Question

(3)y=(x-4)^0+1/((x-1))+(6-x):

Answer 1

问：(3)y=(x-4)^0+1/((x-1))+(6-x):

答：您好，(3)y=(x-4)^0+1/((x-1))+(6-x):（3x-4y)(x+y-1)=03x-4y=0，x+y-1=0y=3x/4,y=1-x分别代入x²+y²=5x=4√5/5,y=3√5/5x=-4√5/5,y=-3√5/5x=4,y=-3x=-3,y=4

答：相关资料：z=y^3=(x-4)^3 (x+1)^2然后求驻点z'=03(x-4)^2(x+1)^2+(x-4)^3*2(x+1)=0(x-4)^2(x+1)[3x+3+2x-8]=0(x-4)^2(x+1)(5x-5)=0x=4,-1,1然后极值的话看二次导数值正负z''=2(x-4)(x+1)(5x-5)+(x-4)^2(5x-5)+(x-4)^2(x+1)*5在x=4,z''=0,所以不是极值点在x=-1,z''=-2500,是局部极小值点所以局部最大为x=-1,y=0局部最小为x=1,y=-3*2^(2/3)