What is the extreme f(x)=(1-cos(x))^2 ?

The extreme f(x)=(1-cos(x))^2 is Minimum(2pin,0),Maximum(pi+2pin,4)

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extreme f(x)=(1-cos(x))^2

extreme points

f (x) = (1 - cos (x))^{2}

Minimum (2 πn, 0), Maximum (π + 2 πn, 4)

Solution steps

Find the critical points:

x = 2 πn, x = π + 2 πn

Domain of

(1 - cos (x))^{2} : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn < x < π + 2 πn,

Decreasing

: π + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = 2 πn

into

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

Minimum (2 πn, 0)

Plug the extreme point

x = π + 2 πn

into

(1 - cos (x))^{2} \Rightarrow y = 4

Maximum (π + 2 πn, 4)

Minimum (2 πn, 0), Maximum (π + 2 πn, 4)

f (x, y) = 400 - 49 x^{2} - 81 y^{2}

f (x) = x^{3} - 4.5 x^{2} - 12 x - 2

f (t) = 3 t + \frac{1}{t}

f (x) = \frac{9}{2} x^{2} - ln (x)

f (x) = 3 x + 1

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