Question

第20题函数求解

Answer 1

(20)
f(x)=aln(1/x) +x +1
a=1
f(x)= ln(1/x) +x +1
1/x >0
x>0

f'(x) = -1/x +1
f'(x) = 0
x=1

f'(x) |x=1+ >0
f'(x) |x=1- <0
x=1 (min)
f(x)= ln(1/x) +x +1
min f(x) = f(1) =2
单调
增加 =(0, 1]
减小=[1, +∞)
(II)
x∈[1,e]
是否存在a 使得 f(x) =1
solution :
f(x)=aln(1/x) +x +1
f'(x) = -a/x +1
f'(x) = 0
x=a
case 1: a≤1
min f(x) = f(1) = 2
case 2: 1<a<e
min f(x)
= f(a)
=aln(1/a) +a +1
=-alna + a +1
>1
case 3: a≥e
min f(x) = f(e) =-a +e +1≤1
a=e
f(e) = -e +e +1 =1
ie
a=e