What is the extreme f(x)= x/(sqrt(2))-2sin(x/2) ?

The extreme f(x)= x/(sqrt(2))-2sin(x/2) is

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extreme f(x)= x/(sqrt(2))-2sin(x/2)

extreme

Minimum (\frac{π}{2} + 4 πn, \frac{π + 8 πn}{2 2} - 2 sin (\frac{π + 8 πn}{4})), Maximum (\frac{7 π}{2} + 4 πn, \frac{7 π + 8 πn}{2 2} - 2 sin (\frac{7 π + 8 πn}{4}))

Solution steps

Find the critical points:

x = \frac{π}{2} + 4 πn, x = \frac{7 π}{2} + 4 πn

Domain of

\frac{x}{2} - 2 sin (\frac{x}{2}) : - \infty < x < \infty

Find intervals:

Decreasing

: 4 πn \leq x < \frac{π}{2} + 4 πn,

Increasing

: \frac{π}{2} + 4 πn < x < \frac{7 π}{2} + 4 πn,

Decreasing

: \frac{7 π}{2} + 4 πn < x < 4 π + 4 πn

Plug the extreme point

x = \frac{π}{2} + 4 πn

into

\frac{x}{2} - 2 sin (\frac{x}{2}) \Rightarrow y = \frac{π + 8 πn}{2 2} - 2 sin (\frac{π + 8 πn}{4})

Minimum (\frac{π}{2} + 4 πn, \frac{π + 8 πn}{2 2} - 2 sin (\frac{π + 8 πn}{4}))

Plug the extreme point

x = \frac{7 π}{2} + 4 πn

into

\frac{x}{2} - 2 sin (\frac{x}{2}) \Rightarrow y = \frac{7 π + 8 πn}{2 2} - 2 sin (\frac{7 π + 8 πn}{4})

Maximum (\frac{7 π}{2} + 4 πn, \frac{7 π + 8 πn}{2 2} - 2 sin (\frac{7 π + 8 πn}{4}))

Minimum (\frac{π}{2} + 4 πn, \frac{π + 8 πn}{2 2} - 2 sin (\frac{π + 8 πn}{4})), Maximum (\frac{7 π}{2} + 4 πn, \frac{7 π + 8 πn}{2 2} - 2 sin (\frac{7 π + 8 πn}{4}))

What is the extreme f(x)= x/(sqrt(2))-2sin(x/2) ?
The extreme f(x)= x/(sqrt(2))-2sin(x/2) is