Question

Помогите пожалуйста.. Очень срочно надо

Answer 1

$y=12tgx-12x+3\pi -7\; \; ,\; \; x\in [-\frac{\pi}{4}\, ;\, \frac{\pi}{4}\, ]\\\\y'=\frac{12}{cos^2x} -12=12\cdot (\frac{1}{cos^2x}-1)=0\; \; \to \; \; cos^2x=1\; \; ,\; \; \frac{1+cos2x}{2}=1\; ,\\\\cos2x=1\; \; ,\; \; 2x=2\pi n\; ,\; \; x=\pi n\; ,\; n\in Z\\\\n=0\; \to \; x=0\in [-\frac{\pi}{4}\, ;\, \frac{\pi}{4}\, ]\\\\x=-\frac{\pi}{4}:\; \; y(-\frac{\pi}{4})=-12+3\pi +3\pi -7=6\pi -19\approx -0,16\\\\x=0:\; \; y(0)=3\pi -7\approx 2,42\\\\x=\frac{\pi}{4}:\; \; y(\frac{\pi}{4})=12-3\pi +3\pi -7=5$

$y_{naibol.}=y(\frac{\pi}{4})=5$