Question

найдите промежутки возрастания и убывания

1) f(x) = 2х^3-3x^2-12x-13

2) f(x) = 5x+x^2

3) f(x) = 8 - x^3

Answer 1

$1)\ \ f(x)=2x^3-3x^2-12x-13\\\\f'(x)=6x^2-6x-12=6\, (x^2-x-2)=6\, (x-2)(x+1)=0\\\\x_1=2\ ,\ x_2=-1\\\\znaki\ f'(x):\ \ \ +++(-1)---(2)+++\\{}\qquad \qquad \qquad \ \, \quad \nearrow \ \ (-1)\ \ \searrow \ \ \ (2)\ \ \nearrow \\{}\qquad \qquad \qquad \qquad \quad \ \ max\ \ \qquad \ min \\\\\\f(x)\ vozrastset:\ \ x\in (-\infty ;-1\ ]\cup [\ 2\, ;+\infty )\\\\f(x)\ ybuvaet:\ \ x\in [-1\ ;\ 2\ ]$

$2)\ \ f(x)=5x+x^2\\\\f'(x)=5+2x=0\ \ ,\ \ \ x=-2,5\\\\znaki\ f'(x):\ \ \ ---(-2,5)+++\\{}\qquad \qquad \qquad \ \, \quad \searrow \ \ \, (-2,5)\ \ \nearrow \\{}\qquad \qquad \qquad \qquad \quad \ \ \ \ \, min\\\\\\f(x)\ vozrastset:\ \ x\in [-2,5\ ;+\infty \, )\\\\f(x)\ ybuvaet:\ \ x\in (-\infty \ ;\ -2,5\ ]$

$3)\ \ f(x)=8-x^3\\\\f'(x)=-3x^2=0\ \ ,\ \ \ x=0\\\\znaki\ f'(x):\ \ \ ---(0)---\\{}\qquad \qquad \qquad \ \, \quad \searrow \ \ (0)\ \ \searrow \\\\f(x)\ ybuvaet:\ \ x\in (-\infty \ ;\ +\infty \, )$