Expert: Paul Klarreich - 4/29/2008

Question
Having trouble figuring out where to start with this problem:
Let f and g be functions that are continuous on [a,b] and differentiable on (a,b). Prove that if f(a)=g(a), and g'(x)>f"(x) for all x in (a,b), then g(b)>f(b).
Thankyou

Answer
Questioner:   Edward Greiner
Category:  Calculus
Private:  No
 
Subject:  Calc
..............................
Hi, Edward,

Your Question:  
Let f and g be functions that are continuous on [a,b] and differentiable on (a,b).

>> Already sounds like Mean Value Theorem stuff.

Prove that if f(a)=g(a), and g'(x)>f"(x) for all x in (a,b), then g(b)>f(b).

Oops -- are you sure you meant to write  g'(x) > f-double-prime(x) ??

I think you meant to write:  g'(x) > f'(x).  If not, let me know.  I am going to assume you mistyped and prove it on that basis.
.......................
Let  h(x) = g(x) - f(x)
Then h(x) has these properties:

I. h is continuous on [a,b] and differentiable on (a,b), just like f,g.

II. h(a) = 0.

III. h'(x) = g'(x) - f'(x) and so h'(x) > 0 for all x in (a,b).

Now you want to prove  g(b) > f(b). Suppose it isn't.
I.E. suppose that  g(b) <=  f(b).  That means  h(b) <= 0.  

The MVT says there exists c in (a,b) such that:

       h(b) - h(a)
h'(c) = -----------
         b - a

But if  h(b) <= 0,  h(a) = 0, and (b-a) > 0, so that fraction is:


       (<=0) - (0)
h'(c) = ----------- <= 0.
          (>0)

But (see item III above) h'(c) > 0 for all c in (a,b), so this is not possible.

That's your proof.