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

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

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

extreme points

f (x) = x^{2} - 4 x + 1, 0 \leq x \leq 3

Maximum (0, 1), Minimum (2, - 3), Maximum (3, - 2)

Solution steps

Find the critical points:

x = 0, x = 2, x = 3

Domain of

x^{2} - 4 x + 1, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = 0

into

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

Maximum (0, 1)

Plug the extreme point

x = 2

into

x^{2} - 4 x + 1 \Rightarrow y = - 3

Minimum (2, - 3)

Plug the extreme point

x = 3

into

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

Maximum (3, - 2)

Maximum (0, 1), Minimum (2, - 3), Maximum (3, - 2)

x - 3

f (x, y) = 3 y^{2} + x^{3} - 6 x y - 9 x

f (x) = 5 sin (x) cos (x)

f (x, y) = x^{4} + y^{2} - 2 x^{2} - 5

f (x) = x + \frac{3}{x}

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