Answer:
[C] [tex]\displaystyle \frac{-3}{250}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Algebra I
- Terms/Coefficients
- Factoring
- Functions
- Function Notation
- Conjugations
Calculus
- Limits
- Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
- Limit Property [Multiplied Constant]: [tex]\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)[/tex]
- Derivatives
- Definition of a Derivative: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle g(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/tex]
[tex]\displaystyle f(x) = \frac{3}{\sqrt{x - 4}}[/tex]
[tex]\displaystyle g(29)[/tex]
Step 2: Differentiate
- Substitute in function [Function g(x)]: [tex]\displaystyle g(x) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{x + h - 4}} - \frac{3}{\sqrt{x - 4}}}{h}[/tex]
- Substitute in x [Function g(x)]: [tex]\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{29 + h - 4}} - \frac{3}{\sqrt{29 - 4}}}{h}[/tex]
- Simplify: [tex]\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{25 + h}} - \frac{3}{5}}{h}[/tex]
- Rewrite: [tex]\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15}{5\sqrt{25 + h}} - \frac{3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}[/tex]
- [Subtraction] Combine like terms: [tex]\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15 - 3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}[/tex]
- Factor: [tex]\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3(5 - \sqrt{25 + h})}{5\sqrt{25 + h}}}{h}[/tex]
- Rewrite: [tex]\displaystyle g(29) = \lim_{h \to 0} \frac{3(5 - \sqrt{25 + h})}{5h\sqrt{25 + h}}[/tex]
- Rewrite [Limit Property - Multiplied Constant]: [tex]\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}}[/tex]
- Root Conjugation: [tex]\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}}[/tex]
- Multiply: [tex]\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{5h\sqrt{25 + h} + h^2 + 25h}[/tex]
- Factor: [tex]\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{h(5\sqrt{25 + h} + h + 25)}[/tex]
- Simplify: [tex]\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + h} + h + 25}[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + 0} + 0 + 25}[/tex]
- Simplify: [tex]\displaystyle g(29) = \frac{3}{5} \cdot \frac{-1}{50}[/tex]
- Multiply: [tex]\displaystyle g(29) = \frac{-3}{250}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
