https://file.aiquickdraw.com/mathbot/file/1729393798447t75ref9s.jpg

Let's tackle the problem step by step. You are asked to find the degree 3 Taylor Polynomial for the function \( f(x) = \sqrt{x} \) centered at \( a = 9 \), and then use it to approximate \( \sqrt{12} \).

### Step 1: Taylor Polynomial Formula
The Taylor polynomial of degree \(n\) for a function \(f(x)\) centered at \(a\) is given by:

\[
T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \dots
\]

### Step 2: Compute Derivatives of \(f(x)\)
The function is \( f(x) = \sqrt{x} = x^{1/2} \).

1. First derivative: 
   \[
   f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
   \]
2. Second derivative:
   \[
   f''(x) = -\frac{1}{4}x^{-3/2} = -\frac{1}{4x^{3/2}}
   \]
3. Third derivative:
   \[
   f'''(x) = \frac{3}{8}x^{-5/2} = \frac{3}{8x^{5/2}}
   \]

### Step 3: Evaluate the Function and Its Derivatives at \(x = 9\)
Now, we evaluate \(f(x)\), \(f'(x)\), \(f''(x)\), and \(f'''(x)\) at \(x = 9\).

1. \( f(9) = \sqrt{9} = 3 \)
2. \( f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6} \)
3. \( f''(9) = -\frac{1}{4 \times 9^{3/2}} = -\frac{1}{4 \times 27} = -\frac{1}{108} \)
4. \( f'''(9) = \frac{3}{8 \times 9^{5/2}} = \frac{3}{8 \times 243} = \frac{1}{648} \)

### Step 4: Construct the Taylor Polynomial
Now we plug these values into the Taylor polynomial formula up to degree 3:

\[
T_3(x) = f(9) + f'(9)(x - 9) + \frac{f''(9)}{2!}(x - 9)^2 + \frac{f'''(9)}{3!}(x - 9)^3
\]

Substitute the computed values:

\[
T_3(x) = 3 + \frac{1}{6}(x - 9) - \frac{1}{108}\frac{(x - 9)^2}{2} + \frac{1}{648}\frac{(x - 9)^3}{6}
\]

Simplifying:

\[
T_3(x) = 3 + \frac{1}{6}(x - 9) - \frac{1}{216}(x - 9)^2 + \frac{1}{3888}(x - 9)^3
\]

### Step 5: Approximate \( \sqrt{12} \)
To approximate \( \sqrt{12} \), substitute \(x = 12\) into the Taylor polynomial \(T_3(x)\).

\[
T_3(12) = 3 + \frac{1}{6}(12 - 9) - \frac{1}{216}(12 - 9)^2 + \frac{1}{3888}(12 - 9)^3
\]

Simplify each term:

\[
T_3(12) = 3 + \frac{1}{6}(3) - \frac{1}{216}(9) + \frac{1}{3888}(27)
\]
\[
T_3(12) = 3 + \frac{1}{2} - \frac{9}{216} + \frac{27}{3888}
\]
\[
T_3(12) = 3 + \frac{1}{2} - \frac{1}{24} + \frac{1}{144}
\]
Now calculate the decimal values:

\[
T_3(12) = 3 + 0.5 - 0.04167 + 0.00694
\]
\[
T_3(12) \approx 3.4653
\]

Thus, the approximation of \( \sqrt{12} \) using the degree 3 Taylor polynomial is approximately \( 3.4653 \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How would increasing the degree of the Taylor polynomial affect the approximation?
2. What is the remainder term in the Taylor series, and how does it affect the approximation accuracy?
3. Can we use the Taylor polynomial to approximate other values near \(x = 9\)?
4. How does the Taylor polynomial differ from a Maclaurin series?
5. What are the potential sources of error when using a Taylor polynomial for approximation?

**Tip**: The more terms (higher degree) in a Taylor polynomial, the closer the approximation will be to the actual function value.

Find the degree 3 Taylor Polynomial for f(x) = sqrt(x) centered at a = 9 and use it to approximate sqrt(12).

This page explains how to find the degree 3 Taylor polynomial for f(x) = sqrt(x), centered at a = 9, and uses it to approximate sqrt(12). Learn step-by-step how to compute derivatives, evaluate at a = 9, and construct the Taylor polynomial.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation