Find the Taylor polynomial
T3(x)
for the function f centered at the number a.
f(x) = 11 tan−1(x),    a = 1

To find the Taylor polynomial \( T_3(x) \) for the function \( f(x) = 11 \tan^{-1}(x) \) centered at \( a = 1 \), we'll need to follow these steps:

1. **Compute the derivatives** of \( f(x) \) up to the third derivative.
2. **Evaluate** these derivatives at \( x = 1 \).
3. **Formulate** the Taylor polynomial using the formula:

\[
T_3(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3
\]

### Step 1: Compute the derivatives

The function is \( f(x) = 11 \tan^{-1}(x) \).

- **First derivative**:
  
  \[
  f'(x) = 11 \cdot \frac{d}{dx} \left(\tan^{-1}(x)\right) = \frac{11}{1 + x^2}
  \]

- **Second derivative**:
  
  \[
  f''(x) = 11 \cdot \frac{d}{dx} \left(\frac{1}{1 + x^2}\right) = 11 \cdot \left(-\frac{2x}{(1 + x^2)^2}\right) = \frac{-22x}{(1 + x^2)^2}
  \]

- **Third derivative**:
  
  \[
  f'''(x) = 11 \cdot \frac{d}{dx} \left(\frac{-2x}{(1 + x^2)^2}\right)
  \]
  To compute this, apply the product and chain rules:

  \[
  f'''(x) = 11 \cdot \left( \frac{-2 \cdot (1 + x^2)^2 - (-2x) \cdot 4x \cdot (1 + x^2)}{(1 + x^2)^4} \right)
  \]

  Simplify:

  \[
  f'''(x) = \frac{-22(1 + x^2)^2 + 88x^2(1 + x^2)}{(1 + x^2)^4}
  \]
  \[
  f'''(x) = \frac{-22(1 + x^2 - 4x^2)}{(1 + x^2)^3}
  \]
  \[
  f'''(x) = \frac{-22(1 - 3x^2)}{(1 + x^2)^3}
  \]

### Step 2: Evaluate at \( x = 1 \)

- \( f(1) = 11 \tan^{-1}(1) = 11 \cdot \frac{\pi}{4} = \frac{11\pi}{4} \)
- \( f'(1) = \frac{11}{1 + 1^2} = \frac{11}{2} \)
- \( f''(1) = \frac{-22 \cdot 1}{(1 + 1)^2} = \frac{-22}{4} = -\frac{11}{2} \)
- \( f'''(1) = \frac{-22(1 - 3 \cdot 1^2)}{(1 + 1^2)^3} = \frac{-22(1 - 3)}{8} = \frac{-22(-2)}{8} = \frac{44}{8} = \frac{11}{2} \)

### Step 3: Formulate the Taylor polynomial

Using the formula:

\[
T_3(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3
\]

Substitute the values:

\[
T_3(x) = \frac{11\pi}{4} + \frac{11}{2}(x - 1) - \frac{11}{4}(x - 1)^2 + \frac{11}{12}(x - 1)^3
\]

This is the third-degree Taylor polynomial for \( f(x) = 11 \tan^{-1}(x) \) centered at \( a = 1 \).

Would you like more details or have any questions?

### Related Questions:
1. What is the Taylor polynomial of degree 4 for the same function?
2. How do you find the radius of convergence for the Taylor series of this function?
3. What is the significance of each term in the Taylor series?
4. Can you explain how the Taylor polynomial approximates the function near \( a = 1 \)?
5. How would the Taylor polynomial change if \( a = 0 \)?
6. What role do higher-order derivatives play in the accuracy of the polynomial?
7. How do you derive the inverse tangent function's Taylor series?
8. Can the Taylor series be used to estimate \( \tan^{-1}(x) \) for large \( x \)?

### Tip:
Remember that the accuracy of a Taylor polynomial increases as you include more terms, but it is most accurate near the center \( a \).

Learn how to find the Taylor polynomial T3(x) for the function f(x) = 11 tan^-1(x) centered at a = 1. This involves computing derivatives, evaluating them, and formulating the polynomial using the Taylor series formula. Suitable for college-level calculus students.

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