What is the extreme y=3x^3-36x^2+7 ?

The extreme y=3x^3-36x^2+7 is Maximum(0,7),Minimum(8,-761)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme y=3x^3-36x^2+7

extreme points

y = 3 x^{3} - 36 x^{2} + 7

Maximum (0, 7), Minimum (8, - 761)

Solution steps

f^{'} (x) = 9 x^{2} - 72 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

3 x^{3} - 36 x^{2} + 7 : 7

Maximum (0, 7)

Plug

x = 8

into

3 x^{3} - 36 x^{2} + 7 : - 761

Minimum (8, - 761)

Maximum (0, 7), Minimum (8, - 761)

f (x, y) = e^{x} ln (x) y

e^{6 x} + e^{- x}

F (x, y) = 12 x^{2} + 5 x y - 17 x + 7 y - 2 y^{2} + 5

5 x y e^{- x^{2} - y^{2}}

f (x) = - 2 x^{2} - 8 x + 5

What is the extreme y=3x^3-36x^2+7 ?
The extreme y=3x^3-36x^2+7 is Maximum(0,7),Minimum(8,-761)