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

The extreme f(x)=2x^3-x^2-4x+2,-1<= x<= 0 is Minimum(-1,3),Maximum(-2/3 , 98/27),Minimum(0,2)

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extreme f(x)=2x^3-x^2-4x+2,-1<= x<= 0

extreme points

f (x) = 2 x^{3} - x^{2} - 4 x + 2, - 1 \leq x \leq 0

Minimum (- 1, 3), Maximum (- \frac{2}{3}, \frac{98}{27}), Minimum (0, 2)

Solution steps

Find the critical points:

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

Domain of

2 x^{3} - x^{2} - 4 x + 2, - 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

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

Minimum (- 1, 3)

Plug the extreme point

x = - \frac{2}{3}

into

2 x^{3} - x^{2} - 4 x + 2 \Rightarrow y = \frac{98}{27}

Maximum (- \frac{2}{3}, \frac{98}{27})

Plug the extreme point

x = 0

into

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

Minimum (0, 2)

Minimum (- 1, 3), Maximum (- \frac{2}{3}, \frac{98}{27}), Minimum (0, 2)

(x - 1)^{3}

f (x) = - x^{2} + 12, - 3 \leq x \leq 4

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 2 x + 10 y

2 x^{3} - 24 x + 1

f (x) = x^{7} e^{2 x} + 2

What is the extreme f(x)=2x^3-x^2-4x+2,-1<= x<= 0 ?
The extreme f(x)=2x^3-x^2-4x+2,-1<= x<= 0 is Minimum(-1,3),Maximum(-2/3 , 98/27),Minimum(0,2)