What is the extreme f(x)=x^3+6x^2-5,-5<= x<= 2 ?

The extreme f(x)=x^3+6x^2-5,-5<= x<= 2 is Minimum(-5,20),Maximum(-4,27),Minimum(0,-5),Maximum(2,27)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x^3+6x^2-5,-5<= x<= 2

extreme points

f (x) = x^{3} + 6 x^{2} - 5, - 5 \leq x \leq 2

Minimum (- 5, 20), Maximum (- 4, 27), Minimum (0, - 5), Maximum (2, 27)

Solution steps

Find the critical points:

x = - 5, x = - 4, x = 0, x = 2

Domain of

x^{3} + 6 x^{2} - 5, - 5 \leq x \leq 2 : - 5 \leq x \leq 2

Find intervals:

Increasing

: - 5 < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 5

into

x^{3} + 6 x^{2} - 5 \Rightarrow y = 20

Minimum (- 5, 20)

Plug the extreme point

x = - 4

into

x^{3} + 6 x^{2} - 5 \Rightarrow y = 27

Maximum (- 4, 27)

Plug the extreme point

x = 0

into

x^{3} + 6 x^{2} - 5 \Rightarrow y = - 5

Minimum (0, - 5)

Plug the extreme point

x = 2

into

x^{3} + 6 x^{2} - 5 \Rightarrow y = 27

Maximum (2, 27)

Minimum (- 5, 20), Maximum (- 4, 27), Minimum (0, - 5), Maximum (2, 27)

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

f (x) = \frac{- x ^{6} - 5 x ^{3} + 5 x}{x ^{2} + 2}

f (x, y) = 200 y^{2} + x^{2} - x^{2} y

2 + x - x^{2}

f (x) = x^{- \frac{1}{3}} (x + 2)

What is the extreme f(x)=x^3+6x^2-5,-5<= x<= 2 ?
The extreme f(x)=x^3+6x^2-5,-5<= x<= 2 is Minimum(-5,20),Maximum(-4,27),Minimum(0,-5),Maximum(2,27)