Functions for Edexcel IGCSE Mathematics Specification B: Methods and Worked Examples

Functions form one of the richest sections of the Edexcel IGCSE Mathematics Specification B specification. Across most European and international curricula, functions are treated as a unifying theme that connects algebra, geometry, and analysis. This exam is no exception.

The edexcel igcse mathematics specification b functions content covers four interconnected areas: Functions and Notation, Variation and Proportion, Graphs and Coordinate Geometry, and Calculus. Together, they take you from the abstract definition of a function to the practical skill of finding gradients and turning points through differentiation. Students taking this course in preparation for A Level Mathematics will find that every topic here connects directly to what comes next.

Functions and notation

What is a function?

A function is a rule that assigns to each input exactly one output. Two notations are used interchangeably in the functions edexcel igcse exam:

f(x) = 2x + 3 (the "f of x equals" form)
f: x → 2x + 3 (the "f maps x to" form)

The domain is the set of all permitted inputs. The range is the set of all possible outputs. For f(x) = 1/(x - 2), the domain excludes x = 2 (because division by zero is undefined), and the range excludes y = 0 (because 1/(x - 2) can never equal zero).

Composite functions

A composite function applies one function after another. If f(x) = 2x + 1 and g(x) = x², then:

Worked example: Find fg(x) and gf(x).

fg(x) = f(g(x)) = f(x²) = 2(x²) + 1 = 2x² + 1.

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

Notice that fg(x) and gf(x) are different. Composition is not commutative. The order matters.

Reading composite notation: In fg(x), apply g first, then f. The function closest to x acts first. This catches many students off guard because the written order (f then g) is the reverse of the application order. Think of it as peeling layers: the outer function waits for the inner one to finish.

Inverse functions

The inverse function f^-1(x) reverses the action of f(x). To find it, swap x and y in the equation y = f(x), then solve for y.

Worked example: Find f^-1(x) for f(x) = (3x - 1)/2.

Let y = (3x - 1)/2.

Swap: x = (3y - 1)/2.

Solve: 2x = 3y - 1, so 3y = 2x + 1, so y = (2x + 1)/3.

Therefore f^-1(x) = (2x + 1)/3.

Check: ff^-1(x) should equal x. f((2x+1)/3) = (3(2x+1)/3 - 1)/2 = (2x + 1 - 1)/2 = 2x/2 = x. Confirmed.

Variation and proportion

Direct and indirect (inverse) proportion are tested both algebraically and in context.

Direct proportion: y is directly proportional to x means y = kx for some constant k. If y is proportional to x², then y = kx².

Inverse proportion: y is inversely proportional to x means y = k/x. If y is inversely proportional to the square root of x, then y = k/sqrt(x).

Worked example: y is directly proportional to x³. When x = 2, y = 40. Find y when x = 3.

y = kx³. Substitute: 40 = k(8), so k = 5.

When x = 3: y = 5(27) = 135.

Worked example: p is inversely proportional to q². When q = 4, p = 3. Find p when q = 6.

p = k/q². Substitute: 3 = k/16, so k = 48.

When q = 6: p = 48/36 = 4/3.

Graphs and coordinate geometry

Straight-line graphs

The equation y = mx + c represents a straight line with gradient m and y-intercept (0, c). The igcse 4mb1 functions content requires you to recognise this form, draw lines from equations, and find equations from graphs.

Worked example: Find the equation of the line passing through (1, 5) and (4, -1).

Gradient m = (-1 - 5)/(4 - 1) = -6/3 = -2.

Using y - y1 = m(x - x1): y - 5 = -2(x - 1), so y = -2x + 7.

Polynomial and reciprocal graphs

The specification requires you to draw and interpret graphs of polynomial functions (quadratics, cubics) and reciprocal functions (y = a/x). Each family has a characteristic shape:

Function type	Shape	Key features
y = ax² + bx + c	Parabola	One turning point (minimum if a > 0, maximum if a < 0)
y = ax³ + ...	S-curve or N-curve	Up to two turning points; passes through the origin if no constant term
y = a/x	Hyperbola	Two branches in opposite quadrants; asymptotes at x = 0 and y = 0

Using graph intersections to solve equations

The edexcel igcse mathematics specification b notes on graphical methods require you to find where two curves meet. The x-coordinates at the intersection points are the solutions to the equation formed by setting the two functions equal.

Gradients from tangents

Before reaching calculus, the specification asks you to determine gradients by drawing tangents to curves at given points. A tangent is a straight line that just touches the curve at that point. You draw it, pick two convenient points on it, and calculate the gradient.

Calculus

This is the topic that sets Specification B apart from Specification A. The Pearson Edexcel IGCSE Mathematics Specification B exam tests introductory differentiation, a topic that many international curricula reserve for pre-university courses.

Differentiating powers of x

The rule: if y = xⁿ, then dy/dx = nx^n-1. This applies term by term to polynomials.

Worked example: Find dy/dx for y = 3x⁴ - 2x³ + 5x - 7.

dy/dx = 12x³ - 6x² + 5.

The constant term (-7) disappears because the derivative of a constant is zero.

Gradients and rates of change

The derivative dy/dx gives the gradient of the curve at any point. Substituting a specific x-value gives the gradient at that point.

Worked example: Find the gradient of y = x³ - 4x + 1 at x = 2.

dy/dx = 3x² - 4.

At x = 2: dy/dx = 3(4) - 4 = 8.

The gradient at x = 2 is 8.

Stationary points and turning points

A stationary point occurs where dy/dx = 0. The nature of the stationary point (maximum, minimum, or point of inflexion) can be determined by examining the sign of the gradient on either side.

Worked example: Find the stationary points of y = 2x³ - 9x² + 12x and determine their nature.

dy/dx = 6x² - 18x + 12 = 6(x² - 3x + 2) = 6(x - 1)(x - 2).

Setting dy/dx = 0: x = 1 or x = 2.

When x = 1: y = 2 - 9 + 12 = 5. Point: (1, 5).

When x = 2: y = 16 - 36 + 24 = 4. Point: (2, 4).

Test the sign of dy/dx around x = 1: at x = 0.5, dy/dx = 6(0.5 - 1)(0.5 - 2) = 6(-0.5)(-1.5) = 4.5 > 0. At x = 1.5, dy/dx = 6(0.5)(-0.5) = -1.5 < 0. Gradient goes from positive to negative, so (1, 5) is a maximum.

Test around x = 2: at x = 1.5, dy/dx < 0. At x = 2.5, dy/dx = 6(1.5)(0.5) = 4.5 > 0. Gradient goes from negative to positive, so (2, 4) is a minimum.

Kinematics and calculus. The specification also applies calculus to distance-time and speed-time problems. If s is displacement and t is time, then ds/dt = velocity and dv/dt = acceleration. The same differentiation rules apply. This is where mathematics meets physics, a connection particularly valued across European education systems.

Common mistakes in Functions

Composition order. fg(x) means apply g first, then f. Writing it the wrong way round is one of the most frequent errors.
Inverse functions. Forgetting to swap x and y before solving, or not rearranging completely, leaves you with a formula that is not actually the inverse.
Proportion constants. After finding k, some students forget to write the final formula and jump straight to substitution. Always state the equation with k substituted before finding the requested value.
Differentiating incorrectly. Common slips: leaving the power unchanged (writing 3x⁴ instead of 12x³), forgetting to reduce the power by 1, or differentiating constants as if they were terms in x.
Confusing dy/dx = 0 with y = 0. Setting the derivative to zero finds stationary points, not roots. Setting y to zero finds where the curve crosses the x-axis. These are different questions.

Self-check questions

Work through these edexcel igcse mathematics specification b practice questions before checking the solutions.

f(x) = 4x - 3 and g(x) = x² + 1. Find fg(2) and gf(2).
Find the inverse of f(x) = (5x + 2)/(x - 1), where x is not equal to 1.
y is inversely proportional to x². When x = 3, y = 4. Find y when x = 6.
Find the coordinates of the stationary points of y = x³ - 6x² + 9x + 1.

Solutions

1. g(2) = 4 + 1 = 5. fg(2) = f(5) = 20 - 3 = 17. f(2) = 8 - 3 = 5. gf(2) = g(5) = 25 + 1 = 26.

2. Let y = (5x + 2)/(x - 1). Swap: x = (5y + 2)/(y - 1). Multiply: x(y - 1) = 5y + 2. xy - x = 5y + 2. xy - 5y = x + 2. y(x - 5) = x + 2. y = (x + 2)/(x - 5). So f^-1(x) = (x + 2)/(x - 5), where x is not equal to 5.

3. y = k/x². 4 = k/9, so k = 36. When x = 6: y = 36/36 = 1.

4. dy/dx = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3). Stationary points at x = 1 and x = 3. When x = 1: y = 1 - 6 + 9 + 1 = 5. When x = 3: y = 27 - 54 + 27 + 1 = 1. Points: (1, 5) and (3, 1).

The edexcel igcse mathematics specification b revision notes for Functions should be revisited regularly, because the skills here reappear throughout the rest of the specification. Every graph question uses coordinate geometry. Every optimisation problem uses calculus. Every proportionality question uses the same constant-finding method. Master these edexcel igcse mathematics specification b explained techniques, and the connections across the whole course become visible.