giải bài tập giải tích 12 nâng cao

515.076 ^BlfCCHI GI-103B NANG r DVL.013442 m Nha xuat ban Dai hoc Quoc gia Ha Noi NGUYEN oCfC CHI ^UJOI bai tdfL I AI TICH 12 XAXG CAO THU VlifJ l\m BINH TH'JAN NHA XUAT BAN DAI HOC QU6c GIA HA LCJl NOI DAU GIAI BAI TAP DAI SO 12, duac bien soan v6i muc dich giiip hoc sinh doi chieu va kiem tra lai cac ket qua khi thtfc hien giai cac bai tap trong sach giao khoa. Muon the, cac em hay danh thcri gian nhat dinh de lam cac bai tap trong sach, sau do doi chieu va kiem tra lai ket qua thiTc hien. GIAI BAI TAP DAI SO 12, phu huynh c6 the suf dung de kiem tra con minh trong viec hoc tap va luyen tap cac kien thufc va ky nang CO ban. GlAl BAI TAP DAI SO 12, cac dong nghiep c6 the suf dung de tham khao. Mong diioc sii gop y chan thanh cua ban doc gan xa. TAC GIA TJTNG DUNG DAO HAM DE KHAO SAT VA VE D6 THI CUA HAM SO §1. TiNH DdN DIEU CUA HAM SO ivdl DUI\ cAi^ I^IOf 1, Dinh li: Gia sil hkm so f c6 dao Mm tren khoang I. • Neu f '(x) > 0, Vx e I thi h^m so f dong bien tren khoang I • N§'u f '(x) < 0, Vx e I thi h^m so' f nghich bien tren khoang I • Neu f '(x) = 0, Vx e I thi h^m so f khong d6i tren khoang I Chii y: Khoang I neu duoc thay bkng mot doan hoac mot niira khoang thi phai bd sung gia thiet "Ham so' lien tuc tren doan hoac niifa khoang do" 2. Vi#c xet chieu bien thien ciia ham so' c6 dao ham c6 the chuyen ve viec xet dau dao ham ciia ham so' do. ^ BAITAP 1. (Bai 1 trang 7 SGK) a) y = 2x^ + 3x^+1 Gidi Ham so' xdc dinh tren R Ta c6; y' = Gx^ + 6x y' = 0 o 6x^ + 6x = 0 o 6x(x + 1) = 0 o Bang bien thien: X = 0 x = -l X —00 -1 0 +0C y' + 0 0 + y —'—* — , — Vay ham so' dong bien tren moi khoang (-oo; -1) va (0; +oo), nghich bien tren khoang (-1; 0) b) y,= x^ - 2x^ + X + 1 Ham so' xdc dinh tren K Ta c6: y' = 3x^ - 4x + 1 y' = 0 <=> 3x^ - 4x + 1 = 0 o X = 1 hoac x = - Bang bien thien 1 3 +00 V§y hkm so d6ng bien trgn moi khodng (-00; ~) vk (1; nghich bien 3 tren khoang 3 3 c) y = X + — X H^m so xdc dinh tren M \1 Ta CO y' = 1 - y'^Oc=>l 5- = 0c^x^ = 3ox = ±73 Bang bien thien: X —00 -73 0 • J3 +00 y' + 0 II - 0 + y -— ' ^ ~—-————" —' ——> Vay ham so' dong bien tren m6i khoang (-00; - 73 ) va (0; 73 ), nghich bien tren moi khoang (- 73 ; 0) va (73 ; +00) 2 d) y = X X Ham so \±c dinh tren R \) Ta CO y' = 1 + ^ > 0, Vx e M \1 x'' Bang bien thien X —00 0 +0C y + II + y Vay hkm so dong bien tren m6i khoang (-00; 0) (0; +00) e) y = x^ - 2x^ - 5 Ham so xac dinh tren M Ta CO y' = 4x^ - 4x ^x = 0 y' = 0 <=> 4x'' - 4x = 0 o 4x(x'' - 1) = 0 o Bdng bien thien x = ±1 Vay ham so dong bien tren m6i khoang (-1; 0) va (1; +00), nghich bien tren moi khoang (-00; -1) va (0; 1). Oy = 74-x' Ham so' xac dinh tren [-2; 2] Tac. y-^^ y' = 0 » X = 0 Bang bien thign X -2 0 +2 y' + 0 y 2. (Bdi 2 trang 7 GSK) • x-2 a) y = —7; • X + 2 Ham so xac dinh tren Gidi 1-2) •Ta c6: y = x + 2 - (x - 2) (x + 2)^ Bdng bien thien: (x + 2) J > 0, Vx e !R\1-21 X —00 -2 +00 y' + + y Vay ham so d6ng bien tren m6i khoang (-00; -2) va (-2; +00) -x'-2x + 3 b) y = X +1 Ham so xac dinh tren M\{-11 (-2x - 2)(x + 1) - (-x' - 2x + 3) -2x' - 2x - 2x - 2 + x' + 2x - 3 y = (x + lf (X + ._ -x' - 2x - 5 (x + • Bang bien thien < 0, Vx e R\{-11 X —00 — +x y' - - y • — Vay ham so nghich bien tren m§i khodng (-00; -1) va (-!;•+«) , (Bdi 3 trang 8 SGK) Gidi a) f(x) = x'^ - Gx"^ + 17x + 4 Ham so xac dinh tren R Ta CO f '(x) = 3x^ - 12x + 17 f '(x) > 0, Vx e R Vay ham so' dong bien tren R b) f{x) = x^ + X - cosx - 4 Ham so xac dinh tren R • Ta c6 f '(x) = 3x^ + 1 + sinx > 0, Vx e R Vay h^m so dong bien tren R . (Bdi 4 trang 8 SGK) Gidi ftx) = ax - x^ Ham so xAc dinh tren R Ta c6: f '(x) = a - Sx^ * N6u a < 0 thi f '(x) < 0, Vx e R. H^m so nghich bi§'n trgn * Ng'u a = 0 thi f '(x) < 0, Vx e R. DAng thiJc chi xay ra khi x = 0 Vay ham so' nghich bien tren R. * Neu a > 0 thi f '(x) = 0 cs- x = ± j- V3 Bang bien thien x —00 fa ~ V3 if +00 f '(x) 0 + 0 f(x) " ———> '—" - /—; I— . Vay a > 0 khong thoa dieu ki§n de toan. H^m so' dong bien tren V£iy vdi a < 0, hkm so nghich bien tren R. 5. (Bdi 5 trang 8 SGK) Gidi fix) = - x^ + ax^ + 4x + 3 3 H^m s6' x^c dinh tren R f '(x) = x^ + 2ax + 4, c6 A' = a^ - 4 * Neu a^ - 4 < 0 hay -2 < a < 2 thi f '(x) > 0, Vx e R. H^m so dong bien tren R • Neu a = 2 thi f '(x) = (x + 2f > 0,\fx* -2. Ham so dong bien tren R. • Neu a = -2 thi f '(x) = (x - 2f > 0, Vx 2. H^m so dong bien tren M. • Neu a > 2 hoac a < -2 thi f '(x) = 0 c6 hai nghiem xi, X2 (xi < X2) Bang bien thien X -_co Xi X2 +00 f'(x) f(x) Ham so nghich bien tren (xj; X2) khong thoa de toan Vay -2 < X < 2 thi ham so da cho dong bien tren M. M LlJYEl^ TAP 6. (Bdi 6 trang 8 GSK) Gidi a) y = -x^ - 2x^ + 4x - 5 3 H^m so' xdc dinh lien tuc tri§n R y' = x^ - 4x + 4 = (x - 2f > 0, Vx e K H^m so' dong bien tren R b) y = -lx^ + 6x2-9x- - 3 3 H^m so xdc dinh va lien tuc tren R y' = -4x^ + 12x - 9 = -(2x - 3)^ < 0, Vx € R H^m so nghich bien tren K c) y = x'' - 8x + 9 x-5 Ham so xac dinh tren R \} (2x - 8)(x - 5) - (x' - 8x + 9) _ 2x' - ISx + 40 - x' + 8x - 9 y = (x-5)^ (X - 5)^ x' -lOx + 31 (x-5)^ Bang bien thien >0,\fx^5 x —00 5 +00 y' + + y Ham so dong bien tren moi khoang (-00; 5) va (5; +00) d)y= N/2X-X' Ham so' xac dinh tren [0; 2] rr. . , 1-X Ta co: y = , V2x - x' y' = 0 <» X = 1 Bang bien thien X 0 1 2 y' + 0 y Ham so dong bien tren khoang (0; 1), nghich bien tren khoang (1; 2) ')y = Vx' -2x + 3 Ham so xac dinh tren R (do x^ - 2x + 3 > 0) , _ 2(x -1) _ x-1 ~ 2N/X' - 2X + 3 Vx' - 2x + 3 y' = 0 O X = 1 Bang bien thien X —00 1 +00 y' 0 + V-— y • -> Ham so dong bien tren khoang (1; +00), nghich bien tren khoang (-00; 1) f)y = - 2x x + 1 Ham so' xac dinh tren R \) -1 - 2x' - 4x - 2 -2x' - 4x - 3 BAng bien thien (x + 1)' (x + D' < 0, Vx G M \I -1 +00 y >• Ham so nghich bien tren m5i khoang (-00; -1) (-1; +00) 7. iBdi 7 trang 8 SGK) f(x) = cos2x - 2x + 3 Ham so' xac dinh tren R Gidi f '(x) = -2sin2x - 2 = -2(sin2x + 1) < 0, Vx e K f '(x) = 0 <=> sin2x = -1 o 2x = - ^ + 2k7t <:=>x = -^ +k7r, ke Ham so nghich bien tren moi doan Vay ham so' nghich bien tren R. . + k7t; - — + (k + 1)71 4 4 , k e Z 8. (Bai 8 trang 8 SGK) Gidi a) X6t ham so' f(x) = x - sinx 0;^ 2 Hkm so lien tuc tren Do do ham so dong bi§'n tren Ta CO fix) > f(0). Vx e : f '(x) = 1 - cosx > 0, Vx e , 0; - V 2. 2] ( Nghia la X - sinx > 0, Vx e 0; - V 2 j hay X > sinx, Vx e 0;-, 2) Mat khdc x > sinx, Vx > — vi sinx < 1 2 Vay x > sinx, Vx > 0 Chufng minh tuong tyf sinx > x, Vx < 0 x^ b) Xet hkm so g(x) = cosx + — - 1 H^m so lien tuc tren nufa khoang [0; +ao), g'(x) = -sinx + x Theo cdu a, g'(x) > 0, Vx > 0, do d6 h^m so g dong big'n tren [0; +«), Va ta CO g(x) > g(0), Vx > 0 x' Nghia la cosx + 1>0, Vx>0 2 TCr d6 suy ra vdi mpi x < 0, ta c6: (-x)^ cos{-x) + 1 > 0 hay cosx + 1>0, Vx<0 2 2 x" Vay cosx > 1 - y , Vx 0 c) Xet h^m so h(x) = x - — - sinx 6 CO h'(x) = 1 - — - cosx Theo cau b) h'(x) < 0, Vx * 0 Do do ham so h nghich bien tren K h(x) < h(0), Vx > 0 x^ x^ ^ X sinx < 0 => X < sinx, Vx > 0 6 6 x^ va h(x) > h(0), Vx < 0 hay x > sinx, Vx < 0 6 J, (Bai 9 trang 9 SGK) Gidi Xet ham so' f(x) = sinx + tanx - 2x lien tuc tren niira khoang 1 0; Co f '(x) = cosx + Do cos^x + cos^x 2 > cos X + cos^x - 2, Vx 6 > 2, Vx e cos'x Do do ham so fix) dong bien tr§n Ta c6: f(x) > f(0;, Vx e Vay sinx + tanx - 2x > 0 hay sinx + tanx > 2x, Vx e 10. (Bai 10 trang 9 SGK) Gidi t + 5 a) Nam 1980, ta c6 t = 10 ^^^^^^ 26.10.10 270 = 18 (nghin) 10 + 5 15 S6 dan cua thi tran nam 1980 la 18 nghin ngif^i Nam 1995, ta CO t = 25 26.25 +10 660 f(25) = = 22 25 + 5 30 So dan cua thi tran nam 1995 la 22 nghin ngKbi b) Ta c6: f '(x) = 26(t + 5) - (26t + 10) 120 (t + 5)^ (t + 5f Vay ham so' dong bien tren [0; +oo) > 0, Vt e (0; +00) c) Toe dp tang dan so vac nam 1990 1^ 120 (20 + 5)' Toe do tSng dan so' v^o n&m 2008 1^ 120 (38 + 5)' f'(20) = -7—=^ ~ 0,192 f'(38)= * 0,065 190 1 00 • Ta CO —= 0,125 o (t + 5)^ = « t + 5 » 31 t = 26 (t + 5)' 0,125 Vao nam 1996 toc dp tang dan so ciia thi tra'n la 0,125. §2. CaC TR! CUA HAM SO 1. Dinh nghia: Gia suT ham so' f xac dinh tren tap c M va XQ e y * XQ dixac gpi la mot diem ciJ^c dai ciia ham so' f neu: Ton tai mot khoang (a; b) chiJa XQ sao cho (a, b) c 7 va f(x) < f(xo) vdi mpi x e (a, b) \. Khi do f(xo) dupe gpi la gia tri cufc dai cua ham so f * Xo difpc gpi la mot diem cxic tieu cua ham so' f neu: Ton tai mot khoang (a; b) chufa XQ sao cho (a; b) e D va f(x) > f(xo) vdi mpi x e (a; b) \. Khi do f(xo) dupe gpi la gia tri cij^c tieu ciia ham so' f * Diem cifc dai va diem cUc tieu gpi chung la diem cu'c tri Gia tri cUe dai va gia tri cUc tieu gpi chung la cu'c tri * Neu Xo la diem CLCC tri cua ham so f thi (XQ; f(xo)) la diem cu'c tri ciia do thi ham so f. 2. Dieu ki^n can de ham so dat ci/c tri Gia sijf ham so f dat CLTC tri tai diem XQ. Khi d6, neu f c6 dao ham tai XQ thi f '(xo) = 0 3. Dieu ki^n dii de ham so dat ci^c tri Gik sijf ham so f lien tuc tren khoang (a; b) chiJa diem xo va c6 dao ham tren cae khoang (a; XQ) (XQ; b). Khi do a) Neu f '(x) < 0 vdi mpi x e (a; XQ) va f '(x) > 0 vdi mpi x e (xo; b) thi ham so' f dat cvtc tieu tai diem XQ. b) Neu f '(x) > 0 vdi mpi x e (a; XQ) va f '(x) < 0 vdi mpi x e (XQ; b) thi ham so dat cue dai tai diem XQ. 4. Qui tSc tim c\ic tri * Qui t^c 1: • Tim f '(x) • Tim cac diem Xi (i = 1, 2, ) tai do dao ham cua ham so' hhng 0 hoac ham so' lien tuc nhung khong c6 dao ham. • Xet da'u f '(x) neu f '(x) ddi da'u khi x qua diem Xj thi ham so' dat cUc tri tai Xj. •f'(X) Xi f'(x) f(x) X, f(X) I f(Xi) cUc tieu - * Qui tac 2 • Tim f '(x) • Tim cac nghi^m Xi (i = 1, 2, ) cua phupng trinh f '(x) = 0 • Tim f "(x) va tinh f "(Xj) + Neu f "(xi) < 0 thi ham s6' dat cUe dai tai diem Xj + Neu f "(Xi) > 0 thi ham s6' dat cUc tieu tai diem Xj JBAI TAP n.(Bai 11 trang 16 - 17 SGK) a)f(x)= ix^ + 2x2 + 3x- 1 Gidi Ham s6' xac dinh tren K f (X) = x^ + 4x + 3; f (X) = 0 o x^ + 4x + 3 = 0 o X = -1 hoac x = Bdng bien thien X 1-00 -3 -1 -3 +0C Vay ham so da cho dat cUc dai tai diem x = -3, gia tri eUe dai la f(-3) : Hkm so dat cUc tieu tai dilm x = -1, gia tri cUc tieu cua ham so la fl-1) = Cdch 2: Ham so xac dinh tren M Ta c6: • f '(x) = x^ + 4x + 3; f '(x) = 0c:.x^ + 4x + 3 = 0ox = -l hofic x = -3; f "(x) = 2x + 4 Vi f "(-1) = 2(-l) + 4 = 2 > 0 nen ham so dat cUc tieu tai diem x 7 f(-l) = 3 f "(-3) = 2(-3) + 4 = -2 < 0 nen ham so dat cUe dai tai diem x f(-3) = -1 b) f(x) = i x^ - x^ + 2x - 10 3 Ham so' xac dinh tren R Ta CO f '(x) = x^ - 2x + 2 > 0, Vx e K . Ham so dong bien tren R, khong c6 cUc tri. -1 7 3 = -1; = -3, c) f(x) = X + — X Ham so xac dinh tren f '(x) = 1 - = R \1 x^-1 2 • f '(X) = 0«x'-l = Oc:>X = ±l Bang bien thign X —oc -1 ( ) 1 + 00 f'fx) + 0 — — 0 + f(x) ^ -2\ ^2 ^ Vay ham so' dat ciJtc dai tai diem x = -1, gid tri ciTc dai la f(-l) = -2 Ham so' dat ciTc tieu tai diem x = 1, gia tri ci/c tieu la f(l) = 2 d) f(x) = IX I (x + 2) Ham so' xac dinh va lien tuc tren R f-x(x + 2) v(Ji X < 0 |x(x + 2) vdi X > 0 " • V6i X < 0, f '(x) = -2x - 2; f '(x) = 0 x = -1 • Vdi X > 0, f'(x) = 2x + 2 > 0 Bang bien thien Ta c6 f(x) = X —00 -1 0 f'(x) + 0 _ fix) Vay ham so' dat cifc dai tai x = -1, gia tri cUc dai fi-1) = 1 va dat ciic tieu tai x = 0, gia tri cifc tieu fiO) = 0 x^ x" e) fix) = —+ 2 5 3 Ham so' xac dinh tren M f '(x) = x^ - x^ = x^ (x^ - 1) f '(x) = 0 <^ x^ (x^ - 1) = 0 X = 0 hoac x = ±1 Bang bien thien X —00 -1 0 1 • +C0 f (x) + 0 0 0 + fix) 32 15 15 Vay ham so dat ciTc dai tai x = -1, gia tri cvtc dai fi-1) 28 32 15 va ham so dat cUc tieu tai x = 1, gia tri ciTc tieu: fil) = 15 f) fix) = x^ - 3x + 3 x-1 Ham so' xac dinh tren M \| (2x - 3)(x - 1) - (x' - 3x + 3) x' - 2x f '(x) = (x -1)^ (x-1)^ f '(x) = 0c:>x-2x = 0<=>x = 0 hoac x = 2 Bang bien thien X —00 0 1 2 + 00 f (x) + 0 0 + fix) ; V$y ham so dat cUc dai tai diem x = 0, gia tri ciTc dai fiO) = -3 va dat cUc tieu tai diem x = 2, gid tri cifc tieu fi2) = 1 I. (Bcii 12 trang 17 SGK) Gidi a) y = x V4-x^ Ham so xac dinh va lien tuc tren [-2; 2] y' = ^ ~ v6i moi X e (-2; 2); y' = 0 « x = ± V2 V4-x' Bang bien thien x -2 -72 72 2 y' 0 + 0 y 0 —*. -2-—- ^2— ~ ^0 Ham so dat cifc tieu tai x = - 72 , gia tri c\ic tieu la -2 H^m so dat c\ic dai tai x = 72 , gia tri cUc dai 1^ 2 b)y = 78-x' - • • Ham so xac dinh va lien tuc tren [-2 72; 2 72 ] y' = , ~^ vdi moi x € (-2 72; 2 72 ); y' = 0 o x = 0 78^ Bang bien thien x -2sl2 0 272 y' + 0 y " 272 • • Ham so dat cifc dai tai x = 0, gia tri cxic dai: 2 72 c) Ap dung qui tAc 2 y = x - sin2x + 2 Ham so xac dinh tren R Ta c6: y' = 1 - 2cos2x y' = 0 <=> 2cos2x = 1 o cos2x = i<=>2x = ±|^ + k27x, k e Z 2 o ox = ±— +k7t, keZ 6 Ngoai ra: y" = 4sin2x * Vi y" — + kn = 4sin = 4sin I 3J = - 4sin = _4^ =-2V3 <0 Do d6 ham so dat ciTc dai tai c6c diem x = -— +k7t, keZ • • 6 Gia tri c\Xc dai y(- — + k7r) = - - + k7i - sin • • 6 6 3 + 2 = — + kn + — + 2 6 2 = 4sin^ =4.^ =273 >0 3 2 Do do h^m so dat cUc tieu tai cAc diem x = - + ku, k e Z 6 * Vi y" — + k7t = 4sin - + k27I k6 U J Gid tri ciTc tieu y — + kn 6 - + k27I = — + k7i - sin 6 7t ,* 73 - = -+k7t +2 6 2 + 2 d) Ap dung qui tic 2 y = 3 - 2cosx - cos2x Ham so xac dinh tren R Ta c6: y' = 2sinx + 2sin2x = 2sinx + 2.2sinxcosx = 2sinx(l + 2cosx) y' = 0 <=> sinx = 0 -1 + 2cosx = 0 sinx = 0 1 o cosx = 2 X = kTI X = ±— + 2k7i 3 Ngoai ra y" = 2cosx + 4cos2x • Vi y"(k7i) = 2cosk7r + 4cos2k7i = 2cosk7i + 4 > 0, Vk s Z. Do do ham so da cho dat cUc tieu tai cac diem x = kn va gid tri cUc tieu y(k7t) = 3 - 2cosk7x - cos2k7r = 2 - 2cosk7r = 2 - 2(-l)'' (k e Z) • Vi y" ±^ + k27: 3 = 2cos ±^ + k27r 3 + 4cos 3 0 271 = 2cos — . 4rr + 4cos — = 3 3 = -2cos — - 4cos — = - 3 3 n ^ n + 4cos ( ^) 7t + — I 3; I 3J IS- =-3 < 0 3 Do do h^m so dat ciic dai tai cac diem x = ± — + 2k7T, k e Z va gia tri 3 cUc dai la: y ^2Vk2; = 3 - 2cos f±2Vk27T^ - cos ^^Vk2.l V 3 , I 3 ) V 3 j = 3 - 2cos^ - cos^ = 3 + 2cos^ + cos- = 3 + 3cos^ = ^ 3 3 3 3 3 2 13. (BM 13 trang 17 SGK) f(x) - ax^ + bx^ + cx + d Giai flam so xac dinh tren K f '(x) = 3ax^ + 2bx + c • Ta CO f(0) = 0 d = 0. Ham so dat cifc tieu tai diem x = 0, nen: f (0) = 0 * Ta CO f(l) = 1 => a + b + c + d = 1 r.r> a + b = 1 Ham so dat cUc dai tai diem x = 1 nSn: f (1) = 0 =; => c = 0 3a + 2b + c = 0 3a + 2b = 0 . Gidi he phiTOng trinh 3a + 2b = 0 a + b = l a? + 2b = 0 -2a - 2b = -2 3a + 2b = 0 a = -2 b = 3 |a = -2 Kiem tra lai ket qua ^ Ta CO ham so f(x) = -2x-' + 3x^ t f '(x) = -6x^ + 6x, f '(x) = 0 c=. X = 0 hoac x = 1 f "(x) = -12x + 6 • f "(0) = 6 > 0. Vay hkm so dat ciTc tieu tai diem x = 0 va f(0) = 0 • f "(1) = - -12 + 6 = -6 < 0. Vay ham so dat cifc dai tai diem x = 1 va f(l) = 1 14. (Bcii 14 trang 17 SGK) Giai f(x) = x"* + ax^ + bx + c Ham so xdc dinh tren K. .f'(x) = 3x^ + 2ax + b Ta CO f(-2) = 0 o -8 + 4a - 2b + c = 0 (1) Hkm so dat cufe tri tai diem x = -2, nen f '(-2) ^- 0 o 12 - 4a + b = 0 (2) D6 thi ham so di qua diem A(l; 0) nen l + a + b + c = 0 (3) 4a - 2b + c - 8 = 0 -4a + b + 12 = 0 a+b+c+l=0 Giai he phUcfng trinh: a+b+c+l=0 o ]-4a + b + 12 = 0 3a - 3b - 9 - 0 a+b+c+l=0 o -12 + b + 12 = 0 a = 3 Kiem tra lai ket qua f(x) = X-' + 3x^ - 4 f '(x) = 3x? + 6x, f '(x) = 0 o X = a+b+c+l=0 -4a + b + 12 = 0 -9a+ 27 = 0 3 + c + 1 = 0 b = 0 a = 3 c = -4 b = 0 a = 3 r\\ / n tt ^ ^ / [...]... = 0 « Bang bien t h i e n X 0 _ b f'(X) ^^^^ + X = 12 12 0 ?ssn f (12) = 480 .12 - 20 .12^ = 2880 +» ^ Vay m a x f(x) = 2880 V5-x' c) c) if ( x ) = V $ y t r e n ( 0 ; +oo) h a m so f d a t g i d t r i I d n n h a t t a i d i e m x = 12 H a m so x d c d i n h t r e n S u y r a t r e n t a p N ' h ^ m so f d a t giA t r i lorn n h a t t a i d i e m n = 12 f'(x) = V a y m u S ' n t h u h o a c h d u o c... x = - 2 ciTc gi^ t r i ciTc t r i 1^ • r ' ^ ^ ^ i t a c tim gia t r i Idn nhat, gia t r i nho n h a t * T i m cac d i e m x,, x-,, x^ thuoc (a; b) t a i do h ^ m so f c6 dao h a m bfing f(-2) = - 8 + 12 - 4 = 0 Do f ( l ) = 1 + 3.1 - 4 = 0 0 hoac k h o n g c6 dao ham , T i n h f(xi), f ( x 2 ) , fix J , f(a), f(b) * So s^nh cdc gia t r i t i m di^cfc _ So Idn n h a t t r o n g cac gia t r i do la gia... t r i lorn n h a t t a i d i e m n = 12 f'(x) = V a y m u S ' n t h u h o a c h d u o c n h i e u n h a t sau m p t vu t h i t r e n m o i d o n v i d i e i , t i c h t r e n m a t h o p h a i t h ^ 12 con cA ^ 33 SGK) = 0 o f'(x) -x^ + 1 = 0 o X *.f5 d) f ( x ) = = ±1 1 0 X 7x^-1 - -oo X 2 - f'(x) ~2 f(x) V a y h ^ m so d a t ciJc t i e u t a i x = - 1 , g i a t r i cifc t i l u l a f ( - l ) = -... u6n U2 '73 hay 3 Do t h i x+ y = —-—x - — 9 9 9 +C0 ^ -1 —00 y + 13 13 873 => y = 9 ^ 9 Gidi 7 X 3 — a) Khdo sat sU bien thien va ve do thi y = x^ - 3x^ + 2 Diem uon: 1- Ham so xac dinh tren R • y " = -12x^ + 4; y " = 0 o x = ± 73 y " = 0 t a i cac diem x = ± — 3 va y " doi dau khi qua cdc diem xi = - — nen X2 = ^ o o r Vi 13) va U la Vay U , 2 ' 9 3 9 hai diem uon cua do t h i Giao diem ciia do t... V2 • y" = 0 tai X = ± — y/2 3 X = ± * Giao diem cua do t h i vdi true tung (0; 1) !^hdn xet: Hkm so da cho Ik hkm so chSn nen nhan true tung Ik true doi xufng + ^ m 3 Do t h i + Diem uo'n: y " = 12x' - 6 = 6(2x' - 1) y" = 0 o 2x^ - 1 = 0 3 D 6 thi ^/2 vk ddi dS'u khi qua c^c diem n^y, nen U i r V2 3^ 2' 4 l a diem uon cua d6 t h i 2 ' 4 + Giao diem ciia do t h i vdi true tung (0; 2) L U Y f t... tai x = ± — 2 = - nen U 1^ diem uon cna d6 thi 3 3 ' 27J • Giao diem eua do thi vdi true tung (0; 1) • y = 0 (x + l)(x - 1)^ = 0 o X = ±1 Do thi c^t true ho^nh tai ( - 1 ; 0) ( 1 ; 0) X 0 • y " = 12x^ - 6 = 6(2x2 _ j ) 0 I 1 vk ddi dS'u khi x qua 3 2 3 Do t h i 3 Do thi y " = 0 tai 1 4 76 0 1 ^ 4 +00 i , nghich bien +00 x/6 —00 y -00 3 J6 • Hkm so dat cifc dai t a i x = 0, gid t r i cUc d a... true do xiJcng x = 0 x^ = m 3 ^ 4 1); y " = 0 o yk diem x = - — 6 >.+00 H ^ m so' c6 m O t cijfc t r i o + x \ +00 — ^ • N l u m < 0 t h i y' = 0 1 x = 0 B d n g bi§'n t h i e n + 0 0 3 D 6 t h i y " = 12x2 _ 2 = 2(6x^ - H ^ m so' C O b a d i e m ciic t r i : +00 2 4 +00 • * Neu m = 0 t h i y' = 0 o • + 0 y' +00 0 2 0 Hkm khodng sl2 -sfm y x = 0; x = ± • H ^ m so d a t ctfc t i e u t a i cac d i e m... cifc t r i : x = l vk ( 1 ; 0) 48 (Bdi sat su biin so xkc J Hktn 31 ^ - 3 6 = ^ 18 x +• hav V •^^ r S 3il 6 ' 31 36 = 36 2^6 9 X + 2 — 9 276 y = 2 X + 31 —+— 13 2^ => y = V i l i m y = +00 X + -— rv6 12 Phi^cfng t r i n 9 t i e p 9 u y e n t a i d i e m 9 h t 36 uo'n U2 6 l i m y = -00 n 6 n dUdng t h i n g x = -— 1^ t i e m 3ii 36 31 v ^-36=^ 31 2V6 2 2s[& 2 31 276 13 ^ - ^ =~ 9 - " " 9 ^ ^ =- ¥ -... u n g (0; 1) Nhan xet: Do t h i nhgn giao d i l m 1(1; - 3 ) cua hai t i ^ m cgn cua do t h i 1km t a m d6i xilng , sat sU bien thiin ham s6 y = 1 1 • 1 \ -2 - 1 A -r \ -2-3- sat sU bii'n thien 1 1 12 x-»+Tr x-)r • x->i' do t h i h k m s6' ( k h i x 1~ va k h i x - » 1*) * l i m [ y - ( - X + 2)] = 0 n e n diidng t h i n g y = - x + 2 I k t i | m c k n xi§n cua I \ 1 do t h i h k m s6' ( k h i x . -40x + 480 f'(x) = 0 » -40x + 480 = 0 « X = 12 Bang bien thien X 0 12 +» _ f'(X) + 0 b ^^^^ ?ssn ^ f (12) = 480 .12 - 20 .12^ = 2880 Vay max f(x) = 2880 V$y tren (0;. n&m 2008 1^ 120 (38 + 5)' f'(20) = -7—=^ ~ 0,192 f'(38)= * 0,065 190 1 00 • Ta CO —= 0 ,125 o (t + 5)^ = « t + 5 » 31 t = 26 (t + 5)' 0 ,125 Vao nam 1996 . 0 o 12 - 4a + b = 0 (2) D6 thi ham so di qua diem A(l; 0) nen l + a + b + c = 0 (3) 4a - 2b + c - 8 = 0 -4a + b + 12 = 0 a+b+c+l=0 Giai he phUcfng trinh: a+b+c+l=0 o ]-4a + b + 12 =