Question-41
Consider a function \(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R}\) that satisfies the following identity for all \(\displaystyle x,y\in \mathbb{R}\):
\[ \begin{equation*} f( x+y) =f( x) \cdot f( y) \end{equation*} \]
If \(\displaystyle f\) is differentiable at \(\displaystyle x=0\) with \(\displaystyle f^{\prime }( 0) =2\) and \(\displaystyle f( 0) \neq 0\), which of the following is true?
Use the definition of derivative.
From the definition of the derivative at \(\displaystyle x\), we have:
\[ \begin{equation*} \begin{aligned} f^{\prime }( x) & =\lim\limits _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h}\\ & \\ & =\lim\limits _{h\rightarrow 0} \ \frac{f( x) f( h) -f( x)}{h}\\ & \\ & =\lim\limits _{h\rightarrow 0} \ f( x)\frac{f( h) -1}{h}\\ & \\ & =f( x) \cdot \lim\limits _{h\rightarrow 0} \ \frac{f( h) -1}{h} \ \end{aligned} \end{equation*} \]
Plugging \(\displaystyle x=y=0\) in the identity, we get:
\[ \begin{equation*} f( 0) =f( 0)^{2} \Longrightarrow f( 0) =1,0 \end{equation*} \]
Since \(\displaystyle f( 0) \neq 0\) as per the question, \(\displaystyle f( 0) =1\). Plugging this back into the equation of the derivative:
\[ \begin{equation*} f^{\prime }( x) =f( x) \cdot \lim\limits _{h\rightarrow 0} \ \frac{f( h) -f( 0)}{h} =f( x) \cdot f^{\prime }( 0) \end{equation*} \]
We are given that \(\displaystyle f^{\prime }( 0) =2\). Therefore:
\[ \begin{equation*} \boxed{f^{\prime }( x) =2f( x)} \end{equation*} \]