Chapter 2.1, Problem 61E

(a)

To determine

To graph: The piecewise function $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ .

Graph:

To graph $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{SIN}}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{(-)}\boxed{2}\boxed{2\text{nd}}\boxed{^}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{\Rightarrow }\boxed{1}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{5}\boxed{0}\boxed{)}$ . Now press Enter after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{\text{COS}}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{0}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{\Rightarrow }\boxed{1}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{6}\boxed{2}\boxed{^}\boxed{)}$ .

The display will show the equation,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\sin \left(\text{X}\right)/\left(\text{X}\ge -2\pi \text{and X}<0\right)\\ {\text{Y}}_2=\cos \left(\text{X}\right)/\left(\text{X}\ge 0\text{and}\text{X}\le 2\pi \right)\end{array}$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

Interpretation: The graph of the piecewise function $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ is the combination of curves $y=\sin x$ defined for $-2\pi \le x<0$ and curve $y=\cos x$ defined for $0\le x\le 2\pi$ .

(b)

To determine

To find: The points $c$ in the domain of $f\left(x\right)$ such that $\underset{x\to c}{\lim }f\left(x\right)$ exist.

Answer to Problem 61E

The limit $\underset{x\to c}{\lim }f\left(x\right)$ exist for every $c$ in the interval $\left(-2\pi ,0\right)\cup \left(0,2\pi \right)$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ .

Calculation:

The function $\sin x$ corresponds for all the values of $x$ in $\left(-2\pi ,0\right)$ . So, the limit $\underset{x\to c}{\lim }f\left(x\right)$ exists for every $x$ in the interval $\left(-2\pi ,0\right)$ .

The function $\cos x$ corresponds for all the values of $x$ in $\left(0,2\pi \right)$ . So, the limit $\underset{x\to c}{\lim }f\left(x\right)$ exists for every $x$ in the interval $\left(0,2\pi \right)$ .

Check the limit of function $f\left(x\right)$ at $x=0$ .

The corresponding function for left hand limit of $f\left(x\right)$ at $x=0$ is $\sin x$ .

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\sin x\\ =\sin 0\\ =0\end{array}$

The corresponding function for right hand limit of $f\left(x\right)$ at $x=0$ is $\cos x$ .

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\cos x\\ =\cos 0\\ =1\end{array}$

As the left hand limit and right hand limit for $f\left(x\right)$ at $x=0$ are not equal. So, the limit of $f\left(x\right)$ at $x=0$ does not exist.

Therefore, the limit $\underset{x\to c}{\lim }f\left(x\right)$ exist for every $c$ in the interval $\left(-2\pi ,0\right)\cup \left(0,2\pi \right)$ .

(c)

To determine

To find: The points $c$ in the domain of $f\left(x\right)$ such that only left hand limit exist at $x=c$ .

Answer to Problem 61E

The points $c$ in the domain of $f\left(x\right)$ such that only left hand limit exist at $x=c$ is $2\pi$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ .

Calculation:

As observed from the graph in part (a), the function is defined for $x=2\pi$ from the left side but not from the right side of $x=2\pi$ .

So, only left hand limit of function $f\left(x\right)$ exist for the point $x=2\pi$ .

Therefore, the points $c$ in the domain of $f\left(x\right)$ such that only left hand limit exist at $x=c$ is $2\pi$ .

(d)

To determine

To find: The points $c$ in the domain of $f\left(x\right)$ such that only right hand limit exist at $x=c$ .

Answer to Problem 61E

The points $c$ in the domain of $f\left(x\right)$ such that only right hand limit exist at $x=c$ is $-2\pi$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\{\begin{array}{l}\sin x,-2\pi \le x<0\\ \cos x,0\le x\le 2\pi \end{array}$ .

Calculation:

As observed from the graph in part (a), the function is defined for $x=-2\pi$ from the right side but not the left side of $x=-2\pi$ .

So, only right hand limit of function $f\left(x\right)$ exist for the point $x=-2\pi$ .

Therefore, the points $c$ in the domain of $f\left(x\right)$ such that only right hand limit exist at $x=c$ is $-2\pi$ .