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The image provided contains a mathematical problem involving the Taylor Series. I will transcribe the problem and provide a solution.

### Transcription of the Image:

**Taylor Series Expansion**

*Express the Taylor series centered at \( x = 0 \) for the function \( f(x) = \frac{1}{\sqrt{1-x}} \) up to and including the \( x^4 \) term.*

**PART B**

*We want to build a bridge across the river where the cross-sectional area of the left support has the equation \( A_L(x) = \frac{1}{\sqrt{1-x}} \) (in units of square feet). Find the cross-sectional area of the left support at \( x = 0.1 \). Then, calculate the area using the Taylor series approximation from Part A.*

### Solution to the Problem:

#### Part A: Taylor Series Expansion

The function given is:
\[
f(x) = \frac{1}{\sqrt{1-x}}
\]

We will expand this function into a Taylor series centered at \( x = 0 \). The general form of the Taylor series for a function \( f(x) \) about \( x = 0 \) is:
\[
f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \dots
\]

Let's calculate the first few derivatives of \( f(x) \):

1. \( f(x) = (1-x)^{-\frac{1}{2}} \)
2. \( f'(x) = \frac{1}{2}(1-x)^{-\frac{3}{2}} \)
3. \( f''(x) = \frac{3}{4}(1-x)^{-\frac{5}{2}} \)
4. \( f'''(x) = \frac{15}{8}(1-x)^{-\frac{7}{2}} \)
5. \( f^{(4)}(x) = \frac{105}{16}(1-x)^{-\frac{9}{2}} \)

Now, evaluating these at \( x = 0 \):

- \( f(0) = 1 \)
- \( f'(0) = \frac{1}{2} \)
- \( f''(0) = \frac{3}{4} \)
- \( f'''(0) = \frac{15}{8} \)
- \( f^{(4)}(0) = \frac{105}{16} \)

Thus, the Taylor series up to the \( x^4 \) term is:

\[
f(x) = 1 + \frac{1}{2}x + \frac{3}{4}x^2 + \frac{15}{8}x^3 + \frac{105}{16}x^4
\]

#### Part B: Area of the Left Support

1. **Exact Calculation**:
   \[
   A_L(0.1) = \frac{1}{\sqrt{1-0.1}} = \frac{1}{\sqrt{0.9}} \approx 1.054
   \]

2. **Taylor Series Approximation**:
   \[
   A_L(0.1) \approx 1 + \frac{1}{2}(0.1) + \frac{3}{4}(0.1^2) + \frac{15}{8}(0.1^3) + \frac{105}{16}(0.1^4)
   \]
   Simplifying this:
   \[
   A_L(0.1) \approx 1 + 0.05 + 0.0075 + 0.001875 + 0.00065625 = 1.06003125
   \]

Thus, using the Taylor series approximation, the area at \( x = 0.1 \) is approximately 1.060, which is slightly higher than the exact value.

Would you like more details on any specific part, or do you have any questions?

### Related Questions
1. What is the general formula for a Taylor series centered at a point \( x = a \)?
2. How would the series change if it were centered at \( x = 1 \) instead of \( x = 0 \)?
3. What is the error bound for the Taylor series approximation in this case?
4. How would you approach the problem if you needed to include terms up to \( x^6 \)?
5. Can you derive the Maclaurin series for a different function, like \( e^x \)?

### Tip
Always check the radius of convergence when working with Taylor series to ensure the series is valid within the interval of interest.

Express the Taylor series centered at x = 0 for the function f(x) = 1/sqrt(1-x) up to and including the x^4 term. We want to build a bridge across the river where the cross-sectional area of the left support has the equation A_L(x) = 1/sqrt(1-x) (in units of square feet). Find the cross-sectional area of the left support at x = 0.1. Then, calculate the area using the Taylor series approximation from Part A.

Explore the Taylor series expansion centered at x = 0 for the function f(x) = 1/sqrt(1-x), and its application in calculating the cross-sectional area of a bridge support. This detailed solution covers series expansion and calculus concepts, suitable for advanced high school students.

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