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  • A function satisfies the equation f(x + y) = f(x) f(y) for all x, y . Suppose that the function is differentiable at x = 0 and f’(0) = 2. Prove that f’(x) = 2f(x).

A function f:R\rightarrow R satisfies the equation f(x + y) = f(x) f(y) for all x, y\in R,f(x)\neq 0. Suppose that the function is differentiable at x = 0 and f’(0) = 2. Prove that f’(x) = 2f(x).

Answers (1)

Given f(x) is differentiable at x = 0 and f(x) ≠ 0

And f(x + y) = f(x)f(y) also f’(0) = 2

To prove: f’(x) = 2f(x)

As we know that,

\\ f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \text { as } f(x+h)=f(x) f(h) \\ \therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x)(f(h)-1\}}{h}=f(x) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}

As

f(x + y) = f(x)f(y)

put x = y = 0

\\ \therefore f(0+0) = f(0)f(0) \\ \Rightarrow $ f(0) = $ \{ $ f(0)$ \} ^2
\therefore f(0) = 1 \quad \{ \because f(x) \neq 0 \ldots \text{given}

\begin{aligned} &f^{\prime}(x)=f(x) \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\\ &\Rightarrow f^{\prime}(x)=f(x) \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\\ &\Rightarrow f^{\prime}(x)=f(x) f^{\prime}(0)\{\text { using formula of derivative }\}\\ &\therefore f^{\prime}(x)=2 f(x) \ldots \text { proved }\left\{\because \text { it is given that } f^{\prime}(0)=2\right\} \end{aligned}

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