f'(x) = x-1/x令f'(x)=0, 得x=1或-1, 所以f(x)在区间[1,e]上单调.f(1) = 1/2, f(e) = 1/2e^2-1 >1/2, 所以f(x)在区间[1,e]上的最大值为1/2e^2-1,最小值1/2
f'x=x 1/x f'x>0f1min,femax