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

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 $2 \pi$.

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

$\sin \left(x\right)$ the period is $2 \pi$ so the graph shows one full period.

Now observe

$\sin \left(x\right) = 0$ at $x = 0$, $x = \pi$ and $x = 2 \pi$
$\sin \left(x\right)$ at $x = \frac{\pi}{2}$ and minimum at $x = \frac{3 \pi}{2}$

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 $\frac{1}{4}$ of the period.

Period of $\sin \left(x\right)$ is $2 \pi$
$\frac{1}{4} \left(2 \pi\right) = \frac{\pi}{2}$

We can see the critical points are at $0 , \frac{\pi}{2} , \frac{3 \pi}{2}$ and $2 \pi$

-

Let us come to our question $f \left(x\right) = \sin \left(\frac{x}{2}\right)$

The period for $\sin \left(B x\right)$ is given by the formula $\frac{2 \pi}{B}$

For $f \left(x\right) = \sin \left(\frac{x}{2}\right)$ the value of $B$ is $\frac{1}{2}$

Period $= \frac{2 \pi}{\frac{1}{2}}$
Period =$4 \pi$

The interval length to find the critical points is $\frac{1}{4}$ the period.

$\frac{1}{4} \left(4 \pi\right) = \pi$

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