Maximum of product of two functions Let's have two (real continuous differentiable) functions such that


*

*$f(x)$ is bounded (from below and from above), positive ($f(x)>0$), and is strictly increasing ($f'(x)>0$, $\forall x$). 

*$g(x)$ is bounded (from below and from above) and has exactly one maximum ($g'(x_0) = 0$ ; $g(x)<g(x_0), \forall x \ne x_0$)


It then follows that function $h(x)=f(x)g(x)$ is also bounded. 
However, additional conditions on either $f(x)$ or $g(x)$ must apply in order for the function $h(x)$ to have again only one maximum.
The question is. What are the additional sufficient conditions on either $f(x)$ or $g(x)$ so that the product $h(x)=f(x)g(x)$ would have again only one maximum?
(Thanks to gammatester for pointing this out).
 A: Dear @pisoir i think you need one easy condition or? read it.
$$h(x)=f(x)g(x)\\
h'(x)=f'(x)g(x)+f(x)g'(x)$$
The function $h(x)$ has a critical point if $h'(x)=0$, so
$$h'(x)=f'(x)g(x)+f(x)g'(x)\\
=f'(x)\big\{ g(x)+\frac{f(x)}{f'(x)}*g'(x)\big\}$$
and we note  that  $f'(x)\ne0$ and that the fraction $\frac{f(x)}{f'(x)}$ is positive. The second derivative is
$$h''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x) \space $$
we have the following cases(i will get the local maximum and discuss conditions to be a unique and hence global maximum):
(1) $x < x_0:$ in this interval both $f$ and $g$ are increasing and $f$ is strictly increase so there is no local maximum of $h$ in this inteval.
(2) $x = x_0$: $g'(x_0)=0$ and ang $g''(x_0)$ is negative, so comparing both $h'(x),h''(x)$, the only case that $h$ has a maximum is that $g(x_0)=0$. notice that in this case it will be a global maximum of $h(x)$ since $g(x)$ will be negative elsewhere
(3) $x \gt x_0$: in this interval $g'(x)$ is negative and the local maximum holds only if
$$g(x)+\frac{f(x)}{f'(x)}*g'(x)=0$$
from this equation we gett that $g$ is positive and hence $h''$ is negative if both $f''$ and $g''$ are negative. this case needs a deep discussion and is rare for general functions  to hold but you have a way to you complete.
A good graphical example for for two functions that have maximum product 10000 at infinity 
A: When you are saying $g$ has just one maximum, I think it means that $g$ has no other point of maximum/minimum.
So the function increases to $g(x_0)$ and then decreases.
For $f$ it just goes on increasing. 
$$ f(x)g(x) < f(x_0)g(x_0) \; \forall \; x \neq x_0 $$
Also check for $(f(x)g(x))'$ to be $0$ at $x = x_0$ we also need $g(x_0) = 0$.
So the conditions would be : 
1) $ f(x)g(x) < f(x_0)g(x_0) \; \forall \; x \neq x_0 $ and 
2) $g(x_0) = 0$.
which can be summarized in 2) only, i.e. $g(x_0) = 0$.
But we do not yet guaranteed that maxima is just one in this case.
If we need that there should be no other minima as well : 
then $$ (f(x)g(x))' = 0 \; iff \;  x = x_0 .$$ 
We analyse it over the 3 regions : 
i) $x < x_0$ where $f'(x) > 0$ and $f(x) > 0$, $g(x) < 0$ and $g'(x) > 0$ ;
ii) $x = x_0$ and 
iii) $x > x_0$ where $f'(x) > 0$ and $f(x) > 0$, $g(x) < 0$ and $g'(x) < 0$ ;
and ensure that in i) and ii) the condition does not hold.
$$ (f(x)g(x))' = f'(x)g(x) + g'(x)f(x) $$ is negative in region iii) so we do not have a min or maxima.
but in region i) the first term is negative while the last term is positive, so we have as all the quantities are positive,
$$f'(x)/g'(x) \neq - f(x)/g(x) \; for \; x < x_0.$$ for making the expression non-zero.
So the final conditions are : 
1) $g(x_0) = 0$ and 
2) $f'(x)/g'(x) \neq - f(x)/g(x) \; for \; x < x_0$.
