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

The extreme f(x)=x^3-3x+1,0<= x<= 3 is Maximum(0,1),Minimum(1,-1),Maximum(3,19)

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

extreme points

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

Maximum (0, 1), Minimum (1, - 1), Maximum (3, 19)

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

x^{3} - 3 x + 1, 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 + 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 = 3

into

x^{3} - 3 x + 1 \Rightarrow y = 19

Maximum (3, 19)

Maximum (0, 1), Minimum (1, - 1), Maximum (3, 19)

f (x) = x^{3} + 2 x^{2} - x + 8

f (x) = \frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5

\frac{x ^{2} - 1}{x ^{2} + 1}

f (x) = x 2 - x

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

What is the extreme f(x)=x^3-3x+1,0<= x<= 3 ?
The extreme f(x)=x^3-3x+1,0<= x<= 3 is Maximum(0,1),Minimum(1,-1),Maximum(3,19)