# Two Functions with the Same Derivative

1. Apr 23, 2015

### Bashyboy

1. The problem statement, all variables and given/known data
I am trying to prove without using the mean value theorem that two different functions with the same derivative differ only by a constant. Is it possible to do this without the mean value theorem? If so, would someone help guide me towards the right solution.

2. Relevant equations

3. The attempt at a solution

Let $f'(x) = h(x)$ and $g'(x) = h(x)$. Define the function $T(x) = h(x) - g(x)$, which has the derivative $T'(x) = 0$ for all $x$. If I could prove that the derivative of a function being identically zero implies that the function is a constant, then I would be finished; however, even this is giving trouble.

2. Apr 23, 2015

### Simon Bridge

What's wrong with the mean value theorem approach?
Use the definition of a derivative on T.

3. Apr 23, 2015

### Bashyboy

Because the mean value theorem isn't available to me.

$T'(x) = \lim_{n \rightarrow \infty} \frac{f(z_n) - f(x)}{z_n -x} = 0$

I am not certain how this would be helpful. I know that $T'(x)$ exists because both $h'(x)$ and $g'(x)$ exist, and therefore the limit exists for every sequence $\{z_n\}$ that converges to $x$. If I could write

$\lim_{n \rightarrow \infty} \frac{f(z_n) - f(x)}{z_n -x} = 0$

as

$\frac{f(z_n) - f(x)}{z_n -x} = 0$

through some argument, then I could arrive at

$f(z_z) = f(x)$.

Because the limit exists for every sequence, I could chose the sequence $\{z_n\}$ so that $z_n \in [a,b]$ for every $n \in \mathbb{N}$ and passes over every point in $[a,b]$. This would mean that, for any two points in $[a,b]$, the function $T(x)$ is equal to itself, which would mean it is a constant function. However, I am not sure to make these arguments precise, nor am I certain it is even possible.

4. Apr 23, 2015

### Simon Bridge

5. Apr 23, 2015

### Bashyboy

The definition of derivative used on that webpage is one I cannot use; I have to use the sequence definition. Furthermore, I don't need the implication that if f(x) is a constant function, then its derivative is zero; rather, I need the converse--namely, if the derivative of a function is zero, then the function must be a constant one. Those are two different statements.

6. Apr 23, 2015

### Simon Bridge

I know that. One of the ways of proving something is to assume something else and disprove it.
If you know how to find the derivative of a function - then go through the process but set the answer to zero ... then find out which function satisfies that answer.

The sequence definition makes it harder ... but the fact that a derivative of zero means a constant function is a consequence of the definition of the derivative, so you should be able to deduce it from that with some work. The alternative is to use a definition of the anti-derivative.

7. Apr 23, 2015

### Bashyboy

So, you are suggesting a proof by contradiction?

8. Apr 23, 2015

### Bashyboy

Well, if we suppose the contrary, then statement becomes $T'(x) = 0$ for all $x \in [a,b]$ but $T(x) \ne k$, where $k$ is some constant. If $T(x) \ne k$, then it must involve at least one variable term, say $x$; hence, $T(x) = x + ...$. The derivative would be $T'(x) = 1 + ... > 0$, which contradicts the fact that $T'(x) = 0$.

Obviously this would not suffice as a proof.

9. Apr 23, 2015

### Ray Vickson

People make suggestions to you and you say you are not allowed to use those results/methods. What ARE you allowed to use? Are you, for example, allowed to look in a book for some results that have not yet been covered in class, and use them? If you are not allowed to use results not yet proved in class, are you allowed to "copy" out the proof and then use the result? I really do not understand why so many restrictions are being placed on you, or even what those restrictions are.

10. Apr 23, 2015

### Simon Bridge

... why is that not sufficient? Can you be explicit?
I think the difficulty you have encountered here is that you have not sufficiently defined your terms.

@Ray: He does not want to use the mean value theorem for derivatives. The definition of the derivative cited is very closely related to the mean value theorem.
I suspect that any attempt at this problem will end up amounting to using the mean value theorem, just starting from a different place. Either that or any useful definition of the derivative will be ruled out, making the problem impossible since what is to be found is no longer defined.

11. Apr 23, 2015

### SammyS

Staff Emeritus
Maybe what you're trying to prove isn't true.

You may have to specify something about the continuity of the functions