The problem asks to approximate the value of $f(1.2)$ using the second-degree Taylor polynomial $T_2(x)$ centered at $a = 1$, where $f(x) = x^2 + \frac{5}{x}$.

To approximate $f(1.2)$ using the Taylor series, we follow these steps:

Step 1: Compute $f(x)$, $f'(x)$, and $f''(x)$

Given $f(x) = x^2 + \frac{5}{x}$, the derivatives are:

• $f'(x) = 2x - \frac{5}{x^2}$
• $f''(x) = 2 + \frac{10}{x^3}$

Step 2: Evaluate the function and its derivatives at $a = 1$

• $f(1) = 1^2 + \frac{5}{1} = 1 + 5 = 6$
• $f'(1) = 2(1) - \frac{5}{1^2} = 2 - 5 = -3$
• $f''(1) = 2 + \frac{10}{1^3} = 2 + 10 = 12$

Step 3: Form the second-degree Taylor polynomial $T_2(x)$

The second-degree Taylor polynomial is: $T_2(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2}(x - 1)^2$ Substitute the values: $T_2(x) = 6 - 3(x - 1) + \frac{12}{2}(x - 1)^2$ Simplifying: $T_2(x) = 6 - 3(x - 1) + 6(x - 1)^2$

Step 4: Approximate $f(1.2)$

Now, substitute $x = 1.2$ into $T_2(x)$: $T_2(1.2) = 6 - 3(1.2 - 1) + 6(1.2 - 1)^2$ Calculate step by step: $T_2(1.2) = 6 - 3(0.2) + 6(0.2)^2$ $T_2(1.2) = 6 - 0.6 + 6(0.04)$ $T_2(1.2) = 6 - 0.6 + 0.24 = 5.64$

Thus, the approximate value of $f(1.2)$ is 5.64.

Would you like further details or have any questions?

Here are some related questions:

1. How is a Taylor series related to the actual function for approximation?
2. Why do we stop at the second degree in the Taylor series?
3. How would the accuracy change if we used a higher-degree Taylor polynomial?
4. Can you approximate $f(x)$ using a Taylor series centered at a different point?
5. What is the geometric interpretation of a Taylor polynomial?

Tip: When approximating using Taylor polynomials, the more terms you include, the more accurate your approximation near the center point.