Chủ đề này có 2 trả lời

[study.maths]

Let f: R ----> R such that:
f(x)x 0, x R,
f is differential and |f'(x)| M, x R.

Prove that |f(x)| M|x|, x R.

Please help me.

Thanhks.

Bài viết đã được chỉnh sửa nội dung bởi study.maths: 20-06-2007 - 18:46

[hoc.toan]

Let f: R ----> R such that:
f(x)x 0, x R,
f is differential and |f'(x)| M, x R.

Prove that |f(x)| M|x|, x R.

Please help me.

Thanhks.

It's clear that f(0)=0 |f(x)| M|x|, x R.

Now we have to prove f(0)=0.

Suppose that x_1,x_2 R such that x_1 x_2.

If f(x_1)=f(x_2) then f(x_1)x_1=f(x_2)x_1, for x_1 0.

Hence, f(x_2)x_1 0 x_2 R, x_1 R\{0}.

This is contradicted. Then f(x_1) f(x_2). It means that kerf={0}. So, f(0)=0.

Applying Lagrangian theorem for f over [0,x], x R, we have the necessary result.

[study.maths]

In my opinion, hoc.toan's proof is very well. Thanks.

I'd like the others to take a care for the solution of this simple problem, if there is anything incorrect then give your comment, please.

Bài viết đã được chỉnh sửa nội dung bởi study.maths: 19-06-2007 - 14:25