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

The extreme f(x)=(x^2-1)^3,-1<= x<= 3 is Maximum(-1,0),Minimum(0,-1),Saddle(1,0),Maximum(3,512)

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

extreme points

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

Maximum (- 1, 0), Minimum (0, - 1), Saddle (1, 0), Maximum (3, 512)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1, x = 3

Domain of

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

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 3

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 = 3

into

(x^{2} - 1)^{3} \Rightarrow y = 512

Maximum (3, 512)

Maximum (- 1, 0), Minimum (0, - 1), Saddle (1, 0), Maximum (3, 512)

g (t) = e^{a t}

f (x) = \frac{8 x}{x ^{2} + 4}

- 3 x^{5} + 5 x^{3} + 1

p (x) = 3 x^{3} - 10 x^{2} + a x - 2

f (x) = \frac{x ^{3} - 4}{x - 1}

What is the extreme f(x)=(x^2-1)^3,-1<= x<= 3 ?
The extreme f(x)=(x^2-1)^3,-1<= x<= 3 is Maximum(-1,0),Minimum(0,-1),Saddle(1,0),Maximum(3,512)