求函数f(xy)=x³-3x-2y y² 1的极值

先对x求偏导u'x=f'(x,xy,xyz)+yf'(x,xy,xyz)+yzf'(x,xy,xyz)所以u'xy=yf''(x,xy,xyz)+xzf''(x,xy,xyz)+f''(x,xy,xy

设a=xy,b=x+y.f(xy,x+y)=x^2+y^2+2xy-2xy=(x+y)^2-2xy把a,b带f(a,b)=b^2-2a所以f(x,y)=y^2-2x同理f(x+y,xy)=x^2+y^

fx=x-y+3=0fy=-x+2y=0解得唯一驻点（-6,-3）A=fxx=1B=fxy=-1C=fyy=2AC-B²=1>0,A>0所以取极小值f(-6,-3)=-9

由fx(x,y)=2x-6y+18=0fy(x,y)=3y^2-6x-39=0解得驻点有(-6,1)(-6,5)(6,1)(6,5)二阶偏导fxx(x,y)=2fxy(x,y)=-6fyy(x,y)=

当x=y=0时f(0+0)+f(0)=2f(0)f(0)f(0)²=f(0)f(0)=1或者f(0)=0当y=0时f(x)+f(0)=2f(x)f(0)若f(0)=0f(x)=0若f（0)=

f'x=3x^2+3y=0-->y=-x^2f'y=-3y^2+3x=0-->y^2=xx=y^2=x^4-->x=0,1,-->y=0,-1f"xx=6x,f"yy=-6y,f"xy=3f(0,0)

令g(x)=f(x)-xg(xy)+xy=x(g(y)+y)+y(g(x)+x)-xyg(xy)=xg(y)+yg(x)令x=0,g(0)=yg(0),g(0)=0若存在|a|>=1使得g(a)不等于

令x=y=1f(1)=f(1)+f(1)f(1)=0

f'x=3x^2-3yf'y=3y^2-3xf'x=0,f'y=0即x^2-y=0y^2-x=0消去yx^4-x=0即x(x-1)(x^2+x+1)=0x=0或1y=0或1x=y=0时f(x,y)=0

(第1步)设x=0,y=2008,代入公式f(x+y)=f(xy)两边,得f(x+y)=f(0+2008)=f(2008);f(xy)=f(0*2008)=f(0);有：f(2008)=f(0);(第

3=1+1+1=f（2）+f(2)+f(2)=f(2*2)+f(2)=f(4*2)=f(8)f(x)+f(x-2)=f(x*(x-2))=f(x^2-2x)结合定义域知识,所以f(x)+f(x-2)0

分别对x,y求偏导数得：f'(x)=2x+y-6f'(y)=2y+x-3令两者都为0,解得驻点为：(3,0)又分别对其求二阶偏导数：f''(x)=2=Af''(y)=2=C用f'(x)再对y求偏导数得

f(xy)=f(x)+f(y)所以,f(1/9)=f(1/3)+f(1/3)=2f(x)+f(2-x)1/9x^2-2x+1/9

分别求X,Y的偏导3x^2-3y=03y^2-3x=0x=0x=1y=0y=1代回去就是

1.f(8)=f(2)+f(4)=f(2)+f(2)+f(2)=3,所以f(2)=1f(2)=f(√2)+f(√2)f(√2)=1/221.原式定义域为R,那么ax^2+4ax+3=0无解a(x+2)

1)令x=y=1,则f(1)=f(1)+f(1)=2f(1),所以,f(1)=0.2）取y=1/x,则f(1)=f(x)+f(1/x),所以,f(x)+f(1/x)=0,因此,f(1/3)+f(1/2

答案是0由f(xy)=f(x)+f(y),得f(1)=0,如f(1/4)=f(1)+f(1/4)推出放f(1)=0所以f(1/3)+f(1/2)+f(1)+f(2)+f(3)=f(1/3)+f(3)+

令x=y=2则xy=4所以f(4)=f(2)+f(2)令x=4,y=2则xy=8所以f(8)=f(4)+f(2)=f(2)+f(2)+f(2)=3f(2)=1令x=y=√2则xy=2所以f(2)=f(

答：f(x,y)=3xy/(x^2+y^2)f(y/x,1)=3*(y/x)*1/[(y/x)^2+1^2]=(3y/x)/[(y^2+x^2)/x^2]=3xy/(x^2+y^2)=f(x,y)x≠