What is the extreme f(x)=x^3+2x^2-7x,0<= x<= 2 ?

The extreme f(x)=x^3+2x^2-7x,0<= x<= 2 is Maximum(0,0),Minimum(1,-4),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+2x^2-7x,0<= x<= 2

extreme points

f (x) = x^{3} + 2 x^{2} - 7 x, 0 \leq x \leq 2

Maximum (0, 0), Minimum (1, - 4), Maximum (2, 2)

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

x^{3} + 2 x^{2} - 7 x, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 0

into

x^{3} + 2 x^{2} - 7 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

x^{3} + 2 x^{2} - 7 x \Rightarrow y = - 4

Minimum (1, - 4)

Plug the extreme point

x = 2

into

x^{3} + 2 x^{2} - 7 x \Rightarrow y = 2

Maximum (2, 2)

Maximum (0, 0), Minimum (1, - 4), Maximum (2, 2)

What is the extreme f(x)=x^3+2x^2-7x,0<= x<= 2 ?
The extreme f(x)=x^3+2x^2-7x,0<= x<= 2 is Maximum(0,0),Minimum(1,-4),Maximum(2,2)