若只
幼苗
共回答了29个问题采纳率:96.6% 举报
(1)证明:对任意x
1,x
2∈R,有
|f(x1)−f(x2)|=|
1+
x21−
1+
x22|=|
x21−
x22
1+
x21+
1+
x22|=
|x1−x2|•|x1+x2|
1+
x21+
1+
x22.…2分
由|f(x
1)-f(x
2)|≤L|x
1-x
2|,即
|x1−x2|•|x1+x2|
1+
x21+
1+
x22≤L|x
1-x
2|.
当x
1≠x
2时,得L≥
|x1+x2|
1+
x21+
1+
x22.
∵
1+
x21>|x1|,
1+
x22>|x2|,且|x
1|+|x
2|≥|x
1+x
2|,
∴
|x1+x2|
1+
x21+
1+
x22<
|x1+x2|
|x1|+|x2|≤1.…4分
∴要使|f(x
1)-f(x
2)|≤L|x
1-x
2|对任意x
1,x
2∈R都成立,只要L≥1.
当x
1=x
2时,|f(x
1)-f(x
2)|≤L|x
1-x
2|恒成立.
∴L的取值范围是[1,+∞).…5分
(2)证明:①∵a
n+1=f(a
n),n=1,2,…,
故当n≥2时,|a
n-a
n+1|=|f(a
n-1)-f(a
n)|≤L|a
n-1-a
n|=L|f(a
n-2)-f(a
n-1)|≤L
2|a
n-2-a
n-1|≤…≤L
n-1|a
1-a
2|
∴
n
![](https://img.yulucn.com/upload/8/90/89088027fba1be3c0df0dc2255073c03_thumb.jpg)
k=1|ak−ak+1|=|a1−a2|+|a2−a3|+|a3−a4|+…+|an−an+1|≤(1+L+L
2+…+L
n-1)|a
1-a
2
1年前
5