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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 1, x = 2

Domain of

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

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 1

into

x^{2} - 2 x + 3 \Rightarrow y = 2

Minimum (1, 2)

Plug the extreme point

x = 2

into

x^{2} - 2 x + 3 \Rightarrow y = 3

Maximum (2, 3)

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

f (x, y) = (y + 204) \cdot 408 + x + 204

f (x) = x + 4 x

f (x) = - 3 sin^{2} (x), 0 \leq x \leq π

f (x, y) = x^{2} + y^{2} + 64

x^{2} + 2

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