Chapter 2.1, Problem 62E

(a)

To determine

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

Explanation of Solution

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

Graph:

To graph $f\left(x\right)=\{\begin{array}{l}\cos x,-\pi \le x<0\\ \sec x,0\le x\le \pi \end{array}$ , 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 $f\left(x\right)=\{\begin{array}{l}\cos x,-\pi \le x<0\\ \sec x,0\le x\le \pi \end{array}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{COS}}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{(-)}\boxed{2\text{nd}}\boxed{^}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{\Rightarrow }\boxed{1}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{5}\boxed{0}\boxed{)}$ . Now, press Enter and write after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(}\boxed{1}\boxed{÷}\boxed{(}\boxed{\text{COS}}\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{0}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{\Rightarrow }\boxed{1}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{6}\boxed{2\text{nd}}\boxed{^}\boxed{)}$ .

The display will show the equations,

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

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 $f\left(x\right)=\{\begin{array}{l}\cos x,-\pi \le x<0\\ \sec x,0\le x\le \pi \end{array}$ is the combination of curves $y=\cos x$ defined for $-\pi \le x<0$ and curve $y=\sec x$ defined for $0\le x\le \pi$ .

(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 62E

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

Explanation of Solution

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

Calculation:

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

The function $\sec x$ corresponds for all the values of $x$ in $\left(0,\pi \right)$ , but the value of the function $\sec x$ does not exist for $x=\frac{\pi }{2}$ . So, the limit $\underset{x\to c}{\lim }f\left(x\right)$ exists for every $x$ in the interval $\left(0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right)$ .

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

(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 62E

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

Explanation of Solution

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

Calculation:

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

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

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

(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 62E

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

Explanation of Solution

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

Calculation:

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

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

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