What is the extreme f(x)=3x^4+4x^3+6x^2-4 ?

The extreme f(x)=3x^4+4x^3+6x^2-4 is Minimum(0,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3x^4+4x^3+6x^2-4

extreme points

f (x) = 3 x^{4} + 4 x^{3} + 6 x^{2} - 4

Minimum (0, - 4)

Solution steps

f^{'} (x) = 12 x^{3} + 12 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

3 x^{4} + 4 x^{3} + 6 x^{2} - 4 : - 4

Minimum (0, - 4)

(- 6.4) (- 9)

f (x) = 3 cos (x)

f (x) = (x - 1)^{2}

f (x) = 2 x^{3} - 4 x^{2} + 8 x + 3

f (x) = \frac{x ^{2}}{1 - x ^{2}}

What is the extreme f(x)=3x^4+4x^3+6x^2-4 ?
The extreme f(x)=3x^4+4x^3+6x^2-4 is Minimum(0,-4)