What is the extreme sin^2(x)-sqrt(2)*sin(x)+2sqrt(2) ?

The extreme sin^2(x)-sqrt(2)*sin(x)+2sqrt(2) is

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extreme sin^2(x)-sqrt(2)*sin(x)+2sqrt(2)

extreme points

sin^{2} (x) - 2 \cdot sin (x) + 22

Minimum (\frac{π}{4} + 2 πn, 22 - \frac{1}{2}), Maximum (\frac{π}{2} + 2 πn, 1 + 2), Minimum (\frac{3 π}{4} + 2 πn, 22 - \frac{1}{2}), Maximum (\frac{3 π}{2} + 2 πn, 1 + 32)

Solution steps

Find the critical points:

x = \frac{π}{4} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{3 π}{2} + 2 πn

Domain of

sin^{2} (x) - 2 \cdot sin (x) + 22 : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{4} + 2 πn,

Increasing

: \frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn,

Decreasing

: \frac{π}{2} + 2 πn < x < \frac{3 π}{4} + 2 πn,

Increasing

: \frac{3 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn,

Decreasing

: \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{4} + 2 πn

into

sin^{2} (x) - 2 \cdot sin (x) + 22 \Rightarrow y = 22 - \frac{1}{2}

Minimum (\frac{π}{4} + 2 πn, 22 - \frac{1}{2})

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

sin^{2} (x) - 2 \cdot sin (x) + 22 \Rightarrow y = 1 + 2

Maximum (\frac{π}{2} + 2 πn, 1 + 2)

Plug the extreme point

x = \frac{3 π}{4} + 2 πn

into

sin^{2} (x) - 2 \cdot sin (x) + 22 \Rightarrow y = 22 - \frac{1}{2}

Minimum (\frac{3 π}{4} + 2 πn, 22 - \frac{1}{2})

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

sin^{2} (x) - 2 \cdot sin (x) + 22 \Rightarrow y = 1 + 32

Maximum (\frac{3 π}{2} + 2 πn, 1 + 32)

Minimum (\frac{π}{4} + 2 πn, 22 - \frac{1}{2}), Maximum (\frac{π}{2} + 2 πn, 1 + 2), Minimum (\frac{3 π}{4} + 2 πn, 22 - \frac{1}{2}), Maximum (\frac{3 π}{2} + 2 πn, 1 + 32)

f (x) = - x + 5

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

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

f (x, y) = x^{2} + y^{2} - 2 x + 6 y + 10

f (x) = 18 x (x - 1)^{3}

What is the extreme sin^2(x)-sqrt(2)*sin(x)+2sqrt(2) ?
The extreme sin^2(x)-sqrt(2)*sin(x)+2sqrt(2) is