Let the function satisfy the equation f(x+y)= | vedclass

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(A) Given $f(x+y)=f(x)f(y)$.Putting $x=0, y=5$,we get $f(5)=f(0)f(5)$.Since $f(5)=3 
eq 0$,we have $f(0)=1$.Now,$f^{\prime}(5) = \lim_{h 	o 0} \frac

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$6$

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(A) Given $f(x+y)=f(x)f(y)$.Putting $x=0, y=5$,we get $f(5)=f(0)f(5)$.Since $f(5)=3 
eq 0$,we have $f(0)=1$.Now,$f^{\prime}(5) = \lim_{h 	o 0} \frac{f(5+h)-f(5)}{h}$.Using the given functional equation,$f(5+h)=f(5)f(h)$.So,$f^{\prime}(5) = \lim_{h 	o 0} \frac{f(5)f(h)-f(5)}{h} = f(5) \lim_{h 	o 0} \frac{f(h)-1}{h}$.Since $f(0)=1$,this is $f(5) \lim_{h 	o 0} \frac{f(h)-f(0)}{h} = f(5)f^{\prime}(0)$.Substituting the values,$f^{\prime}(5) = 3 	imes 2 = 6$.

Let the function satisfy the equation $f(x+y)=f(x)f(y)$ for all $x, y \in \mathbb{R}$,where $f(0) \neq 0$. If $f(5)=3$ and $f^{\prime}(0)=2$,then $f^{\prime}(5)$ is

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