Recall that critical points are simply where the derivative is zero and/or doesn’t exist.

This derivative exists everywhere and so we don’t need to worry about that. Therefore, all we need to do is determine where the derivative is zero. So, all we need to do is solve the equation,

\[\frac{1}{3}\cos \left( \frac{y}{3} \right) + \frac{2}{9} = 0\hspace{0.25in} \to \hspace{0.25in}\cos \left(\frac{y}{3} \right) = - \frac{2}{3}\hspace{0.25in} \to \hspace{0.25in}\frac{y}{3} = {\cos ^{ - 1}}\left( { - \frac{2}{3}} \right) = 2.3005\]

This is the answer we got from a calculator and a quick look at a unit circle gives us a second solution of either -2.3005 or if you want the positive equivalent we could use \(2\pi - 2.3005 = 3.9827\). For this problem we’ll use the positive one, although the negative one could just as easily be used if you wanted to.

All possible solutions to \(\cos \left(\frac{y}{3} \right) = - \frac{2}{3}\) are then,

\[\begin{array}{*{20}{c}}{\displaystyle \frac{y}{3} = 2.3005 + 2\pi n}\\{\displaystyle \frac{y}{3} = 3.9827 + 2\pi n}\end{array}\hspace{0.5in}\,n = 0, \pm 1, \pm 2, \pm , \ldots \]

Finally solving for \(y\) gives all the critical points of the function.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\begin{array}{*{20}{c}}{y = 6.9015 + 6\pi n}\\{y = 11.9481 + 6\pi n}\end{array}\hspace{0.5in}\,n = 0, \pm 1, \pm 2, \pm , \ldots }}\]

If you don’t remember how to solve trig equations you should go back and review those sections in the Review Chapter of the notes.