23 и 24 связано с производными

Решите задачу:

$f(x)= \frac{1}{3}sin3x-\frac{1}{2}x \; ,\; \; f'(x)\ \textless \ 0\\\\f'(x)= \frac{1}{3}\cdot 3\cdot cos3x - \frac{1}{2} \ \textless \ 0\\\\cos3x\ \textless \ \frac{1}{2}\\\\-\frac{\pi}{3}+2\pi n\ \textless \ 3x\ \textless \ \frac{\pi}{3}+2\pi n\; ,\; \; n\in Z\\\\-\frac{\pi }{9}+\frac{2\pi n}{3}\ \textless \ x\ \textless \ \frac{\pi}{9}+\frac{2\pi n}{3}\; ,\; n\in Z\\\\Otvet:\; \; D:\; x\in (-\frac{\pi}{9}+\frac{2\pi n}{3}\; ;\; \frac{\pi}{9}+\frac{2\pi n}{3})\; .$

$2)\; \; f(x)=\Big (\sqrt[4]{x}+ \frac{2}{\sqrt[4]{x}} \Big )\Big (\sqrt[4]{x}- \frac{2}{\sqrt[4]{x}} \Big )=\Big (\sqrt[4]{x}\Big )^2-\Big ( \frac{2}{\sqrt[4]{x}} \Big )^2=\sqrt{x}-\frac{4}{\sqrt{x}}\\\\f'(x)=\frac{1}{2\sqrt{x}}- \frac{-4\cdot \frac{1}{2\sqrt{x}}}{x} =\frac{1}{2\sqrt{x}}+\frac{2}{x\sqrt{x}}\\\\Otvet:\; \; C\; .$