Question

nswered Suppose that a function f satisfies the following conditions for all real values of x and y: 1. f(x+y)=f(x).fl) 2. f(x)= 1+xg(x), where lim g(x)=1. ut of 200 question X +0 Then f is differentiable at all real numbers x and f(x)= f(x). Select one: o True O False

Answer 1

Given that the function f satisfies the below conditions for all real values of x and y.

$1.\, f(x+y)=f(x)\cdot f(y)$

$2.\,f(x)=1+xg(x),\textup{ where }\lim_{x \to 0}g(x)=1$

We want to prove that f is differentiable at all real numbers x and f'(x) = f(x).

According to the definition of derivative, for an arbitrary x

$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

$\Rightarrow f'(x)=\lim_{h \to 0}\frac{f(x)f(h)-f(x)}{h} \,\,\,\,\,\,\, \left ( \textup{using condition 1 on }f(x+h) \right )$

= f'() = lim f(x)(f(n) – 1) h→0

$\Rightarrow f'(x)=\lim_{h \to 0}\frac{f(x)(1+hg(h)-1)}{h} \,\,\,\,\,\,\, \left ( \textup{using condition 2 on }f(h) \right )$

$\Rightarrow f'(x)=\lim_{h \to 0}\frac{f(x)(hg(h))}{h}$

$\Rightarrow f'(x)=\lim_{h \to 0}f(x)g(h)$

$\Rightarrow f'(x)=\lim_{h \to 0}f(x)\cdot \lim_{h \to 0}g(h) \,\,\,\,\,\, \left ( \textup{using limit properties} \right )$

$\Rightarrow f'(x)=\lim_{h \to 0}f(x)\cdot 1 \,\,\,\,\,\,\, \left ( \textup{using condition 2} \right )$

$\Rightarrow f'(x)=\lim_{h \to 0}f(x)$

$\Rightarrow f'(x)=f(x)$

Hence, the given statement is True.