Application of the mean value theorem.

[mtayab1994]

Homework Statement

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)

Homework Equations

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

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??

 

[Robert1986]

Yes, that is correct, but I don't understand why you did the division (also this is dangerous- what if [itex]g'(c) =0[/itex]). Do you see how you can do it without the division?

 

[mtayab1994]

Yes, that is correct, but I don't understand why you did the division (also this is dangerous- what if [itex]g'(c) =0[/itex]). Do you see how you can do it without the division?

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

 

[Robert1986]

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

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 [itex]f(x)=-5x[/itex] and [itex]g(x)=-x^2+1[/itex] and [itex]a=-1[/itex] and [itex]b=1[/itex] the smallest that [itex]g'[/itex] gets is -2. But, [itex]f'(x)=-5 < -2 < g'(x)[/itex]. Right?

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

Now do you see?

 

[mtayab1994]

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 [itex]f(x)=-5x[/itex] and [itex]g(x)=-x^2+1[/itex] and [itex]a=-1[/itex] and [itex]b=1[/itex] the smallest that [itex]g'[/itex] gets is -2. But, [itex]f'(x)=-5 < -2 < g'(x)[/itex]. Right?

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

Now do you see?

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??

 

[Robert1986]

Yep!

 

[mtayab1994]

Yep!

Alright thanx!