
Problem 11.1. Given a function $f(x)$ defined for any real $x$ and $f(\operatorname{tg} x)=\sin 2 x$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Find the minimum and the maximum of the function $f\left(\sin ^{3} x\right) \cdot f\left(\cos ^{3} x\right)$.

Oleg Mushkarov, Nikolai Nikolov


Solution: Let $t=\operatorname{tg} x$. Then $\sin 2 x=\frac{2 t}{1+t^{2}}$ and it follows from the conditions of the problem that $f(t)=\frac{2 t}{1+t^{2}}$ for any $t$. Therefore

$$
\begin{gathered}
f\left(\sin ^{3} x\right) \cdot f\left(\cos ^{3} x\right)=\frac{4 \sin ^{3} x \cdot \cos ^{3} x}{\left(1+\sin ^{6} x\right)\left(1+\cos ^{6} x\right)}= \\
\frac{4(\sin x \cdot \cos x)^{3}}{2-3(\sin x \cdot \cos x)^{2}+(\sin x \cdot \cos x)^{6}}
\end{gathered}
$$

Let $u=\sin x \cdot \cos x=\frac{1}{2} \sin 2 x$. Then $u \in\left[-\frac{1}{2}, \frac{1}{2}\right]$ and we have to find the minimum and the maximum of the function $g(u)=\frac{4 u^{3}}{2-3 u^{3}+u^{6}}$ in the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$. We obtain

$$
g^{\prime}(u)=\frac{12 u^{2}\left(1-u^{2}\right)\left(u^{4}+u^{2}+1\right)}{\left(2-3 u^{2}+u^{6}\right)^{2}}>0
$$

when $u \in\left[-\frac{1}{2}, \frac{1}{2}\right]$ and so $g(u)$ is an increasing function in the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$. It follows now that $\max _{u \in\left[-\frac{1}{2}, \frac{1}{2}\right]} g(u)=g\left(\frac{1}{2}\right)=\frac{32}{81}$ and $\min _{u \in\left[-\frac{1}{2}, \frac{1}{2}\right]} g(u)=g\left(-\frac{1}{2}\right)=-\frac{32}{81}$.
