# Proving if f'=g', then f=g+k for all k constant

1. Oct 27, 2012

### k3k3

1. The problem statement, all variables and given/known data
Let f and g be two continuous functions that are defined on [a,b] and differentiable on (a,b).

Claim: If f'=g' on (a,b), then there exists a constant k such that f(x)=g(x)+k for all x in [a,b]

2. Relevant equations

3. The attempt at a solution

Let h:[a,b]->R be defined by h(x)=g(x)+k-f(x)

h is continuous and differentiable on [a,b] by the assumptions of f and g already made.

Since f'=g' => g'-f'=0

So h'(x) = g'(x) - f'(x) = 0

By the mean value theorem, there exists a c in (a,b) so that

0=(h(b)-h(a))/(b-a) => h(a)=h(b)

I think I showed rolle's theorem. I am trying to use my assumption that f'=g' to show the claim. Any hints on this?

2. Oct 27, 2012

### jgens

Consider the function defined by $h(x) = f(x)-g(x)$ and apply the mean value theorem to show that this is a constant function.

3. Oct 27, 2012

### k3k3

If h(x)=f(x)-g(x), then h'(x)=f'(x)-g'(x) and, since f'=g' by assumption, f'(x)-g'(x)=0 for all x in (a,b). By MVT, there is a c in (a,b) so that h'(c)=h(b)-h(a) = 0 or

[f(b)-g(b)-f(a)+g(a)]/(b-a) = 0 => f(b)-f(a)=g(b)-g(a) ?

If h'(x) = 0 for all x in (a,b), then h is constant.

Is this what you meant?

4. Oct 27, 2012

### jgens

Not exactly. The idea is to suppose $x,y \in (a,b)$ and compute $h(x)-h(y)$ using the mean value theorem. This will tell you that the function is constant and complete the proof.

5. Oct 27, 2012

### k3k3

So choose x,y in (a,b) arbitrary so that h(x)-h(y) = 0?

6. Oct 27, 2012

### Dick

Do a proof by contradiction. Define h(x)=f(x)-g(x) and suppose it's not constant. What does the MVT tell you?

7. Oct 27, 2012

### Ray Vickson

Yes, but you need to prove that, not just write it down.

RGV

8. Oct 27, 2012

### k3k3

MVT tells me that there is a point in the interval (a,b) such that h'(c) = [h(b)-h(a)]/(b-a)

So f'(c)-g'(c) = [h(b)-h(a)]/(b-a) and f'(x) = g'(x) for all x in [a,b]

So 0 = [h(b)-h(a)]/(b-a) => h(b)=h(a) making it a constant function.

9. Oct 27, 2012

### Dick

You aren't really stating that it in "proof by contradiction" form. Suppose h(a) is NOT equal to h(b) what would bother you about that?

10. Oct 27, 2012

### k3k3

What would bother me that is, when f'(x)=g'(x), is that h(a) is h(b). And when h'(x) = 0 for all x in (a,b), then h(x) is constant.

11. Oct 27, 2012

### Dick

You are just stating what you already think you know without proving anything. If h(a)-h(b) is not zero then (h(a)-h(b))/(a-b) is not zero. The MVT should give you a contradiction that would prove that can't happen.

12. Oct 28, 2012

### k3k3

Let me try this again, then...

I want to show that if f'=g' on (a,b), then there exists a constant k such that f(x)=g(x)+k for all x in [a,b]

Let f and g be two continuous functions that are defined on [a,b] and differentiable on (a,b).
Assume f'(x)=g'(x) for all x in [a,b]

Let h:[a,b]->ℝ be defined by h(x)=f(x)-g(x)

Suppose h(a)≠h(b)

Then there is no point c in (a,b) so that h'(c)=0

By the MVT, there is a point d in (x,y)$\subset$[a,b] such that h'(d) = [h(y)-h(x)]/(y-x)

Since h is differentiable on all of [a,b], h'(x)=f'(x)-g'(x)

And f'-g'=0 by assumption.

So h'(d)=0=[h(y)-h(x)]/(y-x) => h(y) = h(x) for all x,y in [a,b]

This means h'(x)=0 for all x in [a,b] and by the monotonicity theorem, h is constant on [a,b].

So f and g must be the same.

13. Oct 28, 2012

### Dick

Things start going wrong about here. That is just plain not true. You are negating statements wrong, I think. What does h(a)≠h(b) really imply?

14. Oct 28, 2012

### k3k3

That the function has different values at the end points of the interval.

15. Oct 28, 2012

### Dick

What does h(a)≠h(b) imply if you use the mean value theorem?

16. Oct 28, 2012

### k3k3

h'(c) = [h(b)-h(a)]/(b-a) ≠ 0

17. Oct 28, 2012

### Dick

Yes! There is c in (a,b) such that h'(c)≠0. But you know h'(x)=0 for all points c in (a,b). So?

18. Oct 28, 2012

### k3k3

h'(c) must be 0 too since c is in (a,b)

19. Oct 28, 2012

### Dick

Yes. So h'(c) must be both zero and nonzero. That's impossible, so your initial assumption h(a)≠h(b) must be wrong. So?