Calculus question

Ok, I know how you feel about homework threads… I’m not looking for
answers, I’m looking for a direction here.

My calc I teacher gave us this project to do. We’re given some
information, and we have to do some things with it, in 3 separate
mini-projects. Here’s what I have, and my ideas on it.

1: f(a+b)=f(a)f(b)
2: f(0)=0
3: f(x) is differentiable at x=0, and f(0)=1

Regardless of my response to project 1, the only things I know about a
function f(x) are properties 1-3, IOW, my answer for project 1 doesn’t
matter when doing the other two projects.
1: Find a function f(x) that satisfies all of the above conditions

So I think I’ve succeded here, with a simple f(x)=e[sup]x[/sup]
function. This isn’t where I’m having the problems.

2: Show that condition 1 implies that either f(0)=0 or f(0)=1. Then
show that if f(0)=0 then f(x)=0 for all x, while if f(0)=1 then f(x)
!=0 for all x.

Huh? Here, the only thing I’ve been able to think of that might work
would be the Mean Value Theorem, using f(0)=0 for my first point, but
I don’t have a 2nd point to define the interval on. Is this even the
right direction to go here? Would saying something like f(0+0)=2f(0)
help there?

  1. Using the limit definition of a derivative, show that
    f’(x)=f(x)f’(0) for all x. What does this say about the continuity of
    f(x) for all x?

I think that here I need to use the product rule, by saying
f(x+0)=f(x)f(0), and then using the product rule to say that
f’(x+0)=f’(x)f(0)+f(x)f’(0), and since information point 2 says that
f(0)=0, then the first term goes away. I think I’m on the right track
there, but I don’t know what to do to say anything about the
continuity. Is it as simple as saying that since f(x) is
differentiable over all x, then it’s continuous over all x?

I think there’s something wrong with you conditions. e^0 is 1, but condition 2 states that it should be equal to 0, while condition 3 states it should be equal to 1. Did you mean to put f’(0) in condition 3?

Oh, my God. I am a superidiot.

Yes, Information 3 should say that f’(x)=1

And I know that e[sup]0[/sup]=1. I… just didn’t remember it.

:smack: :o

I think there’s something wrong with your conditions. exp(x) is the only differentiable function that satisfies f(a + b) = f(a)f(b), but exp(0) = exp’(0) = 1. Did you mean to write f(0) = 1 for condition 2? In that case, condition 3 is correct as-is.

For part 2, you’ve made an error–f(0 + 0) = f(0)[sup]2[/sup], not 2f(0). That should be enough for you to get the rest of the problem.

Part 3 should be pretty straightforward.

The Mean Value Theorem is definitely overkill (and probably useless anyway, since you don’t know anything yet about the derivative of f). You can answer this problem just by applying condition 1 with different values of a and b.

You can’t use the product rule in the manner you describe, I’m afraid, because the second term in the product (namely, f(0)) isn’t a function of x, it’s a constant. When you differentiate both sides of f(x+0)=f(x)f(0), you just get f’(x+0)=f’(x)f(0), which doesn’t tell you anything you didn’t already learn in step two.

Besides, you’re ignoring the first part of the question: “Using the limit definition of a derivative”. This is a hint. Apply the limit definition of f’(x) and think about how you can use condition 1 to simplify/evaluate it.

You guys… remind me never ever to post when I’ve just woken up. I typed this up last night, and could have sworn that I had put an “!=” in there, but now that I’ve copied and pasted it? Maybe I’ve lost it or something.

Condition 2 should read f(0)!=0

Gosh, I am an idiot. :smack: