A function f : R ® R satisfies the equation f ( x + y) = f (x) f (y) for all x, y ÎR, f (x) ¹ 0. Suppose that the function is differentiable at x = 0 and f ¢ (0) = 2. Prove that f ¢(x) = 2 f (x).
Given,
f : R ® R satisfies the equation f ( x + y) = f (x) f (y) for all x, y ÎR, f (x) ¹ 0
Let us take any point x = 0 at which the function f(x) is differentiable.
So, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$2=\lim _{h \rightarrow 0} \frac{f(0) \cdot f(h)-f(0)}{h} \quad[\because f(0)=f(h)]$ .....$\ldots(i)$
$\Rightarrow 2=\lim _{h \rightarrow 0} \frac{f(0)[f(h)-1]}{h}$
Now, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h} \quad[\because f(x+y)=f(x) \cdot f(y)]$
$=\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=2 f(x)$ from eqn. (i)
Therefore, f’(x) = 2f(x).
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