Question

Let $f(x+y+z)=f(x) \cdot f(y) \cdot f(z)$ for all $x, y, z$ where $f$ is a non-zero function i.e. $f(x) \neq 0$ for all $x$. If $f(2)=4, f'(0)=3$, then find $\left|f'(2)\right|$.

Answer 1

$f(x+y+z)=f(x) f(y) f(z)$,
for all $x, y, z\,\,\,\,...(i)$
Substituting $x=y=z=0$ in (i),
we get $f(0)=(f(0))^{3}$
$\Leftrightarrow f(0)\left[(f(0))^{2}-1\right]=0$
$\Leftrightarrow f(0)=\pm 1 \,\,\,\ldots[f(0) \neq 0]$
Substituting $y=2, z=0$ in (i),
we get $f(x+2)=f(x) \cdot f(2) \cdot(\pm 1)$
$\Leftrightarrow f(x+2)=\pm 4 f(x)$, for all $x .$
Differentiating with respect to $x$,
we get $f'(x+2)=\pm 4 f'(x)$, for all $x$
Substituting $x=0$, we get
$f'(2)=\pm 4 f'(0)=\pm 4 \text { (3) }$
$\Rightarrow \left|f'(2)\right|=12$