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. $f(x)\ne 0$ for all $x$. If $f(2)=4,f^{\prime }(0)=3$, then find $\left|f^{\prime }(2)\right|$.

Answer 1

Answer: 12

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