obtain the taylors series of fx= x power 1/3 up to the term x power 2 at a=8

To find the Taylor series of \( f(x) = x^{1/3} \) about \( a = 8 \), we use the formula for the Taylor series expansion:

\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n
\]

### Step 1: Compute the derivatives of \( f(x) = x^{1/3} \).

1. **First derivative** \( f'(x) \):
   \[
   f'(x) = \frac{1}{3}x^{-2/3}
   \]

2. **Second derivative** \( f''(x) \):
   \[
   f''(x) = -\frac{2}{9}x^{-5/3}
   \]

3. **Third derivative** \( f^{(3)}(x) \):
   \[
   f^{(3)}(x) = \frac{10}{27}x^{-8/3}
   \]

### Step 2: Evaluate the function and derivatives at \( x = 8 \).

1. **Evaluate \( f(8) \)**:
   \[
   f(8) = 8^{1/3} = 2
   \]

2. **Evaluate \( f'(8) \)**:
   \[
   f'(8) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}
   \]

3. **Evaluate \( f''(8) \)**:
   \[
   f''(8) = -\frac{2}{9} \cdot 8^{-5/3} = -\frac{2}{9} \cdot \frac{1}{32} = -\frac{1}{144}
   \]

4. **Evaluate \( f^{(3)}(8) \)**:
   \[
   f^{(3)}(8) = \frac{10}{27} \cdot 8^{-8/3} = \frac{10}{27} \cdot \frac{1}{256} = \frac{10}{6912}
   \]

### Step 3: Write out the Taylor series up to \( (x - 8)^2 \).

The Taylor series expansion is:

\[
f(x) \approx f(8) + f'(8)(x - 8) + \frac{f''(8)}{2!}(x - 8)^2
\]

Substitute the values we calculated:

\[
f(x) \approx 2 + \frac{1}{12}(x - 8) - \frac{1}{144} \cdot \frac{(x - 8)^2}{2}
\]

Simplifying:

\[
f(x) \approx 2 + \frac{1}{12}(x - 8) - \frac{1}{288}(x - 8)^2
\]

This is the Taylor series of \( f(x) = x^{1/3} \) about \( x = 8 \) up to the \( (x - 8)^2 \) term.

Would you like any further clarification or more detailed steps?

### Follow-up Questions:
1. What would the next term in the series be if we included the third derivative?
2. How would the series change if we expanded around a different value of \( a \)?
3. How does the accuracy of the approximation depend on the number of terms in the Taylor series?
4. Could this series be used to approximate cube roots for values near \( x = 8 \)?
5. What is the geometric interpretation of the Taylor series approximation?

### Tip:
When expanding functions using Taylor series, it's helpful to calculate a few higher-order derivatives, even if they're not immediately used, as they can show how quickly the series converges.

obtain the Taylor series of f(x) = x^(1/3) up to the term x^2 at a = 8

Discover how to find the Taylor series expansion of f(x) = x^(1/3) around a = 8 up to the (x - 8)^2 term. The solution includes detailed derivative computations and simplification of the series. Ideal for undergraduate calculus students.

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