200 cau-khaosathamso2 (1) 08
- 1. Kh畉o s叩t hm s畛
+ N畉u m 孫 0 , y蔵 続 0, "x (0; m) khi m > 0 ho畉c y蔵 続 0, "x (m; 0) khi m < 0 .
V畉y hm s畛 畛ng bi畉n trong kho畉ng ( x1; x2 ) v畛i x2 - x1 = 1
辿( x ; x ) = (0; m)
v x2 - x1 = 1 棚 m - 0 = 1 m = 賊1 .
辿
棚 1 2
谷( x1; x2 ) = (m;0) 谷0 - m = 1
C但u 9. Cho hm s畛 y = x 4 - 2mx 2 - 3m + 1 (1), (m l tham s畛).
1) Kh畉o s叩t s畛 bi畉n thi棚n v v畉 畛 th畛 c畛a hm s畛 (1) khi m = 1.
2) T狸m m 畛 hm s畛 (1) 畛ng bi畉n tr棚n kho畉ng (1; 2).
揃 Ta c坦 y ' = 4 x 3 - 4mx = 4 x( x 2 - m)
+ m 贈 0 , y 蔵続 0, "x (0; +促) m 贈 0 tho畉 m達n.
+ m > 0 , y 蔵= 0 c坦 3 nghi畛m ph但n bi畛t: - m , 0, m.
Hm s畛 (1) 畛ng bi畉n tr棚n (1; 2) m 贈 1 0 < m 贈 1 . V畉y m ( -促;1湛 .
短
C但u h畛i t動董ng t畛:
a) V畛i y = x 4 - 2(m - 1) x 2 + m - 2 ; y 畛ng bi畉n tr棚n kho畉ng (1;3) . S: m 贈 2 .
mx + 4
C但u 10. Cho hm s畛 y = (1)
x+m
1) Kh畉o s叩t s畛 bi畉n thi棚n v v畉 畛 th畛 c畛a hm s畛 (1) khi m = -1 .
2) T狸m t畉t c畉 c叩c gi叩 tr畛 c畛a tham s畛 m 畛 hm s畛 (1) ngh畛ch bi畉n tr棚n kho畉ng (-促;1) .
m2 - 4
揃 T畉p x叩c 畛nh: D = R {m}. y 蔵= .
( x + m)2
Hm s畛 ngh畛ch bi畉n tr棚n t畛ng kho畉ng x叩c 畛nh y 蔵< 0 -2 < m < 2 (1)
畛 hm s畛 (1) ngh畛ch bi畉n tr棚n kho畉ng (-促;1) th狸 ta ph畉i c坦 - m 続 1 m 贈 -1 (2)
K畉t h畛p (1) v (2) ta 動畛c: -2 < m 贈 -1 .
2 x 2 - 3x + m
C但u 11. Cho hm s畛 y = (2).
x -1
T狸m m 畛 hm s畛 (2) 畛ng bi畉n tr棚n kho畉ng (-促; -1) .
2x2 - 4x + 3 - m f (x)
揃 T畉p x叩c 畛nh: D = R {1} . y ' = 2
= .
( x - 1) ( x - 1)2
Ta c坦: f ( x ) 続 0 m 贈 2 x 2 - 4 x + 3 . 畉t g( x ) = 2 x 2 - 4 x + 3 g '( x ) = 4 x - 4
Hm s畛 (2) 畛ng bi畉n tr棚n (-促; -1) y ' 続 0, "x (-促; -1) m 贈 min g( x )
( -促;-1]
D畛a vo BBT c畛a hm s畛 g( x ), "x (-促; -1] ta suy ra m 贈 9 .
V畉y m 贈 9 th狸 hm s畛 (2) 畛ng bi畉n tr棚n (-促; -1)
2 x 2 - 3x + m
C但u 12. Cho hm s畛 y = (2).
x -1
T狸m m 畛 hm s畛 (2) 畛ng bi畉n tr棚n kho畉ng (2; +促) .
2x2 - 4x + 3 - m f (x)
揃 T畉p x叩c 畛nh: D = R {1} . y ' = 2
= .
( x - 1) ( x - 1)2
Ta c坦: f ( x ) 続 0 m 贈 2 x 2 - 4 x + 3 . 畉t g( x ) = 2 x 2 - 4 x + 3 g '( x ) = 4 x - 4
Hm s畛 (2) 畛ng bi畉n tr棚n (2; +促) y ' 続 0, "x (2; +促) m 贈 min g( x )
[2; +促 )
Trang 7
![Kh畉o s叩t hm s畛
+ N畉u m 孫 0 , y蔵 続 0, "x (0; m) khi m > 0 ho畉c y蔵 続 0, "x (m; 0) khi m < 0 .
V畉y hm s畛 畛ng bi畉n trong kho畉ng ( x1; x2 ) v畛i x2 - x1 = 1
辿( x ; x ) = (0; m)
v x2 - x1 = 1 棚 m - 0 = 1 m = 賊1 .
辿
棚 1 2
谷( x1; x2 ) = (m;0) 谷0 - m = 1
C但u 9. Cho hm s畛 y = x 4 - 2mx 2 - 3m + 1 (1), (m l tham s畛).
1) Kh畉o s叩t s畛 bi畉n thi棚n v v畉 畛 th畛 c畛a hm s畛 (1) khi m = 1.
2) T狸m m 畛 hm s畛 (1) 畛ng bi畉n tr棚n kho畉ng (1; 2).
揃 Ta c坦 y ' = 4 x 3 - 4mx = 4 x( x 2 - m)
+ m 贈 0 , y 蔵続 0, "x (0; +促) m 贈 0 tho畉 m達n.
+ m > 0 , y 蔵= 0 c坦 3 nghi畛m ph但n bi畛t: - m , 0, m.
Hm s畛 (1) 畛ng bi畉n tr棚n (1; 2) m 贈 1 0 < m 贈 1 . V畉y m ( -促;1湛 .
短
C但u h畛i t動董ng t畛:
a) V畛i y = x 4 - 2(m - 1) x 2 + m - 2 ; y 畛ng bi畉n tr棚n kho畉ng (1;3) . S: m 贈 2 .
mx + 4
C但u 10. Cho hm s畛 y = (1)
x+m
1) Kh畉o s叩t s畛 bi畉n thi棚n v v畉 畛 th畛 c畛a hm s畛 (1) khi m = -1 .
2) T狸m t畉t c畉 c叩c gi叩 tr畛 c畛a tham s畛 m 畛 hm s畛 (1) ngh畛ch bi畉n tr棚n kho畉ng (-促;1) .
m2 - 4
揃 T畉p x叩c 畛nh: D = R {m}. y 蔵= .
( x + m)2
Hm s畛 ngh畛ch bi畉n tr棚n t畛ng kho畉ng x叩c 畛nh y 蔵< 0 -2 < m < 2 (1)
畛 hm s畛 (1) ngh畛ch bi畉n tr棚n kho畉ng (-促;1) th狸 ta ph畉i c坦 - m 続 1 m 贈 -1 (2)
K畉t h畛p (1) v (2) ta 動畛c: -2 < m 贈 -1 .
2 x 2 - 3x + m
C但u 11. Cho hm s畛 y = (2).
x -1
T狸m m 畛 hm s畛 (2) 畛ng bi畉n tr棚n kho畉ng (-促; -1) .
2x2 - 4x + 3 - m f (x)
揃 T畉p x叩c 畛nh: D = R {1} . y ' = 2
= .
( x - 1) ( x - 1)2
Ta c坦: f ( x ) 続 0 m 贈 2 x 2 - 4 x + 3 . 畉t g( x ) = 2 x 2 - 4 x + 3 g '( x ) = 4 x - 4
Hm s畛 (2) 畛ng bi畉n tr棚n (-促; -1) y ' 続 0, "x (-促; -1) m 贈 min g( x )
( -促;-1]
D畛a vo BBT c畛a hm s畛 g( x ), "x (-促; -1] ta suy ra m 贈 9 .
V畉y m 贈 9 th狸 hm s畛 (2) 畛ng bi畉n tr棚n (-促; -1)
2 x 2 - 3x + m
C但u 12. Cho hm s畛 y = (2).
x -1
T狸m m 畛 hm s畛 (2) 畛ng bi畉n tr棚n kho畉ng (2; +促) .
2x2 - 4x + 3 - m f (x)
揃 T畉p x叩c 畛nh: D = R {1} . y ' = 2
= .
( x - 1) ( x - 1)2
Ta c坦: f ( x ) 続 0 m 贈 2 x 2 - 4 x + 3 . 畉t g( x ) = 2 x 2 - 4 x + 3 g '( x ) = 4 x - 4
Hm s畛 (2) 畛ng bi畉n tr棚n (2; +促) y ' 続 0, "x (2; +促) m 贈 min g( x )
[2; +促 )
Trang 7](https://image.slidesharecdn.com/200-cau-khaosathamso2108-120710090743-phpapp02/85/200-cau-khaosathamso2-1-08-1-320.jpg)