求函数f(x,y)=x2-y3-6x 12y 15的极端-搜答案-智慧作业帮

从结论上反推即可

y1>y3>y2

#includedoublefun(doublex,doubley){returnx*x*x+y*y*y;}voidmain(){doublea=fun(3.4,2.8);doubleb=fun(10

∵y=-2/x在（负无穷,0）上是增函数∴当y1>y2>y3>0时,0>x1>x2>x3选C

x3+y3=100(x+y)(x^2-xy+y^2)=100因x+y=1所以x^2-xy+y^2=100(x+y)^2-3xy=1001-3xy=100xy=-33x^2+y^2=(x+y)^2-2x

把三条直线画出来求K值其实就是和斜率有关画画图你就能做出来了再问：求f(x,y)的范围我会，但如何理解三角形总是存在的？谢谢！再答：总是存在的就是你把f(x,y)=(x+(k-1)y+4-k)/(x+

要使函数y=lg(x2-4x+3)有意义,则有x^2-4x+3>0即有(x-1)(x-3)>0解得：x3所以函数y=lg(x2-4x+3)求f(x)的定义域为{x|x3}

f(x)=x²-x-5g(x)=1/3x³-5/2x²+4xg'(x)=x²-5x+4y=g'(x)/[f(x)+9]=(x²-5x+4)/(x

(1)当1-2x<0,即x>1/2时,f(x)=1+1-2x=2-2x,求得f(x)与g(x)在[0,1]上的交点为(1,0)由图象可知x∈(1/2,1]时,F(x)=min{f(x),g

f(x+199)=x2-4x+9=(x-2)^2+5=((x+199)-201)^2+5,故f(x)=(x-201)^2+5,故f(x)的最小值=0+5=5

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)

df/dx=3x^2-3y=0df/dy=3y^2-3x=0得驻点(0,0)(1,1)A=d^2f/dx^2=6xC=d^2f/dy^2=6yB=d^2f/dxdy=-3①对驻点(0,0)A=0B=-

f(x,y)=In(1+x2+y2)+1-x3/15-y3/4?应该是f(x,y)=In(1+x²+y²)+1-x³/15-y³/4吧?楼主的问题没法解答.楼主是要

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

求偏导另其等于0即可

x=0或x=整负根号下1-y方

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

y=(1-C1-C2)y1(x)+C1y2(x)+C2y3(x)即y=y1(x)+C1*[y2(x)-y1(x)]+C2*[y3(x)-y1(x)]而y1(x),y2(x),y3(x)都是线性方程y'

x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup