Complete the table of values, rounded to two decimal places.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)=2 e^{-x / 3}$[/tex] \\
\hline
-3 & [tex]$\square$[/tex] \\
\hline
-2 & [tex]$\square$[/tex] \\
\hline
-1 & [tex]$\square$[/tex] \\
\hline
0 & [tex]$\square$[/tex] \\
\hline
1 & [tex]$\square$[/tex] \\
\hline
2 & [tex]$\square$[/tex] \\
\hline
3 & [tex]$\square$[/tex] \\
\hline
\end{tabular}



Answer :

Sure, let's complete the table of values step by step:

Given the function [tex]\( f(x) = 2 e^{-x / 3} \)[/tex], we need to compute the values at specific points [tex]\( x \)[/tex]. We've calculated these values and rounded them to two decimal places.

### Calculating [tex]\( f(x) \)[/tex]:

- For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = 2 e^{-(-3) / 3} = 2 e^{1} = 5.44 \][/tex]

- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2 e^{-(-2) / 3} = 2 e^{2/3} = 3.90 \][/tex]

- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 2 e^{-(-1) / 3} = 2 e^{1/3} = 2.79 \][/tex]

- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2 e^{-0 / 3} = 2 e^{0} = 2.00 \][/tex]

- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2 e^{-1 / 3} = 2 e^{-1/3} = 1.43 \][/tex]

- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2 e^{-2 / 3} = 2 e^{-2/3} = 1.03 \][/tex]

- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 2 e^{-3 / 3} = 2 e^{-1} = 0.74 \][/tex]

### Completing the Table:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x)=2 e^{-x / 3} \\ \hline -3 & 5.44 \\ \hline -2 & 3.90 \\ \hline -1 & 2.79 \\ \hline 0 & 2.00 \\ \hline 1 & 1.43 \\ \hline 2 & 1.03 \\ \hline 3 & 0.74 \\ \hline \end{array} \][/tex]

So, the complete table is:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x)=2 e^{-x / 3} \\ \hline -3 & 5.44 \\ \hline -2 & 3.90 \\ \hline -1 & 2.79 \\ \hline 0 & 2.00 \\ \hline 1 & 1.43 \\ \hline 2 & 1.03 \\ \hline 3 & 0.74 \\ \hline \end{array} \][/tex]