Product of linear and convex function More specific, how many maxima are there for product of these two functions:


*

*$ f(x) = ax + b $, and $ a > 0 $

*$ g(x) $ is (strongly) decreasing convex function, $ \lim_{x\rightarrow\infty} g(x) = 0 $, and it is positive on $ [-\frac{b}{a}, \infty) $


on interval $ [-\frac{b}{a}, \infty) $. We also know that it has x-axis as the horizontal asymptote from the bottom, that is $ \lim_{x\rightarrow\infty} fg = 0 $.
It seems that there is only one maximum (the global one), but I have a hard time to prove that. At the end of the intervals, product is equal to $ 0 $. It is positive. So, there it has the finite maximum. At $ -\frac{b}{a} $ first derivative of $ fg $ is positive. Also, after a particular $ \overline{x}\in [-\frac{b}{a}, \infty) $ first derivative of $ fg $ becomes negative, and it stays like that. Question is, what is going on between $ 0 $ and $ \overline{x} $.
 A: The derivation of the second derivative $g^{''}(x)$ is wrong, unfortunately.
I came back step by step trying to retrieve its starting point and it seems he just made a mistake when applying the quotient rule for derivatives.
Specifically, he applied
$$
g^{''}(x)=-a\bigg(\frac{g^{'}(x)-g(x)f^{'}(x)}{f^2(x)}\bigg)
$$
rather than
$$
g^{''}(x)=-a\bigg(\frac{g^{'}(x)f(x)-g(x)f^{'}(x)}{f^2(x)}\bigg)
$$
This makes the second part of the derivation wrong.
A: Somehow I bumped up again to this question. When I tried with Google to see  if anyone has dealt with it, I found my own question. :) 
Luckily, I got a solution, now.:) It's quite simple and straightforward.
Well, first, assume the opposite, i.e. there are two local maxima. That means there is a local minimum between. So, we have three stationary/critical points $ x_1 $, $ x_2 $, and $ x_3 $ (max, min, max). 
Also, we know that $ f'(x) = a $, $ f''(x) = 0 $, $ g'(x) < 0 $, and $ g''(x) > 0 $, for all $ x > -\frac{b}{a} $. 
Then, we have that $ \frac{\partial}{\partial x} (fg)(x) $ should be less than zero in $ x_1 $ and $ x_3 $, but greater than zero in $ x_2 $, if our assumption is correct. Because those three points are critical, we can derive expression for the $ g'(x) $ in them. Particulary, 
$\qquad\qquad g'(x_i) = -a\frac{g(x_i)}{f(x_i)} $,   for $ i = 1, 2, 3 $.
Then, we can use it to express the second derivative, i.e.
$\qquad\qquad g''(x_i) = a^2\frac{g(x_i)(1 + ax_i + b)}{(ax_i + b)^3} $,   for $ i=1, 2,3 $.
Now, $ (fg)''(x) = \frac{1}{(ax_i + b)^2} a^2 g(x_i)(1 - (ax_i + b)) $. All terms must be positive except $ 1 - (ax_i + b) $. So, we have that


*

*for $ i = 1, 3 $ it must be $ 1 - (ax_i + b) < 0 $, and

*for $ i = 2 $ it should be the opposite, i.e. $ 1 - (ax_2 + b) > 0 $


Well, $ f(x) = ax + b $ is linear increasing function, and $ x_1 < x_2 < x_3 $, which drives us to the contradiction wall. So, only one maximizer exists, i.e. only one global/local maximum.
