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

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Answer 1 · 2016-10-26T10:15:12

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.

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.

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