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

The extreme f(x)=x^3-x^2-8x+8,-2<= x<= 0 is Minimum(-2,12),Maximum(-4/3 , 392/27),Minimum(0,8)

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

extreme points

f (x) = x^{3} - x^{2} - 8 x + 8, - 2 \leq x \leq 0

Minimum (- 2, 12), Maximum (- \frac{4}{3}, \frac{392}{27}), Minimum (0, 8)

Solution steps

Find the critical points:

x = - 2, x = - \frac{4}{3}, x = 0

Domain of

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

Find intervals:

Increasing

: - 2 < x < - \frac{4}{3},

Decreasing

: - \frac{4}{3} < x < 0

Plug the extreme point

x = - 2

into

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

Minimum (- 2, 12)

Plug the extreme point

x = - \frac{4}{3}

into

x^{3} - x^{2} - 8 x + 8 \Rightarrow y = \frac{392}{27}

Maximum (- \frac{4}{3}, \frac{392}{27})

Plug the extreme point

x = 0

into

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

Minimum (0, 8)

Minimum (- 2, 12), Maximum (- \frac{4}{3}, \frac{392}{27}), Minimum (0, 8)

f (x) = \frac{x ^{2} - 11 x + 54}{x - 9}

2 x y

f (x) = \frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9

f (x) = 2 x + ln (x)

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

What is the extreme f(x)=x^3-x^2-8x+8,-2<= x<= 0 ?
The extreme f(x)=x^3-x^2-8x+8,-2<= x<= 0 is Minimum(-2,12),Maximum(-4/3 , 392/27),Minimum(0,8)