What is the extreme f(x)=2x^3-3x^2-12x+4 ?

The extreme f(x)=2x^3-3x^2-12x+4 is Maximum(-1,11),Minimum(2,-16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=2x^3-3x^2-12x+4

extreme

Maximum (- 1, 11), Minimum (2, - 16)

Solution steps

f^{'} (x) = 6 x^{2} - 6 x - 12

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

2 x^{3} - 3 x^{2} - 12 x + 4 : 11

Maximum (- 1, 11)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 4 : - 16

Minimum (2, - 16)

Maximum (- 1, 11), Minimum (2, - 16)

What is the extreme f(x)=2x^3-3x^2-12x+4 ?
The extreme f(x)=2x^3-3x^2-12x+4 is Maximum(-1,11),Minimum(2,-16)