Advanced Placement Calculus AB Pre-requisites Study Notes

Course Assignment: Advanced Placement (AP) Calculus AB.
Instructor/Identifier: Lemes - p03 - sF02.
Source File: calc_pre-requisites_worksheet.pdf (File Size: $1.25\,MB$ ).
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Product of Rational Exponents (Problem 78):
- Expression: $(5a^{2/3})(4a^{3/2})$
- Mathematical Concept: To multiply terms with like bases, multiply the coefficients and add the exponents according to the rule $x^a \times x^b = x^{a+b}$ .
- Coefficient Calculation: $5 \times 4 = 20$ .
- Exponent Addition: $\frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6}$ .
- Simplified Result: $20a^{13/6}$ .
Quotient and Negative Exponents (Problem 79):
- Expression: $\frac{4xy^{-2}}{12x^{-1/3}y^{-5}}$
- Coefficient Simplification: $\frac{4}{12}$ reduces to $\frac{1}{3}$ .
- Variable $x$ Evaluation: Apply the rule $\frac{x^a}{x^b} = x^{a-b}$ . Here, $x^{1 - (-1/3)} = x^{1 + 1/3} = x^{4/3}$ .
- Variable $y$ Evaluation: Apply the same rule: $y^{-2 - (-5)} = y^{-2 + 5} = y^3$ .
- Simplified Result: $\frac{1}{3}x^{4/3}y^3$
Power of a Power Rule (Problem 80):
- Expression: $(4a^{5/3})^{3/2}$
- Rule Application: $(x^a)^b = x^{ab}$ . Both the coefficient and the variable are raised to the power of $\frac{3}{2}$ .
- Coefficient Evaluation: $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$ .
- Variable Evaluation: $a^{(5/3) \times (3/2)} = a^{5/2}$ .
- Simplified Result: $8a^{5/2}$ .

Discrete Function $f(x)$ : Defined as a finite set of ordered pairs: $f(x) = \{(3, 5), (2, 4), (1, 7)\}$ .
Algebraic Function $g(x)$ : A square root function defined as $g(x) = \sqrt{(x - 3)}$ .
Discrete Function $h(x)$ : Defined as a finite set of ordered pairs: $h(x) = \{(3, 2), (4, 3), (1, 6)\}$ .
Algebraic Function $k(x)$ : A quadratic function defined as $k(x) = x^2 + 5$ .

Sum of Functions (Problem 81):
- Expression: $(f + h)(1)$
- Evaluation: Find $f(1)$ and $h(1)$ . From the sets provided, $f(1) = 7$ and $h(1) = 6$ .
- Calculation: $7 + 6 = 13$ .
Difference of Functions (Problem 82):
- Expression: $(k - g)(5)$
- Evaluation of $k(5)$ : $5^2 + 5 = 25 + 5 = 30$ .
- Evaluation of $g(5)$ : $\sqrt{(5 - 3)} = \sqrt{2}$ .
- Calculation: $30 - \sqrt{2}$ .
Single Point Evaluation (Problem 85):
- Expression: $h(3)$
- Evaluation: Locate the ordered pair in set $h$ with an $x$ -coordinate of 3. The pair is $(3, 2)$ .
- Result: 2.

Composite Evaluation (Problem 83):
- Expression: $f(h(3))$
- Inner Function: $h(3) = 2$ .
- Outer Function: $f(2) = 4$ .
- Result: 4.
Algebraic Composition (Problem 84):
- Expression: $g(k(7))$
- Inner Function: $k(7) = 7^2 + 5 = 49 + 5 = 54$ .
- Outer Function: $g(54) = \sqrt{(54 - 3)} = \sqrt{51}$ .
- Result: $\sqrt{51}$ .
Nested Algebraic Composition (Problem 86):
- Expression: $g(g(9))$
- First Evaluation: $g(9) = \sqrt{(9 - 3)} = \sqrt{6}$ .
- Second Evaluation: $g(\sqrt{6}) = \sqrt{(\sqrt{6} - 3)}$ .
- Note: In the real number system, this result involves taking the square root of a negative value, as \sqrt{6} < 3.
Discrete Function Inverse (Problem 87):
- Expression: $f^{-1}(4)$
- Concept: For an inverse function $f^{-1}(y) = x$ , one must find the input $x$ that yields the output $y$ in the original function.
- Evaluation: In $f(x) = \{(3, 5), (2, 4), (1, 7)\}$ , the output 4 corresponds to the input 2.
- Result: 2.
Algebraic Function Inverse (Problem 88):
- Expression: $k^{-1}(x)$
- Procedure: Replace $k(x)$ with $y$ , yielding $y = x^2 + 5$ . Swap $x$ and $y$ to get $x = y^2 + 5$ . Solve for $y$ : $y^2 = x - 5$ .
- Result: $k^{-1}(x) = \sqrt{(x - 5)}$ (assuming the restricted domain of the original function to the positive branch).