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

The extreme f(x)=2+12x^2-8x^3 is Minimum(0,2),Maximum(1,6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Minimum (0, 2), Maximum (1, 6)

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

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

Minimum (0, 2)

Plug

x = 1

into

2 + 12 x^{2} - 8 x^{3} : 6

Maximum (1, 6)

Minimum (0, 2), Maximum (1, 6)

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

f (x, y, z) = z

- 3^{2 x - 4} - 5

f (x) = (8 x - 9) e^{4 x}

y = 6 x e^{- x^{2}}

What is the extreme f(x)=2+12x^2-8x^3 ?
The extreme f(x)=2+12x^2-8x^3 is Minimum(0,2),Maximum(1,6)