Question-23
Consider the function \(\displaystyle f\) defined on the domain \(\displaystyle ( 0,1)\):
\[\begin{equation*} f( x) =x^{m}( x-1)^{n} \end{equation*}\]
We have:
\[\begin{equation*} \begin{aligned} f^{\prime }( x) & =mx^{m-1}( x-1)^{n} +nx^{m}( x-1)^{n-1}\\ & =f( x)\left[\frac{m}{x} +\frac{n}{x-1}\right]\\ & =f( x)\left[\frac{m}{x} -\frac{n}{1-x}\right] \end{aligned} \end{equation*}\]
To find the stationary points, we set \(\displaystyle f^{\prime }( x) =0\). This happens when \(\displaystyle x^{\star } =\frac{m}{m+n}\). This is in \(\displaystyle ( 0,1)\).
\[\begin{equation*} \begin{aligned} f^{\prime \prime }( x) & =f^{\prime }( x)\left[\frac{m}{x} -\frac{n}{1-x}\right] -f( x)\left[\frac{m}{x^{2}} +\frac{n}{( 1-x)^{2}}\right] \end{aligned} \end{equation*}\]
At \(\displaystyle x^{\star } =\frac{m}{m+n}\), we know that \(\displaystyle f^{\prime }\left( x^{\star }\right) =0\). Therefore, \(\displaystyle f^{\prime \prime }\left( x^{\star }\right)\) becomes:
\[\begin{equation*} \begin{aligned} f^{\prime \prime }\left( x^{\star }\right) & =-f\left( x^{\star }\right)\left[\frac{m}{x{^{\star }}^{2}} +\frac{n}{\left( 1-x^{\star }\right)^{2}}\right] \end{aligned} \end{equation*}\]
The sign of \(\displaystyle f^{\prime \prime }\left( x^{\star }\right)\) depends on the sign of \(\displaystyle f\left( x^{\star }\right)\).
\[\begin{equation*} f\left( x^{\star }\right) =x{^{\star }}^{m}\left( x^{\star } -1\right)^{n} \end{equation*}\]
Since \(\displaystyle x^{\star } \in ( 0,1)\), \(\displaystyle f\left( x^{\star }\right)\) is positive when \(\displaystyle n\) is even and negative when \(\displaystyle n\) is odd. It follows that \(\displaystyle x^{\star }\) corresponds to a maximum when \(\displaystyle n\) is even and a minimum when \(\displaystyle n\) is odd.