What is the extreme f(x)=2x^3-6x,0<= x<= 3 ?

The extreme f(x)=2x^3-6x,0<= x<= 3 is Maximum(0,0),Minimum(1,-4),Maximum(3,36)

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extreme f(x)=2x^3-6x,0<= x<= 3

extreme

Maximum (0, 0), Minimum (1, - 4), Maximum (3, 36)

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

2 x^{3} - 6 x, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 3

Plug the extreme point

x = 0

into

2 x^{3} - 6 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

2 x^{3} - 6 x \Rightarrow y = - 4

Minimum (1, - 4)

Plug the extreme point

x = 3

into

2 x^{3} - 6 x \Rightarrow y = 36

Maximum (3, 36)

Maximum (0, 0), Minimum (1, - 4), Maximum (3, 36)

What is the extreme f(x)=2x^3-6x,0<= x<= 3 ?
The extreme f(x)=2x^3-6x,0<= x<= 3 is Maximum(0,0),Minimum(1,-4),Maximum(3,36)