(1)对原函数求导g'(x)=1/2x²-2(a+1)/x³+m ,因为g'(x)过(1,0),g'(1)=1/2-2(a+1)+m=0 ===>m=4a+3/2
(2)a=-1时,m=-1/2 f(x)=1/2x²-1/2-lnx f'(x)=x-1/x 令f'(x)=0 求得驻点x=+1,-1, 函数lnx中x必须大于0,所以我们讨论(0,...
(1)对原函数求导g'(x)=1/2x²-2(a+1)/x³+m ,因为g'(x)过(1,0),g'(1)=1/2-2(a+1)+m=0 ===>m=4a+3/2
(2)a=-1时,m=-1/2 f(x)=1/2x²-1/2-lnx f'(x)=x-1/x 令f'(x)=0 求得驻点x=+1,-1, 函数lnx中x必须大于0,所以我们讨论(0,...