Question

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

Answer 1

Critical points for graphing occurs where the curve is maximum, minimum or has zeros. Let us see a trick to find them.

Explanation:

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`