Stumped.

Calculus Level 4

A function $f$ is defined by $f(x) = |x|^m \cdot |x-1|^n$ for all real numbers $x$ , where $m$ and $n$ are positive integers. What is the local maximum of this function?

Notation: $|\cdot|$ denotes the absolute value function .

\frac {(mn)^{mn}}{(m+n)^{m+n}}

\frac {m^m n^n}{( m+n)^{m+n}}

m^n n^m

1 solution

Chew-Seong Cheong
Sep 5, 2018

Note that $f(x) \ge 0$ for all $x$ , $f(-\infty) = f(\infty) = \infty$ and $f(0) = f(1) = 0$ . Therefore, the local maximum must be within $x \in (0,1)$ . Then we have:

$\begin{aligned} f(x) & = |x|^m \cdot |x-1|^n & \small \color{#3D99F6} \text{For }x \in (0, 1) \\ & = x^m(1-x)^n & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ f'(x) & = mx^{m-1} (1-x)^n - n x^m(1-x)^{n-1} & \small \color{#3D99F6} \text{Putting }f'(x) = 0 \\ mx^{m-1} (1-x)^n & = n x^m(1-x)^{n-1} & \small \color{#3D99F6} \text{For }x \ne 0 \\ m(1-x) & = nx \\ \implies x & = \frac m{m+n} \\ \implies \max (f(x)) & = f\left(\frac m{m+n}\right) \\ & = \left(\frac m{m+n} \right)^m \left(1-\frac m{m+n} \right)^n \\ & = \boxed{ \dfrac {m^mn^n}{(m+n)^{m+n}} } \end{aligned}$

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