📐 Concept diagram

03-01 - Functions

Phase: 3 | Subject: 03-01 Prerequisites: 01-05-linear-functions.md, 01-06-systems-of-linear-equations.md Next subject: 03-02-transformations-of-functions.md

Learning Objectives

By the end of this subject, you will be able to:

1. Define what makes a relation a function
2. Use function notation f(x) correctly
3. Find domain and range from formulas and graphs
4. Identify one-to-one, onto, and bijective functions
5. Compose two functions and evaluate composite functions (f∘g)(x)
6. Identify even and odd functions from equations and graphs
7. Find inverse functions algebraically

Core Content

What is a Function?

A function is a rule that assigns each input (from the domain) exactly ONE output (in the range).

Formal definition: f: A → B means f maps every element of set A to exactly one element of set B.

Vertical line test: On a graph, if any vertical line intersects the curve more than once, it's NOT a function.

Examples: - f(x) = x² is a function (each x gives one y) - x² + y² = 1 (circle) is NOT a function (one x gives two y values)

Function Notation

f(x) means "the output of function f when the input is x."

Example: If f(x) = 2x + 3: - f(2) = 2(2) + 3 = 7 - f(-1) = 2(-1) + 3 = 1 - f(a) = 2a + 3 - f(x + 1) = 2(x + 1) + 3 = 2x + 5

Important: f(x + 1) is NOT f(x) + 1. The input is the entire expression (x + 1).

Domain and Range

Domain: All valid input values (x-values) Range: All possible output values (y-values)

Finding Domain

1. Denominator cannot be zero
2. Expression under square root must be non-negative (for real functions)
3. Expression inside logarithm must be positive

Example: f(x) = √(x - 2) Domain: x - 2 ≥ 0, so x ≥ 2. In interval notation: [2, ∞)

Example: f(x) = 1/(x - 3) Domain: x - 3 ≠ 0, so x ≠ 3. In interval notation: (-∞, 3) ∪ (3, ∞)

Finding Range

Often harder. Look at: - The graph's vertical extent - Limits as x → ±∞ - Minimum/maximum values

Example: f(x) = x² Range: all non-negative reals, [0, ∞)

Types of Functions

One-to-One (Injective)

Every output comes from exactly one input. No two x-values map to the same y-value.

Horizontal line test: If any horizontal line hits the graph more than once, NOT one-to-one.

Example: f(x) = 2x + 1 is one-to-one. f(x) = x² is NOT one-to-one (f(2) = f(-2) = 4).

Onto (Surjective)

Every possible output in the codomain is actually achieved by some input.

Example: f: ℝ → ℝ, f(x) = x³ is onto (hits every real number). f: ℝ → ℝ, f(x) = x² is NOT onto (never outputs negative numbers).

Bijective

Both one-to-one AND onto. Has an inverse function.

Example: f(x) = 2x + 1 is bijective from ℝ to ℝ. f(x) = x² is not bijective from ℝ to ℝ (fails both tests).

Composite Functions

The composition of two functions applies one function AFTER another.

Notation: (f ∘ g)(x) = f(g(x)) — read as "f of g of x" or "f composed with g."

Example: If f(x) = x² and g(x) = 2x + 1: - (f ∘ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)² = 4x² + 4x + 1 - (g ∘ f)(x) = g(f(x)) = g(x²) = 2x² + 1

Important: f ∘ g ≠ g ∘ f in general. Order matters!

Domain of f ∘ g: x must be in the domain of g AND g(x) must be in the domain of f.

Even and Odd Functions

• Even function: f(-x) = f(x) for all x. Graph is symmetric about the y-axis.
• Example: f(x) = x², f(x) = cos(x), f(x) = |x|.
• Odd function: f(-x) = -f(x) for all x. Graph is symmetric about the origin (180° rotation).
• Example: f(x) = x³, f(x) = sin(x), f(x) = 1/x.

Tip: Polynomials with only even powers (x², x⁴, ...) are even; only odd powers (x, x³, ...) are odd. Mixed powers = neither.

Inverse Functions

If f maps x → y, then f⁻¹ maps y → x.

To find f⁻¹: 1. Write y = f(x) 2. Swap x and y 3. Solve for y 4. The result is f⁻¹(x)

Example: Find f⁻¹ if f(x) = 2x + 3

1. y = 2x + 3
2. x = 2y + 3
3. x - 3 = 2y
4. y = (x - 3)/2

f⁻¹(x) = (x - 3)/2

Verification: f(f⁻¹(x)) = 2((x-3)/2) + 3 = x - 3 + 3 = x ✓

Important: Only one-to-one functions have inverses.

Key Terms

• 03 01 Functions
• Bijective
• Composite Functions
• Correct: A)
• Correct: B)
• Correct: C)
• Domain and Range
• Even and Odd Functions
• Example 1: Domain and range
• Example 2: Inverse function
• Example 3: Composite functions and domain
• Finding Domain

Worked Examples

Example 1: Domain and range

f(x) = √(x + 4) / (x - 1)

1. Domain: x + 4 ≥ 0 AND x - 1 ≠ 0
2. x ≥ -4 AND x ≠ 1
3. Domain: [-4, 1) ∪ (1, ∞)

Range: As x → ∞, f(x) → 0 from above. At x = -4, f(-4) = 0/(-5) = 0. Between -4 and 1, f increases from 0 to +∞. After 1, f decreases from +∞ to 0. Range: (0, ∞).

Example 2: Inverse function

f(x) = (x - 1)/(x + 2), find f⁻¹

1. y = (x - 1)/(x + 2)
2. x = (y - 1)/(y + 2)
3. x(y + 2) = y - 1
4. xy + 2x = y - 1
5. xy - y = -1 - 2x
6. y(x - 1) = -(1 + 2x)
7. y = -(1 + 2x)/(x - 1) = (2x + 1)/(1 - x)

f⁻¹(x) = (2x + 1)/(1 - x)

Example 3: Composite functions and domain

Find (f∘g)(x) and its domain if f(x) = √x and g(x) = x - 4.

(f∘g)(x) = f(g(x)) = f(x - 4) = √(x - 4)

Domain: g(x) = x - 4 must be in the domain of f (which requires input ≥ 0). So x - 4 ≥ 0 → x ≥ 4. Domain: [4, ∞).

Quiz

Q1: What does the concept of Bijective primarily refer to in this subject?

A) The definition and application of Bijective B) A visual representation of Bijective C) A historical anecdote about Bijective D) A computational error related to Bijective

Correct: A)

• If you chose A: Bijective is defined as: the definition and application of bijective. The other options describe different aspects that are not the primary focus. Correct!
• If you chose B: This is incorrect. Bijective is defined as: the definition and application of bijective. The other options describe different aspects that are not the primary focus.
• If you chose C: This is incorrect. Bijective is defined as: the definition and application of bijective. The other options describe different aspects that are not the primary focus.
• If you chose D: This is incorrect. Bijective is defined as: the definition and application of bijective. The other options describe different aspects that are not the primary focus.

Q2: What is the primary purpose of Composite Functions?

A) It replaces all other methods in this domain B) It is primarily a historical notation system C) It is used only in advanced research contexts D) It is used to composite functions in mathematical analysis

Correct: D)

• If you chose A: This is incorrect. Composite Functions serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose B: This is incorrect. Composite Functions serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose C: This is incorrect. Composite Functions serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose D: Composite Functions serves the purpose described in the correct answer. The other options misrepresent its role. Correct!

Q3: Which statement about Domain and Range is TRUE?

A) Domain and Range is a fundamental concept covered in this subject B) Domain and Range is not related to this subject C) Domain and Range is mentioned only as a historical footnote D) Domain and Range is an advanced topic beyond this subject's scope

Correct: A)

• If you chose A: Domain and Range is a fundamental concept covered in this subject. This subject covers Domain and Range as part of its core content. Correct!
• If you chose B: This is incorrect. Domain and Range is a fundamental concept covered in this subject. This subject covers Domain and Range as part of its core content.
• If you chose C: This is incorrect. Domain and Range is a fundamental concept covered in this subject. This subject covers Domain and Range as part of its core content.
• If you chose D: This is incorrect. Domain and Range is a fundamental concept covered in this subject. This subject covers Domain and Range as part of its core content.

Q4: Based on the worked examples in this subject, what is the correct result?

A) An unrelated numerical value B) 2x + 3 C) A different result from a common mistake D) The inverse of the correct answer

Correct: B)

• If you chose A: This is incorrect. The worked examples show that the result is 2x + 3. The other options represent common errors.
• If you chose B: The worked examples show that the result is 2x + 3. The other options represent common errors. Correct!
• If you chose C: This is incorrect. The worked examples show that the result is 2x + 3. The other options represent common errors.
• If you chose D: This is incorrect. The worked examples show that the result is 2x + 3. The other options represent common errors.

Q5: How are Domain and Range and Even and Odd Functions related?

A) Domain and Range and Even and Odd Functions are closely related concepts B) Domain and Range and Even and Odd Functions are completely unrelated topics C) Domain and Range is a special case of Even and Odd Functions D) Domain and Range is the inverse of Even and Odd Functions

Correct: A)

• If you chose A: Both Domain and Range and Even and Odd Functions are covered in this subject as interconnected topics. Correct!
• If you chose B: This is incorrect. Both Domain and Range and Even and Odd Functions are covered in this subject as interconnected topics.
• If you chose C: This is incorrect. Both Domain and Range and Even and Odd Functions are covered in this subject as interconnected topics.
• If you chose D: This is incorrect. Both Domain and Range and Even and Odd Functions are covered in this subject as interconnected topics.

Q6: What is a common pitfall when working with Finding Domain?

A) Finding Domain has no common misconceptions B) The main error with Finding Domain is using it when it is not needed C) A common mistake is confusing Finding Domain with a similar concept D) Finding Domain is always computed the same way in all contexts

Correct: C)

• If you chose A: This is incorrect. Students often confuse Finding Domain with similar-sounding or related concepts. Pay attention to the precise definitions.
• If you chose B: This is incorrect. Students often confuse Finding Domain with similar-sounding or related concepts. Pay attention to the precise definitions.
• If you chose C: Students often confuse Finding Domain with similar-sounding or related concepts. Pay attention to the precise definitions. Correct!
• If you chose D: This is incorrect. Students often confuse Finding Domain with similar-sounding or related concepts. Pay attention to the precise definitions.

Q7: When should you apply What Is A Function??

A) What Is A Function? is not practically useful B) Apply What Is A Function? to solve problems in this subject's domain C) Avoid What Is A Function? unless explicitly instructed D) Use What Is A Function? only in pure mathematics contexts

Correct: B)

• If you chose A: This is incorrect. What Is A Function? is a practical tool used throughout this subject to solve relevant problems.
• If you chose B: What Is A Function? is a practical tool used throughout this subject to solve relevant problems. Correct!
• If you chose C: This is incorrect. What Is A Function? is a practical tool used throughout this subject to solve relevant problems.
• If you chose D: This is incorrect. What Is A Function? is a practical tool used throughout this subject to solve relevant problems.

Practice Problems

1. Evaluate f(3) if f(x) = x² - 2x + 1 Answer: f(3) = 9 - 6 + 1 = 4

2. Domain of f(x) = √(x + 5) Answer: x + 5 ≥ 0, so x ≥ -5. Domain: [-5, ∞)

3. Domain of f(x) = 1/(x² - 4) Answer: x² - 4 ≠ 0, so x ≠ ±2. Domain: (-∞, -2) ∪ (-2, 2) ∪ (2, ∞)

4. Is f(x) = x³ one-to-one? Answer: Yes. If x₁³ = x₂³, then x₁ = x₂. Passes horizontal line test.

5. Find f⁻¹ if f(x) = 3x - 7 Answer: y = 3x - 7, swap: x = 3y - 7, solve: y = (x + 7)/3. f⁻¹(x) = (x + 7)/3

6. If f(x) = 2x + 1 and g(x) = x², find (f∘g)(3). Answer: g(3) = 9, f(9) = 2(9) + 1 = 19. Or: (f∘g)(x) = 2x² + 1, so (f∘g)(3) = 2(9) + 1 = 19.

7. Is f(x) = x⁴ - 3x² even, odd, or neither? Answer: f(-x) = (-x)⁴ - 3(-x)² = x⁴ - 3x² = f(x). Even.

Summary

Key takeaways:

• Function: each input has exactly one output
• Vertical line test identifies functions on graphs
• f(x + 1) ≠ f(x) + 1 — substitute the entire input
• Domain: valid inputs; Range: all outputs
• One-to-one: horizontal line test
• Onto: every codomain element is hit
• Bijective: both — has inverse
• Composite functions: (f∘g)(x) = f(g(x)); order matters
• Even: f(-x) = f(x) (y-axis symmetry); Odd: f(-x) = -f(x) (origin symmetry)
• To find inverse: swap x and y, solve for y

Pitfalls

• Thinking f(x + 1) = f(x) + 1: The notation f(x + 1) means substitute (x + 1) into the function definition, not add 1 to the output. These are different operations that only coincide for f(x) = x + c. Always substitute the entire argument expression.
• Confusing domain restrictions: A denominator cannot be zero (x ≠ value), while an even root requires its radicand ≥ 0. With both present (e.g., 1/√(x-3)), you need strict inequality: the radicand must be > 0 because it's also in the denominator.
• Mixing up f∘g and g∘f: (f∘g)(x) = f(g(x)) means apply g FIRST, then f. (g∘f)(x) = g(f(x)) means apply f first, then g. These are almost never equal. The function on the right is applied first.
• Assuming all functions have inverses: Only one-to-one (injective) functions have inverses. f(x) = x² on all reals has no inverse, but restricting the domain to [0, ∞) makes it invertible with f⁻¹(x) = √x.
• Misclassifying even/odd functions: Just because a function has even powers doesn't guarantee it's even — check f(-x) algebraically. For f(x) = x³ - x, compute f(-x) = -x³ + x = -(x³ - x) = -f(x), so it's odd. Mixed even and odd powers usually give neither.

Next Steps

Next up: 03-02-transformations-of-functions.md