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](https://en.wikipedia.org/wiki/Quasiconvex_function)) but apparently it is not that easy.



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


