A calculus problem by Ayushman Chahar

For $x=y=0$ we have:
$f(0)=f^{2}(0)$ ( $f(0)\neq 0$ )
$1=f(0)$
For $y=-x$ we have:
$f(0)=f(x)\cdot f(-x)$
$f(x)\cdot f(-x)=1$
$f(-x)=\frac{1}{f(x)}$

$g(-x)=$
$\frac{f(-x)}{1+(f(-x))^{2}}=$
$\frac{\frac{1}{f(x)}}{1+(\frac{1}{f(x)})^{2}}=$
$\frac{\frac{1}{f(x)}}{\frac{f^{2}(x)+1}{f^{2}(x)}}=$
$\frac{f(x)}{1+f^{2}(x)}=$
$g(x)$
Thus:
$g(-x)=g(x)$
and $g$ is an $\boxed{even}$ function.

Assuming $f(x)$ is differentiable over its entire domain, let us differentiate the above functional equation with respect to x and y respectively:

$f'(x+y) = f'(x)f(y)$ and $f'(x+y) = f(x)f'(y)$

which results in the ODE:

$\frac{f'(x)}{f(x)} = \frac{f'(y)}{f(y)} = A \Rightarrow ln f(x) = Ax + B \Rightarrow f(x) = e^{Ax + B};$ where $A, B \in \mathbb{R}.$

If we now take into account $x = y = 0$ and substitute these values into our functional equation, we now obtain:

$f(0+0) = f(0)f(0) \Rightarrow 0 = f^{2}(0) - f(0) \Rightarrow f(0) = 0, 1$

but since we are given $f(0) \ne 0$ , only $f(0) = 1$ will be permitted. Using this boundary condition gives $1 = e^{B} \Rightarrow B = 0$ , or $f(x) = e^{Ax}.$

We now find that $g(x)$ is an even function, that is $g(-x) = g(x)$ . This can be shown as:

$\boxed{g(-x) = \frac{f(-x)}{1 + f^{2}(-x)} = \frac{e^{-Ax}}{1 + e^{-2Ax}} = \frac{1}{e^{Ax}} \cdot \frac{e^{2Ax}}{1 + e^{2Ax}} = \frac{e^{Ax}}{1 + e^{2Ax}} = g(x)}$