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Fill in the [tex]$y$[/tex] values of the [tex]$t$[/tex]-table for the function [tex]$y=\sqrt[3]{x}$[/tex].

[tex]\[
\begin{tabular}{c|c}
$x$ & $y$ \\
\hline
-8 & \\
-1 & \\
0 & \\
1 & \\
8 & \\
\hline
\end{tabular}
\][/tex]

This is the graph of the function: [tex]$y=\sqrt[3]{x}$[/tex].

Sagot :

GinnyAnswer
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.
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