# Proof f(x)>g(x) in an interval

1. Dec 20, 2016

### Dank2

1. The problem statement, all variables and given/known data
let fx, gx be continuous in [a,b] and differentiable in (a,b). at the end of the interval f(a) >= g(a).

and f'(x) >g'(x) for a<x<b.

proof f(x) > g(x) for a<x<=b

Attempt:
There is a statement says that if the f'x = g'x for x in [a,b] , then there exists k such that f'x - g'x = k for any x in [a,b]

but f'(x) = g'(x) + t(x), where t(x) isn't have to be a line. and i cannot use the statement

2. Dec 20, 2016

### BvU

Shouldn't that be a < sign ?

3. Dec 20, 2016

### Dank2

nope.

4. Dec 20, 2016

### BvU

Might be hard to prove, then:
a=0, b =2, f(x) = x, g(x) = 1 satisifies all criteria, yet f < g on half the interval.

 Ah, the end of the interval is at the beginning. My bad

5. Dec 20, 2016

### Dank2

but f(a) is not equal or or bigger thatn g(a).

6. Dec 20, 2016

### BvU

Yeeah, I get it. Was wrongfooted by the 'end' term.

With f' - g' > 0 and the definition of derivative you should be able to go through the interval without f-g diminishing from its starting value f(a)-g(a).

7. Dec 20, 2016

### Dank2

f'(x) - g'(x) = lim(f(x+h) -f(x) -g(x+h) -g(x))/h > 0

like that? how can i take out f(x) - g(x)

8. Dec 20, 2016

### BvU

Ergo there is a h> 0 such that f(a+h)-g(a+h) > f(a)-g(h)

9. Dec 20, 2016

### Dank2

wait but we need x no?, a is the start of the interval

10. Dec 20, 2016

### BvU

So a+h is the new 'beginning' for the next step. Crux is if we can prove this can be continued all through the a,b interval in a finite number of steps.

11. Dec 20, 2016

### Dank2

but why doesn't the limit and the h at the denominator making problems here?
what i've got is
lim (f(x+h) - g(x+h))/h > lim (f(x) + g(x))/h

12. Dec 20, 2016

### Dank2

ok i think i got it, the limit doesn't change it much

13. Dec 20, 2016

### PeroK

I'm not sure looking at the definition of the derivative is the best way forward. In any case, it's best to introduce a new function, let's call it $k$ such that:

$k(x) = f(x) - g(x)$

What can you say about $k$? And what theorems do you know?

14. Dec 20, 2016

### Dank2

it is continuous and it is differentiable.
However there is a theorem i know regarding k(x) being constant, but i cannot assume it here.

15. Dec 20, 2016

### PeroK

Do you know the Mean Value Theorem?

16. Dec 20, 2016

### Dank2

yes

17. Dec 20, 2016

### PeroK

It might be useful here!

18. Dec 20, 2016

### Dank2

So;
(f(b) - f(a))/b-a > (g(b)-g(a))/b-a

and i can pick any b-a such that the diffrence is very small, so i can use x instead of it ?? so f(b) - f(a) = f(b)

19. Dec 20, 2016

### PeroK

You need to switch your attention to $k(x)$. Also, I suggest a proof by contradiction using the MVT.

20. Dec 20, 2016

### Dank2

ok K'(x) >0 for any x in the interval.