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

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

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

extreme points

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

Maximum (- 1, 4), Minimum (1, 0)

Solution steps

Find the critical points:

x = - 1, x = 1

Domain of

x^{3} - 3 x + 2, (- 2, 2) : - 2 < x < 2

Find intervals:

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = - 1

into

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

Maximum (- 1, 4)

Plug the extreme point

x = 1

into

x^{3} - 3 x + 2 \Rightarrow y = 0

Minimum (1, 0)

Maximum (- 1, 4), Minimum (1, 0)

f (x) = 288 y^{2} + x^{2} - x^{2} y

f (x, y) = 5 x^{2} - x y + 3 y^{2}

p (b, c) = 2 b + 2 c

f (x) = ∣ x - 5 ∣ + 1

g (x) = x 8 - x^{2}

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