What is the extreme f(x)=x^3-3x+1,0<= x<= 2 ?

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

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extreme f(x)=x^3-3x+1,0<= x<= 2

extreme points

f (x) = x^{3} - 3 x + 1, 0 \leq x \leq 2

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

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

x^{3} - 3 x + 1, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 0

into

x^{3} - 3 x + 1 \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 1

into

x^{3} - 3 x + 1 \Rightarrow y = - 1

Minimum (1, - 1)

Plug the extreme point

x = 2

into

x^{3} - 3 x + 1 \Rightarrow y = 3

Maximum (2, 3)

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

f (x, y) = y^{3} + x^{3} - \frac{21}{2} y^{2} - 3 x + 30 y

f (x) = 4 x - 5

+ \frac{x + 1}{1 + \frac{1}{x + 1}}

f (x) = 5 x^{3} e^{- x}

f (x) = 2 x^{3} + 30 x^{2} + 144 x - 1

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