Let f be a function whose domain is the set of all real numbers and f(a+b) = f(a) * f(b) for all a and b.?

Question

furthermore, suppose that f(0) = 1 and f'(0) exists. prove that f'(x) exists for all x and that f'(x) = f'(0) * f(x).

Answer 1

f(a + b) = f(a) * f(b)

Recall the definition of a derivative.

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

Since f(a + b) = f(a)*f(b), then f(x + h) = f(x)*f(h), so

f'(x) =
lim [f(x)f(h) - f(x)]/h
h -> 0

Factoring out f(x), we get

lim { [f(x) [f(h) - 1] } / h
h -> 0

We can pull the f(x) outside of the limit, giving us

f(x) * lim [f(h) - 1]/h
h -> 0

We know that f(h) = f(0 + h), and 1 = f(0). Let's substitute f(h) and f(1) respectively with those values

f(x) * lim [f(0 + h) - f(0)]/h
h -> 0

But, that is precisely the definition of what f'(0) is (see the definition of the limit, above). Therefore,

f'(x) = f(x) * f'(0)

Answer 2

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Answer 3

we have $f'(x)=lim (f(x+h)-f(x))/h$ (where h is going to 0) but $f(x+h)=f(x)f(h)$ then $f'(x)=lim ((f(h)-1)f(x))/h $ but $lim(f(h)-1)/h$ is $f'(0)$ then the question is proved.

Answer 4

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