Thread: proving : tg(x)>x using Lagrange / Cauchy's Lemma Use

I need to prove that when .

I'll use Cauchy's version for Lagrange's lemma:

If f,g are continuous in [a,b] and derivative in (a,b), then , (there exists a 'c' : a<c<b for which...)

So I'll take the range [0,x], while , and f(x):=sin(x), g(x):=xcos(x) :



Now, how can I prove that the last phrase is larger than 1? (so that ) ?

Thank you very much!

Originally Posted by adam63

I need to prove that when .

I'll use Cauchy's version for Lagrange's lemma:

If f,g are continuous in [a,b] and derivative in (a,b), then , (there exists a 'c' : a<c<b for which...)

So I'll take the range [0,x], while , and f(x):=sin(x), g(x):=xcos(x) :



Now, how can I prove that the last phrase is larger than 1? (so that ) ?

Thank you very much!

Don't really think there is any need for Lagrange here...

Let , then

Now, note that:

1.

2. Since for any , we get that .

These two facts, along with the continuity of , give us that and therefore

Originally Posted by Defunkt

Don't really think there is any need for Lagrange here...

Let , then

Now, note that:

1.

2. Since for any , we get that .

These two facts, along with the continuity of , give us that and therefore

That's a great way to solve it, only I haven't 'officially' learned it, and need to use Lagrange \ Roll \ Cauchy's lemmas for functions. Can you help me?

I got it!! I was really close, but I couldn't see it :

I need to prove:









^ Which is right for every