What is the extreme f(x)=x^3-6x^2+9x,0<= x<= 2 ?

The extreme f(x)=x^3-6x^2+9x,0<= x<= 2 is Minimum(0,0),Maximum(1,4),Minimum(2,2)

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extreme f(x)=x^3-6x^2+9x,0<= x<= 2

extreme points

f (x) = x^{3} - 6 x^{2} + 9 x, 0 \leq x \leq 2

Minimum (0, 0), Maximum (1, 4), Minimum (2, 2)

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

x^{3} - 6 x^{2} + 9 x, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2

Plug the extreme point

x = 0

into

x^{3} - 6 x^{2} + 9 x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

x^{3} - 6 x^{2} + 9 x \Rightarrow y = 4

Maximum (1, 4)

Plug the extreme point

x = 2

into

x^{3} - 6 x^{2} + 9 x \Rightarrow y = 2

Minimum (2, 2)

Minimum (0, 0), Maximum (1, 4), Minimum (2, 2)

f (x) = (x - 2) (x + 1)^{3}

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

f (x) = 6 x^{2} + 7 y^{2}

f (x) = - 5 x^{2} + 10 x + 12

14400 + 700 x + x^{2}

What is the extreme f(x)=x^3-6x^2+9x,0<= x<= 2 ?
The extreme f(x)=x^3-6x^2+9x,0<= x<= 2 is Minimum(0,0),Maximum(1,4),Minimum(2,2)