Functions and Graphs

What you'll learn

Functions and graphs form a substantial component of the CIE IGCSE Additional Mathematics syllabus, typically accounting for 10-15% of examination marks. This topic extends beyond basic coordinate geometry to explore function notation, operations on functions, domain and range restrictions, inverse functions, and systematic graph transformations. Mastery of these concepts is essential not only for direct function questions but also for understanding calculus and other advanced topics.

Key terms and definitions

Function — a relation where each input value (from the domain) maps to exactly one output value; written as f(x) or f : x ↦ expression.

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 apply f to the result."

Inverse function — denoted f⁻¹(x), the function that reverses the effect of f(x); if f(a) = b, then f⁻¹(b) = a.

One-to-one function — a function where each output value corresponds to exactly one input value; only one-to-one functions have inverses.

Transformation — a systematic change to a graph's position or shape, including translations, reflections, and stretches.

Asymptote — a line that a graph approaches but never reaches; can be vertical, horizontal, or oblique.

Core concepts

Function notation and evaluation

Function notation provides a precise way to express mathematical relationships. When you see f(x) = 3x² - 5x + 2, the letter f names the function, x represents the input variable, and the expression shows how to calculate outputs.

To evaluate a function at a specific value:

• Substitute the given value for every occurrence of x
• Follow the order of operations carefully
• Simplify to obtain the output

For example, if f(x) = 2x² - 3x + 1 and you need f(4): f(4) = 2(4)² - 3(4) + 1 = 2(16) - 12 + 1 = 32 - 12 + 1 = 21

When the input is an expression rather than a number, such as f(2x), substitute (2x) for every x: f(2x) = 2(2x)² - 3(2x) + 1 = 2(4x²) - 6x + 1 = 8x² - 6x + 1

Domain and range determination

Determining the domain requires identifying values that make the function undefined:

• For fractions: exclude values making the denominator zero
• For square roots: ensure the expression inside remains non-negative (for real functions)
• For logarithms: ensure the argument remains positive

For f(x) = √(x - 3), the domain is x ≥ 3 because (x - 3) must be non-negative.

For g(x) = 1/(x² - 9), the denominator x² - 9 = 0 when x = ±3, so the domain is x ∈ ℝ, x ≠ ±3.

Finding the range often requires considering:

• The minimum or maximum value of quadratic functions (complete the square)
• The behaviour of reciprocal functions
• Restrictions imposed by the domain
• The nature of exponential and logarithmic functions

For f(x) = (x - 2)² + 3, the minimum value occurs at x = 2, giving f(2) = 3, so the range is f(x) ≥ 3.

Composite functions

The composite function fg(x) means "apply g first, then apply f to that result." This is equivalent to f(g(x)).

The order matters critically: fg(x) ≠ gf(x) in most cases.

Process for finding composite functions:

1. Write down the function to be applied second: f( )
2. Replace the input with the entire first function: f(g(x))
3. Substitute g(x) into every occurrence of the variable in f
4. Simplify fully

Example: If f(x) = 2x + 1 and g(x) = x² - 3, find fg(x) and gf(x).

fg(x) = f(g(x)) = f(x² - 3) = 2(x² - 3) + 1 = 2x² - 6 + 1 = 2x² - 5

gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)² - 3 = 4x² + 4x + 1 - 3 = 4x² + 4x - 2

Notice these are different functions, confirming that order matters.

Inverse functions

An inverse function f⁻¹(x) reverses the mapping of f(x). If f maps a → b, then f⁻¹ maps b → a.

Key properties:

• Only one-to-one functions have inverses
• The domain of f becomes the range of f⁻¹
• The range of f becomes the domain of f⁻¹
• ff⁻¹(x) = x and f⁻¹f(x) = x
• The graphs of f and f⁻¹ are reflections in the line y = x

Process for finding inverse functions:

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

Example: Find the inverse of f(x) = (3x - 2)/(x + 1), x ≠ -1

Step 1: y = (3x - 2)/(x + 1)

Step 2: x = (3y - 2)/(y + 1)

Step 3: Rearrange x(y + 1) = 3y - 2 xy + x = 3y - 2 xy - 3y = -x - 2 y(x - 3) = -x - 2 y = (-x - 2)/(x - 3) = -(x + 2)/(x - 3)

Step 4: f⁻¹(x) = -(x + 2)/(x - 3), x ≠ 3

Graph transformations

Transformations systematically alter graphs according to predictable rules. Understanding these allows you to sketch complex functions based on simpler parent functions.

Translations:

• y = f(x) + a: translate a units upward (or downward if a < 0)
• y = f(x - a): translate a units to the right (or left if a < 0)

Note the direction reversal for horizontal translations: f(x - 3) moves right, not left.

Reflections:

• y = -f(x): reflect in the x-axis (flip vertically)
• y = f(-x): reflect in the y-axis (flip horizontally)

Stretches:

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

The horizontal stretch formula is counterintuitive: y = f(2x) compresses the graph horizontally by factor 1/2.

Multiple transformations: When combining transformations, work from the inside out. For y = 2f(x - 3) + 1:

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

Modulus functions

The modulus (absolute value) function |x| gives the non-negative value of x. Graphically, |f(x)| reflects any negative portions of f(x) above the x-axis.

For y = |f(x)|:

• Where f(x) ≥ 0, the graph remains unchanged
• Where f(x) < 0, reflect that portion in the x-axis

For y = f(|x|):

• The graph becomes symmetrical about the y-axis
• The portion for x ≥ 0 remains unchanged
• The portion for x < 0 mirrors the x > 0 portion

Worked examples

Example 1: Composite functions and equations

The functions f and g are defined by f(x) = 2x - 5 and g(x) = x² + 3 for x ∈ ℝ.

(a) Find an expression for gf(x). [2 marks]

(b) Solve the equation gf(x) = 12. [3 marks]

Solution:

(a) gf(x) = g(f(x)) = g(2x - 5) = (2x - 5)² + 3 = 4x² - 20x + 25 + 3 = 4x² - 20x + 28

(b) 4x² - 20x + 28 = 12 4x² - 20x + 16 = 0 x² - 5x + 4 = 0 (x - 4)(x - 1) = 0 x = 4 or x = 1

Example 2: Inverse functions with domain restrictions

The function f is defined by f(x) = x² - 4x + 7 for x ≥ 2.

(a) Express f(x) in the form (x - a)² + b and hence state the range of f. [3 marks]

(b) Find f⁻¹(x), stating its domain. [4 marks]

Solution:

(a) f(x) = x² - 4x + 7 Complete the square: f(x) = (x - 2)² - 4 + 7 = (x - 2)² + 3

Since x ≥ 2, the minimum value occurs at x = 2, giving f(2) = 3. Range of f: f(x) ≥ 3

(b) y = (x - 2)² + 3

Swap variables: x = (y - 2)² + 3

(y - 2)² = x - 3

y - 2 = ±√(x - 3)

Since the domain of f is x ≥ 2, and this is the right branch of the parabola, we take the positive root:

y - 2 = √(x - 3)

y = 2 + √(x - 3)

Therefore f⁻¹(x) = 2 + √(x - 3)

Domain of f⁻¹: x ≥ 3 (this is the range of f)

Example 3: Graph transformations

The diagram shows the graph of y = f(x), which passes through the points (0, 2), (3, 5), and (6, 2).

Sketch the graph of y = 2f(x + 1) - 3, showing clearly the coordinates of the three transformed points. [4 marks]

Solution:

The transformation y = 2f(x + 1) - 3 involves:

• Horizontal translation 1 unit left (x + 1)
• Vertical stretch by factor 2 (coefficient 2)
• Vertical translation 3 units down (-3)

Transform each point:

Point (0, 2):

• Translate left 1: (-1, 2)
• Stretch vertically by 2: (-1, 4)
• Translate down 3: (-1, 1)

Point (3, 5):

• Translate left 1: (2, 5)
• Stretch vertically by 2: (2, 10)
• Translate down 3: (2, 7)

Point (6, 2):

• Translate left 1: (5, 2)
• Stretch vertically by 2: (5, 4)
• Translate down 3: (5, 1)

The transformed points are (-1, 1), (2, 7), and (5, 1).

Common mistakes and how to avoid them

• Confusing fg(x) with gf(x) — Always work from right to left in function notation. The function closest to x is applied first. Write out f(g(x)) explicitly to avoid errors.

• Incorrect domain for inverse functions — The domain of f⁻¹ is the range of f, not the domain of f. After finding the inverse, determine its domain by identifying what the original function's range was.

• Wrong direction for horizontal translations — The transformation f(x - a) shifts the graph a units to the right, not left. The sign appears reversed because you're solving "what input gives the same output."

• Forgetting to restrict the domain when finding inverses — Functions like f(x) = x² don't have inverses over their full domain because they're not one-to-one. You must restrict the domain (e.g., x ≥ 0) before finding the inverse.

• Mixing up horizontal and vertical stretches — A coefficient outside the function, as in 3f(x), stretches vertically. A coefficient inside, as in f(3x), affects horizontally and compresses by factor 1/3, not stretches by 3.

• Incorrect substitution in composite functions — When finding fg(x), you must substitute the entire expression for g(x) into every x in f, including those in denominators, exponents, or under roots. Bracket the substituted expression to avoid sign errors.

Exam technique for Functions and Graphs

• "Find" versus "Solve" command words — "Find an expression for fg(x)" requires algebraic manipulation producing a simplified expression. "Solve fg(x) = k" requires finding specific numerical values of x. Show all working for both, but only the final simplified form for expressions.

• Domain and range questions — Always state restrictions using proper notation: "x ∈ ℝ, x ≠ 3" or "x ≥ -2" or interval notation where appropriate. For range, solve algebraically (completing the square for quadratics) or use calculus where applicable. Sketching helps verify answers.

• Transformation questions worth 3-4 marks — Plot transformed coordinates for given points (1-2 marks) and sketch the curve accurately (1-2 marks). Label axes, mark asymptotes, and show key features like turning points or intercepts.

• Working with restricted domains — When a function has a domain restriction like x ≥ 2, this information is given for a reason. Use it when finding the range (only consider valid x-values) and when determining which square root to take in inverse functions.

Quick revision summary

Functions map inputs to unique outputs using notation f(x) or f : x ↦ expression. The domain comprises valid inputs; the range comprises possible outputs. Composite functions fg(x) mean apply g first, then f. Only one-to-one functions have inverses, found by swapping variables and rearranging; the domain of f⁻¹ equals the range of f. Graph transformations follow systematic rules: f(x) ± a translates vertically, f(x ± a) translates horizontally (direction reversed), -f(x) reflects in x-axis, f(-x) reflects in y-axis, af(x) stretches vertically, f(ax) compresses horizontally. Apply transformations working from inside out.