By considering the third-degree Taylor Polynomial of f(x)=1/(2x+1) about x=0 , which of the following gives the estimated error for f(0.2)

To estimate the error in using the third-degree Taylor polynomial of \( f(x) = \frac{1}{2x+1} \) at \( x = 0 \), we can use the **remainder (error) term** in Taylor's theorem. This remainder term provides an upper bound on the error between the actual value of the function and its Taylor polynomial approximation.

### Steps to determine the error:

1. **Find the third-degree Taylor polynomial of \( f(x) \) about \( x = 0 \):**
   The general form of the Taylor polynomial is given by:
   \[
   P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3
   \]
   We need to compute the derivatives of \( f(x) = \frac{1}{2x+1} \) up to the third order and evaluate them at \( x = 0 \).

2. **Compute the derivatives of \( f(x) \):**
   - \( f(x) = \frac{1}{2x+1} \)
   - First derivative: \( f'(x) = \frac{-2}{(2x+1)^2} \)
   - Second derivative: \( f''(x) = \frac{8}{(2x+1)^3} \)
   - Third derivative: \( f^{(3)}(x) = \frac{-48}{(2x+1)^4} \)

3. **Evaluate each derivative at \( x = 0 \):**
   - \( f(0) = \frac{1}{2(0) + 1} = 1 \)
   - \( f'(0) = \frac{-2}{(2(0) + 1)^2} = -2 \)
   - \( f''(0) = \frac{8}{(2(0) + 1)^3} = 8 \)
   - \( f^{(3)}(0) = \frac{-48}{(2(0) + 1)^4} = -48 \)

4. **Form the third-degree Taylor polynomial:**
   \[
   P_3(x) = 1 - 2x + \frac{8}{2}x^2 - \frac{48}{6}x^3
   \]
   Simplifying:
   \[
   P_3(x) = 1 - 2x + 4x^2 - 8x^3
   \]

5. **Error estimation using the remainder term:**
   The remainder (error) term \( R_3(x) \) for the Taylor polynomial at \( x = 0 \) is given by:
   \[
   R_3(x) = \frac{f^{(4)}(c)}{4!}x^4
   \]
   for some \( c \) between \( 0 \) and \( x \). We need to compute the fourth derivative of \( f(x) \) to find the error term.

6. **Compute the fourth derivative of \( f(x) \):**
   - Fourth derivative: \( f^{(4)}(x) = \frac{384}{(2x+1)^5} \)

7. **Find an upper bound for the error at \( x = 0.2 \):**
   The maximum value of \( f^{(4)}(x) \) in the interval \( [0, 0.2] \) occurs at \( x = 0 \), since the denominator \( (2x+1)^5 \) decreases as \( x \) increases. Thus:
   \[
   f^{(4)}(0) = \frac{384}{(2(0) + 1)^5} = 384
   \]
   Therefore, the error is approximately:
   \[
   R_3(0.2) \approx \frac{384}{4!} (0.2)^4
   \]
   Simplifying:
   \[
   R_3(0.2) \approx \frac{384}{24} \times (0.2)^4 = 16 \times 0.0016 = 0.0256
   \]

Thus, the estimated error in approximating \( f(0.2) \) using the third-degree Taylor polynomial is approximately **0.0256**.

### Do you need further details on any of these steps or any other related questions? 

Here are 5 questions that may help you expand your understanding of this topic:
1. What is the general formula for the remainder term in Taylor’s theorem?
2. How can we find the error for higher-order Taylor polynomials?
3. How can the error estimation change if we use a fifth-degree Taylor polynomial?
4. What happens to the error if we choose a different point for the Taylor series expansion?
5. Can we compute the Taylor series for more complex functions using the same method?

**Tip:** Always verify the behavior of higher derivatives when estimating error, as their growth can significantly impact the accuracy of the approximation!

This page explains how to estimate the error of the third-degree Taylor polynomial for f(x) = 1/(2x+1) about x=0 at x=0.2. It covers step-by-step calculations, including derivatives, the Taylor expansion, and the remainder (error) term. Suitable for high school and college students studying calculus.

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