Question

Find the second degree Taylor polynomial for f(x)= sqrt(x^2+8) at the number x=1

Answer 1

Answer:

$\displaystyle P_2(x) = 3 + \frac{1}{3}(x - 1) + \frac{4}{27}(x - 1)^2$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

Functions

• Function Notation

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                 $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Taylor Polynomials

• Approximating Transcendental and Elementary Functions
• $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^n(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

Step 1: Define

Identify

f(x) = √(x² + 8)

Center: x = 1

n = 2

Step 2: Differentiate

1. [Function] 1st Derivative:                                                                               $\displaystyle f'(x) = \frac{x}{\sqrt{x^2 + 8}}$
2. [Function] 2nd Derivative:                                                                             $\displaystyle f''(x) = \frac{8}{(x^2 + 8)^\bigg{\frac{3}{2}}}$

Step 3: Evaluate

1. Substitute in center x [Function]:                                                                 $\displaystyle f(1) = \sqrt{1^2 + 8}$
2. Simplify:                                                                                                         $\displaystyle f(1) = 3$
3. Substitute in center x [1st Derivative]:                                                         $\displaystyle f'(1) = \frac{1}{\sqrt{1^2 + 8}}$
4. Simplify:                                                                                                         $\displaystyle f'(1) = \frac{1}{3}$
5. Substitute in center x [2nd Derivative]:                                                       $\displaystyle f''(1) = \frac{8}{(1^2 + 8)^\bigg{\frac{3}{2}}}$
6. Simplify:                                                                                                         $\displaystyle f''(1) = \frac{8}{27}$

Step 4: Write Taylor Polynomial

1. Substitute in derivative function values [Taylor Polynomial]:                     $\displaystyle P_2(x) = \frac{3}{0!} + \frac{\frac{1}{3}}{1!}(x - c) + \frac{\frac{8}{27}}{2!}(x - c)^2$
2. Simplify:                                                                                                         $\displaystyle P_2(x) = 3 + \frac{1}{3}(x - c) + \frac{4}{27}(x - c)^2$
3. Substitute in center c:                                                                                   $\displaystyle P_2(x) = 3 + \frac{1}{3}(x - 1) + \frac{4}{27}(x - 1)^2$

Topic: AP Calculus BC (Calculus I + II)  

Unit: Taylor Polynomials and Approximations  

Book: College Calculus 10e