Mat 3 - G2 - Marcela Amorin

65\n\nModalidade: \n\nProva (62)\n\nNome do Professor: \n\nData: 16/09/13\n\n2013\n\nx\n\nx = 0, y = a(b + ex) - 2\n\n(rh lsest)\n\nx = y + z = 0 10\n\n\n\nfx = 30 + 4x - 12y² + 2z\n\nx - 4y = -4...\n\nf0 + x1 + y1 + z1 = 3 6\n\nab = 0\n\n(18x² + 19x + 15) = 0\n\n..\n\n#. = 0\n\nx + 4 + 5x + 30\n\nz -5 = 0\n\n50\n\n0 D\u00e9finition : (f : R2 \u2192 R) x \u2192 f(x) = R x((c)\u00e8te. t \u2192 t2 + 2x + 3)\n\nLaissez faire r\u00e9p\u00e9ter \u2014 \u2248)\n\n2 & f(s) = x^{k} + a2\u00b1 1 + y3 \n\n2 / (1) + t/2 + y\n\n= + (2)\n\n2y = t/2y + t\n\n\u2192 t + r \n\n(1, 0, 0)\n\nP: (0,0) \n\nM(R)= f \u2192 o3\n\nLa: (T) = 6\n\n(M= a, b)\n\nD(0) a= 18 + 42 \n\n\n0[] \n\n[] \n\n7\n\nM(a) \u00f6(1, )\n\nM(a) 6\n\n2\u2014 BUT L= 0 + f l() 0 \n\nE \n\n - 0 + 2 = 0 \n\nA : R = 88 + 16 \n\n= 0 \n\n-0 \n\n-0 1.Tabulation. (1) o = 0 \n\n2. Tabulation(d) e = 0 \n\nZ. P = 0 + 3 \n\nD(P) V 0 \n\nIY \n\nA=(B) 0 + 5= 8. \n\nR (0, 2, 1)\n\n( xy )\n\nA = 2 \n\nAx \u2192 (0, 4) \n\n-0\n\n( p, x^5 + y) \u2192 f(x) \n\nx - 3 + 1= 0 = |3 + 0+ - 2 \n\n= + \n\nM(0) = r\n\nP \n\n(0, 3) = R 0= (4, 3) \n\n= (1) \n\nakt\u00an\n\n(0, 4 )\n\n(1 (3, 2)\n\nu \u2192 0\n\n\n0, - 1 + 1= 0 M+8 + 18 \n\nf(x) \n\nT 3 + O a + b A 3= > M \n\nB B -> W 225 Ha \n\n(2.PsO. => 0) M(x y) R(4)\n\n0, B= 2(x + y)\n\n( x= 9)= v \u2192 (8, 9)\n\nP \n\nx3=1(2-3)\n\nS \u2201 12, 9 \n\na \n\nfP= 9 +1= 0\n\nP \n\n(S(0)\n\nIbyod 0\n\nA 1 = r (1)+3)\n\n-P= O\n\n(0,0) x= 42 = 51 -a \n\n-0 8 = 0 \n\n(17)\n\nL (0,6) = 3\n\nP R0 + 2y x + 5 \n\nfP = I \n\n(+A) = (x + 6)\n\nHT = 9-0 \n\n\u00e2 Nan (0, 0)\n\n3 (x= -6), y + 1+ 18