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

The extreme f(x)=sqrt(x^2-2x+2),-2<= x<= 2 is Maximum(-2,sqrt(10)),Minimum(1,1),Maximum(2,sqrt(2))

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extreme f(x)=sqrt(x^2-2x+2),-2<= x<= 2

extreme

Maximum (- 2, 10), Minimum (1, 1), Maximum (2, 2)

Solution steps

Find the critical points:

x = - 2, x = 1, x = 2

Domain of

x^{2} - 2 x + 2, - 2 \leq x \leq 2 : - 2 \leq x \leq 2

Find intervals:

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = - 2

into

x^{2} - 2 x + 2 \Rightarrow y = 10

Maximum (- 2, 10)

Plug the extreme point

x = 1

into

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

Minimum (1, 1)

Plug the extreme point

x = 2

into

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

Maximum (2, 2)

Maximum (- 2, 10), Minimum (1, 1), Maximum (2, 2)

What is the extreme f(x)=sqrt(x^2-2x+2),-2<= x<= 2 ?
The extreme f(x)=sqrt(x^2-2x+2),-2<= x<= 2 is Maximum(-2,sqrt(10)),Minimum(1,1),Maximum(2,sqrt(2))