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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?

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  • $\begingroup$ Can you tell what the derivative of $f$ is? $\endgroup$ – Arthur Oct 9 '17 at 10:53
  • $\begingroup$ 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 $-n(1-x)^{n-1}$. $\endgroup$ – user228113 Oct 9 '17 at 10:56
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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 $x=1/2$?

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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?

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