2. 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
– Ta có x2 – 2x + 3 = (x – 1)2
2 1
f(x) x 2x 3
|x| 1
VD4:
= HD
0 x m – 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 – 1 (0 ; 2)
m 1 0 m 1
m 1 2 m 3
4. 4
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2
M(x0 ; y0) (G) x0 D và y0 = f(x0)
x
y
x
y
O
TOPPER. Chú ý
0 ; y0 y0 = f(x0)
VD6:
2 ?
= HD
2
1 = 02
O
VD7 0 ; y0 0 ; y0
y = x2
=
0 ; y0
2 – mx + 2 + m khi ta có:
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. Chú ý
y
y = c
O
c
x
1) = f(x2) x1, x2 K
x K (c là
8. 8
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D thì –x D:
VD9: 3 – 4x
= HD
–x và
f(–x) = 2(–x3) – 4(–x) = –2x3 + 4x = –f(x).
VD10:
= HD
[–2 ; 2] thì –x [–2 ; 2] và
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. Chú ý
4
x D thì –x D và f(–x) = f(x)
x D thì –x D và f(–x) = – f(x).
10. 10
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1
(a) (b) {–2 ; 2}
(c) x [–1 ; ) và x 1 (d) x
1
[ ; 2]
2
2
; 1]
(b) f(–2) = –5; f(–1) = 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 có:
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 và( ; 1) (1; )
8
[1; )
A
B
(b) a < 0
(2; )
C m = 1