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

The extreme f(x)=x^3-3x,0<= x<= 3 is Maximum(0,0),Minimum(1,-2),Maximum(3,18)

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

extreme points

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

Maximum (0, 0), Minimum (1, - 2), Maximum (3, 18)

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

x^{3} - 3 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

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 = 3

into

x^{3} - 3 x \Rightarrow y = 18

Maximum (3, 18)

Maximum (0, 0), Minimum (1, - 2), Maximum (3, 18)

f (x, y) = x^{2} + y^{2} - 81

f (x, y) = 2 x^{2} + y^{2} + y + 1

f (x) = 3 sin (2 x)

f (x, y) = 11 x^{2} + 4 x y + 5 y^{2} + 5 x + 5 y

f (x) = ln (\frac{1}{x ^{2} + 2 x + 10})

What is the extreme f(x)=x^3-3x,0<= x<= 3 ?
The extreme f(x)=x^3-3x,0<= x<= 3 is Maximum(0,0),Minimum(1,-2),Maximum(3,18)