Max value of $f_n = x^n(1-x)$?

Question

Here it's claimed that the maximum value for

$$f_n = x^n(1-x)$$

for every $n \in \mathbb{N}$ is at

$$x = n / (n + 1)$$

and is

$$n^n / (n + 1)^{(n + 1)}$$

How is this actually found to be the maximum?

Answer 1

We use calculus. Note that $f_n$ has local extrema when

$$f_n'(x) == 0$$

$$\frac{d}{dx} x^n-x^{n+1} = 0$$

$$nx^{n-1} = (n+1)x^n$$

$$x=0\ \mathrm{or}\ x = \frac{n}{n+1}$$

$x=0$ is easily seen to be a minimum, giving the final result.

Answer 2

If you can't use the calculus technique, and $n$ is a positive integer, you can use AM/GM (assuming $x\geq 0$.)

First, note that for $x>0$, the maximum must be with $x\in [0,1]$ because $x^n(1-x)< 0$ when $x>1$.

Note that the arithmetic mean of $\frac{x}{n},\frac{x}{n},\dots(n\text{ times})\dots,\frac{x}{n},1-x$ is $\frac{1}{n+1}$. The geometric mean is $\sqrt[n+1]{\frac{x^n(1-x)}{n^n}}$.

So we have by AM-GM that $$\frac{1}{n+1}\geq \sqrt[n+1]{\frac{x^n(1-x)}{n^n}}$$

Working this out, we see that $\frac{n^n}{(n+1)^{n+1}}\geq x^n(1-x)$ for all $x\in(0,1)$, with equality only when $\frac{x}{n}=1-x$, or $x=\frac{n}{n+1}$.

Answer 3

Just compute $df/dx$ and look where it is $0$. One has

$$f_n'(x)=nx^{n-1}(1-x)-x^n=x^{n-1}\left(n-(n+1)x\right)$$

And this is $0$ for $x=0$ Corresponding to a minimum or $x=n/n+1$ corresponding to a maximum whose value is

$$f_n\left({n\over n+1}\right)=\left(1+{1\over n}\right)^{-n}{1\over n+1}={n^n\over (n+1)^{n+1}}$$

Answer 4

More generally, the maximum value of function $f_k$ defined by $f_k(x)=x^k(1-x)^{n+1-k}$ on interval $[0,1]$ is easily shown to take place at $x_0:=\frac{k}{n+1}$.

These functions occur naturally, with a convenient normalization factor as the pdfs (probability density functions) of the classical "order statistics" under the name 'Beta distribution" (see also "Bernstein polynomials")