# Differentiability of f(x+y)=f(x)f(y)

Given: f(x+y)=f(x)f(y). f'(0) exists.

Show that f is differentiable on R.

At first, I tried to somehow apply the Mean Value Theorem where f(b)-f(a)=f'(c)(b-a). I ended up lost...

Then I tried showing f(0)=1, because f(x-0)=f(x)f(0) and f(x) isn't equal to 0.
However, with that conclusion, f'(0)=0. This is a problem, because the function is most likely f(x)=e^cx, and f'(x)=ce^cx, and f'(0)=c, which may not be 0....

Am I making this problem harder than it is? Any help would be greatly appreciated!

http://eqworld.ipmnet.ru/en/solutions/fe/fe4101.pdf

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Would I apply f(x+h) to both sides? if so, I got:

f(x+h+y)-f(x+y)=f(x+h)f(y)-f(x)f(y)

which simplifies to:

f(h)=f(h)f(y)

Am I on the right track? What do I do with this?

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vela
Staff Emeritus
Homework Helper
Given: f(x+y)=f(x)f(y). f'(0) exists.

Show that f is differentiable on R.

At first, I tried to somehow apply the Mean Value Theorem where f(b)-f(a)=f'(c)(b-a). I ended up lost...

Then I tried showing f(0)=1, because f(x-0)=f(x)f(0) and f(x) isn't equal to 0.
However, with that conclusion, f'(0)=0. This is a problem, because the function is most likely f(x)=e^cx, and f'(x)=ce^cx, and f'(0)=c, which may not be 0....
Note that f(x)=1 and f(x)=0 both satisfy f(x+y)=f(x)f(y) and have a derivative at x=0.

How did you get from f(0)=1 to the conclusion f'(0)=0?

I don't know how I got f'(0)=0. I just jotted it down, but now I realize that's not a valid conclusion.

vela
Staff Emeritus
Homework Helper
Would I apply f(x+h) to both sides? if so, I got:

f(x+h+y)-f(x+y)=f(x+h)f(y)-f(x)f(y)

which simplifies to:

f(h)=f(h)f(y)
I think it would help if you showed details of your work rather than just showing what you end up with because your errors seem to be in your intermediate steps.

Pinu is suggesting you calculate f'(x) using the definition of the derivative:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Use what you know about f(x) to evaluate the right side. You'll end up with a differential equation you can solve to find f(x) explicitly.

f(x)+f(y)=f(x)f(y)

f(x)=f(x)f(y)-f(y)

f'(x)=limh->0 f(x+h)f(y)-f(y)-[f(x)f(y)-f(y)]/h

f'(x)=limh->0 f(h)f(y)/h.

Is that correct?

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vela
Staff Emeritus
Homework Helper
No. For one thing, x appears on the left but not on the right.

Dang it. I keep working and re-working this problem, and x keeps getting eliminated.
I've also tried substituting f(y)=f(x+y)/f(x), but this doesn't help one bit. I'll have to think about this.

Thank you for your help! I will keep working on it.

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Office_Shredder
Staff Emeritus
Gold Member
Why are there y's?

You should only have x's and h's in your equation for the derivative

$$f'(x) \equiv \lim_{h \rightarrow 0}{\frac{f(x + h) - f(x)}{h}}$$
$$= \lim_{h \rightarrow 0}{\frac{f(x) \, f(h) - f(x)}{h}}$$
$$= f(x) \, \lim_{h \rightarrow 0}{\frac{f(h) - 1}{h}}$$

Next, notice that for arbitrary $x$, taking $y = 0$, we have:
$$f(x + 0) = f(x) = f(x) \, f(0), \; \forall x \in \mathbb{R} \Rightarrow f(0) = 1$$

Therefore, the limit is of the indeterminate form 0/0. Use L'Hospital's rule and another condition you are given in the problem to evaluate it.

Oh, my, goodness.

I've been trying to ride a bike with no steering. Thank you so much for your help.

What a dunce I've been!

Therefore, the limit is of the indeterminate form 0/0. Use L'Hospital's rule and another condition you are given in the problem to evaluate it.

I don't think this is valid. While we do have the continuity of f at 0, L'hopital's rule would basically require the existence of the limit of f'(x) as x approaches 0 as an additional hypothesis, and this does not follow from the existence of f'(0). It's easier to just write out the definition of f'(0), which is a limit we know that exists.

I don't think this is valid. While we do have the continuity of f at 0, L'hopital's rule would basically require the existence of the limit of f'(x) as x approaches 0 as an additional hypothesis, and this does not follow from the existence of f'(0). It's easier to just write out the definition of f'(0), which is a limit we know that exists.

True. But I didn't even have to invoke that. The limit, as written, is:

$$\lim_{h \rightarrow 0}{\frac{f(h) - f(0)}{h}}$$

where, by definition is $f'(0)$.