Chapter 2.1, Problem 57E

(a)

To determine

To graph: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ .

Explanation of Solution

Given information: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ .

Graph:

To graph a function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ , 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 $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{3}\boxed{-}\boxed{\text{X}}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{5}\boxed{2}\boxed{)}\boxed{+}\boxed{(}\boxed{\text{X}}\boxed{÷}\boxed{2}\boxed{+}\boxed{1}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{3}\boxed{2}\boxed{)}$ .

The display will show the equation,

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

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 $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ is a straight line with slope $-1$ and $x$ intercept as $3$ 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 57E

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 $2$ and $1$ respectively.

Explanation of Solution

Given information: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ .

Calculation:

As for the right side of the point $x=2$ , the function corresponds is $f\left(x\right)=\frac{x}{2}+1$ .

Use the function $f\left(x\right)=\frac{x}{2}+1$ for the right hand limit.

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

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

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

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

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

Use the function $f\left(x\right)=3-x$ for 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.

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

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 $2$ and $1$ respectively.

(c)

To determine

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

Answer to Problem 57E

The limit $\underset{x\to c}{\lim }f\left(x\right)$ for $c=2$ does not exist.

Explanation of Solution

Given information: The piecewise function $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ .

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 $2$ and $1$ respectively.

The left hand limit for $f\left(x\right)$ is not equal to right hand limit for $f\left(x\right)$ .

Therefore, the limit $\underset{x\to c}{\lim }f\left(x\right)$ for $c=2$ does not exist.