Product of decreasing function and concave function

Question

I know that a function $f_1(x)$ is concave and another function $f_2(x)$ is decreasing over some $a<x<b$. I want to know is it possible for the product $f_1(x)f_2(x)$ to have two local optimal points in $a<x<b$. Any help will be highly appreciated. Thanks in advance.

Edit:

The functions take positive values for $a<x<b$.

Answer 1

I propose the following example.

$f_1(x)=4-x^2$ it is an inverted parabola thus concave.

$f_2(x)=2-x^3$ since $f'(x)=-3x^2$ this function is $\searrow$.

Both functions are positive over $[-2,b]$ with $1<b<2$ but we don't care too much about the value, $[-2,1]$ is a good interval.

Here is the graph of $f_1,f_2$ and the product : http://tiny.cc/p64jpy

There are $2$ local maxima and a local minimum for the product in $[-2,1]$.

Answer 2

$f_1(x)=\sqrt{x}$ is concave and positive over $\mathbb{R}^+$ and $f_2(x)=\frac{1}{\sqrt{x+\sin x}}$ is positive and decreasing, since $\frac{d}{dx}(x+\sin x)=1+\cos x\geq 0$. On the other hand,

$$ f_1(x)\,f_2(x)=\sqrt{\frac{1}{1+\frac{\sin x}{x}}} $$ has an infinity of stationary points over $\mathbb{R}^+$, since the same holds for $\frac{\sin x}{x}$, which is bounded by $1$ in absolute value.