# Application of the mean value theorem.

1. Nov 21, 2012

### mtayab1994

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

Let f and g be two continuous functions on [a,b] and differentiable on ]a,b[ such that for every x in ]a,b[ : f'(x)<g'(x)

2. Relevant equations

Show that f(b)-f(a)<g(b)-g(a)

3. The attempt at a solution

So I said that there exists a c in ]a,b[ such that f'(c)=(f(b)-f(a))/(b-a) and g'(c)=(g(b)-g(a))/(b-a) and since f'(x)<g'(x) then f'(c)<g'(c) and that implies that f'(c)/g'(c)<1 and when substituting in what we found we get that (f(b)-f(a))/(g(b)-g(a))<1 and that implies that f(b)-f(a)<g(b)-g(a). Is this correct??

Last edited: Nov 21, 2012
2. Nov 21, 2012

### Robert1986

Yes, that is correct, but I don't understand why you did the division (also this is dangerous- what if $g'(c) =0$). Do you see how you can do it with out the division?

3. Nov 21, 2012

### mtayab1994

No I can't see how, but i can suppose that g'(c)≠0 right?

4. Nov 21, 2012

### Robert1986

I don't see how you can suppose that (I might be missing something, but if you can, you will have to prove why it is true, which isn't necessary.) For example, if $f(x)=-5x$ and $g(x)=-x^2+1$ and $a=-1$ and $b=1$ the smallest that $g'$ gets is -2. But, $f'(x)=-5 < -2 < g'(x)$. Right?

So, you know that:
$f'(c) < g'(c)$ and
$f'(c)=\frac{f(b)-f(a)}{b-a}$ and $g'(c)=\frac{g(b)-g(a)}{b-a}$.

Now do you see?

5. Nov 21, 2012

### mtayab1994

So all you are going to do now is say that since f'(c)<g'(c) and when we cancel the b-a we will get f(b)-f(a)<g(b)-g(a) right??

6. Nov 21, 2012

### Robert1986

Yep!

7. Nov 21, 2012

### mtayab1994

Alright thanx!