Approximate sqrt(10) with a Taylor polynomial of degree 2 centered at x=9.

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?

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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.

Learn how to approximate sqrt(10) using a Taylor polynomial of degree 2 centered at x=9. Step-by-step solution including computing derivatives, evaluating at specific points, and applying the Taylor polynomial formula. Suitable for college-level calculus students.

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