Proof of f(b)<g(b) Using the Mean Value Theorem

[tempneff]

1. Suppose that f and g are continuous on [a,b] and differentiable on (a,b). Suppose also that f(a)=g(a) and f '(x)<g '(x) for a<x<b. Prove that f(b)<g(b). [Hint: Apply the Mean Value Theorem to the function h=f-g].
2. {[f(b)-f(a)]\b-a}=f'(c)
3.
I know:
If h(x) = f(x) - g(x) then
h(a) = f(a) - g(a) = 0 and
h is continuous on [a,b] differentiable on (a,b) so Mean Value Theorem applies
[h(b) - h(a)] / (b - a) = h'(c) for some c in (a,b). Therefore
[(f(b) - g(b)) - 0] / (b - a) = h'(c) and
h'(c)=f'(c) - g'(c) which is < 0 because f '(x)<g '(x) for a<x<b.
So [(f(b) - g(b)) - 0] / (b - a) < 0
Now...
I believe that f(b) - g(b) < 0 but I can't prove it. Any tips

 

[micromass]

Everything is correct. So you've got

[tex]\frac{f(b)-g(b)}{b-a}<0[/tex]

Now you need that f(b)-g(b)<0. You know that b-a>0...

 

[tempneff]

I know I am missing something very simple, but how can I say that b-a>0 just because a<x<b??
what is b= -1 and a= -2 then b-a <0?? I know it is something so insignificant ...

 

[micromass]

If b=-1 and a=-2, then b-a=-1-(-2)=-1+2=1>0.
If b>a, then it is always true that b-a>0. Just add -a to both sides...

 

[tempneff]

I am a moron.