Question

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

Answer 1

Critical points are `x={........-9,-6,-3,0,3,6,9,...............}` and

`y=-1/2`, when `x` is odd and is `y=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=-1/2cos(pi/3x)` is all real numbers and

`(dy)/(dx)=-1/2xxsin(pi/3x)xxpi/3=-pi/6sin(pi/3x)`,

it is `0` at `pi/3x=npi`, where `n` is an integer.

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

i.e. at `x={........-9,-6,-3,0,3,6,9,...............}` and `y=-1/2`, when `x` is odd and is `y=1/2`, when `x` is even.