What is the extreme f(x)=x^3-2x^2-4x+8,-1<= x<= 0 ?

The extreme f(x)=x^3-2x^2-4x+8,-1<= x<= 0 is Minimum(-1,9),Maximum(-2/3 , 256/27),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^3-2x^2-4x+8,-1<= x<= 0

extreme

Minimum (- 1, 9), Maximum (- \frac{2}{3}, \frac{256}{27}), Minimum (0, 8)

Solution steps

Find the critical points:

x = - 1, x = - \frac{2}{3}, x = 0

Domain of

x^{3} - 2 x^{2} - 4 x + 8, - 1 \leq x \leq 0 : - 1 \leq x \leq 0

Find intervals:

Increasing

: - 1 < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 0

Plug the extreme point

x = - 1

into

x^{3} - 2 x^{2} - 4 x + 8 \Rightarrow y = 9

Minimum (- 1, 9)

Plug the extreme point

x = - \frac{2}{3}

into

x^{3} - 2 x^{2} - 4 x + 8 \Rightarrow y = \frac{256}{27}

Maximum (- \frac{2}{3}, \frac{256}{27})

Plug the extreme point

x = 0

into

x^{3} - 2 x^{2} - 4 x + 8 \Rightarrow y = 8

Minimum (0, 8)

Minimum (- 1, 9), Maximum (- \frac{2}{3}, \frac{256}{27}), Minimum (0, 8)

What is the extreme f(x)=x^3-2x^2-4x+8,-1<= x<= 0 ?
The extreme f(x)=x^3-2x^2-4x+8,-1<= x<= 0 is Minimum(-1,9),Maximum(-2/3 , 256/27),Minimum(0,8)