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.