B\u00e0i 2 (trang 7 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\nCh\u1ee9ng minh r\u1eb1ng:<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/1-539.png\")
<\/p>\n
\u0111\u1ed3ng bi\u1ebfn tr\u00ean m\u1ed7i kho\u1ea3ng x\u00e1c \u0111\u1ecbnh c\u1ee7a n\u00f3.<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/2-370.png\")
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ngh\u1ecbch bi\u1ebfn tr\u00ean m\u1ed7i kho\u1ea3ng c\u1ee7a n\u00f3.<\/p>\n
L\u1eddi gi\u1ea3i:<\/b><\/p>\n
a) H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R \\ {-2}<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/3-204.png\")
<\/p>\n
N\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean m\u1ed7i kho\u1ea3ng (-\u221e; -2) v\u00e0 (-2 ; +\u221e)<\/p>\n
H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R \\ {-1}<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/4-123.png\")
<\/p>\n
N\u00ean h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean m\u1ed7i kho\u1ea3ng ( -\u221e; -1) v\u00e0 (-1; +\u221e)<\/p>\n
B\u00e0i 3 (trang 8 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\nCh\u1ee9ng minh r\u1eb1ng c\u00e1c h\u00e0m s\u1ed1 sau \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\n
a)f(x)=x3<\/sup>-6x2<\/sup>+17x+4=0;<\/p>\nb) f(x)=x3<\/sup>+x-cos\u2061x-4<\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
H\u00e0m s\u1ed1 f(x)=x3<\/sup>-6x2<\/sup>+17x+4=0 x\u00e1c \u0111\u1ecbnh tr\u00ean R.<\/p>\nTa c\u00f3 f’ (x)=3x2<\/sup>-12x+17=3(x-2)2<\/sup>+5>0 \u2200x \u2208R.<\/p>\nN\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\n
H\u00e0m s\u1ed1 f(x) x\u00e1c \u0111\u1ecbnh tr\u00ean R.<\/p>\n
V\u00e0 f’ (x)=3x2<\/sup>+1+sin\u2061x>0 \u2200x \u2208R (x2<\/sup>\u22650;1+sin2<\/sup>\u2061x\u22650;3x2<\/sup>+1+sin\u2061x=0 v\u00f4 nghi\u1ec7m). n\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\nB\u00e0i 4 (trang 8 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\nV\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a a, h\u00e0m s\u1ed1 y=ax-x2<\/sup>\u00a0ngh\u1ecbch bi\u1ebfn tr\u00ean R?<\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R, y’=a-3x2<\/sup><\/p>\nC\u00e1ch 1. N\u1ebfu a < 0 => y\u2019 < 0 \u2200x \u2208R => h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean R.<\/p>\n
N\u1ebfu a = 0 => y\u2019 = -3x2<\/sup>\u22640,\u2200x \u2208R,y’=0 <=> x = 0<\/p>\nV\u1eady h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean R.<\/p>\n
N\u1ebfu a >0 th\u00ec y’=0<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/1-540.png\")
<\/p>\n
\n
<\/div>\n<\/div>\n
B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/2-371.png\")
<\/p>\n
C\u00e1ch 2. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean R, \u0111i\u1ec1u ki\u1ec7n y’\u22640,\u2200x \u2208R,y’=0 ch\u1ec9 t\u1ea1o m\u1ed9t s\u1ed1 h\u1eefu h\u1ea1n \u0111i\u1ec3m.<\/p>\n
Ta c\u00f3: y’\u22640 <=> a-3x2<\/sup>\u22640 <=> 3x2<\/sup>\u00a0\u2200x \u2208R<\/p>\n<=> a\u2264min\u2061(3x2<\/sup>\u00a0),m\u00e0 3x2<\/sup>\u22650 \u2200x \u2208R<\/p>\n![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/3-205.png\")
<\/p>\n
K\u1ebft lu\u1eadn: v\u1edbi a\u22640 th\u00ec y=ax-3x3<\/sup>\u00a0ngh\u1ecbch bi\u1ebfn tr\u00ean R.<\/p>\nB\u00e0i 5 (trang 8 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\nT\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 a \u0111\u1ec3 h\u00e0m s\u1ed1 f(x) = 1\/3 x3<\/sup>+ax2<\/sup>+4x+3 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
f(x) x\u00e1c \u0111\u1ecbnh tr\u00ean R.<\/p>\n
f’ (x)=x2<\/sup>+2ax+4;\u0394f”=a2<\/sup>-4<\/p>\nC\u00e1ch 1.<\/p>\n
+ n\u1ebfu a2<\/sup>-4<0 hay -2< a < 2 th\u00ec f\u2019(x) > 0, \u2200x \u2208R => h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\n+ N\u1ebfu a2<\/sup>-4=0 hay a=\u00b12<\/p>\nV\u1edbi a = 2 th\u00ec f\u2019(x) = (x+2)2<\/sup>>0 \u2200x \u2260 -2. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\nV\u1edbi a = -2 th\u00ec f\u2019(x) = (x-2)2<\/sup>>0 \u2200x \u2260 2. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\n+ N\u1ebfu a2<\/sup>-4>0 hay a< – 2 ho\u1eb7c a> 2 th\u00ec f\u2019(x) = 0 c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t x1<\/sub>,x2<\/sub>. Gi\u1ea3 s\u1eed x1<\/sub><x2<\/sub>, khi \u0111\u00f3 h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (x1<\/sub>,x2<\/sub>). V\u1eady c\u00e1c gi\u00e1 tr\u1ecb n\u00e0y c\u1ee7a a kh\u00f4ng th\u00f5a m\u00e3n y\u00ean c\u1ea7u b\u00e0i to\u00e1n.<\/p>\nC\u00e1ch 2.<\/p>\n
H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R khi v\u00e0 ch\u1ec9 khi f\u2019(x) > 0 \u2200x \u2208R f\u2019(x) = 0 ch\u1ec9 t\u1ea1i m\u1ed9t s\u1ed1 h\u1eefu h\u1ea1n \u0111i\u1ec3m.<\/p>\n
K\u1ebft lu\u1eadn: h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R khi v\u00e0 ch\u1ec9 khi -2\u2264 a\u22642<\/p>\n
\n
<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
B\u00e0i 1 (trang 7 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 X\u00e9t chi\u1ec1u bi\u1ebfn thi\u00ean c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau: a)y=2×3+3×2+1 b) y=x3-2×2+x+1 c)y=x+3\/x d) y=x-2\/x e) y=x4-2×2-5 L\u1eddi gi\u1ea3i: H\u00e0m s\u1ed1 y=2×3+3×2+1 x\u00e1c \u0111\u1ecbnh tren R. Ta c\u00f3: y’=6×2+6x=0=6x(x+1) y’=0 => x=0 ho\u1eb7c x=-1 Chi\u1ec1u bi\u1ebfn thi\u00ean c\u1ee7a h\u00e0m s\u1ed1 \u0111\u01b0\u1ee3c n\u00eau trong b\u1ea3ng sau: […]<\/p>\n","protected":false},"author":3,"featured_media":24028,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1302],"tags":[1392,1393],"yoast_head":"\n
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