Use the linearization of the function f(x)=x^2/3 at a=27 to estimate the expression (27.09)^2/3.

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.

Learn how to estimate (27.09)^2/3 using the linearization method with the function f(x)=x^{2/3} at a=27. This solution covers key mathematical concepts such as linearization, derivatives, and exponents. Suitable for students in Grades 10-12.

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