Comprehensive Study Guide: Functions, Domain, Range, and Composite Algebra

Course Overview, Policies, and Schedule

Adjustment and Focus:
- Initial focus during the first couple of weeks is typically split due to side jobs and outside commitments.
- Full focus and solid integration into the course routine develop naturally within a few weeks.
Assessment and Expectation Document:
- Students must self-assess and judge their baseline expectations using the distributed assessment sheet.
- The self-assessment document must be kept in the student's course folder and referred to periodically throughout the term.
- The course curriculum covers all topics listed on the assessment sheet plus additional content.
Algebra Preparation Warning:
- Fundamental algebraic concepts recur continuously throughout the course.
- Mastery of algebra is critical; failure to master basic algebraic operations severely impacts calculus performance.
Attendance and Logistics:
- Roster sign-in requires every student to check or write their name for headcount tracking.
- Textbooks are essential; students should copy exercises for early sections (2.1, 2.2, 2.3) immediately to avoid falling behind if textbook delivery is delayed.
- Lack of a physical textbook is not an acceptable excuse for incomplete homework assignments.
Homework and Quiz Schedule:
- Week 1: No homework or quiz is due.
- Week 2 Monday: Class is cancelled / no session.
- Week 2 Wednesday: Section 2.4 coverage begins.
- Week 2 Friday: Quiz #1 covering Sections 2.1, 2.2, and 2.3.
- Homework Submission Rule: Exactly one staple must be used for submitting homework assignments, regardless of how many sections are included in the set (e.g., Sections 2.1, 2.2, and 2.3 turned in together as one stapled entity).
Course Learning Outcomes:
- The syllabus outlines specific learning outcomes serving as a formal academic contract between instructor and student.
- Primary course pillars: Functions, Limits, Derivatives, and Integrals.
- Chapter 2 focus: Functions, Limits, and introduction to Derivatives.
- Chapter 3 focus: Full-blown derivative rules and applications across approximately 30 sub-sections.

Fundamental Concepts of Functions

Definition of a Function:
- A function $f$ is a rule that assigns to each element $x$ in a domain set $A$ one and only one element $f(x)$ in a target set $B$ .
- If an input $x$ maps to multiple outputs in set $B$ (e.g., $f_1$ and $f_2$ ), the rule is a relation, not a function.
Rule Representation:
- Example rule: $f(x) = x^3 + 1$ .
- Input mapping: An input of $x = 2$ yields $f(2) = 2^3 + 1 = 9$ .
Step-by-Step Function Evaluation Example:
- Rule: $g(u) = (3u - 2)^{3/2}$
- Evaluating $g(1)$ :
- $g(1) = (3(1) - 2)^{3/2}$
- $g(1) = (3 - 2)^{3/2}$
- $g(1) = 1^{3/2} = 1$
- Evaluating $g%5Cleft(%5Cfrac{11}{3}%5Cright)$ :
- $g%5Cleft(%5Cfrac{11}{3}%5Cright) = %5Cleft(3%5Cleft(%5Cfrac{11}{3}%5Cright) - 2%5Cright)^{3/2}$
- $g%5Cleft(%5Cfrac{11}{3}%5Cright) = (11 - 2)^{3/2} = 9^{3/2}$
- Express $9$ as $3^2$ : $g%5Cleft(%5Cfrac{11}{3}%5Cright) = (3^2)^{3/2}$
- Apply exponent multiplication: 3^{2 %5Ctimes %5Cfrac{3}{2}} = 3^3 = 27
- Evaluating $g(u + 1)$ :
- $g(u + 1) = (3(u + 1) - 2)^{3/2}$
- Distribute the constant: $g(u + 1) = (3u + 3 - 2)^{3/2}$
- Simplify the inner terms: $g(u + 1) = (3u + 1)^{3/2}$

Algebraic Concepts and Exponent Rules

Base One Exponent Rule:
- The number $1$ raised to any real power is always equal to $1$ :
- $1^p = 1$
- Examples: 1^{%5Csqrt{2}} = 1, 1^{%5Cpi} = 1, $1^{3/2} = 1$
Power of a Power Rule:
- Raising a power to another power involves multiplying the exponents:
- (a^b)^c = a^{b %5Ctimes c}
Perfect Square Trinomial Expansion:
- Square of a sum:
- $(a + b)^2 = a^2 + 2ab + b^2$
- Square of a difference:
- $(a - b)^2 = a^2 - 2ab + b^2$
- Common Error Warning: $(a + b)^2 %5Cneq a^2 + b^2$ . Omitting the middle term ( $2ab$ ) is a critical error that leads to loss of marks on exams.
Calculator Policy:
- Portions of midterms and final exams strictly prohibit calculator usage. Exact algebraic manipulations must be executed by hand.

Domain and Range Analysis

Domain Definition:
- The set of all real values of $x$ for which the function $f(x)$ produces real and well-defined outputs.
Two Primary Sources of Domain Restrictions:
- Even Roots: Radicands under even roots must be non-negative ( $%5Cge 0$ ) to prevent imaginary/complex values.
- Division by Zero: Denominators must never equal zero, as division by zero produces undefined/infinite results:
- $%5Cfrac{1}{0} = %5Ctext{undefined}$
Interval Notation and Symbols:
- Open Parenthesis ( or ) indicates exclusion of the endpoint.
- Square Bracket [ or ] indicates inclusion of the endpoint.
- All Real Numbers Notation: $(-%5Cinfty, %5Cinfty)$ .
- Infinity Exclusion Rule: Positive and negative infinity must ALWAYS use parentheses, never brackets (e.g., [-%5Cinfty, %5Cinfty] is mathematically invalid because infinity is not a bounded real number).
Domain Example 1: Function with No Restrictions
- $f(x) = %5Csqrt{x^2 + 1}$
- Analysis: Since $x^2 %5Cge 0$ for all real $x$ , $x^2 + 1 %5Cge 1 > 0$ .
- Conclusion: The expression under the square root is strictly positive for all real numbers.
- Domain: $(-%5Cinfty, %5Cinfty)$
Domain Example 2: Combined Restrictions (Midterm Caliber)
- $f(x) = %5Cfrac{%5Csqrt{1 - x}}{x^2 - 4}$
- Step 1: Numerator Restriction (Even Root)
- The radicand must be non-negative: $1 - x %5Cge 0 %5Cimplies x %5Cle 1$
- Interval form of numerator constraint: $(-%5Cinfty, 1]$
- Step 2: Denominator Restriction (Non-Zero Division)
- Set denominator to zero to locate undefined points: $x^2 - 4 = 0$
- Solve for $x$ : $x^2 = 4 %5Cimplies x = %5Cpm 2$
- Critical Warning: Taking the square root of a squared term yields both positive and negative roots ( $%5Cpm$ ). Forgetting the negative root ( $-2$ ) loses valid solutions and causes downstream errors in min/max and inflection point calculations.
- Excluded points: $x %5Cneq 2$ and $x %5Cneq -2$
- Step 3: Intersecting Constraints
- Primary constraint: $x %5Cle 1$ (this automatically excludes $x = 2$ ).
- Secondary constraint: Exclude $x = -2$ from the interval $(-%5Cinfty, 1]$ .
- Domain Interval: $(-%5Cinfty, -2) %5Ccup (-2, 1]$

Graphing Functions and the Vertical Line Test

Graphing via Point Plotting:
- Function: $g(x) = 4 - %5Csqrt{x}$
- Domain Analysis: Radicand $x %5Cge 0 %5Cimplies [0, %5Cinfty)$ .
- Selected Table of Values:
- $x = 0 %5Cimplies g(0) = 4 - %5Csqrt{0} = 4$
- $x = 1 %5Cimplies g(1) = 4 - %5Csqrt{1} = 3$
- $x = 2 %5Cimplies g(2) = 4 - %5Csqrt{2} %5Capprox 4 - 1.4 = 2.6$
- $x = 4 %5Cimplies g(4) = 4 - %5Csqrt{4} = 2$
- $x = 9 %5Cimplies g(9) = 4 - %5Csqrt{9} = 1$
- $x = 16 %5Cimplies g(16) = 4 - %5Csqrt{16} = 0$
- Domain and Range Determination:
- Domain: $[0, %5Cinfty)$
- Range: Maximum value of $g(x)$ is $4$ (at $x = 0$ ), decreasing toward negative infinity as $x$ grows: $(-%5Cinfty, 4]$ .
Vertical Line Test:
- A visual test used to determine if a graph represents a valid function.
- Rule: If a vertical line intersects a graph at more than one point, the graph does NOT represent a function.
- Case Studies:
- Vertical Ellipse / Closed Curve: Fails (intersects line at two points) %5Crightarrow NOT a function.
- Standard Single-Valued Curves: Passes (intersects line at exactly one point) %5Crightarrow Function.
- Circle centered at origin: Fails (intersects line at two points) %5Crightarrow NOT a function.

Algebra of Functions and Composite Functions

Basic Combination Rules:
- Addition: $(f + g)(x) = f(x) + g(x)$
- Multiplication: $(f %5Ccdot g)(x) = f(x) %5Ccdot g(x)$
- Division: $%5Cleft(%5Cfrac{f}{g}%5Cright)(x) = %5Cfrac{f(x)}{g(x)}$ , provided $g(x) %5Cneq 0$
Definition of Composite Functions:
- Symbol notation: $(g %5Ccirc f)(x) = g(f(x))$
- Conceptual definition: The function $f(x)$ acts on $x$ , and the resulting output becomes the input domain for function $g$ .
- Substitution Rule: Every occurrence of $x$ in the outer function $g(x)$ is entirely replaced by the expression for $f(x)$ .
- Non-Commutativity Property: In general, $(f %5Ccirc g)(x) %5Cneq (g %5Ccirc f)(x)$ in 99% of cases (exceptions include inverse functions such as logarithmic and exponential pairings).
Comprehensive Composite Example 1:
- Given Functions: $f(x) = 2%5Csqrt{x} + 3$ and $g(x) = x^2 + 1$
- Part 1: Determine $(f %5Ccirc g)(x) = f(g(x))$
- Replace variable $x$ in $f(x)$ with $g(x)$ :
- $f(g(x)) = 2%5Csqrt{g(x)} + 3$
- $f(g(x)) = 2%5Csqrt{x^2 + 1} + 3$
- No further algebraic simplification is possible.
- Part 2: Determine $(g %5Ccirc f)(x) = g(f(x))$
- Replace variable $x$ in $g(x)$ with $f(x)$ :
- $g(f(x)) = (f(x))^2 + 1$
- $g(f(x)) = (2%5Csqrt{x} + 3)^2 + 1$
- Expand using $(a + b)^2 = a^2 + 2ab + b^2$ :
- $(2%5Csqrt{x} + 3)^2 = (2%5Csqrt{x})^2 + 2(2%5Csqrt{x})(3) + (3)^2$
- $= 4x + 12%5Csqrt{x} + 9$
- Add the remaining term $+ 1$ :
- $g(f(x)) = 4x + 12%5Csqrt{x} + 9 + 1 = 4x + 12%5Csqrt{x} + 10$
Comprehensive Composite Example 2 (Direct Point Evaluation):
- Given Functions: $f(x) = %5Cfrac{1}{x - 1}$ and $g(x) = x^2 + 1$
- Task: Evaluate $h(2)$ where $h(x) = (g %5Ccirc f)(x) = g(f(x))$
- Strategy: Substitute the evaluation point directly into the algebraic expressions without performing full symbolic expansion, reducing execution time and algebraic error risk.
- Symbolic expression setup:
- $h(x) = g(f(x)) = %5Cleft(%5Cfrac{1}{x - 1}%5Cright)^2 + 1$
- Direct evaluation at $x = 2$ :
- $f(2) = %5Cfrac{1}{2 - 1} = %5Cfrac{1}{1} = 1$
- $h(2) = g(f(2)) = g(1) = (1)^2 + 1 = 2$