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

The extreme f(x)=sin(x/2) is Maximum(pi+4pin,1),Minimum(3pi+4pin,-1)

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

extreme points

f (x) = sin (\frac{x}{2})

Maximum (π + 4 πn, 1), Minimum (3 π + 4 πn, - 1)

Solution steps

Find the critical points:

x = π + 4 πn, x = 3 π + 4 πn

Domain of

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

Find intervals:

Increasing

: 4 πn \leq x < π + 4 πn,

Decreasing

: π + 4 πn < x < 3 π + 4 πn,

Increasing

: 3 π + 4 πn < x < 4 π + 4 πn

Plug the extreme point

x = π + 4 πn

into

sin (\frac{x}{2}) \Rightarrow y = 1

Maximum (π + 4 πn, 1)

Plug the extreme point

x = 3 π + 4 πn

into

sin (\frac{x}{2}) \Rightarrow y = - 1

Minimum (3 π + 4 πn, - 1)

Maximum (π + 4 πn, 1), Minimum (3 π + 4 πn, - 1)

f (x, y) = x^{3} + y^{3} - 3 x - 3 y + 20

f (x, y) = x^{3} + y^{2} - 4 x y + 17 x - 10 y + 8

f (x) = 2 x^{3} + 45 x^{2} + 84 x + 29

x^{3} - 12 x^{2} + 36 x

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

What is the extreme f(x)=sin(x/2) ?
The extreme f(x)=sin(x/2) is Maximum(pi+4pin,1),Minimum(3pi+4pin,-1)