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 points

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

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)

f (x, y) = 2 x^{2} - 4 x + y^{2} - 4 y + 1

3 x^{2} - 24 ln (x)

f (x) = \frac{4 x}{x ^{2} + 1}, - 4 \leq x \leq 0

f (x) = ln (27 + x^{3})

f (x) = x^{3} - 9 x^{2} + 27 x - 3

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)