Chapter 3, Problem 77RE

To determine

To graph : The function $y=\sin \left(x-\sin x\right)$ and verify that if this has horizontal tangents at the $x-\text{axis}$

Explanation of Solution

Given information

  $y=\sin \left(x-\sin x\right)$

Graph:

Graph of $y=\sin \left(x-\sin x\right)$ is below:

  

Interpretation :

Substitute $0$ for $y$ :

  $\begin{array}{l}\sin \left(x-\sin x\right)=0\\ x-\sin \left(x\right)=k\pi ,\text{where},k=-2,-1,0,1,2\end{array}$

Differentiate $y=\sin \left(x-\sin x\right)$ with respect to $x$ :

  $\begin{array}{c}\frac{d}{dx}\left[y\right]=\frac{d}{dx}\left[\sin \left(x-\sin x\right)\right]\\ =\cos \left(x-\sin x\right)\cdot \frac{d}{dx}\left[x-\sin x\right]\\ =\cos \left(x-\sin x\right)\left(1-\cos x\right)\end{array}$

To find the value of the function, substitute, $k\pi \text{for}\left(x-\sin x\right)$

  $\begin{array}{c}\cos \left(x-\sin x\right)=\cos \left(k\pi \right)\\ =\pm 1\end{array}$

The value of $y\text{and}\frac{dy}{dx}$ are $0$ when $1-\cos x=0\text{and}x=k\pi$

In the graph, the curve lies in the interval $-2\pi \le x\le 2\pi$ .

Also, the value of:

  $\begin{array}{c}\cos \left(-\pi \right)=\cos \pi \\ =-1\end{array}$

The given equation is valid for $k=-2,0\text{and}2.$

Therefore, the curve contains horizontal tangent lines along $x-\text{axis}$ for the values of $x=-2\pi ,0\text{and}2\pi$ .

Thus, the curve contains infinite numbers of horizontal lines.