Question

Помогите ,срочно пожалуйста!

Answer 1

$1)\; \; f(x)=sin2x-2x\; ,\; \; x\in [-\frac{\pi}{2},\frac{\pi}{2}]\\\\f'(x)=2cos2x-2=0\; \; \to \; \; cos2x=1\\\\2x=2\pi n\; ,\; \; x=\pi n\, ,\; n\in Z\\\\x\in [-\frac{\pi }{2},\frac{\pi }{2}]\; \; \to \; \; x=\pi \cdot 0=0\; \\\\f(0)=sin0-2\cdot 0=0\\\\f(-\frac{\pi}{2})=sin(-\pi )+\pi =0+\pi =\pi \\\\f(\frac{\pi}{2})=sin\pi -\pi =-\pi \\\\-\pi \ \textless \ 0\ \textless \ \pi \; \; \Rightarrow \; \; f_{naimen.}=-\pi \; \; ,\; \; f_{naibol.}=\pi$

$2)\; \; f(x)=\sqrt3cosx+sin\frac{\pi}{6}+\frac{x^2}{\pi }\; ,\; \; x_0= \frac{\pi }{6}\\\\f'(x)=-\sqrt3sinx+\frac{2x}{\pi }\\\\f'( \frac{\pi }{6})=-\sqrt3\cdot sin \frac{\pi }{6}+ \frac{1}{3}=-\sqrt3\cdot \frac{1}{2}+\frac{1}{3}=\frac{2-3\sqrt3}{6}$