Question

最速下降法求min f(x1,x2)=x1^6+x2^6+3*x1^2*x2^2-x1^2-x2^2-2*x1*x2

Answer 1

证：
x>0，x³>0，e^x>0，令g(x)=ln[f(x)]
底数e>1，对数值随真数f(x)的增大而单调递增。即：g(x)与f(x)有相同的单调性。
g(x)=ln(x³/e^x)=ln(x³)-ln(e^x)=3lnx -x
x1+x2=6，x2=6-x1
g(x2)-g(x1)=3ln(x2)-x2 -[3ln(x1)-x1]
=3ln(x2)-6+x1 -3ln(x1)
=3ln(6/x1 -1)+2x1-6
令h(x)=3ln(6/x -1) +2x -6，(0<x<3)
h'(x)=[3/(6/x -1)](6/x -1)' +2
=[3x/(6-x)]·(-6/x²) +2
=-6[3/x(6-x)] +2
=18/[x(x-6)] +2
=[18+2x(x-6)]/[x(x-6)]
=2(x²-6x+9)/[x(x-6)]
=2(x-3)²/[x(x-6)]
0<x<3，(x-3)²>0，x-6<0，2(x-3)²/[x(x-6)]<0
h'(x)<0，即0<x<3时，h(x)=3ln(6/x -1) +2x -6单调递减
令x=3，得h(3)=3ln(6/3 -1)+2×3-6=3ln1 +6-6=0
0<x<3时，h(x)>0
g(x2)-g(x1)>0
f(x2)-f(x1)>0
f(x1)<f(x2)