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If $f: R \rightarrow R$ is a differentiable function such that $f(x+y)=f(x) . f(y)$ for all $x, y \in R$ and if $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$, then $f(4)=$
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$\begin{aligned} & \text { Let } x=4, y=0 \Rightarrow f(4+0)=f(4) \cdot f(0) \\ & \Rightarrow \quad \begin{aligned} f(4) & =f(4) f(0) \Rightarrow f(0)=1 \\ f^{\prime}(4) & =\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(4) f(h)-f(4)}{h}=\lim _{h \rightarrow 0}\left(\frac{f(h)-1}{h}\right) f(4)\end{aligned}\end{aligned}$
$\begin{aligned} & f^{\prime}(4)=f(4) \lim _{h \rightarrow 0} \frac{(f(h)-f(0)}{h} \\ & f^{\prime}(4)=f(4) \cdot f^{\prime}(0) \\ & f^{\prime}(4)=24, f^{\prime}(0)=3 \Rightarrow \quad f(4)=\frac{f^{\prime}(4)}{f^{\prime}(0)}=\frac{24}{3}=8 .\end{aligned}$
$\begin{aligned} & f^{\prime}(4)=f(4) \lim _{h \rightarrow 0} \frac{(f(h)-f(0)}{h} \\ & f^{\prime}(4)=f(4) \cdot f^{\prime}(0) \\ & f^{\prime}(4)=24, f^{\prime}(0)=3 \Rightarrow \quad f(4)=\frac{f^{\prime}(4)}{f^{\prime}(0)}=\frac{24}{3}=8 .\end{aligned}$
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