# How do you find the critical points to graph y = -½ cos(π/3 x)?

Oct 26, 2016

Critical points are $x = \left\{\ldots \ldots . . - 9 , - 6 , - 3 , 0 , 3 , 6 , 9 , \ldots \ldots \ldots \ldots \ldots\right\}$ and

$y = - \frac{1}{2}$, when $x$ is odd and is $y = \frac{1}{2}$, when $x$ is even.

#### Explanation:

A critical point of a differentiable function is any point on the curve in its domain, where its derivative is $0$ or undefined.

Now as domain of $y = - \frac{1}{2} \cos \left(\frac{\pi}{3} x\right)$ is all real numbers and

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \times \sin \left(\frac{\pi}{3} x\right) \times \frac{\pi}{3} = - \frac{\pi}{6} \sin \left(\frac{\pi}{3} x\right)$,

it is $0$ at $\frac{\pi}{3} x = n \pi$, where $n$ is an integer.

or $x = 3 n$
graph{-1/2cos(pi/3x) [-20, 20, -2, 2]}

i.e. at $x = \left\{\ldots \ldots . . - 9 , - 6 , - 3 , 0 , 3 , 6 , 9 , \ldots \ldots \ldots \ldots \ldots\right\}$ and $y = - \frac{1}{2}$, when $x$ is odd and is $y = \frac{1}{2}$, when $x$ is even.