If f(x + y) = f(x)f(y) for all x and y and f( | vedclass

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(C) Given the functional equation $f(x + y) = f(x)f(y)$.Setting $y = 0$,we get $f(x + 0) = f(x)f(0)$,which implies $f(x) = f(x)f(0)$. Since $f(5) = 2$

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$6$

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(C) Given the functional equation $f(x + y) = f(x)f(y)$.Setting $y = 0$,we get $f(x + 0) = f(x)f(0)$,which implies $f(x) = f(x)f(0)$. Since $f(5) = 2$,$f(x)$ is not identically zero,so $f(0) = 1$.By the definition of the derivative,$f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$.Substituting $f(x + h) = f(x)f(h)$,we get $f'(x) = \lim_{h 	o 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h 	o 0} \frac{f(h) - 1}{h}$.Since $f(0) = 1$,this is $f'(x) = f(x) \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = f(x)f'(0)$.Given $f(5) = 2$ and $f'(0) = 3$,we have $f'(5) = f(5)f'(0) = 2 	imes 3 = 6$.

If $f(x + y) = f(x)f(y)$ for all $x$ and $y$ and $f(5) = 2$,$f'(0) = 3$,then $f'(5)$ will be

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