Functions, Domain & Range, Composite/Inverse & Modulus Graphs
Functions are the structural backbone of Year 2 Pure Mathematics. Examiners test subtle distinctions: why a many-to-one quadratic must have its domain restricted at its vertex before an inverse function can exist, why the domain of f⁻¹(x) is always determined by finding the range of f(x) first, and how to solve modulus equations and inequalities without introducing extraneous roots. You must also sketch and combine multi-step graph transformations, distinguishing clearly between the vertical reflection y = |f(x)| and the horizontal reflection y = f(|x|).
- Mappings, Functions, Domain, and Range
- A function is a rule that maps each input x in the domain to a single unique output f(x) in the range (so one-to-many mappings like y² = x are not functions). A function is one-TO-one if every distinct input gives a distinct output (passing the horizontal line test, such as strictly increasing or strictly decreasing functions), and many-TO-one if two or more inputs give the same output (such as f(x) = x² or sin x). To find the range of a quadratic f(x) on a restricted domain, always complete the square first to check whether the vertex lies inside the domain.
- Composite Functions fg(x) and Domain Compatibility
- The composite function fg(x) means f(g(x)): apply the inner function g first, then apply the outer function f to the output of g. In general, fg(x) ≠ gf(x). For the composite function fg(x) to exist for all x in the domain of g, the range of the inner function g must lie completely inside the domain of the outer function f. The range of fg(x) is found by feeding the range of g(x) as the input domain into f.
- Inverse Functions f⁻¹(x) and Domain/Range Swap
- An inverse function f⁻¹(x) exists if and only if f(x) is one-TO-one. If a function is many-to-one (such as a quadratic with vertex at x = k), the largest possible domain of the form x ≥ a for which f⁻¹ exists is x ≥ k (the x-coordinate of the turning point). The golden rule of inverse functions is that the domain and range swap completely: Domain of f⁻¹(x) = Range of f(x), and Range of f⁻¹(x) = Domain of f(x). Geometrically, y = f⁻¹(x) is the reflection of y = f(x) in the line y = x, so any intersection of y = f(x) and y = f⁻¹(x) on the line of symmetry can be found simply by solving f(x) = x!
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Mappings, Functions, Domain, and Range
- A function is a rule that maps each input x in the domain to a single unique output f(x) in the range (so one-to-many mappings like y² = x are not functions). A function is one-TO-one if every distinct input gives a distinct output (passing the horizontal line test, such as strictly increasing or strictly decreasing functions), and many-TO-one if two or more inputs give the same output (such as f(x) = x² or sin x). To find the range of a quadratic f(x) on a restricted domain, always complete the square first to check whether the vertex lies inside the domain.
- Composite Functions fg(x) and Domain Compatibility
- The composite function fg(x) means f(g(x)): apply the inner function g first, then apply the outer function f to the output of g. In general, fg(x) ≠ gf(x). For the composite function fg(x) to exist for all x in the domain of g, the range of the inner function g must lie completely inside the domain of the outer function f. The range of fg(x) is found by feeding the range of g(x) as the input domain into f.
- Inverse Functions f⁻¹(x) and Domain/Range Swap
- An inverse function f⁻¹(x) exists if and only if f(x) is one-TO-one. If a function is many-to-one (such as a quadratic with vertex at x = k), the largest possible domain of the form x ≥ a for which f⁻¹ exists is x ≥ k (the x-coordinate of the turning point). The golden rule of inverse functions is that the domain and range swap completely: Domain of f⁻¹(x) = Range of f(x), and Range of f⁻¹(x) = Domain of f(x). Geometrically, y = f⁻¹(x) is the reflection of y = f(x) in the line y = x, so any intersection of y = f(x) and y = f⁻¹(x) on the line of symmetry can be found simply by solving f(x) = x!
- The Modulus Function |x| and Solving Modulus Equations
- For any real number a, the modulus |a| is its non-negative magnitude: |a| = a if a ≥ 0, and |a| = -a if a < 0. To solve an equation or inequality involving |ax + b| = cx + d or |ax + b| < |cx + d|: 1. Always sketch both graphs on the same axes first to see how many intersections exist and which branch (positive +(ax + b) or negative -(ax + b)) is involved at each intersection. 2. Solve +(ax + b) = cx + d and -(ax + b) = cx + d (or square both sides when both sides are moduli: (ax + b)² = (cx + d)²), and check every solution in the original equation to reject any extraneous roots where cx + d < 0.
- Sketching y = |f(x)| vs y = f(|x|)
- 1. To sketch y = |f(x)|: first sketch y = f(x); keep everything on or above the x-axis (y ≥ 0) unchanged, and REFLECT any part of the curve that lies below the x-axis (y < 0) upwards across the x-axis (multiplying its y-coordinates and equation by -1). 2. To sketch y = f(|x|): first sketch y = f(x) for x ≥ 0 only (to the right of the y-axis); delete everything to the left of the y-axis, and REFLECT the right-hand half (x ≥ 0) across the y-axis into the left-hand half (x < 0) so the graph is symmetric about the y-axis.
- Combining Graph Transformations
- Changes outside the function f(.) affect y-coordinates vertically in the expected direction: y = f(x) + a translates by (0, a)^T; y = k f(x) stretches vertically by scale factor k; y = -f(x) reflects in the x-axis. Changes inside the function f(.) affect x-coordinates horizontally in the opposite way: y = f(x + a) translates by (-a, 0)^T; y = f(kx) stretches horizontally by scale factor 1 / k; y = f(-x) reflects in the y-axis. For y = a|bx + c| + d, the sharp vertex ('V' point) occurs where the inside of the modulus is zero (bx + c = 0 → x = -c/b) and has y-coordinate y = d.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
How do you find the domain of an inverse function f⁻¹(x)?
Why does f(x) = x² - 6x + 11 for x ∈ ℝ not have an inverse, and what is the minimum value of k such that f(x) has an inverse for x ≥ k?
What is the quickest way to solve f(x) = f⁻¹(x) when f(x) is a quadratic with a restricted domain?
What are the coordinates of the vertex of y = 5 - 2|3x - 6|, and is it a minimum or a maximum?
How does the graph of y = f(|x|) differ from y = |f(x)|?
What sequence of two transformations maps y = f(x) onto y = f(2x - 6)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
The function f is defined by f(x) = x² - 8x + 20 for x ≥ 4. (a) State the range of f. (b) Find an expression for the inverse function f⁻¹(x) and state its domain and range. (c) Find the exact coordinate where the graphs of y = f(x) and y = f⁻¹(x) intersect. (7 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Complete the square for f(x): f(x) = (x - 4)² + 4 [1 mark]. 2. Since x ≥ 4, (x - 4)² ≥ 0, so the range of f is f(x) ≥ 4 [1 mark]. 3. (b) Let y = (x - 4)² + 4, swap x and y (or rearrange for x): (x - 4)² = y - 4 → x = 4 ± √(y - 4) [1 mark]. 4. Since the domain of f is x ≥ 4, choose the positive square root: f⁻¹(x) = 4 + √(x - 4) [1 mark]. 5. State the domain and range of f⁻¹: Domain is x ≥ 4 (matching the range of f), and Range is f⁻¹(x) ≥ 4 (matching the domain of f) [1 mark]. 6. (c) Since y = f(x) and y = f⁻¹(x) intersect on the line y = x, solve f(x) = x: x² - 8x + 20 = x → x² - 9x + 20 = 0 [1 mark]. 7. Factorise (x - 4)(x - 5) = 0, giving intersection points (4, 4) and (5, 5) (both satisfy x ≥ 4) [1 mark].
- Q2
Solve the inequality |2x - 5| > 3 - x. Show all algebraic steps clearly. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Find the two critical intersection points where |2x - 5| = 3 - x by solving both the positive and negative branches [1 mark]. 2. Positive branch (where 2x - 5 ≥ 0): 2x - 5 = 3 - x → 3x = 8 → x = 8/3 [1 mark]. 3. Negative branch (where 2x - 5 < 0): -(2x - 5) = 3 - x → -2x + 5 = 3 - x → x = 2 [1 mark]. 4. Check both critical values in |2x - 5| = 3 - x: at x = 2, |4 - 5| = 1 and 3 - 2 = 1 (valid); at x = 8/3, |16/3 - 5| = 1/3 and 3 - 8/3 = 1/3 (valid) [1 mark]. 5. From a sketch of the V-graph y = |2x - 5| (steep gradient ±2) and the line y = 3 - x (gradient -1), the V-graph lies strictly above the line outside the two intersection points: x < 2 or x > 8/3 [1 mark].
- Q3
The functions f and g are defined by f(x) = 3e^(2x) - 1 for x ∈ ℝ, and g(x) = ln(x - 2) for x > 2. (a) State the range of f and the range of g. (b) Show that fg(x) = 3(x - 2)² - 1 and solve the equation fg(x) = 26. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Since e^(2x) > 0 for all x ∈ ℝ, 3e^(2x) - 1 > -1, so the range of f is f(x) > -1; since ln(x - 2) takes all real values for x > 2, the range of g is g(x) ∈ ℝ [2 marks]. 2. (b) Form the composite function fg(x) = f(g(x)) = 3e^(2 ln(x - 2)) - 1 [1 mark]. 3. Use the logarithm power law 2 ln(x - 2) = ln[(x - 2)²] and the inverse property e^(ln u) = u: fg(x) = 3e^(ln[(x - 2)²]) - 1 = 3(x - 2)² - 1 [1 mark]. 4. Set fg(x) = 26 and solve: 3(x - 2)² - 1 = 26 → 3(x - 2)² = 27 → (x - 2)² = 9 → x - 2 = ±3 → x = 5 or x = -1 [1 mark]. 5. Check compatibility with the domain of g (x > 2): reject x = -1 since -1 ≤ 2, leaving the unique solution x = 5 [1 mark].
- Q4
A function f(x) has a single minimum turning point at P(-2, -6) and crosses the axes at (-5, 0), (1, 0), and (0, -4). Write down the coordinates of the image of P under the following transformations: (a) y = |f(x)|, (b) y = f(|x|), (c) y = 3f(x - 2) + 5, and (d) y = f(-2x). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Under y = |f(x)|, points below the x-axis (y < 0) are reflected across the x-axis (y → -y), so P(-2, -6) maps to (-2, +6) [1 mark]. 2. (b) Under y = f(|x|), the left half of the graph (x < 0) is deleted and replaced by the reflection of the right half (x ≥ 0); since P(-2, -6) has x < 0, it disappears entirely from the graph (and (0, -4) becomes a local vertex) [1 mark]. 3. (c) Under y = 3f(x - 2) + 5, x increases by 2 (-2 + 2 = 0) and y is multiplied by 3 then increased by 5 (3(-6) + 5 = -13), mapping P to (0, -13) [1 mark]. 4. (d) Under y = f(-2x), x is divided by -2 (-2 ÷ (-2) = +1) while y is unchanged, mapping P to (1, -6) [1 mark].
Common Exam Questions (4)
Why did I lose a mark when choosing between the + and, square root for the inverse of a quadratic?
When you complete the square as y = (x - a)² + b and rearrange to x = a ± √(y - b), look at the restricted domain of f(x): if the domain of f is x ≥ a (the right half of the parabola), you must choose the + sign (f⁻¹(x) = a + √(x - b)); if the domain of f is x ≤ a (the left half of the parabola), you must choose the, sign (f⁻¹(x) = a - √(x - b)).
Why did squaring both sides of |2x - 1| = x - 3 give two answers that were both wrong?
Squaring both sides actually solves |2x - 1| = |x - 3|, which includes points where x - 3 is negative. Since a modulus |2x - 1| can never be negative, any solution where the right-hand side (x - 3) < 0 (i.e. x < 3) is an extraneous root and must be rejected.
When combining two horizontal transformations like y = f(2x + 6), which order do I apply them in?
Horizontal transformations inside the bracket work in reverse order of operations (BIDMAS backwards): first apply the translation by (-6, 0)^T (to get f(x + 6)), and second apply the horizontal stretch by scale factor 1/2 (replacing x with 2x to get f(2x + 6)).
What does it mean if a function is self-inverse?
A function is self-inverse if f⁻¹(x) = f(x) for all x in its domain (which is equivalent to ff(x) = x), meaning its graph is already symmetrical about the line y = x (for example, f(x) = k / x or f(x) = (ax + b) / (cx - a)).
Need more questions on Functions, Domain & Range, Composite/Inverse & Modulus Graphs, or want to organise your own notes?
Type a specific exam question you are stuck on, pick one of the prompts below, or paste your class notes to build a custom set of flashcards and practice questions on this page.
Related study guides
Exponentials, Logarithms & Linearising Non-Linear Data
Exponential functions and logarithms appear across Pure Mathematics, Differential Equations, and Statistics. At A-Level, you must move fluently between exponential and logarithmic forms, solve disguised quadratic equations in e^x or a^x, and differentiate e^(kx) to find instantaneous rates of growth or decay. A guaranteed exam question tests your ability to reduce non-linear relationships, specifically power laws y = ax^n and exponential laws y = ab^x - into straight-line graphs by plotting log(y) against log(x) or log(y) against x.
Proof by Contradiction, Algebraic Division & Partial Fractions
Mathematical proof and algebraic manipulation underpin every topic in Pure Mathematics across AQA, Edexcel, and OCR A-Level Maths. In Year 2 Pure, you must construct rigorous proofs by contradiction, including proving the irrationality of √2 and Euclid's proof that there are infinitely many prime numbers. Alongside formal proof, decomposing rational functions into partial fractions is an essential gateway skill: you cannot integrate complex rational expressions or expand them using the general binomial theorem until you have split them into simpler fractions, including cases with repeated linear factors and improper algebraic fractions.
Differentiation: Rules & Applications
Differentiation measures the instantaneous rate of change of a function and gives the gradient of the tangent to a curve at any point. At A-Level Maths, extending differentiation with the chain rule, product rule, quotient rule, and implicit differentiation lets you locate stationary points and solve optimisation problems.