Functions and Graphs

What you'll learn

This revision guide covers all testable content on Functions and Graphs for CXC CSEC Additional Mathematics. You will master function notation, domain and range, composite and inverse functions, function transformations, and graphing techniques including asymptotes and curve sketching. These topics typically account for 15-20% of Paper 2 marks.

Key terms and definitions

Function — a relation where each element in the domain maps to exactly one element in the codomain, written as f: x → y or f(x) = y

Domain — the complete set of possible input values (x-values) for which a function is defined

Range — the complete set of possible output values (y-values) that a function can produce

Composite function — a function formed by applying one function to the result of another, written as fg(x) or f(g(x)), meaning "apply g first, then f"

Inverse function — a function f⁻¹ that reverses the effect of function f, so that f⁻¹(f(x)) = x for all x in the domain of f

One-to-one function — a function where each element in the range corresponds to exactly one element in the domain; only one-to-one functions have inverses

Asymptote — a line that a curve approaches but never touches as x or y approaches infinity or a specific value

Transformation — a change in the position, size, or orientation of a graph through translation, reflection, or stretch

Core concepts

Function notation and evaluation

Function notation f(x) indicates "the value of function f at x". To evaluate a function, substitute the given value into the expression.

For f(x) = 3x² - 5x + 2:

• f(2) = 3(2)² - 5(2) + 2 = 12 - 10 + 2 = 4
• f(-1) = 3(-1)² - 5(-1) + 2 = 3 + 5 + 2 = 10
• f(a+1) = 3(a+1)² - 5(a+1) + 2 requires expansion of brackets

When finding x for a given f(x) value, form an equation and solve algebraically. If f(x) = 7, then 3x² - 5x + 2 = 7, leading to 3x² - 5x - 5 = 0.

Domain and range restrictions

The domain may be restricted by:

• Division by zero: f(x) = 1/(x-3) has domain x ∈ ℝ, x ≠ 3
• Square roots of negative numbers: f(x) = √(x-2) has domain x ≥ 2
• Logarithms of non-positive numbers: f(x) = ln(4-x) has domain x < 4
• Explicit restrictions given in the question

The range is found by:

• Analyzing the function's minimum/maximum values
• Completing the square for quadratics
• Considering transformations of standard graphs
• Examining the behaviour as x approaches domain boundaries

For f(x) = (x-2)² + 3 with domain x ≥ 2:

• Minimum occurs at x = 2, where f(2) = 3
• As x increases, f(x) increases without bound
• Range: f(x) ≥ 3 or [3, ∞)

Composite functions

A composite function fg(x) means apply g first, then apply f to the result. The order matters: fg(x) ≠ gf(x) in general.

Given f(x) = 2x + 1 and g(x) = x²:

• fg(x) = f(g(x)) = f(x²) = 2(x²) + 1 = 2x² + 1
• gf(x) = g(f(x)) = g(2x+1) = (2x+1)²

To evaluate fg(3):

• Method 1: Find fg(x) first, then substitute x = 3
• Method 2: Find g(3) = 9, then find f(9)

The domain of fg(x) requires:

• x must be in the domain of g
• g(x) must be in the domain of f

For f(x) = √x and g(x) = x - 5:

• fg(x) = √(x-5) has domain x ≥ 5
• gf(x) = √x - 5 has domain x ≥ 0

Inverse functions

To find f⁻¹(x):

Method 1: Algebraic substitution

1. Write y = f(x)
2. Rearrange to make x the subject
3. Swap x and y
4. Write as f⁻¹(x)

For f(x) = (3x - 2)/5:

1. y = (3x - 2)/5
2. 5y = 3x - 2, so 3x = 5y + 2, giving x = (5y + 2)/3
3. Swap: y = (5x + 2)/3
4. f⁻¹(x) = (5x + 2)/3

Method 2: Function composition Use ff⁻¹(x) = x to form an equation and solve for f⁻¹(x).

Key properties:

• The domain of f⁻¹ equals the range of f
• The range of f⁻¹ equals the domain of f
• The graph of f⁻¹ is the reflection of f in the line y = x
• Only one-to-one functions have inverses over their entire domain

Transformations of graphs

Given the graph of y = f(x), these transformations produce:

Translations:

• y = f(x) + a: vertical translation by a units (positive = upward)
• y = f(x - a): horizontal translation by a units (positive = rightward)

Reflections:

• y = -f(x): reflection in the x-axis
• y = f(-x): reflection in the y-axis

Stretches:

• y = af(x), a > 1: vertical stretch by factor a; 0 < a < 1: vertical compression
• y = f(ax), a > 1: horizontal compression by factor 1/a; 0 < a < 1: horizontal stretch

Combined transformations must be applied in the correct order. For y = 2f(x-3) + 1:

1. Translate 3 units right (x-3)
2. Stretch vertically by factor 2
3. Translate 1 unit up

Caribbean example: A biologist models the population of Leatherback turtles nesting in Trinidad with P(t) = 500 + 150sin(πt/6), where t is months after January. The function P(t-3) + 100 represents the nesting pattern delayed by 3 months with 100 additional turtles.

Graphing techniques and asymptotes

Vertical asymptotes occur where the denominator equals zero in rational functions:

• f(x) = 3/(x-2) has vertical asymptote x = 2
• As x approaches 2, f(x) approaches ±∞

Horizontal asymptotes describe end behaviour as x → ±∞:

• f(x) = 3/(x-2): as x → ±∞, f(x) → 0, so y = 0 is the horizontal asymptote
• f(x) = (2x+1)/(x-3): as x → ±∞, f(x) → 2, so y = 2 is the horizontal asymptote

Oblique asymptotes occur when the numerator's degree exceeds the denominator's degree by 1. Use polynomial division.

Curve sketching checklist:

1. Find x-intercepts (set y = 0)
2. Find y-intercept (set x = 0)
3. Identify asymptotes
4. Determine behaviour near asymptotes
5. Check for symmetry
6. Plot key points
7. Sketch the curve showing all features

For f(x) = (x+1)/(x-2):

• x-intercept: x + 1 = 0, so x = -1
• y-intercept: f(0) = 1/(-2) = -0.5
• Vertical asymptote: x = 2
• Horizontal asymptote: y = 1
• Test regions: x < -1, -1 < x < 2, x > 2

Worked examples

Example 1: The functions f and g are defined as f(x) = 2x - 3 and g(x) = x² + 1.

(a) Find fg(x) in its simplest form. [2 marks] (b) Calculate fg(4). [2 marks] (c) Solve the equation fg(x) = 15. [3 marks]

Solution:

(a) fg(x) = f(g(x)) = f(x² + 1) = 2(x² + 1) - 3 = 2x² + 2 - 3 = 2x² - 1 ✓✓

(b) Method 1: fg(4) = 2(4)² - 1 = 2(16) - 1 = 31 ✓✓

Method 2: g(4) = 4² + 1 = 17 ✓
          fg(4) = f(17) = 2(17) - 3 = 31 ✓

(c) fg(x) = 15 2x² - 1 = 15 ✓ 2x² = 16 x² = 8 ✓ x = ±√8 = ±2√2 ✓

Example 2: A function is defined by f(x) = (4x + 3)/(2x - 1), x ≠ ½.

(a) Find f⁻¹(x). [4 marks] (b) State the domain of f⁻¹. [1 mark]

Solution:

(a) Let y = (4x + 3)/(2x - 1) ✓ y(2x - 1) = 4x + 3 2xy - y = 4x + 3 2xy - 4x = y + 3 x(2y - 4) = y + 3 ✓ x = (y + 3)/(2y - 4) ✓ Therefore f⁻¹(x) = (x + 3)/(2x - 4) or (x + 3)/(2(x - 2)) ✓

(b) Domain of f⁻¹ = Range of f The range of f excludes the horizontal asymptote y = 2 Domain of f⁻¹: x ∈ ℝ, x ≠ 2 ✓

Example 3: The diagram shows the graph of y = f(x) passing through points A(0, 2), B(3, 5), and C(6, 2).

Sketch the graph of y = f(x - 2) - 1, clearly showing the coordinates of the images of points A, B, and C. [4 marks]

Solution:

The transformation y = f(x - 2) - 1 represents:

• Translation 2 units right (x - 2) ✓
• Translation 1 unit down (- 1) ✓

Apply to each point:

• A(0, 2) → A'(0+2, 2-1) = A'(2, 1) ✓
• B(3, 5) → B'(3+2, 5-1) = B'(5, 4) ✓
• C(6, 2) → C'(6+2, 2-1) = C'(8, 1) ✓

[Sketch showing same curve shape through these new points]

Common mistakes and how to avoid them

• Confusing fg(x) and gf(x): Remember fg(x) means "g first, then f". Always work from right to left when reading composite functions. Write out the full substitution to avoid errors.

• Incorrect transformation directions: y = f(x - 3) moves right, not left. The sign seems opposite to the direction. Remember: "do the opposite of what you see" for horizontal translations.

• Forgetting domain restrictions in composite functions: When finding fg(x), check that g(x) lies within the domain of f. A common error is finding the expression without considering whether all values are valid.

• Swapping x and y too early when finding inverses: Complete all algebraic manipulation to isolate x before swapping variables. Swapping too early leads to incorrect rearrangement.

• Misidentifying horizontal asymptotes: For y = (2x² + 1)/(x² - 4), the horizontal asymptote is y = 2 (ratio of leading coefficients), not y = 0. When numerator and denominator have the same degree, divide the leading coefficients.

• Not showing approach to asymptotes on sketches: Your sketch must clearly show whether the curve approaches the asymptote from above or below. Test values on either side of vertical asymptotes to determine this.

Exam technique for Functions and Graphs

• Command words matter: "Find" requires an exact answer with full working. "Sketch" needs key features labelled (intercepts, asymptotes, turning points) but not precise plotting. "State" requires a direct answer, often with minimal working.

• Show substitution clearly: When evaluating f(3), write f(3) = ... with the substitution visible, then simplify. This earns method marks even if your final answer is incorrect. Examiners cannot award marks for working done "in your head".

• Domain and range notation: Use correct notation: x ∈ ℝ, x ≠ 2 or x > -3 or [-3, 5]. Avoid vague descriptions like "all real numbers except 2" when formal notation is expected. Both interval and inequality notation are accepted.

• Verify inverse functions: If time permits, check your inverse by showing ff⁻¹(x) = x or f⁻¹f(x) = x. This verification often earns the final mark and catches algebraic errors.

Quick revision summary

Functions map each input to exactly one output. Master function notation f(x), evaluation, and finding x for given f(x). Composite functions fg(x) apply g first, then f. Find inverse functions by swapping x and y after making x the subject. Transformations: f(x)±a is vertical translation, f(x±a) is horizontal translation (opposite sign), af(x) is vertical stretch, f(ax) is horizontal compression. Rational functions have vertical asymptotes where denominators equal zero and horizontal asymptotes determined by end behaviour. Always show full working and label key features on graphs.