What is the extreme f(x)=3x^4+16x^3+24x^2+32 ?

The extreme f(x)=3x^4+16x^3+24x^2+32 is Saddle(-2,48),Minimum(0,32)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=3x^4+16x^3+24x^2+32

extreme points

f (x) = 3 x^{4} + 16 x^{3} + 24 x^{2} + 32

Saddle (- 2, 48), Minimum (0, 32)

Solution steps

f^{'} (x) = 12 x^{3} + 48 x^{2} + 48 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

3 x^{4} + 16 x^{3} + 24 x^{2} + 32 : 48

Saddle (- 2, 48)

Plug

x = 0

into

3 x^{4} + 16 x^{3} + 24 x^{2} + 32 : 32

Minimum (0, 32)

Saddle (- 2, 48), Minimum (0, 32)

What is the extreme f(x)=3x^4+16x^3+24x^2+32 ?
The extreme f(x)=3x^4+16x^3+24x^2+32 is Saddle(-2,48),Minimum(0,32)