Question

Use the function f(x)=3x
to find the approximation 33√
, correct to three decimal places, by creating a table with x-values closer and closer to 3–√
and the corresponding f(x)
values.(4 points)

Answer 1

To approximate \( \sqrt[3]{33} \) using the function \( f(x) = 3x \), we first need to establish that we are looking for the value of \( x \) such that \( f(x) = \sqrt[3]{33} \). In other words, we want to find \( x \) such that:

\[
3x = \sqrt[3]{33}
\]

To find this, we can rewrite it as:

\[
x = \frac{\sqrt[3]{33}}{3}
\]

Next, we'll create a table where we can evaluate \( f(x) \) for various values of \( x \) that are close to \( \frac{\sqrt[3]{33}}{3} \).

First, let's calculate \( \sqrt[3]{33} \):

\[
\sqrt[3]{33} \approx 3.208 \quad (\text{to three decimal places})
\]

Now, we divide by 3:

\[
x \approx \frac{3.208}{3} \approx 1.0693
\]

Now we will evaluate \( f(x) \) for \( x \) values near \( 1.0693 \). We will use \( x \) values of \( 1.06, 1.07, 1.08, 1.069, 1.070, \) and \( 1.071 \).

### Table of Values

| \( x \) | \( f(x) = 3x \) |
|-----------|-------------------|
| 1.060 | \( 3 \times 1.060 = 3.180 \) |
| 1.067 | \( 3 \times 1.067 = 3.201 \) |
| 1.069 | \( 3 \times 1.069 = 3.207 \) |
| 1.070 | \( 3 \times 1.070 = 3.210 \) |
| 1.071 | \( 3 \times 1.071 = 3.213 \) |
| 1.080 | \( 3 \times 1.080 = 3.240 \) |

### Summarized Results

Based on the table:

- For \( x = 1.060 \), \( f(x) = 3.180 \)
- For \( x = 1.067 \), \( f(x) = 3.201 \)
- For \( x = 1.069 \), \( f(x) = 3.207 \)
- For \( x = 1.070 \), \( f(x) = 3.210 \)
- For \( x = 1.071 \), \( f(x) = 3.213 \)
- For \( x = 1.080 \), \( f(x) = 3.240 \)

From the values calculated, we can see that \( f(1.069) = 3.207 \) is very close to \( \sqrt[3]{33} \approx 3.208 \).

### Conclusion
Thus, based on the values obtained from the function \( f(x) \), and observing the output of the function as we get closer to \( x = 1.0693 \), we approximate:

\[
\sqrt[3]{33} \approx 3.208 \quad \text{(to three decimal places)}
\]

The value of \( x \) that gives this output is roughly \( 1.0693 \).