Chapter 2.3, Problem 5E

To determine

To find:The points of continuity and discontinuity of the function $y=\sqrt{2x+3}$ . Also write the type of each discontinuity.

Answer to Problem 5E

The function $y=\sqrt{2x+3}$ is continuous for all the values of $x$ in the domain. The points of discontinuity are $x<-\frac{3}{2}$ .

Explanation of Solution

Given information:The functionis $y=\sqrt{2x+3}$ .

Calculation:

To graph a function $y=\sqrt{2x+3}$ , 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 $y=\sqrt{2x+3}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{2\text{nd}}\boxed{x^2}\boxed{2}\boxed{\text{X}}\boxed{+}\boxed{3}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\sqrt{2\text{X}+3}$

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 $y=\sqrt{2x+3}$ is continuous at every point of $x$ in the domain.

The function $y=\sqrt{2x+3}$ is not defined for all the values of $x<-\frac{3}{2}$ . So, the function $y=\sqrt{2x+3}$ is discontinuous for all the values of $x<-\frac{3}{2}$ .

Therefore, the function $y=\sqrt{2x+3}$ is continuous for all the values of $x$ in the domain. The points of discontinuity are $x<-\frac{3}{2}$ .