What is the extreme f(x)=x^3+12x^2-27x+8,-10<= x<= 0 ?

The extreme f(x)=x^3+12x^2-27x+8,-10<= x<= 0 is Minimum(-10,478),Maximum(-9,494),Minimum(0,8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^3+12x^2-27x+8,-10<= x<= 0

extreme

Minimum (- 10, 478), Maximum (- 9, 494), Minimum (0, 8)

Solution steps

Find the critical points:

x = - 10, x = - 9, x = 0

Domain of

x^{3} + 12 x^{2} - 27 x + 8, - 10 \leq x \leq 0 : - 10 \leq x \leq 0

Find intervals:

Increasing

: - 10 < x < - 9,

Decreasing

: - 9 < x < 0

Plug the extreme point

x = - 10

into

x^{3} + 12 x^{2} - 27 x + 8 \Rightarrow y = 478

Minimum (- 10, 478)

Plug the extreme point

x = - 9

into

x^{3} + 12 x^{2} - 27 x + 8 \Rightarrow y = 494

Maximum (- 9, 494)

Plug the extreme point

x = 0

into

x^{3} + 12 x^{2} - 27 x + 8 \Rightarrow y = 8

Minimum (0, 8)

Minimum (- 10, 478), Maximum (- 9, 494), Minimum (0, 8)

What is the extreme f(x)=x^3+12x^2-27x+8,-10<= x<= 0 ?
The extreme f(x)=x^3+12x^2-27x+8,-10<= x<= 0 is Minimum(-10,478),Maximum(-9,494),Minimum(0,8)