I found an online answer to this question, but I think there is something wrong with it
Question: $f(x) = |x|^m |x - 1|^n, \forall x \in \mathbb{R}$
(m) and (n) are natural numbers (positive integers). The local maximum value for the function is:
a. $(1)$ b. $(m^n n^m)$ c. $(\frac{m^m n^n}{(m + n)^{m+n}})$ d. $(-\frac{(mn)^{mn}}{(m + n)^{m+n}})$
Answer and Explanation:
We have $f(X)= \begin{cases}(-1)^{m+n} x^m(x-1)^n & \text { if } x<0, \\ \\ \\ \\ (-1)^n x^m(x-1)^n & \text { if } 0 \leq x<1, \\ \\ \\ \\ x^m(x-1)^n & \text { if } x \geq 1\end{cases}$ Let $g(X)=x^m(x-1)^n$ Then $$ \begin{aligned} & g'(x)=m x^{m-1}(x-1)^n+n x^m(x-1)^{n-1} \\ & =x^{m-1}(x-1)^{n-1}\{(m x-m+n x)\} \end{aligned} $$ $$ \text { Now } f(X)=0 \text { or } g(x)=0 \text { i.e } \\ x=0,1 \text { or } \frac{m}{m+n} $$ $$ \mathrm{F}(\mathrm{0})=0, \\ \mathrm{f}(1)=0 \text { and } $$ $$ f\left(\frac{m}{m+n}\right)=(-1)^n \frac{m^{m n^n(-1)^n}}{(m+n)^{m+n}}=\frac{m^m n^n}{(m+n)^{m+n}}>0 $$
thus the answer is C
I wonder from the beginning how $\frac{m}{m+n}$ is the x value at which the function is maximum because $\frac{m}{m+n}$ is smaller than 1
and $f(x) = x^m(x-1)^n \text { only if } x \geq 1$