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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).
Solution
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).
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