Question

find points on the curve ( y = 1cos^{4}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

Use the chain rule. If \(y = \cos^{4}x\), let \(u=\cos x\), then \(y = u^{4}\).
The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=4u^{3}\), and the derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx}=-\sin x\).
By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), so \(\frac{dy}{dx}=4\cos^{3}x(-\sin x)=- 4\cos^{3}x\sin x\).

Step2: Find when the derivative is zero

A horizontal tangent occurs when \(\frac{dy}{dx} = 0\).
Set \(-4\cos^{3}x\sin x=0\).
This equation is satisfied when \(\cos x = 0\) or \(\sin x=0\).
If \(\cos x=0\), then \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\). When \(x = \frac{\pi}{2}+k\pi\), \(y=\cos^{4}(\frac{\pi}{2}+k\pi)=0\).
If \(\sin x = 0\), then \(x = k\pi,k\in\mathbb{Z}\). When \(x=k\pi\), \(y=\cos^{4}(k\pi)=1\).

Answer:

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