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

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

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

extreme

Maximum (0, 1), Minimum (1, 0), Maximum (3, 4)

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

x^{2} - 2 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^{2} - 2 x + 1 \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 1

into

x^{2} - 2 x + 1 \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = 3

into

x^{2} - 2 x + 1 \Rightarrow y = 4

Maximum (3, 4)

Maximum (0, 1), Minimum (1, 0), Maximum (3, 4)

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