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Xét tính đơn điệu của hàm số dựa vào công thức Toán 12
Xét tính đơn điệu của hàm số dựa vào công thức Toán 12 được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn học sinh cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Toán 12 3.9 K tài liệu
Xét tính đơn điệu của hàm số dựa vào công thức Toán 12
Xét tính đơn điệu của hàm số dựa vào công thức Toán 12 được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn học sinh cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Chủ đề: Chương 1: Ứng dụng đạo hàm để khảo sát và vẽ đồ thị của hàm số 375 tài liệu
Môn: Toán 12 3.9 K tài liệu
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3x 6x 9 0 3 x 1 w8 1 0 10
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3x 8x 5 0 1 x 3 3 3 – – Email: w8 1 0 10
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3 10 3; 0 x y x x y 3 x 3 3 2 y x 5x 3x 1 1 ; 1 3; ;3 3 3 3 6 – – Email: 3 2 y x 5x 3x 1 D – y 3 sin52x D 3;3 y 6cos5 2x; 10 k2 y 0 5 2x k x ,k 2 4 3 7 – – Email: y 3 sin5 2x 10 3 10 k ; k 4 4 103 10 k ; k 4 4 T 3;3 D – x e y xe 1 D \ 0 x e y 0, x D xe 21 x e y xe 1 x e y ; 0 0; x e 1 D \ 0 D ; 0 0; 3 8 – – Email: – y f x 4 2 x 2x 0;1 ; 1 ; 1 0; 1 y f x 4 2 x 2x x 1 f x 3 4x 4x 0 x 0 x 1 x 1 0 1 f 'x 0 0 0 1 1 f x 0 – – 0; 1 x 0.5 x 0.5 qyp[^4$+ 3 9 – – Email: 2[d$p0.5= – 3 y x 3x 4 ; 1 1; 1; 1 ; 1 1; D x 1 2 y ' 3x 3,y ' 0 x 1 1; 1 x 2, x 0, x 2 x 2 qyp[^3$+ 3[p4$p2= 4 0 – – Email: x 0 qyp[^3$ +3[p4$0= x 2 qyp[^3$ +3[p4$2= – 2022 y x 1 ; 1 ; 0; 0 ; 2 2 2021 y 2022x x 0 y 0 x 0 y 0 ( ; 0) x 0.4, x 1, x 2 4 1 – – Email: x 1 qy[^2022$p1= x 0.4 qy[^2022$0.4= x 0.4 2021 2021 y 2022 x y 2021 2022x y0.4 2 y 2 2022 0 0 5 5 x 0.4 qy[^2022$2= y 2022 x y 2021 x y 2021 2022 2 2022.2 CASIO CAPACITY x 2 4 2 – – Email: – 2x 3 y x 1 ; 2 ; 0; 1 y 0, x 1 x 21 ; 1 1; D \ 1 – ; 4 2 y x 3x 2x 1. x 1 y . 2x 2 3 2 y x x 2x 1. 3 y x 3. ; 3 2 y x x 2x 1 D 2 2 1 5
y 3x 2x 2 3 x 0, x 3 9 4 3 – – Email: ; 3 y x 3 D 2 y 3x 0 ; y 4 ax 3 bx 2 cx dx ,e a 0 y 3 ax 2 4 3bx 2cx d a 0 lim y 0 x x lim y ; y x 0 0 0 0 x x x0 2 x x x x x x 0 0 0 0 x x x x 0 0 a 0 lim y 0 x x lim y ; y x 0 0 0 0 x x x0 2 x x x x x x 0 0 0 0 x x x x 0 0 – 4 4 – – Email: – x x y 3 log x 2 y log x e y 2 y 3 4 5 y 3 log x D 0; y log 2x D \ 0 3 2 x 5 x y 5 1 5 2 2 x e y e 1 4 4 – x 1 y 2x 1 y x 2 y x 3 y x 1 x 3 2x 1 x 2 x 1 y 2 y 0, x 1 x 1 x 21 x 1 y x 1 2x 1 y 7 y 0 x 3 x 3 x 2 3 2x 1 y x 3 x 2 y 3 1 y 0, x 2x 1 x 2 2 2 1 4 5 – – Email: x 2 y 2x 1 x 3 y 1 y 0, x 2 x 2 x 2 2 x 3 y x 2 – f x x 1 g x x 1 ; ; h x 4 2 x 3x 1; k x 3 x 3x 1 x 1 x 2 0 1 2 3 f x x 1 x 2 g x x 1 x 2 h x 4 2 x 3x 1 k x 3 x 3x 1 k x 3 x 3x 1 d D \ c d d ; ; c c – f x x 2 ; g x 3 2 x 1 2 ; h x 3 log x ; k x 2 x 2 x 3 4 6 – – Email: 0 1 2 3 f x x 2 f x 5 0, x 3 x 3 x 2 3 2 1 2x f x f x 2 x 1 2x.2 .ln2 2 x 1 x 0, 2x.2 .ln2 0 2 1 2x f x ; 0 f x 3 log x 3 log x D 0; 2 2 f x 3 0, x D x.ln2 32 f x x D 0; f x 1 3 2 x 0, x D 2 – y log 2 2x x 1 . 2 1 ; 1; 2 1 ; 1; 2 1 ; 1; 2 4 7 – – Email: 1 ; 1; 2 1 D ; 1; . 2 4x 1 1 y ;y 0 x 2 2x x 1ln2 4 x 1; y 0 1 x ; y 0 2 1 ; 1; . 2 – 3 2 y x 2x x 1 1; 1 ;1 3 1 ; 3 1 ;1 3 2 y 3x 4x 1 x 1 y 0 1 x 3 4 8 – – Email: 1 ;1 3 – 4 2 y x 2x 2 2; 2; ; 0 ; 0 x 1 3 y 4x 4x 0 x 0 x 1 a 1 0 ab 20 W ; 1 0; 1 1 ; 0 1; 2; 1; 2; – y f x f x 2 x 5x 4 1;4 3; ; 3 4 9 – – Email: 1;4 f x 2
0 x 5x 4 0 1 x 4 1;4 – 5 3 y 3x 5x ; 1 1; ; ; 1 1; 1; 1 4 2 2 y x x x 2 15 15 15 x 1 x 1 2 y 0 x 1 0 x 1 ; 1 1; 1; 1 – m 4 3 2 y x 4x mx 2 ; 0 9 m 9 m 9 m 9 m 4 2 2 4 ; 0 3 2 y 4x 12x 2mx ; 0 y 0, x ; 0 5 0 – – Email: 3 2 2
4x 12x 2mx 0 2x 6x m 0, x ; 0 2
m 2x 6x, x ; 0 m min f x f x 2 2x 6x ; 0 f x 2 2x 6x f x 3
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g x f 2x gx x f 2 3 2 . 3 x 0 . x 0 x 0 x 0 2 2 3 x 6 x 9 x 3 2 2 3 x 1 x 4 x 2 2 2 3 x 2 x 1 x 1 – y f x f x m
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4 3m 2 0 m a 3 0 3 – 6 6 – – Email: y f x. y f x g x f 12x 1; 0. ; 0. 0; 1. 1; . f x x 1 0 . 1 x 2
gx 2f 12x. x 1 gx 1 2x 1 0 f 12x 0 1 12x 2 1 x 0 2 g x 1 ;0 1; 2 – f x . f x 1;3 1 x g x f x 2 4; 2. 2; 0. 0; 2. 2;4. 6 7 – – Email: gx 1 x f 1 1 2 2 gx x 0 f 1 . 2 2 x x f 1
2 2 1 3 4 x . 2 2 2 4; 2 x x f 1 2 1
1 a 0 2 2 2a x 4 2 2 22a;4 2;4. 1 x g x f x 4; 2. 2 – m 5;5 x e m x 1 5 6 7 10 x e 0 m 0 x e m x 1 1 x 1 1 1 x x e f x x m e m 1 x f x x e ' x f x xe 0 x 0 6 8 – – Email: 1 1 m 0 m 1 m 1 0 m m
m 5;4;3;2;1;1 m5;5 x y e x y e 0 x C m 0 y m x 1 1;0 : x C y e m 0 x y e 0 m 0 : y m x 1 x e m x 1 x 0 C m 1 x e m m 1 – 3 4 2 3 2 2
x x m x x 2 2 1 x 1 1m m m x 1 . 6 9 – – Email: m 1 . m 1 . m 5 m 5 . 4 4 a 3 4x 2x m 3 a 3 b 4 x 2 x m . 1 b 3 2 2x 1 3 3 3 3
ab a b 0 a a b b a b . 3 a 3 b 4 x 2 x 1 m . f x 4 2 x x 1 1; . f x 3 x x x 2 4 2
2 2x 1 0, x 1 . f x 1; 1; . m
min f x f 1 1 m 1 . 1; – y f x g x f 3 2x x 1m . m max g x 10 . 0; 1 m 13 . m 12 . m 1. m 3 . t x 3 2x x 1 x 0; 1 . tx 2 6x 1 0, x 0; 1. t x x 0;1 t 1; 2 . 7 0 – – Email:
max f t 3 max f t m 3 m . 1;2 1;2
3 m 10 m 13 . – y f x. y f x f 1 1, 1 f 1 2. f x ln x m x 1; e e m 2 m 2 m 3 m 3 m f x ln x 1 x 1 ; e m max g x
g x f x lnx 1 1 ; e 1 g x f x x f 1 x 0, x 1; e 1 g x x 1 1 0, 1; e 0, x 1; x e g x 1 1 ; 1 1 1; max g x g 3. e e 1 1 ; e e – 7 1 – – Email: 2 2
3 x 6 x 18 3x x m m 1 m m 5 ;5 x 3;6 3 . 5 . 9 . 10 . t x 3 6 x . 2 t x 2 9 2 18 3 x . t 1 1 t 1 1 x 3 ; 0 0 3; 6 . 2 x 3 2 6 x 2 x 3 2 6 x 2 t 3;3 2 . 2 t 9 2 2 t 2t t m m 9 1 2 m m 1. 2 2 f t 2 t 2t 9 t 3;3 2 . 2 f t t 1 0 t 3;3 2 . f t 3;3 2 max f t f 3 3 . 3;3 2 m 2 2 2 2
m m 1 max f t m m 1 3 m m 2 0 . 3;3 2 m 1 – y f x f 0 0
f x 0,x 1; 2 7 2 – – Email:
f x 0,x 1; 2 y f x 1; 2 y f x x f x f 1 f x f 2 f x f
0 0, x 1; 0 y f x 1; 0 f x f 0 0, x 0; 2 0; 2 x 0 y f x K x h;x h o o K K \ x h 0 0 7 3 – – Email: f x 0, x x h;x
f x 0,x x ;x h x o o o o o y f x
f x 0,x x h;x
f x 0,x x ;x h x o o o o o y f x – f x
f x x 2 2 1 x 2x x .
g x f 2x 2x 2 2 3 . 1 . . 3 2
gx x f 2 2 1 x 2x 2
2x 1 x 2x 2 21 x 2x 2 2 2
2 2 2x 2x 2 2x 5 1 x 4 1 1 . x 5 0 x 1 2 1 x 4 1 1 0 . x 2 0; 1, 2; 3 g x. 7 4 – – Email: 7 5