一、选择填空 (x2 1、已知lim-ax-b=0,则( ) x→x+1 (A)a=1,b=1(B)a=-1,b=-1(C)a=-1,b=1(d)a=1b=-1 2、函数f(x)=x(x2-3x+2)(x+2)有()个不可导点。 (A)1(B)2(C)3(D)4 3、设f(x)=x(x-1)(x-2)…(x-2004),则f(0)=() (A)-2003(B)-2004(C)2003(D)2004 4、设f(x)={sin≠0

9-1两小球处于如题9-1图所示的平衡位置时每小球受到张力T,重力mg以及库仑力 F的作用,则有Tcos=mg和Tsin=F,∴F=mgtg,由于很小,故 F=1q2 =mgtg0 mg sin=mg

then there exists AE R\ such that (Kuhn-Tucker condition) G(s') =0 and 1. Lagrange Method for Constrained Optimization FOC: D.L(,\)=0. The following classical theorem is from Takayama(1993, p.114). Theorem A-4 (Sufficieney). Let f and, i= ,..m, be quasi-concave, where Theorem A-1. (Lagrange). For f: and G\\, consider the following G=(.8 ) Let r' satisfy the Kuhn-Tucker condition and the FOC for (A.2). Then, x' problem is a global maximum point if max f() (1)Df(x') =0, and f is locally twice continuously differentiable,or

5.1.3线性空间上的对称双线性函数、二次型函数的定义 定义若f为V上的双线性函数且f(a,B)=f(B,a),则称f为V上的对称双线性函数

作业9:设函数f和g在[a,b]上连续,单调,有xn∈[ab使得 g(xn)=f(xn+1)(n=1,2,)证明:3x0∈[a,b],使得f(x0)=g(x)。 证:不妨设f(x)单调递增

定理5.2.1(levi定理)若n(x)为可测集E上的非负可测函数列, 且满足中(x)≤中+1(x),中n(x)→f(x)(n→+∞),则 fdx= lim 中dx n-JE 证明G(f,E)={(x,y)0≤y