Prove that for all we have

Before proceeding with the proof, we recall the second mean-value theorem for integrals (Theorem 5.5 on p. 219 of Apostol). For a continuous function on the interval if has a continuous derivative which never changes sign on the interval then there exists a such that

* Proof. * Now, we want to apply the mean-value theorem above with and . Since is continuous everywhere , it is continuous on any interval . Then,

is continuous for all (so, in particular, for all ). Furthermore, since for all we have that for all . Thus, is continuous and never changes sign in any interval . Therefore, we can apply the mean-value theorem to conclude there exists a for any such that

But since for all we know for any and . Hence,

At the end, following the same procedure, you get (x(1-cosc)+1-cosc)/(x+1) which is always greater or equal to 0 for any choice of c or x, such that 0<=c<=x.

I’m sorry it should be (x(1-cosc)+1-cosx)/(x+1)

I agree with Batman…

i got 1 – (x/(x+1)) * cos (c) – cos(x) / (x + 1) > or = 0, with equality only holding when cos(c) = cos(x) = 1

The integral limits were from 0 to x. this one is ok: f(a)=f(0) = 1/(1+0), but wasn’t the other supposed to be f(b) = f(x) = 1/(1+x)? You’ve written 1/(1+1), why?