if f(x+y)=f(x).f(y) forr all real x,y and f'(0)=3,f(5)=2 then how can find f'(5)???

Question

michele_laino · Answer

we have:
$$\begin{gathered}
  f(0) = f\left( {0 + 0} \right) = {\left( {f(0)} \right)^2} \hfill \
  f\left( 0 \right)\left( {1 - f\left( 0 \right)} \right) = 0 \hfill \
  f(0) = 1 \hfill \
  f(0) = 0 \hfill \ 
\end{gathered} $$

anonymous · Answer

then??

anonymous · Answer

Setting $y=0$ gives
$$f(x+0)=f(x)f(0)~~\iff~~f(x)=f(x)f(0)$$
and differentiating gives
$$f'(x)=f'(x)f(0)$$
for all $x$. This would suggest that $f(0)=1$, since we know that $f'(0)\neq0$.

michele_laino · Answer

we have:
\[\begin{gathered}
  \frac{{f\left( {h + 5} \right) - f\left( 5 \right)}}{h} = \frac{{f\left( h \right)f\left( 5 \right) - f\left( 5 \right)}}{h} = f\left( 5 \right)\frac{{f\left( h \right) - 1}}{h} \hfill \
  \frac{{f\left( h \right) - f\left( 0 \right)}}{h} = \frac{{f\left( {h + 0} \right) - f\left( 0 \right)}}{h} = f\left( 0 \right)\frac{{f\left( h \right) - 1}}{h} \hfill \
  f'\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h + 5} \right) - f\left( 5 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)f\left( 5 \right) - f\left( 5 \right)}}{h} = f\left( 5 \right)\mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - 1}}{h} \hfill \
  f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h + 0} \right) - f\left( 0 \right)}}{h} = f\left( 0 \right)\mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - 1}}{h} \hfill \
  f'\left( 5 \right) = f\left( 5 \right)\frac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} \hfill \ 
\end{gathered} \]

michele_laino · Answer

with:
$$f\left( 0 \right)=1$$

mathmath333 · Answer

On differentiating the equation we get
$\large \color{black}{\begin{align} 

f(x+y)=f(x)f(y)
\end{align}}$ 

with respect to x.
$\large \color{black}{ \begin{align} 
f'(x+y)\dfrac{d}{dx}(x+y)&=f'(x)f(y)+f'(y)f(x)\dfrac{dy}{dx}\~\
f'(x+y)(1+y')&= f'(x)f(y)+f(x)f'(y)(y') \~\
\end{align}}$ 

put $\large x=0$ and $\large y=5$

$\large \color{black}{ \begin{align}
f'(x+y)(1+y')&= f'(x)f(y)+f(x)f'(y)(y') \~\
f'(0+5)(1+5')&= f'(0)f(5)+f(0)f'(5)(5') \~\
f'(5)(1)&= f'(0)f(5)+f(0)f'(5)0 \~\
f'(5)(1)&= f'(0)f(5)\~\
f'(5)(1)&= 3\cdot 2\~\
\end{align}}$

anonymous · Answer

Can we generalize and arrive at $\dfrac{f'(x)}{f(x)}=3$? We can tell that an exponential function will satisfy the required conditions. From here we can solve for $f(x)$ and find $f'(5)$.

anonymous · Answer

thank u