To approximate \( \sqrt[3]{33} \) using the function \( f(x) = 3x \), we want to find the value of \( x \) such that \( f(x) = 33 \). This means we need to solve the equation \( 3x = 33 \), leading us to find \( x = \frac{33}{3} = 11 \). Therefore, \( \sqrt[3]{33} \approx 11 \).
However, since you're looking for a table with \( x \)-values that get closer to \( \sqrt[3]{33} \), let's evaluate \( f(x) \) for values of \( x \) that will give us a good approximation of \( \sqrt[3]{33} \).
To find the approximation, we will evaluate \( f(x) \) at values of \( x \) around \( 3 \) since \( \sqrt[3]{27} = 3 \). Therefore:
Table of \( x \) values and \( f(x) \)
\[ \begin{array}{|c|c|} \hline x & f(x) = 3x \ \hline 2.5 & 3(2.5) = 7.5 \ 2.8 & 3(2.8) = 8.4 \ 2.9 & 3(2.9) = 8.7 \ 3.0 & 3(3.0) = 9.0 \ 3.1 & 3(3.1) = 9.3 \ 3.2 & 3(3.2) = 9.6 \ 3.3 & 3(3.3) = 9.9 \ 3.4 & 3(3.4) = 10.2 \ 3.5 & 3(3.5) = 10.5 \ 3.6 & 3(3.6) = 10.8 \ 3.7 & 3(3.7) = 11.1 \ 3.8 & 3(3.8) = 11.4 \ \hline \end{array} \]
We see that as \( x \) increases from \( 3.4 \) to \( 3.6 \), \( f(x) \) gets close to \( 11 \).
Since we want values that approach a final approximation for ·\( \sqrt[3]{33} \), we can narrow down the range further:
Narrowed down table
\[ \begin{array}{|c|c|} \hline x & f(x) = 3x \ \hline 3.0 & 9.0 \ 3.4 & 10.2 \ 3.5 & 10.5 \ 3.6 & 10.8 \ 3.7 & 11.1 \ \hline \end{array} \]
To refine it even more:
Further Refined table
\[ \begin{array}{|c|c|} \hline x & f(x) = 3x \ \hline 3.3 & 9.9 \ 3.4 & 10.2 \ 3.45 & 10.35 \ 3.5 & 10.5 \ 3.6 & 10.8 \ 3.7 & 11.1 \ 3.65 & 10.95 \ \hline \end{array} \]
By interpolation, we can estimate that \( \sqrt[3]{33} \) is close to \( 3.2 \) to \( 3.5 \), which gives us a better approximation:
Final Approximation:
After looking at the values, we see that \( f(3.4) \) gives \( 10.2 \) and \( f(3.7) \) gives \( 11.1 \). Continuing to narrow down shows us that \( \sqrt[3]{33} \) should be around \( 3.59 \) because:
\[ f(3.59) = 3(3.59) = 10.77 \]
The value of \( \sqrt[3]{33} \) is approximately \( 3.2 \).
Thus, rounding to three decimal places:
\[ \sqrt[3]{33} \approx 3.2. \]
Therefore, the final answer is:
\[ \sqrt[3]{33} \approx 3.207 \] to three decimal places.
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