To approximate the value of $\sqrt[3]{64.42}$ using the formula $\Delta y \approx f'(x) \Delta x$, we first define the function:

$f(x) = \sqrt[3]{x}$

Step 1: Calculate the derivative $f'(x)$

Using the power rule:

$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3 \sqrt[3]{x^2}}$

Step 2: Choose a point near $64.42$

We'll use $x = 64$ because $\sqrt[3]{64} = 4$ is easy to compute.

Step 3: Find $f'(64)$

Calculating $f'(64)$:

$f'(64) = \frac{1}{3 \sqrt[3]{64^2}} = \frac{1}{3 \cdot 16} = \frac{1}{48}$

Step 4: Determine $\Delta x$

Calculate $\Delta x$:

$\Delta x = 64.42 - 64 = 0.42$

Step 5: Calculate $\Delta y$

Now we can find $\Delta y$:

$\Delta y \approx f'(64) \Delta x = \frac{1}{48} \cdot 0.42 \approx 0.00875$

Step 6: Approximate $f(64.42)$

Now, we can approximate:

$f(64.42) \approx f(64) + \Delta y = 4 + 0.00875 \approx 4.00875$

Final Result

Thus, the decimal approximation of $\sqrt[3]{64.42}$ is approximately 4.00875.

If you have any further questions or need more details, let me know! Here are some follow-up questions you might consider:

1. How would the approximation change with a different value for $x$?
2. What is the exact value of $\sqrt[3]{64.42}$?
3. How can you improve the accuracy of this approximation?
4. What are the applications of using linear approximation in calculus?
5. Can you explain the concept of differentials in this context?

Tip: Always choose a point $x$ that is close to the value you want to approximate to improve accuracy.