# Exercise with lagrange and derivatives

1. Dec 22, 2012

### Felafel

1. The problem statement, all variables and given/known data

Being a>0 and f:[a,b]--->R continuos and differentiable in (a,b), show that there exists a t $\in$ (a,b) such that:
$\frac{bf(a)-af(b)}{b-a}=f(t)-tf'(f)$

3. The attempt at a solution
For lagrange's theorem, we have:
$\frac{f(a)-f(b)}{b-a}= -f'(t)$

thought i could find f(t) from the equation $f(x) = f(t) + f'(t)(x-t)+R_1(x-t)$ ignoring R.
but then a "x" would appear and I don't know how to deal with it.

Last edited by a moderator: Dec 22, 2012
2. Dec 22, 2012

### Staff: Mentor

To simplify the equation, consider g:[a,b]->R with
$$g(x)=f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}$$
This is chosen to get g(a)=g(b)=0 and $g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$

Is there some t ∈ (a,b) such that $0=g(t)-tg'(t)$?
Working backwards, you should be able to get the original equation.

3. Dec 23, 2012

### Felafel

sounds good, but I'm explicitly asked to use Lagrange's theorem

4. Dec 23, 2012

### Staff: Mentor

I think you can use that theorem to solve the reduced problem ;).

5. Dec 23, 2012

### Felafel

another hint? i don't get it, i always end up in a loop :(