B\u00e0i 1 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0\u00c1p d\u1ee5ng Quy t\u1eafc 1, h\u00e3y t\u00ecm c\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/strong><\/span><\/p>\n a) y = 2x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 36x – 10 ; b) y = x4<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 3;<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n a)<\/b>\u00a0TX\u0110: D = R<\/p>\n y’ = 6x2<\/sup>\u00a0+ 6x – 36 = 6(x2<\/sup>\u00a0+ x – 6)<\/p>\n y’ = 0 => x = -3 ho\u1eb7c x = 2<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u1eady \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i l\u00e0 (-3; 71) v\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u l\u00e0 (2; -54).<\/p>\n b)<\/b>\u00a0TX\u0110: D = R<\/p>\n y’= 4x3<\/sup>\u00a0+ 4x = 4x(x2<\/sup>\u00a0+ 1) = 0; y’ = 0 => x = 0<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u1eady h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u l\u00e0 (0; -3).<\/p>\n c)<\/b>\u00a0TX\u0110: D = R \\ {0}<\/p>\n <\/p>\n y’ = 0 => x = \u00b11<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u1eady h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i l\u00e0 xC\u0110<\/sub>\u00a0= -1 v\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u l\u00e0 xCT<\/sub>\u00a0= 1.<\/p>\n d)<\/b>\u00a0TX\u0110: D = R<\/p>\n y’= 3x2<\/sup>(1 – x)2<\/sup>\u00a0– 2x3<\/sup>(1 – x) = x2<\/sup>(5x2<\/sup>\u00a0\u2013 8x + 3)<\/p>\n y’ = 0 => x = 0; x = 1 ho\u1eb7c x = 3\/5<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u1eady h\u00e0m s\u1ed1 c\u1ef1c \u0111\u1ea1i xC\u0110<\/sub>\u00a0= 3\/5 v\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u xCT<\/sub>\u00a0= 1<\/p>\n (L\u01b0u \u00fd:<\/b>\u00a0x= 0 kh\u00f4ng ph\u1ea3i l\u00e0 c\u1ef1c tr\u1ecb v\u00ec t\u1ea1i \u0111i\u1ec3m \u0111\u00f3 \u0111\u1ea1o h\u00e0m b\u1eb1ng 0 nh\u01b0ng \u0111\u1ea1o h\u00e0m kh\u00f4ng \u0111\u1ed5i d\u1ea5u khi \u0111i qua x = 0.)<\/p>\n <\/p>\n V\u1eady D = R.<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u1eady h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u xCT<\/sub>\u00a0= 1\/2.<\/p>\n B\u00e0i 2 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0\u00c1p d\u1ee5ng Quy t\u1eafc 2, h\u00e3y t\u00ecm c\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 sau:<\/strong><\/span><\/p>\n a) y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 1 ; \u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 b) y = sin2x \u2013 x<\/strong><\/span><\/p>\n c) y = sinx + cosx ; \u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 d) y = x5<\/sup>\u00a0– x3<\/sup>\u00a0– 2x + 1<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n a)<\/b>\u00a0TX\u0110: D = R.<\/p>\n y’ = 4x3<\/sup>\u00a0– 4x<\/p>\n y’= 0 => x = 0; x = \u00b11.<\/p>\n y” = 12x2<\/sup>\u00a0– 4<\/p>\n y”(0) = -1 < 0 => x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1.<\/p>\n y”(\u00b11) = 8 > 0 > x = -1 v\u00e0 x = 1 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1.<\/p>\n b)<\/b>\u00a0TX\u0110: D = R<\/p>\n y’ = 2cos2x \u2013 1;<\/p>\n <\/p>\n c)<\/b>\u00a0TX\u0110: D = R<\/p>\n <\/p>\n d)<\/b>\u00a0TX\u0110: D = R<\/p>\n y’= 5x4<\/sup>\u00a0– 3x2<\/sup>\u00a0– 2<\/p>\n y’ = 0 => x \u00b11.<\/p>\n y” = 20x3<\/sup>\u00a0– 6x<\/p>\n y”(-1) = -20 + 6 = -14 < 0 => x = -1 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1.<\/p>\n y”(1) = 20 \u2013 6 = 14 > 0 => x = 1 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1.<\/p>\n B\u00e0i 3 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0Ch\u1ee9ng minh h\u00e0m s\u1ed1 y = \u221a|x| kh\u00f4ng c\u00f3 \u0111\u1ea1o h\u00e0m t\u1ea1i x = 0 nh\u01b0ng v\u1eabn \u0111\u1ea1t \u0111\u01b0\u1ee3c c\u1ef1c ti\u1ec3u t\u1ea1i \u0111i\u1ec3m \u0111\u00f3.<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n T\u00ednh theo\u00a0\u0111\u1ecbnh ngh\u0129a \u0111\u1ea1o h\u00e0m<\/b>\u00a0t\u1ea1i xo<\/sub>\u00a0= 0 ta c\u00f3:<\/p>\n <\/p>\n Ngh\u0129a l\u00e0 h\u00e0m s\u1ed1 y = \u221a|x| kh\u00f4ng c\u00f3 \u0111\u1ea1o h\u00e0m t\u1ea1i x = 0. (1)<\/p>\n M\u1eb7t kh\u00e1c ta c\u00f3: \u221a|x| \u2265 0 \u2200 x. D\u1ea5u “=” x\u1ea3y ra khi x = 0.<\/p>\n Do \u0111\u00f3 h\u00e0m s\u1ed1 y = \u221a|x| \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 0. (2)<\/p>\n T\u1eeb (1) v\u00e0 (2) suy ra \u0111i\u1ec1u ph\u1ea3i ch\u1ee9ng minh.<\/p>\n B\u00e0i 4 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0Ch\u1ee9ng minh r\u1eb1ng v\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 m, h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n y = x3<\/sup>\u00a0– mx2<\/sup>\u00a0– 2x + 1<\/strong><\/span><\/p>\n lu\u00f4n lu\u00f4n c\u00f3 m\u1ed9t c\u1ef1c \u0111\u1ea1i v\u00e0 m\u1ed9t \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– mx2<\/sup>\u00a0– 2x + 1 ta c\u00f3:<\/p>\n TX\u0110: D = R<\/p>\n y’ = 3x2<\/sup>\u00a0– 2mx – 2<\/p>\n <\/p>\n V\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb c\u1ee7a m ta \u0111\u1ec1u c\u00f3 x1<\/sub>\u00a0< 0 < x2<\/sub>.<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n T\u1eeb b\u1ea3ng tr\u00ean ta th\u1ea5y h\u00e0m s\u1ed1 lu\u00f4n c\u00f3 m\u1ed9t \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i xC\u0110<\/sub>\u00a0= x1<\/sub>\u00a0v\u00e0 m\u1ed9t \u0111i\u1ec3m c\u1ef1c ti\u1ec3u xCT<\/sub>\u00a0= x2<\/sub>\u00a0v\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb c\u1ee7a m (\u0111pcm).<\/p>\n B\u00e0i 5 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0T\u00ecm a v\u00e0 b \u0111\u1ec3 c\u00e1c c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n <\/p>\n \u0111\u1ec1u l\u00e0 nh\u01b0ng s\u1ed1 d\u01b0\u01a1ng v\u00e0 xo<\/sub>\u00a0= -5\/9 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n – N\u1ebfu a = 0 th\u00ec y = \u20139x + b. V\u1eady h\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/p>\n – N\u1ebfu a \u2260 0. Ta c\u00f3: y’= 5a2<\/sup>x2<\/sup>\u00a0+ 2ax – 9<\/p>\n y’= 0 => x = 1\/a ho\u1eb7c x = -9\/5a<\/p>\n +\u00a0V\u1edbi a > 0<\/b>\u00a0ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u00ec xo<\/sub>\u00a0= -5\/9 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i n\u00ean<\/p>\n <\/p>\n Theo \u0111\u1ec1 b\u00e0i th\u00ec yCT<\/sub>\u00a0d\u01b0\u01a1ng n\u00ean v\u1edbi a = 81\/25 th\u00ec khi \u0111\u00f3:<\/p>\n <\/p>\n \u00a0\u00a0\u00a0 +\u00a0V\u1edbi a < 0<\/b>\u00a0ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n V\u00ec xo<\/sub>\u00a0= -5\/9 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i n\u00ean<\/p>\n <\/p>\n Theo \u0111\u1ec1 b\u00e0i th\u00ec yCT<\/sub>\u00a0d\u01b0\u01a1ng n\u00ean v\u1edbi a = -9\/5 th\u00ec khi \u0111\u00f3:<\/p>\n <\/p>\n V\u1eady c\u00e1c gi\u00e1 tr\u1ecb a, b c\u1ea7n t\u00ecm l\u00e0:<\/p>\n <\/p>\n B\u00e0i 6 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n <\/p>\n \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 2.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n TX\u0110: D = R \\ {-m}<\/p>\n <\/p>\n y’ = 0 => x1<\/sub>\u00a0= -m – 1; x2<\/sub>\u00a0= -m + 1<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 2 \u21d4 – m \u2013 1 = 2 =>\u00a0m = \u20133<\/b>.<\/p>\n B\u00e0i 1 (trang 18 SGK Gi\u1ea3i t\u00edch 12):\u00a0\u00c1p d\u1ee5ng Quy t\u1eafc 1, h\u00e3y t\u00ecm c\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau: a) y = 2×3\u00a0+ 3×2\u00a0– 36x – 10 ; b) y = x4\u00a0+ 2×2\u00a0– 3; L\u1eddi gi\u1ea3i a)\u00a0TX\u0110: D = R y’ = 6×2\u00a0+ 6x – 36 = 6(x2\u00a0+ x – 6) […]<\/p>\n","protected":false},"author":3,"featured_media":21502,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1298],"tags":[1377,1356,1355],"yoast_head":"\n