Chapter 2.1, Problem 58E

(a)

To determine

To graph: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ .

Explanation of Solution

Given information: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ .

Graph:

To graph a function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ , 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 after the symbol ${\text{Y}}_1$ . Enter the keystrokes as given below:

  $\begin{array}{l}\boxed{(}\boxed{3}\boxed{-}\boxed{\text{X}}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{5}\boxed{2}\boxed{)}\boxed{+}\boxed{2}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{1}\boxed{2}\boxed{)}\\ \boxed{+}\boxed{(}\boxed{\text{X}}\boxed{÷}\boxed{2}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{3}\boxed{2}\boxed{)}\end{array}$

Now, enter the key $\boxed{\text{MODE}}$ and set the graph at dot mode.

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(3-\text{X}\right)\left(\text{X}<2\right)+2\left(\text{X}=2\right)+\left(\text{X}/2\right)\left(\text{X}>2\right)$

Adjust the window at $\left[-2,6\right]$ by $\left[-1,5\right]$ .

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 $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ is a straight line with slope $-1$ and $x$ intercept as $3$ for $x<2$ , value of $f\left(x\right)$ is $2$ for $x=2$ and a straight line with slope $\frac{1}{2}$ for $x>2$ .

(b)

To determine

To find: The value of limits $\underset{x\to c^{+}}{\lim }f\left(x\right)$ and $\underset{x\to c^{-}}{\lim }f\left(x\right)$ for $c=2$ .

Answer to Problem 58E

The value of limits $\underset{x\to c^{+}}{\lim }f\left(x\right)$ and $\underset{x\to c^{-}}{\lim }f\left(x\right)$ for $c=2$ are both equal to $1$ .

Explanation of Solution

Given information:

The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ .

Calculation:

As for the right side of the point $x=2$ , the function corresponds is $f\left(x\right)=\frac{x}{2}$ .

Use the function $f\left(x\right)=\frac{x}{2}$ for the right hand limit.

  $\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\left(\frac{x}{2}\right)$

Substitute $2$ for $x$ in the above expression.

  $\begin{array}{c}\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\left(\frac{x}{2}\right)\\ =\frac{2}{2}\\ =1\end{array}$

So, the value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ is $1$ .

Now for the left side of the point $x=2$ , the function corresponds is $f\left(x\right)=3-x$ .

Use the function $f\left(x\right)=3-x$ to find the left hand limit.

  $\begin{array}{c}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\left(3-x\right)\\ =\underset{x\to 2^{-}}{\lim }\left(3\right)+\underset{x\to 2^{-}}{\lim }x\\ =3-\underset{x\to 2^{-}}{\lim }x\end{array}$

Substitute $2$ for $x$ in the above expression.

  $\begin{array}{c}\underset{x\to 2^{-}}{\lim }f\left(x\right)=3-\underset{x\to 2^{-}}{\lim }x\\ =3-2\\ =1\end{array}$

So, the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ is $1$ .

Therefore, the value of limits $\underset{x\to c^{+}}{\lim }f\left(x\right)$ and $\underset{x\to c^{-}}{\lim }f\left(x\right)$ for $c=2$ are both equal to $1$ .

(c)

To determine

To check: Whether the limits $\underset{x\to c}{\lim }f\left(x\right)$ for $c=2$ exists or not.

Answer to Problem 58E

The limit $\underset{x\to c}{\lim }f\left(x\right)$ for $c=2$ exist and is equal to $1$ .

Explanation of Solution

Given information: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ .

As calculated in part (b), the value of limits $\underset{x\to c^{+}}{\lim }f\left(x\right)$ and $\underset{x\to c^{-}}{\lim }f\left(x\right)$ for $c=2$ are both equal to $1$ .

The left hand limit for $f\left(x\right)$ is equal to right hand limit for $f\left(x\right)$ at $x=2$ .

Therefore, the limit $\underset{x\to c}{\lim }f\left(x\right)$ for $c=2$ exist and is equal to $1$ .