Chapter 2.1, Problem 59E

(a)

To determine

To graph: The piecewise function $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

Explanation of Solution

Given information: The piecewise function $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

Graph:

To graph a function $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ , 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 $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes as given below:

  $\begin{array}{l}\boxed{(}\boxed{1}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{-}\boxed{1}\boxed{)}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{5}\boxed{1}\boxed{)}\\ \boxed{+}\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{3}\boxed{-}\boxed{2}\boxed{\text{X}}\boxed{+}\boxed{5}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{1}\boxed{)}\end{array}$

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

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(1/\left(\text{X}-1\right)\right)\left(\text{X}<1\right)+\left(\text{X}^3-2\text{X}+5\right)\left(\text{X}\ge 1\right)$

Adjust the window at $\left[-3,5\right]$ by $\left[-4,10\right]$ .

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 $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ is the combination of curves $y=\frac{1}{x-1}$ defined for $x<1$ and curve $x^3-2x+5$ defined for $x\ge 1$ .

(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 $c=1$ .

Answer to Problem 59E

The value of limit $\underset{x\to c^{+}}{\lim }f\left(x\right)$ is $4$ and the limit $\underset{x\to c^{-}}{\lim }f\left(x\right)$ does not exist for $c=1$ .

Explanation of Solution

Given information: The piecewise function is $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

Calculation:

As for the right side of the point $x=1$ , the function corresponds is $f\left(x\right)=x^3-2x+5$ .

Use the function $f\left(x\right)=x^3-2x+5$ for the right hand limit.

  $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(x^3-2x+5\right)$

Substitute $1$ for $x$ in the above expression.

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=1^3-2\times 1+5\\ =1-2+5\\ =4\end{array}$

So, the value of $\underset{x\to 1^{+}}{\lim }f\left(x\right)$ is $4$ .

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

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

  $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(\frac{1}{x-1}\right)$

This does not exist as it forms $\frac{0}{0}$ .

So, the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ doesn’t exist.

Therefore, the value of limit $\underset{x\to c^{+}}{\lim }f\left(x\right)$ is $4$ and the limit $\underset{x\to c^{-}}{\lim }f\left(x\right)$ does not exist for $c=1$ .

(c)

To determine

To check: Whether the limits $\underset{x\to c}{\lim }f\left(x\right)$ for $c=1$ exists or not.

Answer to Problem 59E

The limit $\underset{x\to c}{\lim }f\left(x\right)$ for $c=1$ does not exist as the left hand limit of the function does not exist.

Explanation of Solution

Given information: The piecewise function is $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

As calculated in part (b), the value of limit $\underset{x\to c^{-}}{\lim }f\left(x\right)$ for $c=1$ does not exist.

Therefore, the limit $\underset{x\to c}{\lim }f\left(x\right)$ for $c=1$ does not exist.