17数分AI期末考试

E? 2017*2018c1Y 2 A 2 B AI“MATH120014 m?X?/4 6 ; K 8?no8o ? (30 )! WK(zK 5 ): (1) ? f(x) = 3x4 16x3 42x2+ 120 x, K f(x) 3m 3,3 ?65. 5: f0(x) = 12x3 48x2 84x + 120 = 12(x 5)(x 1)(x + 2). ? f(1) = 65,f(2) = 232,f(3) = 63,f(3) = 207. (2) Z ? ex?lnx + 1 x ? + lnx + 1 ? dx =(ex+ x)lnx + C. (3) ? f(x) 3 x = 1 kn?, v f0(x) = f ? f(x) ? 9 f(1) = 1. K f(x) 3 x = 1 ? Peano .?n? Taylor : 1 + (x 1) + (x1)2 2 + (x1)3 3 + o(x 1)3),(x 1). 5:d f(x) 3 x = 1 kn?, ? f 3 x = 1 NCY?, udX f0(x) = f ? f(x) ? q? f kY?n?. l? f00(x) = f0 ? f(x) ? f0(x), f000(x) = f00 ? f(x) ? f0(x) ?2 + f0 ? f(x) ? f00(x). g? f0(1) = f ? f(1) ? f(1) = 1, f00(1) = 1, f000(1) = 2. (4) u n 1, ? n ?0,1 n ? v sin 1 n = 1 n cosn, K lim n nn= 3 3 . 5: n= arccos ? nsin 1 n ? . d lim nnn =lim x0+ arccos sin x x x =lim x0+ sinarccos sin x x x =lim x0+ q 1 sin2x x2 x =lim x0+ r x + sinx x r x sinx x3 = 3 3 . ?| LHospital K Taylor , k lim nnn =lim x0+ arccos sin x x x =lim x0+ sin xx cos x x2 q 1 sin2x x2 =lim x0+ tanx x p2x3(x sinx)=lim x0+ x3 3 + o(x3) q 2x3(x3 6 + o(x3) = 3 3 . 11 ( ? 4) (5) ?- y = x23: A(x0,x2 0) ? x ?u P :. e AP ? 1, K x0= s 65 1 8 . 5: y = x2 0+ 2x0(x x0). )? P :I ( x0 2 ,0). ) (x0 x0 2 )2+ (x2 0) 2 = 1 ?(J. (6) ? f(x) = x1999exsinx, K f(2018)(0) = 2018! 19! 29. 5:? g(x) exsinx = 19 X n=0 anxn+ o(x19)(x 0), K f(x) = 19 X n=0 anxn+1999+ o(x2018)(x 0). d f(2018)(0) = 2018!a19= 2018! 19! g(19)(0) = 2018! 19! Im ? (1 + i)19e(1+i)x ? ? ? x=0 ? = 2018! 19! (2)19sin 19 4 = 2018! 19! 29. ?(10 )! ? f(x) 3 1,1 Y, 3 (1,0) (0,1) S?. e lim x0 f0(x) 3, y: f(x) 3 x = 0 ?. y: du f 3 1,1 Y, d, lim x0 f(x) = 0. d, d LHospital K, = f0(0) = lim x0 f(x) f(0) x = lim x0 f0(x) 3. K: 1. ?ur LHospital K-#y?H. 2. k?rm?m?. 3. k? f0(x) Y. c7, 1n. n(20 )! ? y = y(x) d x = et+ 2t, y = sint t, (. (1) O d2y dx2 .(2) O Z y(x)dx. (3) ? x0 R, xn+1= y(xn) (n 0). y: xn . ): (1) dy dx = dy/dt dx/dt = cost 1 e2+ 2 . d2y dx2 = d dt ?dy dx ? 1 dx/dt = et(1 sint cost) 2sint (e2+ 2)3 . (2) Z y(x)dx = Z (sint t)(et+ 2)dt = (sint 2 cost 2 t + 1)et 2cost t2+ C. (3) x t A, ?=? t + ( ) x + ( ). k lim x+(y(x)x) = limt+(sintte t+2t) = , lim x(y(x)x) = limt+(sintte t+2t) = +. ud0n3 x R ? y( x) x = 0. , d dx(y(x) x) = dy(x) dx 1 1, d y(x) x , l?k :. ( x y(x) x ?:, ? x 7 y(x) k?: x. 5? ? ?dy(x) dx ? ? 1, k |xn+1 x| |xn x| . |x0 x|. d, xn k. ? M = supn|xn|, k = max |x|M ? ?dy(x) dx ? ?. K k 1, |xn+1 x| k|xn x| . kn+1|x0 x|. ? ?lim n+ xn= x. 12 ( ? 4) 5: 1. (3) ?ykCz, X, Ley: x2n x2n+1 ONk., ?4O a,b. K b = y(a), a = y(b). |y(b) y(a)| = |b a|. % a = b, l? xn . 2. k? ? ? ? dy(x) dx ? ? ? 0 (x (1,0) (0,+) y0(x) 3 (1,+) O. d 3 (1,+) . 13 ( ? 4) 8(10 )! m (0,1) S? 0.a1a2a3.an. L, ? 9 !. f(0.a1a2a3.an.) := 0.a10a20a30.an0., f(x) 3 (0,1) S?=?:Y, =?:Y. ): ? x0 (0,1) k?, f(x) 3 x0Y, ? x0 (0,1) ?, f(x) 3 x0Y. N/, e x0= 0.a1a2.an, an 1,2,.,9. K f(a0) = n X k=1 ak 102k1 , ? x (0,x0) , f(x) n1 X k=1 ak 102k1 + an 1 102n1 + 10 102n+1 . d f(x0) f(x) 9 102n. L f(x) 3 x0Y?, l?Y. e x0?, = x0= 0.a1a2.ak., ak 0,1,2,.,9, akk ?. K5K, 7k k ? ak,ak+1, 9, = 1 10ak+ ak+1 98. ? 0, - k = kv? 1 10ak+ak+1 98, 1 102k , K? |xx0| 1 10k+1 x ?:?c k 1 7, x0?c k 1 , d |f(x) f(x0)| 10 102k+1 . d, f(x) 3 x0:Y. ?, 3k?, f(x) E,kmY5. 5: krkn?uk?. (10 )! ? f(x) 3 0,+) k?Y?, v f(0) = f0(0) = 0 9 f00(x) + 3f0(x) + 2f(x) 0. y: f(x) 0. y: 5?f00(x)+3f0(x)+2f(x) = (D2+3D+2)f(x) = (D+1)(D+2)f(x), g(x) = ex(f0(x) + 2f(x), K g(x) 3 0,+) Y?, g0(x) = ex?f00(x) + 3f0(x) + 2f(x)? 0. d g(x) NO, g(x) g(0) = 0 (x 0). l?