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

The extreme f(x)=(x^2-1)^3,-1<= x<= 2 is Maximum(-1,0),Minimum(0,-1),Saddle(1,0),Maximum(2,27)

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

extreme points

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

Maximum (- 1, 0), Minimum (0, - 1), Saddle (1, 0), Maximum (2, 27)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1, x = 2

Domain of

(x^{2} - 1)^{3}, - 1 \leq x \leq 2 : - 1 \leq x \leq 2

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = - 1

into

(x^{2} - 1)^{3} \Rightarrow y = 0

Maximum (- 1, 0)

Plug the extreme point

x = 0

into

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

Minimum (0, - 1)

Plug the extreme point

x = 1

into

(x^{2} - 1)^{3} \Rightarrow y = 0

Saddle (1, 0)

Plug the extreme point

x = 2

into

(x^{2} - 1)^{3} \Rightarrow y = 27

Maximum (2, 27)

Maximum (- 1, 0), Minimum (0, - 1), Saddle (1, 0), Maximum (2, 27)

f (x) = x (1 - x^{2})^{2}

f (x) = x^{3} - 12 x^{2} - 27 x - 26

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

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

f (x) = 2 x^{3} - 9 x^{2} + 2

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