# How to find the maximum of the given function?

Find the maximum of the function $f(x)=x^{n}(1-x)^{n}$, $n\in \mathbb{N}$, $x\in [0,1]$?

I tried to use the Second derivative test, I am not getting the maximum.

$f'(x)=0 \tag{1}$

$\implies$ $x=0,1$

Can anyone help me to find the Maximum?

• Can you tell what the derivative of $f$ is? – Arthur Oct 9 '17 at 10:53
• Your "$\implies$" is wrong. Probably due to a mistake of algebra where you thought $((1-x)^n)'$ to be $n(1-x)^{n-1}$ rather than $-n(1-x)^{n-1}$. – user228113 Oct 9 '17 at 10:56

You could start by noting that $$f(x) = \left( x (1-x) \right)^n.$$ Now think about $x \mapsto x(1-x)$ on $[0,1]$. In particular, $f(0)=0=f(1)$, while $f$ is symmetric with respect to the line $x=1/2$. What happens at $x=1/2$?
The derivative of $f(x) = (x(1-x))^n$ is
$$f'(x) = n(x(1-x))^{n-1}(1-2x)$$
and so the critical points are at $x = 0, 1$ and $x = \frac{1}{2}$. Can you apply the second derivative test to these points?