What is the extreme f(x)=(x^2-1)^3,-1<= x<= 6 ?

The extreme f(x)=(x^2-1)^3,-1<= x<= 6 is Maximum(-1,0),Minimum(0,-1),Saddle(1,0),Maximum(6,42875)

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extreme f(x)=(x^2-1)^3,-1<= x<= 6

extreme

Maximum (- 1, 0), Minimum (0, - 1), Saddle (1, 0), Maximum (6, 42875)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1, x = 6

Domain of

(x^{2} - 1)^{3}, - 1 \leq x \leq 6 : - 1 \leq x \leq 6

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 6

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 = 6

into

(x^{2} - 1)^{3} \Rightarrow y = 42875

Maximum (6, 42875)

Maximum (- 1, 0), Minimum (0, - 1), Saddle (1, 0), Maximum (6, 42875)

What is the extreme f(x)=(x^2-1)^3,-1<= x<= 6 ?
The extreme f(x)=(x^2-1)^3,-1<= x<= 6 is Maximum(-1,0),Minimum(0,-1),Saddle(1,0),Maximum(6,42875)