What is the extreme y=2x^3-3x^2-9 ?

The extreme y=2x^3-3x^2-9 is Maximum(0,-9),Minimum(1,-10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme y=2x^3-3x^2-9

extreme points

y = 2 x^{3} - 3 x^{2} - 9

Maximum (0, - 9), Minimum (1, - 10)

Solution steps

f^{'} (x) = 6 x^{2} - 6 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

2 x^{3} - 3 x^{2} - 9 : - 9

Maximum (0, - 9)

Plug

x = 1

into

2 x^{3} - 3 x^{2} - 9 : - 10

Minimum (1, - 10)

Maximum (0, - 9), Minimum (1, - 10)

- (x - 4)^{2} + 1

f (x) = 1. 5^{x} + 4

(- 2, - 1), (- 5, 8)

f (x) = \frac{5 x}{6 x + 7}

f (x) = \frac{- 2 x + 2}{x + 7}

What is the extreme y=2x^3-3x^2-9 ?
The extreme y=2x^3-3x^2-9 is Maximum(0,-9),Minimum(1,-10)