Let f(x+y)=f(x) cdot f(y) for all x, y in R. | vedclass

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(D) We are given the functional equation $f(x+y)=f(x) \cdot f(y)$.By the definition of the derivative,$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3

Vedclass · Accepted Answer

$33$

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(D) We are given the functional equation $f(x+y)=f(x) \cdot f(y)$.By the definition of the derivative,$f^{\prime}(3)=\lim _{h ightarrow 0} \frac{f(3+h)-f(3)}{h}$.Using the given functional equation,$f(3+h)=f(3) \cdot f(h)$.Substituting this into the limit expression:$f^{\prime}(3)=\lim _{h ightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} = f(3) \lim _{h ightarrow 0} \frac{f(h)-1}{h}$.Since $f(0+0)=f(0) \cdot f(0)$,we have $f(0)=f(0)^2$,which implies $f(0)=1$ (assuming $f(x)$ is not identically zero).Thus,$f^{\prime}(0)=\lim _{h ightarrow 0} \frac{f(h)-f(0)}{h} = \lim _{h ightarrow 0} \frac{f(h)-1}{h} = 11$.Substituting $f(3)=3$ and $\lim _{h ightarrow 0} \frac{f(h)-1}{h} = 11$ into the expression for $f^{\prime}(3)$:$f^{\prime}(3) = 3 	imes 11 = 33$.

Let $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. Suppose that $f(3)=3$ and $f^{\prime}(0)=11$,then $f^{\prime}(3)$ is given by

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