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

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

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

extreme points

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

Minimum (1, 0), Maximum (2, 9)

Solution steps

Find the critical points:

x = 1, x = 2

Domain of

(3 x - 3)^{2}, - \infty < x \leq 2 : x \leq 2

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 1

into

(3 x - 3)^{2} \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = 2

into

(3 x - 3)^{2} \Rightarrow y = 9

Maximum (2, 9)

Minimum (1, 0), Maximum (2, 9)

f (x, y, z) = x ln (y)

f (x, y) = 200 - x - y

f (x) = x^{2} + 49

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

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

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