f(x)=x^3－(9x^2)*(cosα)+48x(cosβ)+18*(sinα)^2rng(x)=3x^2-18(cosα)x+48(cosβ)rn对任意的实数t均有g(1+cost)≥0,g(3+sint)≤0rncost＝1时,g(1+1)=g(2)>=0rnsint=-1时,g(3-1)=g(2)<=0rng(2)=0rn也就是rn0＝12－36cosα＋48(cosβ)rng(x)=3x^2-18(cosα)x+36cosα－12rng(3+sint)≤0rng(4)=36-36cosα≤0rncosα=1rnf(x)=x^3-(9x^2)*(cosα)+x(36cosα－12)=x^3-9x^2+24xrng(x)=3x^2-18x+24=3(x-2)(x-4)rnrnf(x)在区间（a-1,a)rna<＝2时,值域〔f(a-1),f(a)〕rn2<a<3时,值域〔{f(a-1),f(a)}min,f(2)〕rn3=<a<4时,值域〔f(a),f(a-1)〕rn5>a>4时,值域〔f(4),{f(a),f(a-1)}max〕rna>=5时,值域〔f(a-1),f(a)〕rnrn3）若对任意的m∈[-26,6],恒有-9x^2+24x＋11≥-mx,求x的取值范围rn你用数形结合,-9x^2+24x＋11是抛物线,-mx是绕原点转的直线rn-9x^2+24x＋11=－6x的小根为prn-9x^2+24x＋11=26x的大根为qrnx的范围是〔p,q〕rn自己解一元2次方程吧rnrnf(x)=x^3-xrnf(x)'=(3x^2-1)rny-f(t)=(3t^2-1)(x-t)rn直线y=(3t^2-1)(x-t)+t^3-trnrny=(3t^2-1)(x-t)+t^3-t过(a,b),有三条切线rn所以b＝(3t^2-1)(a-t)+t^3-t有3个不同实数根rng(t)=(3t^2-1)(a-t)+t^3-t=-2t^3+3at^2-arng(t)'=-6t^2+6at=6t(a-t)rn所以g(t)在(-无穷,0)和(a,+无穷)是减的rn在(0,a)是增的rn画草图,数形结合rny=b与g(t)=-2t^3+3at^2-a有三个交点rn-a=g(0)<b<g(a)=a^3-a=f(a)rnrn（3）设函数g(x)=(1+1/n)^x(n∈N,且n>1,x∈N),对任意实数x,证明[g(2x)+g(2)]/2>g'(x)rnm=1+1/nrn1<m<=2rng(x)=(1+1/n)^x＝m^xrng(2x)+g(2)=(m^x)^2+m^2rn2g(x)'=2m^x(lnm)rn(m^x)^2+m^2>2m^x(lnm)rn等价于rn(m^x)+m^2/m^x>2(lnm)rn(m^x)+m^2/m^x>=2mrn要证2m>2(lnm)rn即m>lnmrnlnm-m<0rnP(x)=lnx-xrn(1=<x<=2)rnP(x)'=1/x-1<0rn所以P(x)=lnx-xrn(1<x<=2)MAX=<P(1)=-1<0rnlnm-m<0rn得证