已知函数f(x)=(1/2)x^2+Inx求证当x>1时,1/2x^2+Inx
问题描述：
已知函数f(x)=(1/2)x^2+Inx求证当x>1时,1/2x^2+Inx
答
设g（x）=f(x)-2/3 x³=1/2 x²+lnx-2/3 x³
即证明当x＞1时,g（x）＞0
g(x)′=x+1/x -2x²
=(x²+1-2x³)/x
=(x²-2x+1+2x-2x³)/x
=[(x-1)²-2x(x+1)(x-1)]/x
=-(2x²+x+1)(x-1)/x
显然当x≥1时,g(x)′＞0,g(x)递增
所以当x＞1时,g(x)＞g(1)=1+1-2=0