Mathematics, 11.07.2019 00:00 2023greenlanden

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

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Answer from: kahmir15

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

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra I

Functions

Function Notation

Calculus

Differentiation

DerivativesDerivative 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:

f(x) = cxⁿf’(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

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

Step 3: Evaluate

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

Step 4: Write Taylor Polynomial

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

Answer from: Quest

Idon’t understand sorry

Answer from: Quest

8121 because

130,000 divided by 16 is 8121

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

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