Question

Please help. the packet is due tonight

Answer 1

[C]
`\displaystyle (-3)/(250)`

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

• Terms/Coefficients
• Factoring
• Functions
• Function Notation
• Conjugations

Calculus

• Limits
• Limit 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)`
• Derivatives
• Definition of a Derivative:
  `\displaystyle f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)`

Explanation:

Step 1: Define

Identify

`\displaystyle g(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)`

`\displaystyle f(x) = (3)/(√(x - 4))`

`\displaystyle g(29)`

Step 2: Differentiate

1. Substitute in function [Function g(x)]:
   `\displaystyle g(x) = \lim_(h \to 0) ((3)/(√(x + h - 4)) - (3)/(√(x - 4)))/(h)`
2. Substitute in x [Function g(x)]:
   `\displaystyle g(29) = \lim_(h \to 0) ((3)/(√(29 + h - 4)) - (3)/(√(29 - 4)))/(h)`
3. Simplify:
   `\displaystyle g(29) = \lim_(h \to 0) ((3)/(√(25 + h)) - (3)/(5))/(h)`
4. Rewrite:
   `\displaystyle g(29) = \lim_(h \to 0) ((15)/(5√(25 + h)) - (3√(25 + h))/(5√(25 + h)))/(h)`
5. [Subtraction] Combine like terms:
   `\displaystyle g(29) = \lim_(h \to 0) ((15 - 3√(25 + h))/(5√(25 + h)))/(h)`
6. Factor:
   `\displaystyle g(29) = \lim_(h \to 0) ((3(5 - √(25 + h)))/(5√(25 + h)))/(h)`
7. Rewrite:
   `\displaystyle g(29) = \lim_(h \to 0) (3(5 - √(25 + h)))/(5h√(25 + h))`
8. Rewrite [Limit Property - Multiplied Constant]:
   `\displaystyle g(29) = (3)/(5) \lim_(h \to 0) (5 - √(25 + h))/(h√(25 + h))`
9. Root Conjugation:
   `\displaystyle g(29) = (3)/(5) \lim_(h \to 0) (5 - √(25 + h))/(h√(25 + h)) \cdot (5 + √(25 + h))/(5 + √(25 + h))`
10. Multiply:
    `\displaystyle g(29) = (3)/(5) \lim_(h \to 0) (-h)/(5h√(25 + h) + h^2 + 25h)`
11. Factor:
    `\displaystyle g(29) = (3)/(5) \lim_(h \to 0) (-h)/(h(5√(25 + h) + h + 25))`
12. Simplify:
    `\displaystyle g(29) = (3)/(5) \lim_(h \to 0) (-1)/(5√(25 + h) + h + 25)`
13. Evaluate limit [Limit Rule - Variable Direct Substitution]:
    `\displaystyle g(29) = (3)/(5) \lim_(h \to 0) (-1)/(5√(25 + 0) + 0 + 25)`
14. Simplify:
    `\displaystyle g(29) = (3)/(5) \cdot (-1)/(50)`
15. Multiply:
    `\displaystyle g(29) = (-3)/(250)`

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

Unit: Derivatives

Book: College Calculus 10e