已知函数f(x)=x的平方+alnx,[1]当a=-2时,求函数f(x)的单调区间;[2]若g(x)=f(x)+2/x在[1,正无穷大)上是增

(1) f(x) = x^2 - 2lnx ==> f'(x) = 2(x^2-1)/x ==> 递减区间为(0,1),递增区间为(1,正无穷)(2) g(x) = x^2 - alnx + 2/x ==> g'(x) = (2x^3 - ax - 2)/x^2 因为g(x)=f(x)+2/x在[1,正无穷大)上是增函数,所以2x^3 - ax...