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

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Homework Statement


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]


Homework Equations





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?
 

Answers and Replies

  • #2
jgens
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Consider the function defined by [itex]h(x) = f(x)-g(x)[/itex] and apply the mean value theorem to show that this is a constant function.
 
  • #3
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Consider the function defined by [itex]h(x) = f(x)-g(x)[/itex] and apply the mean value theorem to show that this is a constant function.
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
jgens
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Is this what you meant?
Not exactly. The idea is to suppose [itex]x,y \in (a,b)[/itex] and compute [itex]h(x)-h(y)[/itex] using the mean value theorem. This will tell you that the function is constant and complete the proof.
 
  • #5
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Not exactly. The idea is to suppose [itex]x,y \in (a,b)[/itex] and compute [itex]h(x)-h(y)[/itex] using the mean value theorem. This will tell you that the function is constant and complete the proof.
So choose x,y in (a,b) arbitrary so that h(x)-h(y) = 0?
 
  • #6
Dick
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So choose x,y in (a,b) arbitrary so that h(x)-h(y) = 0?
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
Ray Vickson
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So choose x,y in (a,b) arbitrary so that h(x)-h(y) = 0?
Yes, but you need to prove that, not just write it down.

RGV
 
  • #8
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Do a proof by contradiction. Define h(x)=f(x)-g(x) and suppose it's not constant. What does the MVT tell you?
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
Dick
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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.
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
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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?
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
Dick
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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.
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
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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.
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)[itex]\subset[/itex][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
Dick
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Suppose h(a)≠h(b)

Then there is no point c in (a,b) so that h'(c)=0
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
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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?
That the function has different values at the end points of the interval.
 
  • #15
Dick
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That the function has different values at the end points of the interval.
What does h(a)≠h(b) imply if you use the mean value theorem?
 
  • #16
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What does h(a)≠h(b) imply if you use the mean value theorem?
h'(c) = [h(b)-h(a)]/(b-a) ≠ 0
 
  • #17
Dick
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h'(c) = [h(b)-h(a)]/(b-a) ≠ 0
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
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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?
h'(c) must be 0 too since c is in (a,b)
 
  • #19
Dick
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h'(c) must be 0 too since c is in (a,b)
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?
 

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