What is the extreme f(x)=8+81x-3x^3,0<= x<= 4 ?

The extreme f(x)=8+81x-3x^3,0<= x<= 4 is Minimum(0,8),Maximum(3,170),Minimum(4,140)

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extreme f(x)=8+81x-3x^3,0<= x<= 4

extreme

Minimum (0, 8), Maximum (3, 170), Minimum (4, 140)

Solution steps

Find the critical points:

x = 0, x = 3, x = 4

Domain of

8 + 81 x - 3 x^{3}, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < 4

Plug the extreme point

x = 0

into

8 + 81 x - 3 x^{3} \Rightarrow y = 8

Minimum (0, 8)

Plug the extreme point

x = 3

into

8 + 81 x - 3 x^{3} \Rightarrow y = 170

Maximum (3, 170)

Plug the extreme point

x = 4

into

8 + 81 x - 3 x^{3} \Rightarrow y = 140

Minimum (4, 140)

Minimum (0, 8), Maximum (3, 170), Minimum (4, 140)

What is the extreme f(x)=8+81x-3x^3,0<= x<= 4 ?
The extreme f(x)=8+81x-3x^3,0<= x<= 4 is Minimum(0,8),Maximum(3,170),Minimum(4,140)