f(x)=(a-1/2)x^2+lnx求导和当a=1时,f(x)在区间[1,e]的最大值和最小值

Question

f(x)=(a-1/2)x^2+lnx 求导和当a=1时,f(x)在区间[1,e]的最大值和最小值

Answer 1

f'（x）=2（a-1/2）x+1/x
当a=1时,f'(x)=x+1/x在[1,e]上恒大于0,所以单调递增.
f(x)max=f(e)=1/2*e^2+1
f(x)min=f(1)=1/2