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2

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SGK: Sách giáo khoa.


4

I.
Môn Toán

giúp

HS)

môn
Toán,

HS

.
trong hai
p chúng ta

:

lý do nêu trên

.


5



f '( x0 )

lim
x x0

f ( x) f ( x0 )
x x0

1.1.1.2.

f '( x0 )

lim
x x0

f ( x) f ( x0 )
x x0
x0

f

x, tính y

x0
x
y

f '( x0 )



x

f x

a; b

a;b)

f' x

x

y'

y'

f' x

y

f x

x và kí

f ''( x).

y ''

Chú ý:

x

n

f ( x) , kí

n

x
n

,n 4 .

f

n

x .

f

n

x

f

n 1

x '


1.2.1.2. Bài

liên quan:
M x0 ; y0 .

A

1.2.2.1.

.
.

t0
t0 : v t0

s ' t0 .

Q Q t

Q

àm

Q Q t
Q Q t

t0

t 0 : I t0

v
t

(t ),

nên


9

,...
quan

,
có

.

Q(t
Q'(t
t1, t2
Q(t 2) Q(t1 )


10

I tb

I(t
sau: I

Tuy nhiê

2.2
2.2

.

.

2.2

.

2.2

.
2.2

2.2

.


12

[2].

g

+


A. f (3) 6

C. f (3) 3

B. f (3) 0

D. f (3)

y

Câu 2:
A. x0

C. y0

B. x x x0

D. y

Câu 3:

x0 là?

f x0

C. y

f x



y
C. lim

f ( x) f ( x0 )
x x0

D. lim

f ( x) f ( x0 )
x x0

x

B.

y

y y0

f ( x)

f x0

Câu 4:

y0

y



.

Câu 1:

y

f ( x)

x 2 3x

Câu 2:

y

f ( x)

x 2 3x

y

f (3

x)

x0
2

f (3) 3 x



x2 2 x 1

x0

f (1) 0
...(1)...

lim

x2 2 x 1

0

x 1

x 1

( x 1) 2
1
x 1

lim
x

lim( x 1) 0
x 1

.


0 . Ta có:

3

y
=......?
x

x0 1

x2 2 x 1

f x

f '(3) lim

x) 0

lim(3
x

f x

3

x

x 2 3x

f ( x)

(3) :

x

x

1

0

1.3.1.

toán 1:
x0

sau:
Cách 1:
x0 . Tính

x

y

f x0

x

f x0 .

y


2 x2 4x 1

x0 1.


17

x0 1 . Ta có:

x
f (1)

1

y

f 1

y
x

2

lim
x

0

x

y
x

f 1

2 x

lim 2 x
x

0

0

Cách 2:
2 x2 4x 1

f x
f (1)

x0 1 . Ta có:

1

2x2 4x 1
f ( x) f (1)
lim
= lim
x 1
x 1

y

x

x

2

f

4 2 x 7
( 3)
3
2 x

2

1

lim

0

2 . Ta có:

3

f

y

3
5 x

x
5

x

2.


18

Cách 2:
4x 7
3 x

f x
f ( 2)

2 . Ta có:

x0

3
4x 7
( 3)
lim 3 x
x
2


1
.
5

f '( 2)

f x

x2

x 1
x 1

HD:
Cách 1:
x

3 . Ta có:

x0

13
4

f (3)
y

f 3


2

x
15 x
x.(4 x 16)

y
lim
x 0 x

lim
x

f '(3)

4

0

2

x
15 x
x.(5 x 20)

4 x 15
0 4 x 16

lim
x

f (3)

x2
f ( x) f (3)
lim
x 3
x 3

lim
x

3

x 1 13
x 1
4
x 3

4 x2 9 x 9
3 4 x 1 x 3

lim
x

4x 3
2 4( x 1)

15
16


f '( x0 )

lim
x

f '( x0 )
f '( x0 )

x0

f ( x) f ( x0 )
.
x x0

lim
x

x0

f ( x) f ( x0 )
.
x x0

f '( x0 )
f x

x 2 3x
x 1

x 1


x 1 ( 2)
x 1

x 2 3x ( 2)
lim
x 1
x 1

1 nên

y

f ( x)

lim
x 1

lim
x 1

( x 1)
x 1

lim( 1)

1

x 1


x

f '(0 )

f ( x) f (0)
x 0

lim
0

f ( x) f (0)
x 0

lim
x

0

f '(0 )

y

sin x 0
x

lim
x

0


0 và

x0

f x

2 x2 7 x 3
2x 1

x

a

x

1
2.
1
2
1
.
2

x0

HD:
1
2

f


f ( x)

1
,d
2

x0

f

1
2

lim f ( x)
x

1
2

x0

lim x 3
x

1
2

5
2

0.


21

y

x2
;x 1
.
ax b ; x 1

f ( x)

x0 1.

Tìm
HD:
f (1)

lim f ( x) 1
x 1

lim f ( x)

a b

x 1

y


x 1

lim
x 1

lim
x 1

x2 1
x 1

lim
x 1

ax b 1
x 1

f ( x)

a b 1

x 1 x 1
x 1

lim
x 1

b


b 1 a

lim( x 1) 2

toán 3:

y

f ( x)

x 1

x0 1

Thay a 2 vào b 1 a
y

y

x0 1

y
.
x

x

a( x 1)
x 1



x 3x 2 3 x x
y
x

3x 2 3x. x

lim
x

0

y
x

2

x
x

2

lim 3x 2 3x. x
x

x

0

2


x

x. 3 x 2 3 x. x

x

3x 2 3x. x

lim
x

x

f x

y
x

x

2

lim 3 x 2 3 x. x
x

0

y


2. Bài

.
.

Câu 1

y

f ( x)

f ( x)

A. f '( x) 0, x K

C. f '( x) 0, x K

B. f '( x) 0, x K

D. f '( x) 0, x K

Câu 2

y

f ( x)

f ( x)

A. f '( x) 0, x K

B.

D.

Câu 5:

A.

C.

B.

D.

Hoàn thành các

.

Câu 1: Cho hàm s y

f ( x)

x3 2 x 2 4 . T

f ( x)

x3 2 x 2 4

nh c a hàm s


4
0

f '( x)

+

f ( x)

1

(
(4;

;4) .

)

Câu 5:
y

T

f ( x) 2 x 3 6 x 1

nh D

y ' 6 x2 6 ; y ' 0

...(1)... 0

1,1 .