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

The extreme f(x)=x^3-x^2-x+8,-1<= x<= 0 is Minimum(-1,7),Maximum(-1/3 , 221/27),Minimum(0,8)

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extreme f(x)=x^3-x^2-x+8,-1<= x<= 0

extreme points

f (x) = x^{3} - x^{2} - x + 8, - 1 \leq x \leq 0

Minimum (- 1, 7), Maximum (- \frac{1}{3}, \frac{221}{27}), Minimum (0, 8)

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

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

Decreasing

: - \frac{1}{3} < x < 0

Plug the extreme point

x = - 1

into

x^{3} - x^{2} - x + 8 \Rightarrow y = 7

Minimum (- 1, 7)

Plug the extreme point

x = - \frac{1}{3}

into

x^{3} - x^{2} - x + 8 \Rightarrow y = \frac{221}{27}

Maximum (- \frac{1}{3}, \frac{221}{27})

Plug the extreme point

x = 0

into

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

Minimum (0, 8)

Minimum (- 1, 7), Maximum (- \frac{1}{3}, \frac{221}{27}), Minimum (0, 8)

f (x) = \frac{12 x}{9 + x ^{2}}

f (x, y) = 4 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = 2 x^{\frac{5}{3}} - 5 x^{\frac{4}{3}}

f (x) = 2 (3 x + 2)^{2} - 3 x^{4}

y = x (x + 4)^{3}

What is the extreme f(x)=x^3-x^2-x+8,-1<= x<= 0 ?
The extreme f(x)=x^3-x^2-x+8,-1<= x<= 0 is Minimum(-1,7),Maximum(-1/3 , 221/27),Minimum(0,8)