If f: R rightarrow R is a differentiable func | vedclass

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(D) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.Setting $x=4$ and $y=0$,we get $f(4+0)=f(4) \cdot f(0)$,which implies $f(4)=f(4) \cdot f(0)

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$8$

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(D) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.Setting $x=4$ and $y=0$,we get $f(4+0)=f(4) \cdot f(0)$,which implies $f(4)=f(4) \cdot f(0)$. Since $f(x)$ is differentiable,$f(x)$ is not identically zero,so $f(0)=1$.By the definition of the derivative,$f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.Substituting $f(x+h)=f(x) \cdot f(h)$,we have $f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h \rightarrow 0} \frac{f(h)-1}{h}$.Since $f(0)=1$,this is $f^{\prime}(x) = f(x) \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h} = f(x) \cdot f^{\prime}(0)$.Given $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$,we substitute these into the equation $f^{\prime}(4) = f(4) \cdot f^{\prime}(0)$.$24 = f(4) \cdot 3$.Therefore,$f(4) = \frac{24}{3} = 8$.

If $f: R \rightarrow R$ is a differentiable function such that $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$ and if $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$,then $f(4)=$

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