To estimate $(27.09)^{2/3}$ using the linearization of the function $f(x) = x^{2/3}$ at $a = 27$, follow these steps:

1. Find the function and its derivative

Given the function $f(x) = x^{2/3}$, the linearization of the function at $x = a$ is given by:

$L(x) = f(a) + f'(a)(x - a)$

First, calculate $f(27)$:

$f(27) = 27^{2/3} = 9$

Now, find the derivative of $f(x)$:

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

At $x = 27$, the derivative is:

$f'(27) = \frac{2}{3} \cdot 27^{-1/3} = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$

2. Write the linearization

Now, substitute $a = 27$ into the linearization formula:

$L(x) = f(27) + f'(27)(x - 27)$

$L(x) = 9 + \frac{2}{9}(x - 27)$

3. Estimate $(27.09)^{2/3}$

To estimate $(27.09)^{2/3}$, substitute $x = 27.09$ into the linearization:

$L(27.09) = 9 + \frac{2}{9}(27.09 - 27)$

$L(27.09) = 9 + \frac{2}{9}(0.09)$

$L(27.09) = 9 + \frac{2}{9} \cdot 0.09 = 9 + 0.02$

$L(27.09) = 9.02$

Thus, the estimated value of $(27.09)^{2/3}$ is approximately 9.02.

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Would you like more details on any of the steps? Here are some follow-up questions:

1. How is linearization generally used for estimating values of functions?
2. Can the linearization method be applied to higher-order approximations?
3. What are other techniques for approximating functions near a specific point?
4. How does the derivative impact the accuracy of the linear approximation?
5. How would the estimation change if we chose a different base point $a$?

Tip: Linearization works best for small deviations from the base point. For larger deviations, higher-order approximations (like using the second derivative) improve accuracy.