TITLE: Is it a concave function?
QUESTION [1 upvotes]: What kind of function is this $$f(x)=x^{n}(1-x)$$ where $0<x<1$ and $n>1$. I have plotted it for different values of $n$ the function looks to be concave. But when I solve the second derivative it is not negative always. So anybody who knows what type of function is the above function.

REPLY [4 votes]: No, for $n>1$, $f_n(x)=x^n(1-x)$ is not concave in $[0,1]$. The second derivative is
$$f''(x)=x^{n-2}n((n-1)-(n+1)x)$$
which is $\geq 0$ in $[0,(n-1)/(n+1)]$ and $\leq 0$ in $[(n-1)/(n+1),1]$.
Hence $f_n$ is concave just in $[(n-1)/(n+1),1]$ (and convex in $[0,(n-1)/(n+1)]$), an interval which becomes smaller and smaller as $n$ grows. 
P.S. On the other hand, for $n>1$, $f_n(x)=x^n(1-x)$ is quasi-concave in $[0,1]$. The first derivative is
$$f'_n(x)=x^{n-1}(n-(n+1)x)$$
and therefore $f_n$ is increasing in $[0,n/(n+1)]$ and it is decreasing in $[n/(n+1),1]$, that is $f_n$ is unimodal which implies quasi-concavity: for all $x,y,t\in [0,1]$,
$$f(tx+(1-t)y)\geq \min(f(x),f(y)).$$
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