Calculus & Differentiation Problem!?

Question

Consider a function f which satisfies the following properties:
(i) f(x+y)= f(x)f(y)
(ii) f(0)≠0
(iii) f(0)=1

1. Use the definition of the derivative to show that f'(x)=f(x) for all real numbers x.

2. Let g be another function that satisfies properties (i)-(iii) and let k(x)=f(x)/g(x). Show that k is defined for all x and find k'(x). Use this to discover the relationship between f and g.

3. Can you think of a function which satisfies (i)-(iii). Can there be more than one such function?

Please help! I don't understand how to approach this problem at all!

Answer 1

First (iii) should state that f'(0), NOT f(0) is equal to 1. This is because the fact that f(0)=1 can be derived from (i) and (ii) as follows:

f(0) = f(0+0) = f(0)f(0) by (i).

Because f(0)≠0 by (ii), we can divide the left and right sides of the equation by f(0), obtaining 1 = f(0).

And now to the problem!

1) Fix any real value x.
Since f is defined for all x, f(x) is some constant.

By the def. of derivative

f'(x) = lim y-->0 [f(x+y) - f(x)]/(y) = lim y-->0 [f(x)f(y) - f(x)]/(y)=
=lim y-->0 f(x)[f(y)-1]/y = f(x) lim y-->0 [f(0+y) - f(0)]/y **this follows because f(x) is constant for a particular value of x (and so could be moved outside of the limit), f(y) = f(0+y), and 1 = f(0)**.

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

This last statement follows from the fact that f'(0)=1 (iii).

2) Any function g with the above properties cannot be equal to 0 for any x. Assume to the contrary that there is an x such that g(x)=0. But then g(0) = g(x - x) = g(x+(-x))= g(x)g(-x) = 0*g(-x) = 0, contradicting the fact that g(0)=1.

Hence, the ratio f(x)/g(x) = k(x) is defined for all x.

Now k'(x) = (d/dx)[f(x)/g(x)] = [g(x)f'(x)-g'(x)f(x)]/[g(x)]^2 = [g(x)f(x)-g(x)f(x)]/[g(x)]^2 **by 1)**. Hence k'(x) = 0.

It follows by a corollary to the mean-value theorem, that k(x) = c for some nonzero constant c. Hence f(x)=cg(x).

3) The function f(x) = e^(kx) for any nonzero k satisfies these 3 properties. If g(x) is any other function of which (i)-(iii) hold, g(x) = cf(x) by exercice 2. It should be noted that c cannot be negative, because, since f(x)>0, if c<0, g(x)<0 for all x. In particular g(0) is negative, which is a contradiction. Thus c>0 and therefore c=e^r for some real r and g(x) = cf(x) = e^r*e^(kx). Since g(0)=1 we have e^r*e^(k*0) = e^r = c = 1.

Thus all functions satisfying these 3 properties are of the form e^(kx).

Answer 2

Let a be f(1)

If n is a natural number f(n) = a^n because
f(n + 1) = f(n)*f(1) = a^n * a = a^(n + 1)

a = f(1) = f(0 + 1) = f(0)*f(1) = a*f(0)

Thus, a[f(0) - 1] = 0 and either f(1)=0 or f(0)=1

If x=1/n for a natural number n then a=f(1) = f(n * 1/n)

a = f(1/n)^n

Thus f(1/n) = a^(1/n)

Now, be q a rational number of form (m/n)

f(q) = f(m * 1/n) = f(1/n)^m = [a^(1/n)] ^m = a^(m/n) = a^q

I do not have enough details to prove for irrational numbers.

But if f(z) = a^z for a real number, z, then

1 = f(0) = f(z - z) = f(z)*f(-z), thus

f(-z) = 1/f(z) = 1/a^z = a^(-z)

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1. We haven't proven that a=e