Question

Допоможіть будь ласка 11 і 13 завдання.

Answer 1

Ответ:

$11)\ \ f(x)=4x-\dfrac{1}{3}\, x^3\ \ ,\ \ x_0=3\\\\f(3)=4\cdot 3-\dfrac{1}{3}\cdot 27=12-9=3\\\\f'(x)=4-x^2\ \ ,\ \ \ f'(3)=4-9=-5\\\\y=3-5(x-3)\ \ ,\ \ \ \ \boxed{\ y=-5x+18\ }$

$13)\ \ sin\Big(4x-\dfrac{\pi}{6}\Big)=-1\ \ ,\ \ x\in [-2\pi \, ;\ \pi \ ]\\\\\\a)\ \ 4x-\dfrac{\pi}{6}=-\dfrac{\pi}{2}+2\pi n\ \ ,\ \ \ 4x=-\dfrac{\pi}{3}+2\pi n\ \ ,\ \ \boxed{\ x=-\dfrac{\pi }{12}+\dfrac{\pi n}{2}\ ,\ n\in Z\ }\\\\\\\\b)\ \ x\in [-2\pi \, ;\ \pi \ ]:\ \ -2\pi \leq -\dfrac{\pi}{12}+\dfrac{\pi n}{2}\leq \pi \ \ ,\ \ -2+\dfrac{1}{12}\leq \dfrac{n}{2}\leq \pi +\dfrac{\pi }{12}\ \ ,\\\\\\-\dfrac{23}{6}\leq n\leq \dfrac{13}{6}\ \ \Rightarrow \ \ n=-3,-2,-1\, ,0\ ,\ 1\ ,\ 2$

$x_1=-\dfrac{19\pi }{12}\ ,\ x_2=-\dfrac{13\pi }{12}\ ,\ x_3=-\dfrac{7\pi}{12}\ ,\ \ ,\ x_4=-\dfrac{\pi}{12}\ ,\ x_5=\dfrac{5\pi}{12}\ ,\ x_6=\dfrac{11\pi}{12}\ .$