About this chapter
Functions and graphs is a regular on the WMA13 paper. The June 2024 paper opened with a modulus graph, its transformation and an inequality, and later asked for an inverse function and the range of a composite, which was one of the hardest parts of the whole paper.
This chapter covers the modulus function and its equations and inequalities, domain and range, composite and inverse functions, the graphs of |f(x)| and f(|x|), combined transformations, and modulus graphs with unknown constants. Every question asks for exact answers, and most need a sketch as well as the algebra.
The ten questions
- Q1 (6 marks): from a sketch of y = f(x), draw y = |f(x)| and y = f(|x|) with every key point.
- Q2 (5 marks): solve |3x − 5| = x + 7, then the matching inequality.
- Q3 (5 marks): sketch |3x + a| + a in terms of a, then solve an equation in terms of a.
- Q4 (5 marks): where the point (3, −4) moves under four different transformations.
- Q5 (5 marks): find fg(x), then solve fg(x) = gf(x) in surds.
- Q6 (10 marks): the range of a rational function, its inverse with domain, a composite value, and an equation with no solutions.
- Q7 (10 marks): a modulus graph with two constants: key points in terms of a and b, then their values from two intersections.
- Q8 (10 marks): sketch y = f(|x|) for a quadratic, count solutions of f(|x|) = k, then solve |f(x)| = 3.
- Q9 (10 marks): where y = |2x + 3| meets y = x², an inequality from the sketch, and an equation with a rejected root.
- Q10 (10 marks): the range of f, a composite gf(x), its range, and an equation.
Key skills tested
The modulus function. |x| is x when x is positive or zero, and −x otherwise. The graph of y = |ax + b| is a V shape with its vertex on the x-axis.
Modulus equations. Solve the positive and negative versions separately, then check each answer in the original equation. A sketch shows how many solutions to expect.
Domain and range. The range is the set of outputs for the given domain. Sketch the function to find both ends.
Composite functions. fg(x) means f(g(x)): apply g first, and its outputs must lie in the domain of f.
Inverse functions. Only one-to-one functions have inverses. Swap x and y, rearrange, and write the result as the inverse function of x. Its domain is the range of f.
The graphs of |f(x)| and f(|x|). For |f(x)|, reflect the parts below the x-axis upwards. For f(|x|), reflect the part for positive x in the y-axis.
Combined transformations. For y = a f(bx + c) + d, change x by the c first and then the b, and y by the a first and then the d.
Points under transformations. Track one point at a time. For example, (3, −4) on y = f(x) moves to (5, 1) on y = f(x−2) + 5.
Parameters in modulus graphs. Write the key points in terms of the constants, then substitute known intersections into the correct branch.
Worked example
Question 2 from this chapter. (a) Solve |3x − 5| = x + 7. (b) Hence find the set of values of x for which |3x − 5| < x + 7.
(a) Positive case: 3x − 5 = x + 7, so 2x = 12 and x = 6. Negative case: −(3x − 5) = x + 7, so −3x + 5 = x + 7, giving −4x = 2 and x = −½. Both check in the original equation: at x = 6 each side is 13, and at x = −½ each side is 6.5.
(b) The line y = x + 7 is above the V shape between the two meeting points, so −½ < x < 6.
A quick sketch of the V and the line shows which region is wanted and stops you giving two separate ranges.
Seen on real papers
Every question in our booklets is original. These are the recent papers where each question type has appeared.
- June 2024, Q1: a transformed modulus graph, an inequality with only one solution, and where a point moves under a transformation, as in Q4 and Q9.
- June 2024, Q5(a), (c) and (d): solving an equation involving f, an inverse function with its domain, and the range of a composite, as in Q6 and Q10.
Where marks are lost
- A solution that does not exist. In June 2024 many gave two answers to a modulus inequality that had only one. Drawing the line on the graph showed where it really crossed.
- No domain for the inverse. Leaving out the domain of an inverse function was very common in June 2024, even though this type of question appears regularly on WMA13.
- Stopping at x and y. Many rearranged correctly but never rewrote the answer as the inverse function of x, and lost the accuracy mark.
- The range of a composite. Fully correct ranges for fg were very rare in June 2024. Link the range of the inner function to the outer one.
Common questions
How do I find the domain of an inverse function?
It is the range of the original function. Find that range first, from a sketch if needed.
Why check modulus solutions?
Solving each case separately can produce a value that does not satisfy the original equation, so substitute each one back in.
What is the difference between |f(x)| and f(|x|)?
|f(x)| reflects the negative parts of the graph upwards. f(|x|) copies the right-hand half of the graph onto the left.
Where this chapter leads
- Pure 3 Chapter 5: Exponentials and Logarithms: inverse functions of e to the power x and ln x.
- Pure 3 Chapter 3: Trigonometric Functions: ranges and inverses of the trig functions.