Mathematics, 15.06.2021 19:20 prettygirl1053480

Please help. the packet is due tonight

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Answer from: brainewashed11123

[C] $\displaystyle \frac{-3}{250}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra I

Terms/CoefficientsFactoringFunctionsFunction NotationConjugations

Calculus

LimitsLimit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$ Limit Property [Multiplied Constant]: $\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$ DerivativesDefinition of a Derivative: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle g(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$\displaystyle f(x) = \frac{3}{\sqrt{x - 4}}$

$\displaystyle g(29)$

Step 2: Differentiate

Substitute in function [Function g(x)]: $\displaystyle g(x) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{x + h - 4}} - \frac{3}{\sqrt{x - 4}}}{h}$ Substitute in x [Function g(x)]: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{29 + h - 4}} - \frac{3}{\sqrt{29 - 4}}}{h}$ Simplify: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{25 + h}} - \frac{3}{5}}{h}$ Rewrite: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15}{5\sqrt{25 + h}} - \frac{3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}$ [Subtraction] Combine like terms: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15 - 3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}$ Factor: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3(5 - \sqrt{25 + h})}{5\sqrt{25 + h}}}{h}$ Rewrite: $\displaystyle g(29) = \lim_{h \to 0} \frac{3(5 - \sqrt{25 + h})}{5h\sqrt{25 + h}}$ Rewrite [Limit Property - Multiplied Constant]: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}}$ Root Conjugation: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}} \cdot \frac{5 + \sqrt{25 + h}}{5 + \sqrt{25 + h}}$ Multiply: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{5h\sqrt{25 + h} + h^2 + 25h}$ Factor: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{h(5\sqrt{25 + h} + h + 25)}$ Simplify: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + h} + h + 25}$ Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + 0} + 0 + 25}$ Simplify: $\displaystyle g(29) = \frac{3}{5} \cdot \frac{-1}{50}$ Multiply: $\displaystyle g(29) = \frac{-3}{250}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

The answer is 1 because you have to flip the sign when dividing by negative number and 3

Answer from: Quest

The answer to your question is 1/9