Functions not differing by a constant whose derivatives do not equal h(x)

[kingstrick]

Homework Statement

If h(x):=0 for x<0 and h(x):=1 for x≥0. Prove there dne f:ℝ→ℝ such that f'(x) = h(x) for all x in ℝ. Give examples of two functions not differing by a constant whose derivatives equal h(x) for all x ≠ 0

Homework Equations

The Attempt at a Solution

I don't know how to begin. Am I trying to prove that h(x) has no antiderivative. If so...can someone give me a hint as to which theorem I am supposed to use.

 

[HallsofIvy]

Do you know the theorem that says that, while a derivative is not necessarily continuous, every derivative has the "intermediate value property"? Having said that there does not exist a function whose derivative is equal to h, there certainly cannot exist two, even if you do not specify the value at x= 0 so you will need to find functions that are not differentiable (perhaps not even continuous at x= 0). I would try functions that are constant for x< 0 then some constant plus x for x> 0.

 

[kingstrick]

Do you know the theorem that says that, while a derivative is not necessarily continuous, every derivative has the "intermediate value property"?

Are you referencing Darboux's Theorem? If so, then yes... Should I try to prove this by contradiction then and show that h(x) is a derivative of some other function?

 

[Citan Uzuki]

Should I try to prove this by contradiction then

Yes.

and show that h(x) is a derivative of some other function?

No, you don't show it's the derivative of a different function. You show it's not a derivative at all.

Look, you know that derivatives have the intermediate value property. Does h have the intermediate value property?

 

[kingstrick]

Yes.



No, you don't show it's the derivative of a different function. You show it's not a derivative at all.

Look, you know that derivatives have the intermediate value property. Does h have the intermediate value property?

No because there is no intermediate value it is only 0 or 1...

 

[Citan Uzuki]

Exactly. And that proves the theorem. Derivatives have the IVP, h does not, so h is not a derivative. That's all you need to say for that part of the problem.

As for finding two functions whose derivatives match h at every nonzero point, I'd look back to HallsofIvy's suggestion.

 

[kingstrick]

Exactly. And that proves the theorem. Derivatives have the IVP, h does not, so h is not a derivative. That's all you need to say for that part of the problem.

As for finding two functions whose derivatives match h at every nonzero point, I'd look back to HallsofIvy's suggestion.

Thank you for everything. So one eq. Was obvious but to make another that isn't only different by a constant does this imply that I need a function that is composition which reduces to 1 as a derivative?

 

[Citan Uzuki]

I'm not sure what you mean. You need a function whose derivative is 0 for x<0 and 1 for x>0 (this function will not be differentiable at zero). Then you need a second function of the same type that differs from the first by something that isn't a constant.

You really have to go with HallsofIvy's suggestion here, as there aren't any other functions with the correct derivative on $\mathbb{R} \setminus \{ 0 \}$. This means that you need two functions of the form:
$$f(x) = \begin{cases} A & \text{if } x<0 \\ B+x & \text{if } x \geq 0 \end{cases}$$

Note that you may choose the constants A and B independently.