A function f : R ® R satisfies the equation

Question:

A function : R ® R satisfies the equation y) = (x(y) for all xÎR, (x) ¹ 0. Suppose that the function is differentiable at = 0 and ¢ (0) = 2. Prove that ¢(x) = 2 (x).

Solution:

Given,

: R ® R satisfies the equation y) = (x(y) for all xÎR, (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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