To fill in the [tex]\( y \)[/tex] values of the [tex]\( t \)[/tex]-table for the function [tex]\( y = \sqrt[3]{x} \)[/tex], we need to evaluate this function at each given [tex]\( x \)[/tex] value. Here are the steps:
1. For [tex]\( x = -8 \)[/tex]:
[tex]\[
y = \sqrt[3]{-8}
\][/tex]
The cube root of [tex]\(-8\)[/tex] results in a complex number:
[tex]\[
y = (1.0000000000000002 + 1.7320508075688772j)
\][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = \sqrt[3]{-1}
\][/tex]
The cube root of [tex]\(-1\)[/tex] also results in a complex number:
[tex]\[
y = (0.5000000000000001 + 0.8660254037844386j)
\][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = \sqrt[3]{0}
\][/tex]
The cube root of [tex]\(0\)[/tex] is straightforward:
[tex]\[
y = 0.0
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = \sqrt[3]{1}
\][/tex]
The cube root of [tex]\(1\)[/tex] is simply:
[tex]\[
y = 1.0
\][/tex]
5. For [tex]\( x = 8 \)[/tex]:
[tex]\[
y = \sqrt[3]{8}
\][/tex]
The cube root of [tex]\(8\)[/tex] is:
[tex]\[
y = 2.0
\][/tex]
Thus, you can now fill in the [tex]\( t \)[/tex]-table with these values:
[tex]\[
\begin{tabular}{c|c}
$x$ & $y$ \\
\hline
-8 & (1.0000000000000002 + 1.7320508075688772j) \\
-1 & (0.5000000000000001 + 0.8660254037844386j) \\
0 & 0.0 \\
1 & 1.0 \\
8 & 2.0 \\
\hline
\end{tabular}
\][/tex]
This represents the values of [tex]\( y \)[/tex] for the function [tex]\( y = \sqrt[3]{x} \)[/tex] at the given [tex]\( x \)[/tex] points.
