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.