What is the extreme f(x)=-x^3+3x^2-6 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=-x^3+3x^2-6

extreme points

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

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

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} - 6 : - 6

Minimum (0, - 6)

Plug

x = 2

into

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

Maximum (2, - 2)

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

f (x) = \frac{1}{x - 4} + \frac{1}{6 - x}

f (x) = - x^{3} - 3 x^{4}

f (x) = \frac{( x + 1 )}{x ^{2} - 9}

f (x) = 4 + 4 + x

f (x) = - (4 x + 10 y) = - 9

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