Question

find points on the curve ( y = 1cos^{2}x ) that have a horizontal tangent.
select the correct answer below and, if necessary, fill in the answer box(es) to complete your choice.
○ a. there are a finite number of points that have a horizontal tangent line.
the point(s) is/are
(type an ordered pair. use a comma to separate answers as needed. type an exact answer using ( pi ) as needed.)
○ b. there are an infinite number of points that have a horizontal tangent line.
these points have the form(s)
, for any integer ( k ).
(type an ordered pair. use a comma to separate answers as needed. type an exact answer using ( pi ) as needed.)
○ c. there are no points that have a horizontal tangent line.

Explanation:

Step1: Differentiate the function

We know that if \(y = \cos^{2}x\), then using the chain - rule \((u^{2})^\prime=2u\cdot u^\prime\) where \(u = \cos x\) and \(u^\prime=-\sin x\). So \(y^\prime=2\cos x(-\sin x)=-\sin(2x)\) (since \(\sin(2x) = 2\sin x\cos x\)).

Step2: Set the derivative equal to zero

A horizontal tangent occurs when \(y^\prime = 0\). So we set \(-\sin(2x)=0\). Then \(\sin(2x)=0\).
We know that \(\sin\theta=0\) when \(\theta = k\pi\), \(k\in\mathbb{Z}\). Let \(\theta = 2x\), then \(2x=k\pi\), \(x=\frac{k\pi}{2}\), \(k\in\mathbb{Z}\).

Step3: Find the \(y\) - values

When \(x = \frac{k\pi}{2}\), \(y=\cos^{2}(\frac{k\pi}{2})\).
If \(k = 2n\) (even, \(n\in\mathbb{Z}\)), \(x=n\pi\) and \(y=\cos^{2}(n\pi)=1\).
If \(k = 2n + 1\) (odd, \(n\in\mathbb{Z}\)), \(x=(n+\frac{1}{2})\pi\) and \(y=\cos^{2}((n + \frac{1}{2})\pi)=0\).

Answer:

B. There are an infinite number of points that have a horizontal tangent line. These points have the form(s) \((\frac{k\pi}{2},0)\) and \((\frac{k\pi}{2},1)\) for any integer \(k\).