Chapter 2.3, Problem 2E

To determine

To find:The points of continuity and discontinuity of the function $y=\frac{x+1}{x^2-4x+3}$ and also identify the type of each discontinuity.

Answer to Problem 2E

The function $y=\frac{x+1}{x^2-4x+3}$ is continuous at all the values of $x$ except $x=1$ and $x=3$ with infinite discontinuity.

Explanation of Solution

Given information:The functionis $y=\frac{x+1}{x^2-4x+3}$ .

Calculation:

To graph a function $y=\frac{x+1}{x^2-4x+3}$ , 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 $y=\frac{x+1}{x^2-4x+3}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{4}\boxed{\text{X}}\boxed{+}\boxed{3}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(\text{X}+1\right)/\left(\text{X}^2-4\text{X}+3\right)$

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)

As observed from graph, the function $y=\frac{x+1}{x^2-4x+3}$ is continuous at every point of $x$ except $x=1$ and $x=3$ . Also, the type of discontinuity at $x=1$ and $x=3$ is of infinite type of the function $y=\frac{x+1}{x^2-4x+3}$ .

Therefore, the function $y=\frac{x+1}{x^2-4x+3}$ is continuous at all the values of $x$ except $x=1$ and $x=3$ with infinite discontinuity.