Question

已知函数f(x)=ln x+a/x，若函数f(x)在[1,e]上最小值是3/2，求a

Answer 1

f'(x)=1/x-a/x^2=(x-a)/x^2=0
x<=a时，f(x)减x>a时，f(x)增
a<=1时在[1,e]增fmin=f(1)=a=3/2与a<=1矛盾
1<a<=e在[1,e]先减后增，f(a)=ln(a)+1=3/2,a=根号e
a>e在[1,e]减fmin=f(e)=1+a/e=3/2 a=1/2e与a>e矛盾

Answer 2

f'(x)=1/x-a/x^2=0,x=a极值点
f''(x)=-1/x^2+2a/x^3,f''(a)=1/a^2>0,x=a极小值点
if a属于[1,e],f(a)=lna+1=3/2,a=√e
if a<1,f(x)单增,最小值f(1)=a=3/2,矛盾
if a>e,f(x)单减,最小值f(e)=1+a/e=3/2 a=e/2,矛盾

Answer 3

f'(x)=1/x-a/x^2=(x-a)/x^2
当x=a时，f'(a)=0
当 1<x<a时，f'(x)<0 f(x)在(1,a)内单调递减
当 a<x<e时，f'(x)>0 f(x)在（a,e)内单调递增 则x=a时，f(x)有最小值3/2
f(a)=lna+1=3/2
lna=1/2
a=e^(1/2)

Answer 4

令一阶导数f'(x)=1/x-a/x^2=0得x=a
1.当a<1时f(x)单调递增，最小值为f(1)=a=3/2与a<1矛盾舍去
2.当a>e时f(x)单调递减，最小值为f(e)=1+a/e=3/2得a=e/2,舍去
3.当1<=a<=e时最小值为f(a)=ln(a)+1=3/2,a=根号e合题意

Answer 5

a=根号e