Question

СРОЧНО ПОМОГИТЕ ХОТЬ С ЧЕМ ТО ДО 10:45

Answer 1

Ответ:

$2)\ \ f(x)=4x-\dfrac{1}{3}\, x^3\\\\f'(x)=4-x^2<0\ \ ,\ \ \ (2-x)(2+x)<0\ \ ,\ \ \ znaki:\ ---(-2)+++(2)---\\\\\boxed{\ x\in (-\infty ;-2)\cup (\, 2;+\infty )\ }\\\\\\4a)\ \ f(x)=x^3-x^2+6x-2\ \ ,\ \ x_0=1\ ,\\\\f(1)=1-1+6-2=4\\\\f'(x)=3x^2-2x+6\ \ ,\ \ f'(1)=3-2+6=7\\\\y=7+4(x-1)\\\\\boxed{y=4x+3}$

$b)\ \ f(x)=sin^4x\ \ ,\ \ \ x_0=\dfrac{\pi}{2}\ \ ,\\\\f(\dfrac{\pi}{2})=sin^4\dfrac{\pi}{2}=1\\\\f'(x)=4sin^3x\cdot cosx\ \ ,\ \ \ f'(\dfrac{\pi}{2})=4sin^3\dfrac{\pi}{2}\cdot cos\dfrac{\pi}{2}=0\\\\y=1+0\cdot (x-\dfrac{\pi}{2})\\\\\boxed {\ y=1\ }$