Question

In Exercises 13-20, use a grapher to (a) identify the domain and range and (b) draw the graph of the function. $$y=\sqrt[3]{1-x^{2}}$$

Short answer

The domain of the function \(y=\sqrt[3]{1-x^{2}}\) is \(x\) is element of \(-\infty, +\infty\) and the range is \( y \) is element of \([0, +1]\). The graph of the function is symmetric with respect to the y-axis.

Step-by-step solution

Determine the Domain

Recall that the domain of a function is the set of all valid input values (x-values) for which the function is defined. For the function \(y=\sqrt[3]{1-x^{2}}\), it's visible that any real value of x is permissible. Within the cube root, regardless of the value of x (be it positive or negative or even zero), it will always be valid, as the square of any real number is a positive number or zero, and 1 minus a positive number or zero is valid under the cube root. Thus, the domain of the function is \(x\) is element of \(-\infty , +\infty\).

Determine the Range

When it comes to the range, it is the set of all possible output values (y-values) after the x-values are plugged into the function. For the function \(y=\sqrt[3]{1-x^{2}}\), the minimum value is reached when \(x^2\) is at its maximum, which is 1, resulting in \(y=\sqrt[3]{1-1} = 0\). The maximum value of y can reach \(+1\) because y cannot be more than \(1\) as it will exceeds the real number limit. As a result, it's the interval from \([0,+1]\). Therefore, the range is \( y \) is element of \([0, +1]\).

Sketch the Graph

Once the domain and range are known, a sketch of the function can be created using a graphing tool. According to the domain and range findings, the graph extends indefinitely along the x-axis (both negative and positive) and rises from 0 to 1 along the y-axis. Because the function is not explicitly defined in terms of x, the approach of this function is to get a mirror reflection along y-axis as it's symmetrical for both positive and negative values of x. Use these guidelines to assist with the design of the graph.