Question

A function f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f′(0) = 2. Prove that f′(x) = 2f(x).

Answer 1

Given that, f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0.

Let us take any point x = 0 at which the function f(x) is differentiable.

∴ f'(0) = `lim_("h" -> 0) ("f"(0 + "h") - "f"(0))/"h"`

2 = `lim_("h" -> 0) ("f"(0) * "f"("h") - "f"(0))/"h"`   ......[∵ f(0) = f(h)]  ....(i)

⇒ 2 = `lim_("h" -> 0) ("f"(0)["f"("h") - 1])/"h"`

Now f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`

= `lim_("h" -> 0) ("f"(x) * "f"("h") - "f"(x))/"h"`  .....[∵  f(x + y) = f(x) . f(y)]

= `lim_("h" -> 0) ("f"(x)["f"("h") - 1])/"h"`

= 2f(x)

From equation (i)

Hence, f'(x) = 2f(x).