Chapter 2.3, Problem 6E

To determine

To find:The points of continuity and discontinuity of the function $y=\sqrt[3]{2x-1}$ . Also write the type of each discontinuity.

Answer to Problem 6E

The function $y=\sqrt[3]{2x-1}$ is continuous for all the values of $x$ and point of discontinuity does not exist.

Explanation of Solution

Given information:The functionis $y=\sqrt[3]{2x-1}$ .

Calculation:

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

The display will show the equation,

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

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 $y=\sqrt[3]{2x-1}$ is continuous at every point of $x$ .So, the function has no point of discontinuity of the function $y=\sqrt[3]{2x-1}$ .

Therefore, the function $y=\sqrt[3]{2x-1}$ is continuous for all the values of $x$ and point of discontinuity does not exist.