Chapter 1.2, Problem 15E

(a)

To determine

To find:The domain and range of the function $y=\sqrt[3]{1-x^2}$ .

Answer to Problem 15E

The domain and range of the function $y=\sqrt[3]{1-x^2}$ is $\left(-\infty ,\infty \right)$ and $\left(-\infty ,1\right]$ respectively.

Explanation of Solution

Given information:The given function is $y=\sqrt[3]{1-x^2}$ .

Calculation:

Domain is the set of all input values for which function is defined. The function $y=\sqrt[3]{1-x^2}$ is defined for all the real values of $x$ . So, the domain of the function is the set of real numbers or the interval $\left(-\infty ,\infty \right)$ .

The range is the set of all output values for which function is defined. The function is $y=\sqrt[3]{1-x^2}$ defined for all input values less than or equal to $1$ .So, the range of the function is the interval $\left(-\infty ,1\right]$ .

Therefore, the domain and range of the function $y=\sqrt[3]{1-x^2}$ is $\left(-\infty ,\infty \right)$ and $\left(-\infty ,1\right]$ respectively.

(b)

To determine

To graph:The function $y=\sqrt[3]{1-x^2}$ with the help of graphing calculator.

Explanation of Solution

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

Graph:

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

The display will show the equation,

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

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given quadratic equation in standard window as shown in Figure (1).

  

Figure (1)

Interpretation:It can be observed that the graph of the function $y=\sqrt[3]{1-x^2}$ is symmetric about $y-$ axis and intersects it at point $\left(0,1\right)$ .