# AA x,y,z in RR f(x+y+z) = f(x)f(y)f(z)!=0 & f(2)=5 , f'(0)=2 then find the value of f'(2) ?

Oct 13, 2016

$f ' \left(2\right) = \pm 10$

#### Explanation:

$f \left(x + 0 + 0\right) = f \left(x\right) f {\left(0\right)}^{2} = f \left(x\right) \to f {\left(0\right)}^{2} = 1$
$f \left(x + \delta + 0\right) = f \left(x\right) f \left(\delta\right) f \left(0\right)$

${\lim}_{\delta \to 0} \frac{f \left(x + \delta\right) - f \left(x - \delta\right)}{2 \delta} = {\lim}_{\delta \to 0} \frac{f \left(x\right) f \left(\delta\right) f \left(0\right) - f \left(x\right) f \left(- \delta\right) f \left(0\right)}{2 \delta} =$
$= f \left(x\right) f \left(0\right) {\lim}_{\delta \to 0} \frac{f \left(\delta\right) - f \left(- \delta\right)}{2 \delta} = f \left(x\right) f \left(0\right) f ' \left(0\right)$

but $f \left(0\right) = \pm 1$ and $f ' \left(0\right) = 2$ so

$f ' \left(x\right) = f \left(0\right) f ' \left(0\right) f \left(x\right) = \pm 2 f \left(x\right)$ and finally

$f ' \left(2\right) = \pm 10$

Note:

$\frac{f ' \left(x\right)}{f} \left(x\right) = \pm \frac{1}{2}$ then $f \left(x\right) = {C}_{0} {e}^{\pm \frac{x}{2}}$