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

The extreme f(x)=3x^3-3x^2-3x+2,-1<= x<= 0 is Minimum(-1,-1),Maximum(-1/3 , 23/9),Minimum(0,2)

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

extreme points

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

Minimum (- 1, - 1), Maximum (- \frac{1}{3}, \frac{23}{9}), Minimum (0, 2)

Solution steps

Find the critical points:

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

Domain of

3 x^{3} - 3 x^{2} - 3 x + 2, - 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

3 x^{3} - 3 x^{2} - 3 x + 2 \Rightarrow y = - 1

Minimum (- 1, - 1)

Plug the extreme point

x = - \frac{1}{3}

into

3 x^{3} - 3 x^{2} - 3 x + 2 \Rightarrow y = \frac{23}{9}

Maximum (- \frac{1}{3}, \frac{23}{9})

Plug the extreme point

x = 0

into

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

Minimum (0, 2)

Minimum (- 1, - 1), Maximum (- \frac{1}{3}, \frac{23}{9}), Minimum (0, 2)

f (x, y) = x^{2} + y^{2} - 4 x + 17 x - 10 y + 2021

f (x) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

f (x, y) = - 3 x y + x^{3} + y^{3}

f (x) = 16 - x^{2}, - 4 \leq x \leq 4

f (x) = 6 + 3 x - 3 x^{2}

What is the extreme f(x)=3x^3-3x^2-3x+2,-1<= x<= 0 ?
The extreme f(x)=3x^3-3x^2-3x+2,-1<= x<= 0 is Minimum(-1,-1),Maximum(-1/3 , 23/9),Minimum(0,2)