# Thread: interesting problem with a function (not defined) - quite intimidating

1. ## interesting problem with a function (not defined) - quite intimidating

I found this question to be quite intimidating:

Suppose a function f has the following properties:
• f is differentiable everywhere.
• For all values of x and y, .
• .
• .

a) Show that:

b) Show that for all values of x by first showing that for all x.

c) Use the definition of the derivative to prove that
for all x.

I'd assume f would be a constant function. It seems to have similar properties as such a function. Anyway, thanks for help!

2. ## Re: interesting problem with a function (not defined) - quite intimidating

Originally Posted by zachd77
I found this question to be quite intimidating:

Suppose a function f has the following properties:
• f is differentiable everywhere.
• For all values of x and y, .
• .
• .

a) Show that:

b) Show that for all values of x by first showing that for all x.

c) Use the definition of the derivative to prove that
for all x.

I'd assume f would be a constant function. It seems to have similar properties as such a function. Anyway, thanks for help!
Hi zachd77!

Have you tried fiddling around with f(x+y)=f(x)f(y)?
What if you fill in a couple of special values?
Like x=0 and y=0?
Or like x=x and y=-x?

3. ## Re: interesting problem with a function (not defined) - quite intimidating

yeah... the book I got this from said to try plugging in values.. I'll try it and see what happens.

4. ## Re: interesting problem with a function (not defined) - quite intimidating

Ok so I got (a) down so far, but I got stuck on (b) and on (c) I ended up with an indeterminate form. Probably made a mistake or I'm just over thinking this problem a lot.

5. ## Re: interesting problem with a function (not defined) - quite intimidating

So what did you get from filling in x=y=0 respectively y=-x?

The first should take care of (a), the second in combination with (a) should take care of (b).

6. ## Re: interesting problem with a function (not defined) - quite intimidating

Originally Posted by zachd77
I found this question to be quite intimidating:

Suppose a function f has the following properties:
• f is differentiable everywhere.
• For all values of x and y, .
• .
• .

a) Show that:

b) Show that for all values of x by first showing that for all x.

c) Use the definition of the derivative to prove that
for all x.

I'd assume f would be a constant function. It seems to have similar properties as such a function. Anyway, thanks for help!
a)f(0)=f2(0)=f3(0)=... and so on. It's possible when f(0)=1 only. f(0)=f(x)f(-x)=f(x)f(-x)f(0) => f(x)f(-x)=1
b)if f(x)<0 then f(x)=f(x/2+x/2)=f2(x/2)<0 it's absurd. As f(x)f(-x)=1 then .

7. ## Re: interesting problem with a function (not defined) - quite intimidating

well since you KNOW f(0) = 1, then:

1 = f(0) = f(a+(-a)) = f(a)f(-a).

this means that f(a) ≠ 0 (or else we would get 1 = 0, right?). this works for ANY a.

note that since f(a)f(-a) = 1, f(a) and f(-a) always have the same sign. so on the interval (-a,a) = (0-a,0+a) f is either entirely positive, or entirely negative, and thus positive (since f(0) = 1 > 0).

(this is where continuity comes in: suppose for some b > 0, we had f(b) < 0. by continuity, at some point in (0,b), say c, we must have f(c) = 0 (the intermediate value theorem). but there IS no such c. a similar argument works for any b < 0).

but if f is positive on any interval (-a,a), it must be positive on all of R.

c) is relatively easy: by definition-

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} f(x)\cdot \frac{f(h) - 1}{h}$

$= f(x)\cdot \lim_{h \to 0} \frac{f(0)f(h) - f(0)}{h} = f(x)\cdot f'(0)$, so....

also: your initial guess that f is constant is wrong: if f was constant, then f'(0) would always be 0 (constant functions are "flat").

one class of (continuous) functions for which f(x+y) = f(x)f(y) are THESE: f(x) = ax, for a > 0. the trouble with this is, what does ax even MEAN for an irrational x?

one way of doing this is by imagining a continuous function ax that "connects the dots" where "the dots" are ax for RATIONAL x. it would be helpful if we could produce a specific "a" that we KNEW existed (besides 1, which is boring). that is what this problem is trying to do: figure out what such a function might be like.

what's coming next is: for what function f(x) = ax (if it exists, and is differentiable) might we have f(x) = f'(x)?

8. ## Re: interesting problem with a function (not defined) - quite intimidating

e^x..... that's quite obvious

9. ## Re: interesting problem with a function (not defined) - quite intimidating

Since we are given that f is differentiable, it is continuous and it is relatively easy to show that $f(x)= e^{ax}$ (equivalently [/tex]f(x)= r^x[/tex]) are the only continuous functions satisfying f(x+y)= f(x)f(y). There are many NON-continous functions satisfying that but they have to be really discontinuous! In fact, they must be unbounded on any interval.

10. ## Re: interesting problem with a function (not defined) - quite intimidating

my point was: how do you define the function f(x) = ex if one doesn't already know what it is? what's so special about "e" that (ex)' = ex, but not for other positive real numbers?

but, we have something we've learned about f(x) (just ignore e for a bit, pretend you've never heard about it). it equals its own derivative, AND...this derivative (itself) is always positive. that is: f(x) is a strictly increasing function.

and THAT means: for every f(x) in the range of f, we have an inverse function g:f(x)--->x.

here's the cool thing about inverse functions:

suppose we have for a,b in R: [f(b) - f(a)]/(b - a), and f has an inverse on its entire domain (so f-1 is defined for all f(x) in the range of f).

suppose that f(a) = c, and f(b) = d. we can re-write the fraction above as:

(c - d)/[f-1(c) - f-1(d)]

in particular, if b = a+h:

[f(a+h) - f(a)]/h = (c - d)/[f-1(c) - f-1(d)]

suppose that we call d-c, k, so that d = c+k.

then:

[f(a+h) - f(a)]/h = -k/[f-1(c) - f-1(c+k)] = k/[f-1(c+k) - f-1(c)]

if f is continuous, then as h-->0, f(a+h) - f(a) -->0, which means that d-c -->0, that is: c+k - c = k -->0.

so:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{k \to 0} \frac{k}{f^{-1}(c+k) - f^{-1}(c)} = \frac{1}{(f^{-1})'(c)}$

if the derivative of f is never 0 for any of the x's we're considering:

$(f^{-1})'(c) = (f^{-1})'(f(a)) = \frac{1}{f'(a)}$

we can also write this as:

$(f^{-1})'(c) = \frac{1}{f'(f^{-1}(c))}$

which will be more useful for us.

what does this mean for the PARTICULAR f(x) we've been discussing in this thread? to simplify the notation, let's use g(y) instead of f-1(y).

it says that:

$g'(y) = \frac{1}{f'(g(y))}$

but, in this case: f'(g(y)) = f(g(y)) = y (since f and g are inverses, and since f'(x) = f(x)). that is:

g'(y) = 1/y.

note that since f takes x+y to f(x+y) = f(x)f(y)

we have g(f(x)f(y)) = g(f(x+y)) = x + y = g(f(x)) + g(f(y))., so writing f(x) = a, f(y) = b:

g(ab) = g(a) + g(b). holy canoli! we've discovered LOGARITHMS!

also, since f(0) = 1, g(1) = 0.

if you know the fundamental theorem of calculus, you know that:

$\int_a^b f'(t)\ dt = f(b) - f(a)$

so for our newly-discovered function g:

$\int_1^x g'(t)\ dt = g(x) - g(1) = g(x)$

that is:

$g(x) = \int_1^x \frac{1}{t}\ dt$.

now, all along we're thinking f(x) = ax for some positive number a. now we have a way of finding a, it is the real number a such that:

$\int_1^a \frac{1}{t}\ dt = 1$ <-of course we have to have some means of approximating this integral, which might be tough, but we CAN do it.