To approximate $\sqrt{10}$ using a Taylor polynomial of degree 2 centered at $x = 9$, we follow these steps:

Step 1: Define the function

Let $f(x) = \sqrt{x} = x^{1/2}$.

Step 2: Compute the derivatives

We need to find the first and second derivatives of $f(x)$:

1. First derivative: $f'(x) = \frac{d}{dx} \left( x^{1/2} \right) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$

2. Second derivative: $f''(x) = \frac{d}{dx} \left( \frac{1}{2\sqrt{x}} \right) = \frac{-1}{4} x^{-3/2} = \frac{-1}{4x^{3/2}}$

Step 3: Evaluate the function and derivatives at $x = 9$

Now, evaluate $f(x)$, $f'(x)$, and $f''(x)$ at $x = 9$:

1. $f(9) = \sqrt{9} = 3$
2. $f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}$
3. $f''(9) = \frac{-1}{4 \times 9^{3/2}} = \frac{-1}{108}$

Step 4: Write the Taylor polynomial

The Taylor polynomial of degree 2 centered at $x = 9$ is given by:

$P_2(x) = f(9) + f'(9)(x - 9) + \frac{f''(9)}{2!}(x - 9)^2$

Substituting the values we calculated:

$P_2(x) = 3 + \frac{1}{6}(x - 9) + \frac{-1/108}{2}(x - 9)^2$

$P_2(x) = 3 + \frac{1}{6}(x - 9) - \frac{1}{216}(x - 9)^2$

Step 5: Approximate $\sqrt{10}$

Now, approximate $\sqrt{10}$ by substituting $x = 10$ into the Taylor polynomial:

$P_2(10) = 3 + \frac{1}{6}(10 - 9) - \frac{1}{216}(10 - 9)^2$

Simplifying:

$P_2(10) = 3 + \frac{1}{6} \times 1 - \frac{1}{216} \times 1^2$ $P_2(10) = 3 + \frac{1}{6} - \frac{1}{216}$ $P_2(10) = 3 + \frac{36}{216} - \frac{1}{216} = 3 + \frac{35}{216}$ $P_2(10) = 3 + 0.1620 \approx 3.1620$

So, the Taylor polynomial approximation of $\sqrt{10}$ is approximately $3.1620$.

Would you like any further details or have any questions?

---

Here are 5 related questions you might find interesting:

1. How would the approximation change if we used a Taylor polynomial of degree 3 centered at $x = 9$?
2. What is the actual value of $\sqrt{10}$, and how does it compare to the Taylor approximation?
3. Can you generalize the method to approximate $\sqrt{x}$ at any point near $x = a$?
4. What is the error bound for this Taylor approximation?
5. How does the approximation improve with higher-degree Taylor polynomials?

Tip: When using Taylor series approximations, the accuracy increases as the degree of the polynomial increases, especially when the point of interest is close to the center of the expansion.