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

The extreme f(x)=x^3-3x,0<= x<= 4 is Maximum(0,0),Minimum(1,-2),Maximum(4,52)

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

extreme points

f (x) = x^{3} - 3 x, 0 \leq x \leq 4

Maximum (0, 0), Minimum (1, - 2), Maximum (4, 52)

Solution steps

Find the critical points:

x = 0, x = 1, x = 4

Domain of

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

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 4

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = 1

into

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

Minimum (1, - 2)

Plug the extreme point

x = 4

into

x^{3} - 3 x \Rightarrow y = 52

Maximum (4, 52)

Maximum (0, 0), Minimum (1, - 2), Maximum (4, 52)

What is the extreme f(x)=x^3-3x,0<= x<= 4 ?
The extreme f(x)=x^3-3x,0<= x<= 4 is Maximum(0,0),Minimum(1,-2),Maximum(4,52)