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To solve this problem, we will use the Taylor series expansion for the given function \( y = 5x^5 + 4x^4 + 3x^3 + 2x^2 + x + 1 \), around \( x = 1 \), and calculate the value at \( x = 2 \). The Taylor series expansion of a function \( f(x) \) around \( a \) is given by:

\[
f(x) \approx f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4 + \frac{f^{(5)}(a)}{5!}(x - a)^5 + \ldots
\]

We need to find the derivatives of the function up to the fifth order and evaluate them at \( x = 1 \).

1. **Calculate \( f(1) \)**:
   Given that \( f(1) = 16 \).

2. **Find the derivatives of \( f(x) \)**:
   \[
   f(x) = 5x^5 + 4x^4 + 3x^3 + 2x^2 + x + 1
   \]

   - First derivative, \( f'(x) = 25x^4 + 16x^3 + 9x^2 + 4x + 1 \)
   - Second derivative, \( f''(x) = 100x^3 + 48x^2 + 18x + 4 \)
   - Third derivative, \( f'''(x) = 300x^2 + 96x + 18 \)
   - Fourth derivative, \( f^{(4)}(x) = 600x + 96 \)
   - Fifth derivative, \( f^{(5)}(x) = 600 \)

3. **Evaluate the derivatives at \( x = 1 \)**:
   - \( f'(1) = 25(1)^4 + 16(1)^3 + 9(1)^2 + 4(1) + 1 = 25 + 16 + 9 + 4 + 1 = 55 \)
   - \( f''(1) = 100(1)^3 + 48(1)^2 + 18(1) + 4 = 100 + 48 + 18 + 4 = 170 \)
   - \( f'''(1) = 300(1)^2 + 96(1) + 18 = 300 + 96 + 18 = 414 \)
   - \( f^{(4)}(1) = 600(1) + 96 = 696 \)
   - \( f^{(5)}(1) = 600 \)

4. **Substitute these values into the Taylor series expansion up to the fifth term**:
   \[
   f(2) \approx 16 + 55(2-1) + \frac{170}{2!}(2-1)^2 + \frac{414}{3!}(2-1)^3 + \frac{696}{4!}(2-1)^4 + \frac{600}{5!}(2-1)^5
   \]
   \[
   f(2) \approx 16 + 55 + \frac{170}{2} + \frac{414}{6} + \frac{696}{24} + \frac{600}{120}
   \]
   \[
   f(2) \approx 16 + 55 + 85 + 69 + 29 + 5
   \]
   \[
   f(2) \approx 259
   \]

Hence, the estimated value of the function at \( x = 2 \) using the Taylor series expansion up to the fifth term is 259.

Would you like more details on the steps? 

Here are some related questions:
1. What happens if we include higher-order terms in the Taylor series?
2. How does the accuracy of the approximation change if we expand around a different point?
3. Can Taylor series be used for functions with discontinuities?
4. What is the geometric interpretation of a Taylor series?
5. How does the remainder term in the Taylor series affect the accuracy of approximation?

**Tip:** When approximating a function using Taylor series, more terms generally provide better accuracy, especially near the point of expansion.

Given function y = 5x^5 + 4x^4 + 3x^3 + 2x^2 + x + 1 and that the value of the function at x = 1 is 16, calculate the function value at x = 2 using the Taylor series expansion (up to the fifth term only).

Learn how to calculate the value of a function using the Taylor series expansion for y = 5x^5 + 4x^4 + 3x^3 + 2x^2 + x + 1, up to the fifth term, evaluated at x = 2. Includes derivatives and function values at x = 1.

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