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Section 4.2 : Critical Points

11. Determine the critical points of \(s\left( z \right) = 4\cos \left( z \right) - z\).

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We’ll need the first derivative to get the answer to this problem so let’s get that.

\[s'\left( z \right) = - 4\sin \left( z \right) - 1\] Show Step 2

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,

\[ - 4\sin \left( z \right) - 1 = 0 \hspace{0.25in} \to \hspace{0.25in} \sin \left( z \right) = - {\textstyle{1 \over 4}} \hspace{0.25in} \to \hspace{0.25in} z = {\sin ^{ - 1}}\left( { - {\textstyle{1 \over 4}}} \right) = - 0.2527\]

This is the answer we got from a calculator and we could use this or we could use the equivalent positive angle : \(2\pi - 0.2527 = 6.0305\). Either can be used, but we’ll use the positive one for this problem.

Now, a quick look at a unit circle gives us a second solution of \(\pi + 0.2527 = 3.3943\).

Finally, all possible solutions to this equation, and hence, all the critical points of the original function are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\begin{array}{*{20}{c}}{z = 6.0305 + 2\pi n}\\{z = 3.3943 + 2\pi n}\end{array}\,\,\,\,\,\,\,\,n = 0, \pm 1, \pm 2, \pm , \ldots }}\]

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.