�ݺ�ߣ

1
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htttp://tuhoc.edu.vn/blog
1
.
�C D
�C
�C hay
�C
x y = f(x)
VD1:
= HD
3
f(x) 2 x 1
|x| 2
x 1 0 x 1 x 1
x 1 v? x 2
|x| 2 0 |x| 2 x 2
TOPPER. Ch�� ?
0
0
1
f(x)
f(x)
f(x)
g(x)
g(x) 0

2
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VD2:
= HD
x 2
f(x) 2x 3
3 x
3
2x 3 0 x 3
x 32
23 x 0
x 3
VD3:
= HD
�C Ta c�� x2 �C 2x + 3 = (x �C 1)2
2 1
f(x) x 2x 3
|x| 1
VD4:
= HD
0 x m �C 1
2x
f(x)
x m 1
VD5:
= HD
y x m 1 2x m
x m 1
x m 1 0
m
2x m 0 x
2
m 1 0
m 0m
0
2
m �C 1 (0 ; 2)
m 1 0 m 1
m 1 2 m 3

3
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2
(b) T��nh f(�C2), f(�C1), , f(1).
3
2
2x 1 khi x < 1
f(x)
1 x khi 1 x 1
2
f( )
2
y x m 2x m 1
1
(a) (b)
(c) (d)
2x 1
y
|x| 1
x
y
|x| 2
x 1
y
x 1
y 2x 1 2 x

4
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2
M(x0 ; y0) (G) x0 D v�� y0 = f(x0)
x
y
x
y
O
TOPPER. Ch�� ?
0 ; y0 y0 = f(x0)
VD6:
2 ?
= HD
2
1 = 02
O
VD7 0 ; y0 0 ; y0
y = x2
=
0 ; y0
2 �C mx + 2 + m khi ta c��:
hay2
0 0 0
y x mx 2 m 2
0 0 0
y x 2 m(1 x )
0 0
2
00 0
1 x 0 x 1
y 3y x 2
TOPPER. Ch�� ?
y
y = c
O
c
x
1) = f(x2) x1, x2 K
x K (c l��

5
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4
(C): A(�C1 ; 1), B(�C1 ; �C1), C(1 ; �C1), D(1 ; 0)
5 0 ; y0
mx 1
y
x m

6
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3
x1, x2 K,
x1 > x2 f(x1) > f(x2)
x1, x2 K,
x1 > x2 f(x1) < f(x2)
x1, x2 K, x1 x2,
x1, x2 K, x1 x2,
2 1
2 1
f(x ) f(x )
0
x x
2 1
2 1
f(x ) f(x )
0
x x
VD8
tr��n
= HD
1 v�� x2 1 x2 ta c��:
.
2 1 2 1
2 1 2 1
f(x ) f(x ) (2x 3) (2x 3)
2 0
x x x x

7
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6
y = f(x) = x2 v��( ; 1) (1 ; ).
7
v��
2
x 1
( ; 1) (1 ; ).
8
(a) v��
(b)
3
y
x
( ; 0) (0 ; ).
y x 1 [1 ; ).
ax
y f(x) .
x 2
(2; ).
(2; ).

8
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D th�� Cx D:
VD9: 3 �C 4x
= HD
�Cx v��
f(�Cx) = 2(�Cx3) �C 4(�Cx) = �C2x3 + 4x = �Cf(x).
VD10:
= HD
[�C2 ; 2] th�� Cx [�C2 ; 2] v��
y f(x) 2 x 2 x
2 x 0
2 x 2
2 x 0
f( x) 2 x 2 x f(x)
TOPPER. Ch�� ?
4
x D th�� Cx D v�� f(�Cx) = f(x)
x D th�� Cx D v�� f(�Cx) = �C f(x).

9
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A
(a) f(x) = �C2x + 5 (b) f(x) = �Cx3 + 2x
(c) (d)
3
f(x)
x 2
2
f(x) x 2|x|
B 3
y f(x) 2x|x |
C 3 + (m2 �C 1)x2
VD11:
=
0 hay x �C1
.
y x 1
y x 1
[ 1; )
[ 1; ).

10
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1
(a) (b) {�C2 ; 2}
(c) x [�C1 ; ) v�� x 1 (d) x
1
[ ; 2]
2
2
; 1]
(b) f(�C2) = �C5; f(�C1) = 0; ; f(1) = 0
2 2
f( )
2 2
3
0
x m
x m 0
m 1
2x m 1 0 x
2
m 0
m 1
0
2
4
5 0 ; y0 khi ta c��:
hay 0 m
x0y0 + 1 = m(x0 + y0)
x0 khi:
mx 1
x m
0
0
0
mx 1
y
x m 0 0 0 0
x y my mx 1
0 0
0 0
x y 0
x y 1 0
m 1, m 1
7 v��( ; 1) (1; )
8
[1; )
A
B
(b) a < 0
(2; )
C m = 1

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M.2a.01. dai cuong ve ham so

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