √2)5 ii (√5 √2)10 iii Å 1 √x + 3 √xã12 (√2x 1 + 3 √2 x )n 5 1

i n ii n x 20 n

i 2x =2 x x

3x = x(3 4 2 x)

2x = 2 x 2 x

3x = x(4 2 x 3) v 4x = x(8 2 x 4 x) vi 4x =8 x 8 2 x +1 n 0 , n 1 , n 2 ,..., n k ,..., n n

n 1 11 121 1331 14641 15101051 1615201561 172135352171 n (a + b)n (a + b)7 = a 7 +7a 6 b +21a 5 b2 +35a 4 b3 +35a 3 b4 +21a 2 b5 +7ab6 + b7 . (a + b)n = n 0 an + n 1 an 1 b + ··· + n n bn a = b =1 a =1 b = 1 n 0 + n 1 + + n n =2n ; n 0 n 1 + ··· +( 1)n n n =0. Ä n +2 k ä = Ä n k ä +2Ä n k 1 ä + Ä n k 2 ä. (x +1)n+2 =(x +1)n (x +1)2 x x k 0 k n +2 x k Ä n +2 k ä x k x n + Ä n 1 äx n 1 + +1 (x 2 +2x +1). x k x k 2 x 2 x k 1 x x k x k Ä n k 2 ä +2Ä n k 1 ä + Ä n k ä,

Ä n +3 k ä = Ä n k ä +3Ä n k 1 ä +3Ä n k 2 ä + Ä n k 3 ä

Ä n +4 k ä = Ä n k ä +4Ä n k 1 ä +6Ä n k 2 ä +4Ä n k 3 ä + Ä n k 4 ä

1+2Ä n 1 ä + ··· +2n Ä n n ä =3n

Ä 2n 0 ä + Ä 2n 2 ä + ··· + Ä 2n 2n ä =2n 1

Ä 2n 1 0 ä + Ä 2n 1 2 ä + + Ä 2n 1 2n 2 ä =2n 1

Ä n 0 äÄ n k ä + Ä n 1 äÄ n k +1 ä + ··· + Ä n n k äÄ n n ä = Ä 2n n + k ä

Ä m + n m äÄ n 0 ä + Ä m + n m +1 äÄ n 1 ä + ··· + Ä m + n m + n äÄ n n ä = Ä 2n + m n ä

Ä m + n 0 äÄ n 0 ä + Ä m + n 1 äÄ n 1 ä + + Ä m + n n äÄ n n ä = Ä 2n + m n ä f gn

(f + g ) = f + g ;(f + g ) = f + g ,...

(f + g )(n) = f (n) + g (n) (n =1, 2,... ). (fg ) = f g + fg , (1) (fg ) =(f g + fg ) = f g + f g + f g + fg (fg ) = f g +2f g + fg , (2)

(fg )(3) =(f g +2f g + fg ) = f (3) g + f g +2f g +2f g + f g + fg (3) ,

(fg )(3) = f (3) g +3f g +3f g + fg (3) , (3)

1 1 1 2 1 2 1 3

3 3 1 1 2 3 n f g (fg )(n) = f (n) g + n 1 f (n 1) g + + n k f (n k ) g (k ) + + ··· + n n 1 f g (n 1) + fg (n) . (4) 4 n ∈ N n =1 4 1 4 n ∈ N 4 (fg )(n+1) = f (n+1) g + f (n) g + n 1 Äf (n) g + f (n 1) g ä + ··· + n k Äf (n k +1) g (k ) + f (n k ) g (k +1) ä + + n n 1 Äf g (n 1) + f g (n) ä + f g (n) + fg (n+1) = f (n+1) g + 1+ n 1 f (n) g + + n k + n k 1 f (n k +1) g (k ) + ··· + n n 1 +1 f g (n) + fg (n+1) (fg )(n+1) = f (n+1) g + n +1 1 f (n) g + ··· + n +1 k f (n k +1) g (k ) + ··· + n +1 n f g (n) + fg (n+1) , 4 n n +1 4 n ∈ N 4 n (fg )(n) = n k =0 n k f (n k ) g (k ) , f (0) = f g (0) = g F F (x)= x x F (6) (x)=(x x)(6) = x( x)(6) + Ä 6 1 ä( x)(5) . ( x)(n) =( 1)n 1 (n 1)! xn ,

(2k +1) (0)=( 1)k (2k )!(k =1, 2,... ).

(4) (0)= 3 2 f (0)=0,f (6) (0)= 5 4 f (4) (0)=0,

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

i (x1 +1)(x2 +1)(x3 +1)(x4 +1) ii (x1 +1)(x2 +1) (xn +1)

p n np 1 ≡ 1( p). (2) p n np ≡ n( p) (3) n =1 3 n ∈ N (n +1)p = np + Ä p 1 änp 1 + + Ä p p 1 än +1 (4) Ä p k ä k =1, 2,...,p 1 p p 4 (n +1)p ≡ np +1( p), (n +1)p (n +1)= np n ( p). (5)

S = {P,G}

S = {P,G,R}

{1} {2} {3} {4} {5} {6} {2, 4, 6}

= {1, 2, 3, 4, 5, 6}

i E1 S1 = {C,P,B } E1

23 =8 S1 S1 = {∅, {C }, {P }, {B }, {C,P }, {C,B }, {P,B }, {C,P,B }}. {C,P,B }

= {P,C }

E2 S2 = {BP,BC,PC }

S3 = {BP,PB,BC,CB,PC,CP }

iv E4 S4 = {BP,PB,BC,CB,PC,CP,BB,PP,CC } {1} {2}

A1 ∩ A2 ∩···∩ An Ai 1 i n

A1 ∪ A2 ∪···∪ An Ai 1 i n A1 ∪ A2 ∪···∪ An = S A1 ,A2 ,...,An S

S = {e1 ,e2 ,...,en } A1 ,A2 ,...,An Ai = ei i = 1, 2,...,n {e1 }∪{e2 }∪···∪{en } = {e1 ,e2 ,...,en } S A ⊂ S S = A ∪ A S A A A B A B e A B A B A B A B = A ∩ B

A ∩ B = ∅ A B A B A B S {i,j } i j i ∈{1, 2, 3, 4, 5, 6} j ∈{P,G}

S = {(1,P ), (1,G), (2,P ), (2,G), (3,P ), (3,G), (4,P ), (4,G), (5,P ), (5,G), (6,P ), (6,G)}

A = {(2,P ), (2,G), (4,P ), (4,G), (6,P ), (6,G)} B B = {(1,P ), (2,P ), (3,P ), (4,P ), (5,P ), (6,P )} A ∪ B

A ∪ B = {(1,P ), (2,P ), (2,G), (3,P ), (4,P ), (4,G), (5,P ), (6,P ), (6,G)}, A ∩ B

A ∩ B = {(2,P ), (4,P ), (6,P )}.

A = {(1,P ), (1,G), (3,P ), (3,G), (5,P ), (5,G)}. A B A B = {(2,G), (4,G), (6,G)} C 4 C = {(4,P ) (4,G)} C A C ⊂ A

S 42 =16

S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

4 C A = {(1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4)}, B = {(1, 1), (1, 2), (2, 1)}, C = {(1, 1), (2, 2), (3, 3), (4, 4)}. A ∩ B

= {(1, 1)}

⊂ A

∩ C B C 4 B C = {(1, 2), (2, 1)} A C A ∩ C A C = {(1, 3), (2, 4), (3, 1), (4, 2)} S A ∪ A = A A ∩ A = A

∪ B = B ∪ A A ∩ B = B ∩ A A ∪ (B ∪ C )=(A ∪ B ) ∪ C

∩ (B ∩ C )=(A ∩ B ) ∩ C A ∪ (B ∩ C )=(A ∪ B ) ∩ (A ∪ C ) A ∩ (B ∪ C )=(A ∩ B ) ∪ (A ∩ C )

∪ A = S

∩ A = ∅ A ∪ S = S

∩ S = A A ∪ B = A ∩ B A ∩ B = A ∪ B

A ∩ B A ∪ B A ∪ (A ∩ B ) ∪ (A ∪ B )= S A ∪ (A ∩ B ) ∪ (A ∪ B )= A ∪ (A ∩ B ) ∪ (A ∩ B ) = A ∪ (A ∩ (B ∪ B )) = A ∪ (A ∩ S ) = A ∪ A = S.

P (A)

(A)

f1 (A) f2 (A) f3 (A) ...

(A)

k (A) k =1, 2,...

, 5

20 30 40

3 1 3 S = {e1 ,e2 ,...,en } A A ∈ S

(A)

S R P (A)

i ei Ni Ni N f (ei )= Ni /N 0 f (ei ) 1 P

0 P (ei ) 1. ii f (e1 ),f (e2 ),...,f (en ) e1 ,e2 ,...,en

f (e1 )+f (e2 )+ + f (en )= N1 N + N2 N + + Nn N = N1 + N2 + + Nn N = N N =1. P P (e1 )+ P (e2 )+ + P (en )=1. iii e1 ,e2 ,...,er r n N1 ,N2 ,...,Nr A = {e1 ,e2 ,...,er } A A = {e1 }∪{e2 }∪···∪{er }

f (A)= N1 + N2 + ··· + Nr N = N1 N + N2 N + ··· + Nr N = f (e1 )+ f (e2 )+ ··· + f (er ) {ei } ei

P (A)= P (e1 )+ P (e2 )+ + P (er )

iv ∅ f (∅)= 0 N =0

P (∅)=0

S = {e1 ,e2 ,...,en } A ∈ S P (A) P S R

i 0 P (ei ) 1 1 i n ii n i=1 P (ei )=1

iii A = {e1 ,e2 ,...,er } P (A)= r i=1 P (ei )

iv P (∅)=0 P (S ) 1

S = {e1 }∪{e2 }∪···∪{en } iii ii

P (S )= n i=1 P (ei )=1

A ⊂ B

P (A) P (B )

A ⊂ B A

B A = B P (A)= P (B ) A = B B A ek A = {e1 ,e2 ,...,er }

B = {e1 ,e2 ,...,er ,ek } iii P (A)= P (e1 )+ P (e2 )+ ··· + P (er )

P (B )= P (e1 )+ P (e2 )+ ··· + P (er )+ P (ek ) P (B )= P (A)+ P (ek ) i P (ek ) 0 P (B ) P (A)

A 0 P (A) 1

A ∅⊂ A ⊂ S i 0 P (A) 1

S [0, 1] A A P (A)+ P (A)=1 A A

A P (A)+ P (A)

P (A)+ P (A)= P (S )=1

S = {e1 ,e2 ,...,en }

P (e1 )+ P (e2 )+ ··· + P (en )=1 P (e1 )= P (e2 )= ··· = P (en )

P (ei )= 1 n (1 i n). A = {e1 ,e2 ,...,er } = {e1 }∪{e2 }∪···∪{er }

P (A)= P (e1 )+ P (e2 )+ ··· + P (er )= 1 n + 1 n + ··· + 1 n = r n . 1/2 A = {e1 ,e2 ,...,er } e1 ,e2 ,...,er A A A = {2, 4, 6} A {2} {4} {6} A 5

S = {1, 2, 3, 4, 5, 6} A = {1, 2, 3, 4} A P (A)= 4 6 = 2 3 7 (i,j ) i,j ∈{1, 2, 3, 4, 5, 6} i j S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

)= 28 91

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

P (C )= 15 91 C

Ä 47 2 ä Ä 50 2 ä 1 1960 5 i ii i Ä 5 3 ä =10 1 5 1 P (A)= Ä 4 2 ä¿10= 6 10 = 3 5 ii A P (A)=1 3 5 = 2 5 1, 2, 3, 4, 5, 6 64 6 (1, 1, 1, 1) (2, 2, 2, 2) (3, 3, 3, 3) (4, 4, 4, 4) (5, 5, 5, 5) (6, 6, 6, 6)

6 64 = 1 63 = 1 216 i 4 ii 2 4!=24 i 4 4 32 36 52 56 2 3 5 6 4 2=8 8 24 1 3 ii 2 2 2 6 3!=6 2 · 6=12 12 24 1 2 6! 2!4! =15 5! 2!3! =10 10 15 = 2 3 i 4 ii 5 iii iv

(A1 ∪ A2 ∪ A3 )= P (A1 )+ P (A2 )+ P (A3 )= 14 100 + 6 100 + 4 100 = 6 25 . 32 1 8 9 16 17 24

(A ∪ B )= P (A)+ P (B ) P (A ∩ B )

P (∅)=0 5 1

1 A ∩ B =

∪ B = A ∪ (A ∩ B ),B =(A ∩ B ) ∪ (A ∩ B ), (6)

1 P (A ∪ B )= P (A)+ P (A ∩ B ),P (B )= P (A ∩ B )+ P (A ∩ B ). P (

P (A ∩ B )= 1 6 P (A ∪ B )= P (A)+ P (B ) P (A ∩ B )= 3 6 + 2 6 1 6 = 2 3 3 4

(i,j ) i,j ∈{1, 2, 3, 4, 5, 6} 36 A 3 3 6

A = {(1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6), (4, 3), (4, 6), (5, 3), (5, 6), (6, 3), (6, 6), (3, 1), (6, 1), (3, 2), (6, 2), (6, 5), (3, 4), (6, 4), (3, 5)}. A 20 P (A)= 20 36 = 5 9 B 4 4

B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)} 11 P (B )= 11 36

A ∩ B = {(3, 4), (4, 3), (4, 6), (6, 4)},

P (A ∩ B )= 4 36 5 P (A ∪ B )= 20 36 + 11 36 4 36 =

= 3 4 . A

C P (A ∪ B ∪ C ) 23

A = {(P,G,G), (G,P,G), (G,G,P )},

B = {(G,P,P ), (P,G,P ), (P,P,G), (G,G,P ), (G,P,G), (P,G,G), (G,G,G)}, C = {(P,G,G)}, P (A)= 3 8 P (B )= 7 8 P (C )= 1 8

A ∩ B = A,A ∩ C = B ∩ C = C,A ∩ B ∩ C = C,

P (A ∩ B )= 3 8 ,P (A ∩ C )= P (B ∩ C )= P (A ∩ B ∩ C )= 1 8 ,

P (A ∪ B ∪ C )= 3 8 + 7 8 + 1 8 3 8 1 8 1 8 + 1 8 = 7 8

(

)

P (B )= 2 5 5 A

(B )= 1 4

P (B |A) 1 5 S = {(i,j ) | i = j,i j ∈{1, 2, 3, 4, 5}}

5 · 4=20 A (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 3) (2, 4) (2, 5) 8 A P (A) 8 20 2 5

B (1, 2) (2, 1) A B P (B |A)= 2 8 = 1 4 S = {e1 ,e2 ,...,en } A B k =0 A r k r k B B A r k P (B |A)= r k P (A)= k n P (A ∩ B )= r n P (A ∩ B ) P (A) = r n k n = r k

(A ∩ B )

(B |A)=

(A) (1)

(B |A)

P (B |A)=

(A) , (1)

(B |A)

(A) =0

(A)=0

P (B |A)=

(A ∩ B )

(A) .

P (B |A)= 1 2 ,P (B )= 1 5 , P (B |A) = P (B ), A B P (B |A) = P (B ) A B S A ⊂ S S 3 1 2 3 4 5 6 P (A|S )

(A)

(A ∩ S )

∩ S = A

(S )=1 P (A|S )=

(S ) P (A|S )= P (A), (3) P (A|S ) P (A) 3 A S 5 (i,j ) i,j ∈{1, 2, 3, 4} 16 A

(B |A)= P (B ) A B

(A ∩ B )= P (A)P (B ). (1) 1 A B 1 2 3 4 5

A B S = {1C, 2C, 3C, 4C, 5P, 6P }, C P A = {2C, 4C, 6P },B = {1C, 2C, 3C, 4C },

(A)= 3 6 = 1 2 ,P (B )= 4 6 = 2 3

∩ B = {2C, 4C }, P (A ∩ B )= 2 6 = 1 3

(A ∩ B )= P (A)P (B ) A B 1 2 3 4 5 6 A B S {1C, 2C, 3C, 4P, 5P, 6P } A = {2C, 4P, 6P },B = {1C, 2C, 3C },A ∩ B = {2C }.

(A)= P (B )= 1 2 P (A ∩ B )= 1 6 P (A ∩ B ) = P (A)P (B ) A B 1

(A ∩ B ∩ C )= P (A)P (B )P (C ). (2) 2 P (A ∩ B )= P (A)P (B ),P (B ∩ C )= P (B )P (C ),P (A ∩ C )= P (A)P (C ). (3) A B C

(A)= P (B )= P (C )= 2 4 = 1 2 . A ∩ B B ∩ C C ∩ A

(A ∩ B )= P (B ∩ C )= P (C ∩ A)= 1 4 , 3 A ∩ B ∩ C P (A ∩ B ∩ C )= 1 4 2

1 2 0, 10 0, 20 0, 25 i ii iii iv

B C A B C A B C A B C A

A B C A ∩ B ∩ C A ∩ B ∩ C (A ∩ B ∩ C ) ∩ (A ∩ B ∩ C )= ∅ (A ∩ B ∩ C ) (A ∩ B ∩ C ) (A ∩ B ∩ C ) (A ∩ B ∩ C )

i A ∩ B ∩ C P (A ∩ B ∩ C )= P (A)P (B )P (C )=0, 1 · 0, 2 · 0, 25=0, 005

ii (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C )

P ((A ∩ B ∩ C ) ∪ (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C ))= P (A ∩ B ∩ C )+ P (A ∩ B ∩ C )

+ P (A ∩ B ∩ C )= P (A)P (B )P (C )+ P (A)P (B )P (C )+ P (A)P (B )P (C )

=0, 1 0, 2 0, 75+0, 1 0, 8 0, 25+0, 9 0, 2 0, 25=0, 08 iii (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C )

P ((A ∩ B ∩ C ) ∪ (A ∩ B ∩ C ) ∪ (A ∩ B ∩ C ))= P (A ∩ B ∩ C )+ P (A ∩ B ∩ C )

+ P (A ∩ B ∩ C )= P (A)P (B )P (C )+ P (A)P (B )P (C )+ P (A)P (B )P (C )

=0, 1 · 0, 8 · 0, 75+0, 9 · 0, 2 · 0, 75+0, 9 · 0, 8 · 0, 25=0, 375.

iv A ∩ B ∩ C

P (A ∩ B ∩ C )= P (A)P (B )P (C )=0, 9 0, 8 0, 75=0, 54. i ii iii iv 1 0, 6 i ii iii iv v vi vii A B C A B C

⊂ S B = S ∩

=(A1

∪···∪ An ) ∩ B

=(A1 ∩ B ) ∪ (A2 ∩ B ) ∪···∪ (An ∩ B ). A1 ,A2 ,...,An A1 ∩ B A2 ∩ B An ∩ B P (B )= P (A1 ∩ B )+ P (A2 ∩ B )+ + P (An ∩ B ). k 1 k n P (Ak ∩ B )= P (Ak )P (B |Ak ),

P (B )= P (A1 )P (B |A1 )+ P (A2 )P (B |A2 )+ ··· + P (An )P (B |An ), (1)

P (B )= n k =1

P (Ak )P (B |Ak ). (2) 1 2 B X Y Z X 50 Y 20 Z 30 X Y 2 Z 5 A1 X A2 Y A3 Z P (A1 )=0, 5 P (A2 )=0, 2 P (A3 )=0, 3 B

P (B )= P (A1 )P (B |A1 )+ P (A2 )P (B |A2 )+ P (A3 )P (B |A3 ).

P (B |A1 )=0, 02,P (B |A2 )=0, 02,P (B |A3 )=0, 05, P (B )=0, 5 0, 02+0, 2 0, 02+0, 3 0, 05=0, 029. A1 A2 ... An B

P (Ak ∩ B )= P (Ak )P (B |Ak )= P (B )P (Ak |B )(k =1, 2,...,n)

P (Ak |B )= P (Ak )P (B |Ak ) P (B ) ,

P (Ak |B )= P (Ak )P (B |Ak ) n k =1

P (Ak )P (B |Ak ) (k =1, 2,...,n).

... An

1 A2 An

X 1 1

S = {G,P } X (S )= {0, 1} S (i,j ) i,j =1, 2, 3, 4, 5, 6 S 36 X X (i,j )= i + j

X (1, 1)=2,X (1, 2)= X (2, 1)=3,X (1, 3)= X (2, 2)= X (3, 1)=4, X (S )= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, 11

X (S )= {x1 ,x2 ,... } X x1 x2 X (S ) X

X (S ) X (X = xk X xk ∈ X (S ) P (X = xk )

xk ∈ X (S ) p(xk )= P (X = xk ) {p(xk ) | xk ∈ X (S )} X X

x1 x2 ...xn ... p(x1 ) p(x2 ) ...p(xn ) ...

. S = {G,P } X X (G)=0 X (P )=1 X (S )= {0, 1} 0 p(0)= P (X =0) 1 p(1)= P (X =1) X 0 1 p(0)= p(1)= 1 2 01 1 2 1 2 X (i,j )= i + j X p(2)= P (X =2)= 1 36 , X 2 {(1, 1)} p(3)= P (X =3)= 2 36 , X 3 {(1, 2), (2, 1)}

p(k ) k =2, 3,..., 12 p(8)= P (X =8)= 5 36 , X 8 {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

(X = x1 ) ∪ (X = x2 ) ∪ ..., (X ∈ X (S )) p(x1 )+ p(x2 )+ ··· =1 . (xk ,p(xk ))

p(0)= P (X =0)= 1 3

p(1)= P (X =1)= 2 3 · 1 3

p(2)= P (X =2)= 2 3 2 3 1 3

p(3)= P (X =3)= 2 3 2 3 2 3

1 3 2 9 4 27 8

(A,A)= X1 ∩ X2 , (A, A)= X1 ∩ Y2 , (A,A)= X2 ∩ Y1 , (A, A)= X2 ∩ Y2 .

P (A,A) = P (X1 ∩ X2 )= P (X1 )P (X2 )= p · p = p 2 , P (A, A)= P (X1 ∩ Y2 )= P (X1 )P (Y2 )= p q,

P (A,A)= P (X2 ∩ Y1 )= P (X2 )P (Y1 )= q · p = p · q, P (A, A)= P (X2 ∩ Y2 )= P (X2 )P (Y2 )= q · q = q 2 .

E n P (B )= p k q n k n B A k A n k

P (A)= p P (A)= q =1 p n A k 0 k n

S = {A, A}

S n S n Xn A p(k ) Xn k k ∈{0, 1,...,n} S n n A A n k A n k A p k q n k n k k n p(k )= P (Xn = k )= n k p k q n k . Xn 0, 1, 2,...,n p(0)+ p(1)+ + p(n)= q n + n 1 pq n 1 + + n k p k q n k + + pn =1, p + q =1

P ((A,A)) P (A,A)

p = 2 8 = 1 4 q = 3 4 {0, 1, 2, 3, 4, 5}

p(0)= P (X =0)= 3 4 5 = 243 1024 ,

p(1)= P (X =1)=5 1 4 3 4 4 = 405 1024 ,

p(2)= P (X =2)= Ä 5 2 ä 1 4 2 3 4 3 = 270 1024 ,

p(3)= P (X =3)= Ä 5 3 ä 1 4 3 3 4 2 = 90 1024 ,

p(4)= P (X =4)= Ä 5 4 ä 1 4 4 3 4 = 15 1024 ,

p(5)= P (X =5)= 1 4 5 = 1 1024 . X X 3 X 3 P (X 3) P (X 3)= P (X =3)+ P (X =4)+ P (X =5) = p(3)+ p(4)+ p(5), P (X 3)= 90 1024 + 15

100 0, 03

X P (X =4)= 100 4 (0, 03)4 (0, 97)96 . P (X =4) 0, 171 100 4 n k n k n→∞ pn =0, n→∞ npn = λ> 0, (1) k =0, 1, 2,... n→∞ n k p k n (1 pn )n k = 1 k ! λk e λ . (2) Ä n k äp k n (1 pn )n k = n(n 1) (n k +1) k ! p k n (1 pn )n (1 pn ) k = 1 k ! (1 pn ) k n(n 1) (n k +1) nk (npn )k Å(1 pn ) 1 pn ã npn . (3) 1 n→∞ 1 k ! (1 pn ) k = 1 k ! , n

Å(1 pn ) 1 pn ã npn = e λ , n →∞ 3 2 2 n p Ä n k äp k (1 p)n k = 1 k ! (np)k e np (4) p< 0, 1 n> 50 np< 5 4 p =0, 03(< 0, 1) n =100(> 50) np =3(< 5) Ä 100 4 ä(0, 03)4 (0, 97)96 1 4! 34 e 3 0, 168

X 0, 1, 2,...

P (X = k )= 1 k ! λk e λ (λ> 0; k =0, 1, 2,... ) (5) 5

i P (X = k ) > 0 k =0, 1, 2,...

ii P (X =0)+ P (X =1)+ ··· + P (X = k )+ ··· = e λ + 1 1! λe λ + ··· + 1

= e λ e λ =1. t Xt [0,t) [0,t) n t/n pn t/n n t/n pn pn = a t n a> 0 n→∞ pn =0, n→∞ npn = at. Ai i =1, 2,...,n i t/n Xt A1 An

P (Ai )= pn A1 An Xt n ∈ N P (Xt = k )= Ä n k äp k n (1 pn )n k n →∞ t k P (Xt = k )= n→∞ Ä n k äp k n (1 pn )n k = 1 k ! e at (at)k S

]

a,b]

x ∈ [a,b]

P (X = x)=0. X {x1 ,x2 ,... } p(xk )= 1 2k p(x1 )+ p(x2 )+ ··· = 1 2 + 1 4 + 1 8 + ··· =1

x ∈ [a,b] 0

0 P (X = x)=0

(xk ,p(xk ))

k X

xk p(xk )

P (X = xk )= p(xk ). y = p(x) (x,p(x)) P (X = x)= p(x) P (X = x)=0

{x1 x2 ,...,xn } [a,b] x1 x2 ... xn 1 xk +1 xk =1 k =1, 2,...,n 1

k =1, 2,...,n 1 Ak =(xk , 0) Mk =(xk ,p(xk )) Nk =(xk +1

p(xk )) Π(Ak Ak +1 Nk Mk )

Ak Ak +1 Nk Mk

xn [a,b]

p(xk )=Π(Ak Ak +1 Nk Mk ) X x1 x2

P (a X b)

P (a X b)= n k =1 p(xk )p(xk )= n k =1 Π(Ak Ak +1 Nk Mk ),

[a,b]

P (a X b) x1 x2 xn ∈ [a,b] y = p(x) x x = a x = b X

P (a X b)

P (X = x)=0 P (a X b) b a

P (a X b)= P (X = x)=0 X P (X = x)

(a X b) X

[0, 30]

P (X =3)=0

P (2, 97 X 3, 03)

i p(x) 0 x ∈ R

ii +∞

p(x) x =1

iii P (a X b)= b a p(x) x a −∞ b +∞ X p p(x)= ß

(X< 1)=

(X< 3)=

) x

v P Å(1 <X< 2) X> 1 2 ã = P (1 <X< 2) ∩ X> 1 2 P X> 1 2 = P (1 <X< 2) P X> 1 2 (1, 2) ∩ 1 2 ,

E (X )=0 E (X ) > 0

E (X ) < 0 X p(x)= ® 3 8 (5 4x + x 2 ), 1 x 3 0, x E (X )= +∞ −∞ xp(x) x = 3 8 3 1 (5x 4x 2 + x 3 ) x =2

X p(k )= Ä n k äp k (1 p)n k , (4) E (X )= n k =0 kp(k )= n k =0 k Ä n k äp k (1 p)n k np E (X )= np. ii 1

E (X )= +∞ −∞ xp(x) x = μ ( ) X

D (X )

D (X )= E Ä(X E (X ))2 ä . (5) 5 X E (X ) X E (X ) X

S = {x1 ,x2 ,... } X a a {x1 a,x2 a,... } f S f (X ) {f (x1 ),f (x2 ),... } X X E (X ) (X E (X ))2 (X E (X ))2

D (X )= E (X 2 ) (E (X ))2 , 5

(X )

(S )

x1 x2 ... xk x1 < x2 < ··· <xk N

f1 f2 fk f1 f2 fk

f1 + f2 + + fk = N x1 x2 ... xk

fr1 = f1 N ,fr2 = f2 N ,...,frk = fk N ,

fr1 + fr2 + ··· + frk =1 x1 ,x2 ,...,xk

X x1 x2 x3 ... xk Σ

fi f1 f2 f3 ... fk N

fri fr1 fr2 fr3 ... frk 1

fi fri 32 4 9 10 3 6 5 4 2

X 1 2 3 4 5 Σ

fi 6 3 10 9 4 32

fri 3 16 3 32 5 16 9 32 1 8 1 [a,b] k

[a,a1 ) [a1 ,a2 ) ... [ak 1 ,b)

X fi fri

[a,a1 ) f1 fr1

[a1 ,a2 ) f2 fr2

[a2 ,a3 ) f3 fr3

[ak 1 ,b) fk frk

(x1 ,f1 ) (x2 ,f2 ) ... (xk ,fk ) x1 ,x2 ,...,xk (a,a1 )

(a1 ,a2 ) ... (ak 1 ,b) f1 ,f2 ,...,fk

[a,b) a1 ,a2 ,...,ak 1 [a,a1 ) [a1 ,a2 ) [ak 1 ,b) a1 ,a2 ,...,am 1 m>k

Mo = a1 + f2 f1 (f2 f1 )+(f2 f3 ) d, a1 d f1 f2 f3 [165, 170) a1 =165 f1 =7 f2 =8 f3 =5 d =5 Mo =165+ 1 1+3 · 5=165+1, 25=166, 25 x1 x2 xk {21, 25, 27, 30, 32} 27

{17, 19, 21, 23, 26, 28} Me = 21+23 2 =22

18 2 0 2 14 0 14

= |x1 x| + |x2 x| + + |xN x| N = 1 N N i=1 |xi x| X : x1 x2 xN x X : x1 ,x2 ,...,xN f1 f2 ... fk f1 + f2 + ··· + fk = N

= f1 |x1 x| + f2 |x2 x| + ··· + fk |xk x| N = 1 N k i=1 fi |xi x| 16 18 20 4 18 32 θ = |− 2| +2 3 = 4 3 θ = |− 14| +14 3 = 28 3 x X x x1 x2 xN x σ 2 = (x1 x)2 +(x2 x)2 + ··· +(xN x)2 N = 1 N N i=1 (xi x)2 X x1 x2 ... xk f1 f2 ... fk xi i =1, 2,...,k x X

2 = f1 (x1 x)2 + f2 (x2 x)2 + ··· + fk (xk x)2 N = 1 N k i=1 fi (xi x)2 ,

) (0, 32, 5, 68)

x = 5375 32 =167, 97; Mo =165+ 8 7 (8 7)+(8 5) · 5=166, 25;

σ 2 = 2817, 94 32 =88, 06; σ =9, 38 Mo< x (x σ, x + σ )=(158, 59 177, 35) 20 62, 50 (x 2σ, x +2σ )= (149, 21, 186, 73) 30 93, 75 X Y (X,Y ) S

2 X Y (X,Y )

Y (x1 ,y1 ) (x2 ,y2 ) ... (xn ,yn ) x1 x2 ... xn X

y1 y2 ... yn Y

(xi ,yi ) i =1, 2,...,n y

x y = ax + b

y = ax + b (1)

xi i =1, 2,...,n X 1 yi (= axi + b)

y ∗ i yi Y X Y

r 2 = n i=1 (y ∗ i y )2 n i=1 (yi y )2 r = œ n i=1 (y ∗ i y )2 n i=1 (yi y )2 1 X Y 0 1 X Y Y = aX + b

r 2 = Å n i=1 (xi x)(yi y )ã2 n i=1 (xi x)2 n i=1 (yi y )2 = Å 1 n n i=1 (xi x)(yi y )ã σ 2 X σ 2 Y σX σY X Y r r = CXY σX σY , CXY = 1 n n i=1 (xi x)(yi y )= 1 n n i=1 xi yi xy 10 X Y X 65887610497 Y 877105810686 X Y

xi yi xi x yi y (xi x)2 (yi y )2 (xi x)(yi y ) 6 8 1 0, 5 1 0, 25 0, 5 5 7 2 0, 5 4 0, 25 1 8 7 1 0, 5 1 0, 25 0, 5 8 10 1 2, 5 1 6, 25 2, 5 7 5 0 2, 5 0 6, 25 0 6 8 1 0, 5 1 0, 25 0, 5 10 10 3 2, 5 9 6, 25 7, 5

6 3 1, 5 9 2, 25 4, 5 9 8 2 0, 5 4 0, 25 1 7 6 0 1, 5 0 2, 25 0 Σ 70 75 30 24, 50 15 x =7 y =7, 5 r = 10 i=1 (xi x)(yi y ) 10 i=1 (xi x)2 10 i=1 (yi y )2 = 15 √30 24, 5 =0, 5533 X Y X Y

i ii iii iv

550

f (x,y )=0

y x f (x,y )=0 y (x,y )

f (x,y )=0 f (x,y )=0 (x,y ) x 2 + y 2 =1 (x,y ) 1 xy x y yy y 2 3

(x,y,z ) =(0, 0, 0)

xn + y n = z n n

f (x + h) f (x) h =0, (1) h h =0 f 1

f (x)=0

(2a x)y 2 = x 3 a> 0

x y y =3x x 2 3x x 2 y =0.

x x +˙xO y y +˙ yO 3x +3˙xO x 2 2x(˙xO ) (˙xO )2 y ˙ yO =0 (˙xO )2 2

18 e ix = x + i x,

(1) Å x n + i x n ãn = x + i x (n ∈ N)

(2) x n x n x n 0 1 x n x n n 2 Å1+ ix n ãn = x + i x.

(3) n Å1+ z n ãn ez n 3 e ix 1

f (x) f (x + y )

f : x f :(x + y )

fx f (x + y )

f (x)= ®x 1, 1 x< 0 x 2 +1, 0 x 1 [ 1, 1]

limites

(A,B )

A,B ) ∅ = A ⊂ Q, ∅ = B ⊂ Q,A ∪ B = Q,a ∈ A ∧ b ∈ B ⇒ a<b. A B (A,B )

1 1 f A B

f : A → B A B N K = {n 2 | n ∈ N}

f : N → K f (n)= n 2 K N N K N

N K

Q N R N Q R A A A A N N ( N) ( ( N)) ... 0 1

i {(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)} ii 512 iii 27 iv {(1,a)} {(1,a), (2,a)} {(1,a), (2,b)} {(1,a) (1,b) (2,c)} {(1,a), (2,b) (2,c)} v {(1,a), (2,a), (3,a)} {(1,a), (2,a), (3,b)} {(1,a) (2,b) (3,b)} {(1,b), (2,c) (3,c)} vi {(1,a), (2,b), (3,c)} {(1,a) (2,c) (3,b)} {(1,b) (2,a) (3,c)} vii {(1,a) (2,b) (3,c)} {(1,b) (2,a) (3,c)} {(1,c), (2,a), (3,b)} {(1,a), (2,c), (3,b)} {(1,b), (2,c), (3,a)} {(1,c), (2,b), (3,c)}

Df +g = Df g = Dfg = Df /g =(0, +∞) Df +g = Df g = Dfg = Df /g = R \{0}

Df +g = Df g = Dfg = Df /g =(0, +∞) Df +g = Df g = Dfg = R Df /g = R \{kπ | k ∈ Z} Df +g = Df g = Dfg = Df /g = R \{1} Df +g = Df g = Df g = (0, +∞) Df /g =(0, 1) ∪ (1, +∞) Df +g = Df g = Dfg = R \ ¶ 1 2 kπ | k ∈ Z© Df /g = R \ Ä¶ 1 2 kπ | k ∈ Z© ∪{−1, 1}ä Df +g = Df g = Dfg =(0, +∞) Df /g =(0, 1) ∪ (1, +∞) R R îkπ, π 2 + kπ ä k ∈ Z {x | x 0 x = 1 4 k 2 π 2 ,k ∈ Z} [0, +∞) (−∞, 2] ∪ [2, +∞) (0, +∞) (−∞, 2) ∪ (2, +∞) (−∞, 3) ∪ Ä 2 3 , +∞ä [0, +∞) R R (−∞, 1) ∪ ( 1, 1) ∪ (1, +∞) ( 4, 1 2√2) ∪ ( 1+2√2, 2) Ä 7 5 , +∞ä Ä−∞, 3 2 ä∪Ä 3 2 , 2ä∪(2, +∞) R R\{2} (−∞, 2)∪ Ä 5 2 , 4ä ∪ (4, +∞) Ä 3 5 , 4 5 ä ∪ Ä 4 5 , 13 2 ó (−∞, 4) ∪ (4, 5) (0, +∞) (−∞, 0) ∪ (0, +∞) ( 3, 2√2) ∪ (2√2, 3) ∪ (3, +∞) (−∞, √2) ∪ ( √2, 1) ∪ (1, √2) ∪ (√2, +∞) (2kπ,π +2kπ ) k ∈ Z x ∈ R x =0 x = 2 (1+2k )π k ∈ Z x ∈ R x = kπ x = π 4 + kπ k ∈ Z Äkπ, π 2 + kπ ä k ∈ Z x ∈ R x = kπ x = π 2 + kπ k ∈ Z i t2 3t +1 ii t 2 3t 1 +1 iii x 2 3x + t2 t iv 4t2 6t +1 v x 4 x 2 1 vi t4 3t2 +2 x 3(x +1) 1 x 3 x 2 x +1 3(x2 +1) 2x +1 x +4 2x 2 +4x 1 x(x 1) x 2x +1 x 2 x 1 x x 23x 2x 3 i xy x + y

x + y 1+

f Ä 1 x ä = f (x) i ii 1 2 iii a ∈ [1, 4] f Ä 1 x ä = 1 −∞ <x 1 2 f Ä 1 x ä = 1 x 1 <x< +∞ f Ä 1 x ä =1 1 4 x< 1 î 2, 1 2 ó ∪{1}

A = R A+ =(−∞, 1) ∪ Ä 4 5 , +∞ä A0 = ¶ 1, 4 5 © A = Ä 1, 4 5 ä A = R \{3} A+ = Ä−∞, 1 3 ä ∪ (3, +∞) A0 = ¶ 1 3 © A = Ä 1 3 , 3ä A = R A+ =(0, 1) ∪ (2, +∞) A0 = {0, 1, 2} A =(−∞, 0) ∪ (1, 2) A = R \{1} A+ =( 2, 0) ∪ (1, +∞) A0 = {0, 2}

A =(−∞, 2) ∪ (0, 1) A = R \{−1, 1} A+ =(−∞, 2) ∪ ( 1, 1) ∪ (2, +∞) A0 = {−2, 2} A =( 2, 1) ∪ (1, 2) A = R \ ¶ 1 2 , 2© A+ =(−∞, 5) ∪ Ä 1 2 , 2ä ∪ (3, +∞) A0 = {−5, 3} A = Ä 5, 1 2 ä ∪ (2, 3) A = R \{−1, 5} A+ = 1, 2 3 ∪ (2, 5) A0 = 2 3 , 2 A =(−∞, 1) ∪ Ä 2 3 , 2ä ∪ (5, +∞) A = Ä 7 5 , +∞ä A+ = Ä 5 3 , +∞ä

A0 = ¶ 5 3 © A = Ä 7 5 , 5 3 ä A =( 2, +∞) A+ =( 2, 3) A0 = {3} A =(3, +∞) A =( 3, +∞) A+ =( 2, +∞) A0 = {2} A =( 3, 2) x> 0 x< 0 x> 2 x< 2 Ä π 2 +2kπ, π 2 +2kπ ä Ä π 2 +2kπ, 3π 2 +2kπ ä k ∈ Z

i a> 0 ii a< 0

(12x 8x 3 )e x 2 421 0 m n m(m 1) (m n +1)x m n m<n 0 m n m(m 1) (m n +1)(x a)m n m<n 0 n! (1+ x)n+1 +( 1)n (1 x)n+1 2(1 x2 )n+1 ( 1)n+1 n! xn Äx + nπ 2 ä a n Äax + nπ 2 ä e x n k =0 n k 2 k !x n k e x n k =0 n k g (k ) (x) y 8x +13=0 8y + x 91=0 y +6x =16 6y x =22 12y 6√3x =9 2√3π 12√3y +24x =9√3+8π 4x 5y +12=0 5x +4y 26=0 3y +2x 30=0 2y 3x +19=0 (2, 12) ( 2, 20) Äa, a 2 ä Ä a, a 2 ä (2, 4) (3/11) x 3y +4=0 y +3x =28 7x 10y +44=0

, 3) (

t a + y t b =1

x x

Äx + nπ 2 ä x n ( 1)n+1 n! x n xn 12, 042 63, 04 5, 72 0, 11 1 1 1 3 (√7 1) 0 2 1 ± 19 3 (e 1) Ä 2

3) x ∈ Ä π 2 +2kπ, π 2 +2kπ ä x ∈ Ä π 2 +2kπ, 3π 2 +2kπ ä k ∈ Z R ad bc> 0 ad bc< 0 f (

i abcd abdc acbd acdb adbc adcb bacd badc bcad bcda bdac bdca cabd cadb cbad cbda cdab cdba dabc dacb dbac dbca dcab dcba ii 5679 5697

5796 5967

6597 6759 6795 6957 6975 7569 7596 7659 7695 7956 7965 9567 9576 9657 9675 9756 9765 i badc bdac ii cabd cadb iii bdca dbca i 5! ii 5! 4! i

ii 5!3! 2 · 3! 21543 43151 43251 i 6! ii 5! i 593 ii 510 iii 13990 i 20 ii 10 2(5!)2 9! 4!3!2! (4!3!2!)3! 10! 5! i 4! 2! ii 3!

49 6 4

C D

ab ac ad ba bc bd ca cb cd da db dc; abc abd acb acd adb adc bac bad bca bcd bda bdc cab cad cba cbd cda cdb dab dac dba dbc dca dcb i 60 ii 37 iii 756 30 29 28 2970 480 30 29 28 i 20 ii 10 iii 5 iv

15 6 24

ab ac ad ae af bc bd be bf cd ce cf de df ef abc abd abe abf acd ace acf ade adf aef bcd bce bcf bde bdf bef cde cdf def abcd abce abcf abde abdf abef acde acdf acef adef bcde bcdf bcef bdef cdef 12345 12456 12457 12458 23456 23457 23458 24567

10! 2 · 9!

4+4+4+4

4+3+2+2

4+5+5+4

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

y = k 2 y

Јован Д.

МАТЕМАТИКА ЗА IV РАЗРЕД ГИМНАЗИЈЕ

о издање, 20 . година

Издавач Завод за уџбенике Београд, Обилићев венац 5 www.zavod.co.rs

Ликовни уредник Тијана Павлов

Лектор

Графички уредник Александар

Дизајн и прелом Жељко Хрчек

Корице Александар Радовановић

Коректор Добрило Тошић

Обим: 20,75 штампарских табака

Формат: 16,5×23,5 cm

Тираж: 00 примерака

Штампање завршено 20 . године.

Штампа

Штампа , Београд