How do you find the critical points to graph #y=sin (x/2)#?

Since the question is about the sine curve, let me put a figure of a sin(x) curve between #0# and #2pi#.

The red arrows show where the curve has zero or #x-#intercept.
The green arrows indicate where the curve got maximum,.

#sin(x)# the period is #2pi# so the graph shows one full period.

Now observe

#sin(x) = 0# at #x=0#, #x=pi# and #x=2pi#
#sin(x)# at #x=pi/2# and minimum at #x=(3pi)/2#

We can see how the curve moves from Zero, max, zero, min and zero.

Each happens at the same interval, if you see carefully it is #1/4# of the period.

Period of #sin(x)# is #2pi#
#1/4 (2pi) = pi/2#

We can see the critical points are at #0, pi/2, (3pi)/2# and #2pi#

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Let us come to our question #f(x)=sin(x/2)#

The period for #sin(Bx)# is given by the formula #(2pi)/B#

For #f(x)=sin(x/2)# the value of #B# is #1/2#

Period #=(2pi)/(1/2)#
Period =#4pi#

The interval length to find the critical points is #1/4# the period.

#1/4 (4pi) = pi#

The critical points would be at #0,pi, 2pi, 3pi# and #4pi#
The zeros would be at #0,2pi# and #4pi#
The maximum would be at #pi#
The minimum would be at #3pi#