What is the extreme y=-x^3+3x^2-5 ?

The extreme y=-x^3+3x^2-5 is Minimum(0,-5),Maximum(2,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme y=-x^3+3x^2-5

extreme points

y = - x^{3} + 3 x^{2} - 5

Minimum (0, - 5), Maximum (2, - 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

- x^{3} + 3 x^{2} - 5 : - 5

Minimum (0, - 5)

Plug

x = 2

into

- x^{3} + 3 x^{2} - 5 : - 1

Maximum (2, - 1)

Minimum (0, - 5), Maximum (2, - 1)

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

f (x) = 3 x^{2} + 4 y^{3} - 12 x y + 37

f (x) = 15 x + 18 y - 2 x^{2} + 2 x y + 3 y^{2}

f (x, y) = x^{3} - 12 x + y^{2} + 6 y + 14

f (x) = x^{- 7} ln (x)

What is the extreme y=-x^3+3x^2-5 ?
The extreme y=-x^3+3x^2-5 is Minimum(0,-5),Maximum(2,-1)