Question

Suppose f : R → R is differentiable at x = 0, that f (a + b) = f

(a)f (b) for all a, b, ∈ R, and that f (0) = 1. Prove that f is
differentiable and f ′(x) = f ′(0)f (x) Real Analysis. Please use def of derivative and provide all justifications

Answer 1

The differentiability of f(x) and the relationship f'(x) = f'(0)f(x) are proven using the limit definition of the derivative and the given functional equation of f.

To prove that f is differentiable and that f'(x) = f'(0)f(x), let's start by using the fact that f is differentiable at x = 0 and f(0) = 1. The definition of the derivative gives us f'(0) = lim(h->0)(f(h) - f(0))/h. Because f is given by f(a + b) = f(a)f(b), we can look at the limit definition of the derivative at any point x.

Applying the functional equation to the definition of derivative: f'(x) = lim(h->0)(f(x+h) - f(x))/h

= lim(h->0)(f(x)f(h) - f(x))/h

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

= f(x)f'(0).

Thus, we have proven that the derivative of f at any point x equals f'(0) multiplied by f(x).