Let f(x+y)=f(x)f(y) for all x, y where f(0) n | vedclass

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(A) Given the functional equation $f(x+y) = f(x)f(y)$.To find $f'(x)$,we use the definition of the derivative:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f

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$6$

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(A) Given the functional equation $f(x+y) = f(x)f(y)$.To find $f'(x)$,we use the definition of the derivative:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$Using the given functional equation $f(x+h) = f(x)f(h)$:$f'(x) = \lim_{h 	o 0} \frac{f(x)f(h) - f(x)}{h}$$f'(x) = f(x) \lim_{h 	o 0} \frac{f(h) - 1}{h}$Since $f(0+0) = f(0)f(0)$,we have $f(0) = f(0)^2$. Given $f(0) 
eq 0$,it follows that $f(0) = 1$.Thus,$f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0} \frac{f(h) - 1}{h} = 3$.Substituting this back into the expression for $f'(x)$:$f'(x) = f(x) \cdot f'(0) = 3f(x)$.At $x = 5$:$f'(5) = 3f(5) = 3 	imes 2 = 6$.

Let $f(x+y)=f(x)f(y)$ for all $x, y$ where $f(0) \neq 0$. If $f(5) = 2$ and $f'(0) = 3$,then $f'(5)$ is equal to

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