Let $f_n(x) = nx(1-x)^n$ for $x \in [0,1]$. I need to get sup$\lbrace \mid f_n(x) \mid \rbrace$ for $x\in [0,1]$. So im not sure if its correct how i did it:
$f_n(x)$ is continous on $[0,1]$ and thats an compact interval. That means $f_n(x)$ gets to his maximum on this interval. So i got the first derivative: $f'_n(x) = n(1-x)^n-n^2x(1-x)^{n-1}$.
Then $f'_n(x) == 0 \Leftrightarrow x=\dfrac{1}{n+1}$. Since $f_n$ is continous and $f_n(0) = 0 < \dfrac{1}{n+1}$ for every $n \in \mathbb{N}$ this has to be the maximum. (Because if $x$ is the only max/min every value of $f_n$ should be smaller or greater than $f_n(x)$ with $x = \dfrac{1}{n+1}$.)
That follows, that sup$\lbrace \mid f_n(x) \mid \rbrace = f_n(\dfrac{1}{n+1})$. Is this right? How would you show the supremum?
I could proof that this is the supremum of this function showing that it is the smallest upper boundary for this function. Is this still necessary after doing the work above?
There is one minor (notational?) mistake: you wrote
You should write
But indeed, since $f'$ has only one zero in $(0, 1)$ and $f(0) < f\left(\frac{1}{n + 1}\right)$, $f$ attains a (local) maximum in $x = \frac{1}{n + 1}$. You should however also check whether $f$ attains a maximum on the boundary of the interval, since you can only find extrema in the open interval $(0, 1)$ using the first derivative.
In this case, the supremum of $f_n$ is the maximum of $f_n$, as we work with a continuous function on a compact set.