Quasi-concavity of $f(x)=(1-\frac{x}{1000})\log_2(1+2^x)$ on $[0~1000]$

Question

I want to prove that function $f:[0~1000]\rightarrow R$, $$f(x)=(1-\frac{x}{1000})\log_2(1+2^x)$$ is quasi-concave. Any idea how to do the proof? I already tried to prove that any super-level set is convex (see this) but apparently it is not that easy.

Answer 1

We have \begin{equation} f_2(x):=f''(x)\Big/\frac{2^{x-3}}{125 \left(2^x+1\right)^2}= -x \ln2-2 \left(2^x+1-500 \ln2\right) \end{equation} and \begin{equation} f''_2(x)=-2^{1 + x} \ln^2 2<0, \end{equation} so that $f_2$ is concave. Also, $f_2(0)>0$ and $f'_2(0)<0$. So, $f_2$ decreases on $[0,1000]$ from $f_2(0)>0$ to $f_2(1000)<0$. Thus, for some $c\in(0,1000)$ we have $f_2>0$ and hence $f''>0$ on $[0,c)$, and $f_2<0$ and hence $f''<0$ on $(c,1000]$. So, the function $f$ is convex on $[0,c]$ and concave on $[c,1000]$. Moreover, $f(0)=1>0$, $f'(0)>0$, and $f(1000)=0$. So, $f$ increases on $[0,c]$ and then continues to increase on $[c,d]$ for some $d\in[c,1000]$, then switching to decrease on $[d,1000]$. Thus, $f$ is indeed quasi concave.