# If f(a)=g(a) and f'(x)>g'(x) for all x, use MVT to prove f(x)>g(x)

1. Dec 8, 2011

### NWeid1

1. The problem statement, all variables and given/known data
Give a graphical argument that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. Use the Mean Value Theorem to prove it.

2. Relevant equations

3. The attempt at a solution
I have sketched a graphical argument to show that f(x)>g(x). This is what I got:

Let h(x) = f(x) - g(x).
Then h'(x) = f'(x) - g'(x) > 0 for any x > a.
Now apply the MVT on the interval [a, x].

So.. h'(c) = (h(x) - h(a))/(x -a)

And then
$$h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0$$
$$h'(c) =f(x) - f(a) - g(x) + g(a) > 0$$
$$h'(c) =f(x) - f(a) > g(x) - g(a)$$
$$h'(c) =\frac{f(x) - f(a)}{x-a} > \frac{g(x) - g(a)}{x-a}$$
==> h'(c) = f(x) > g(x)
==> f(x) > f(x) for all x>a

But I feel like this is wrong?!

2. Dec 8, 2011

### Fredrik

Staff Emeritus
The equalities that all begin with h'(c)= look very wrong. They're saying that a single number h'(c) is equal to several different numbers. That clearly can't be true. I also don't see why you conclude that h'(c)>0.

3. Dec 8, 2011

### NWeid1

Well, I know to set h(x) = f(x) - g(x) but I don't know where to go from there. lol

4. Dec 8, 2011

### Fredrik

Staff Emeritus
The good news is that that's the right way to start.

Can you tell me exactly what the MVT says about h on the interval [a,x]?

5. Dec 8, 2011