Let's assume that $ f: V \rightarrow \mathbb{R}_{>0} $ is a monotonic, continuous and smooth function, i.e. the $n$-th derivative $ f^{(n)} $ exists for all integers $ n $, where $ V $ is a real interval. I am interested in the monotonicity of the following function: $$ g(x) = \frac{f(x)}{f(c \times x)}, \quad c\in\mathbb{R}_{>0} $$ One could answer this by differentiating $ g(x) $ and check whether $ g'(x) \geq 0 $ (or $ \leq 0 $) for all $ x \in V $. By applying the quotient rule one finds that it is sufficient to show that $ c \times f(x) f'(c \times x) - f(c \times x)f'(x) \geq 0 $ (or $ \leq 0 $) for all $ x \in V $. However, this can get rather messy at times.
I was wondering if there are any properties of $ f(x) $ that imply monotonicity of $ g(x) $.
Suppose for example that $ V = [0, \infty) $ and that $ f^{(n)}(x) > 0 $ for all $ x \in V $ and all integers $ n $. Further suppose that $ f(0) = 1 $. I could not find an example were in such a case $ g(x) $ was not monotonically increasing for $ c < 1 $ or not monotonically decreasing for $ c > 1 $.
Any reference, counter example, proof or proof idea regarding this issue would be appreciated.
Counter example:
Consider function $f(x) = e^{sin(x)+x}$ and $c=\frac{1}{2}$.
$f'(x) = (cos(x)+1)e^{sin(x)+x}$
$f'(x) \geq 0$ for $x \in \mathbb{R}$ so $f$ is monotonic.
$g(x) = \frac{f(x)}{f(x/2)} = \frac{e^{sin(x)+x}}{e^{sin(x/2)+x/2}} = e^{sin(x)-sin(x/2)+x/2 } $
$g'(x) = (cos(x) - \frac{1}{2}cos(x/2) + \frac{1}{2} ) e^{sin(x)-sin(x/2)+x/2}$
And function $g$ is not monotonic in $\mathbb{R}$.