What is the extreme x^3-3x+1,-3/2 <= x<= 3 ?

The extreme x^3-3x+1,-3/2 <= x<= 3 is Minimum(-3/2 , 17/8),Maximum(-1,3),Minimum(1,-1),Maximum(3,19)

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extreme x^3-3x+1,-3/2 <= x<= 3

extreme points

x^{3} - 3 x + 1, - \frac{3}{2} \leq x \leq 3

Minimum (- \frac{3}{2}, \frac{17}{8}), Maximum (- 1, 3), Minimum (1, - 1), Maximum (3, 19)

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

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

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < 3

Plug the extreme point

x = - \frac{3}{2}

into

x^{3} - 3 x + 1 \Rightarrow y = \frac{17}{8}

Minimum (- \frac{3}{2}, \frac{17}{8})

Plug the extreme point

x = - 1

into

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

Maximum (- 1, 3)

Plug the extreme point

x = 1

into

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

Minimum (1, - 1)

Plug the extreme point

x = 3

into

x^{3} - 3 x + 1 \Rightarrow y = 19

Maximum (3, 19)

Minimum (- \frac{3}{2}, \frac{17}{8}), Maximum (- 1, 3), Minimum (1, - 1), Maximum (3, 19)

F (X, Y) = X (X + Y) + X (X + Y)

- x^{2} - 3.7 x - 6.5

f (x) = - 2 x^{3} + 33 x^{2} - 168 x + 10

f (x) = 4 x^{3} - 6 x^{2} - 144 x + 1

f (x, y) = \frac{20480}{x} + \frac{20480}{y} + 5 x y

What is the extreme x^3-3x+1,-3/2 <= x<= 3 ?
The extreme x^3-3x+1,-3/2 <= x<= 3 is Minimum(-3/2 , 17/8),Maximum(-1,3),Minimum(1,-1),Maximum(3,19)