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

1 Answer
Jan 2, 2016

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.

enter image source here

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