What is the extreme f(x)=8+12x^2-8x^3 ?

The extreme f(x)=8+12x^2-8x^3 is Minimum(0,8),Maximum(1,12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=8+12x^2-8x^3

extreme points

f (x) = 8 + 12 x^{2} - 8 x^{3}

Minimum (0, 8), Maximum (1, 12)

Solution steps

f^{'} (x) = - 24 x^{2} + 24 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

8 + 12 x^{2} - 8 x^{3} : 8

Minimum (0, 8)

Plug

x = 1

into

8 + 12 x^{2} - 8 x^{3} : 12

Maximum (1, 12)

Minimum (0, 8), Maximum (1, 12)

f (x) = x^{3} - 15 x^{2} + 78 x + 5

x^{2} - 2 x y + 2 y^{2} - 2 x + 4 y - 1

f (x) = - 2 sin (x) cos (x)

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

f (x) = 14 x^{2} - x^{3}

What is the extreme f(x)=8+12x^2-8x^3 ?
The extreme f(x)=8+12x^2-8x^3 is Minimum(0,8),Maximum(1,12)