AQA A-level Mathematics: Graphs of Functions, Composite Functions and Inverse Functions (7357) – Revision Notes

These notes condense AQA A-level Mathematics 7357 content B7 (graphs of functions) and B8 (composite and inverse functions), from the specification version 1.3 dated 31 January 2018. Longer explanations are in the study guide first.

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Key definitions

Term	Meaning
Root	A value of x where the curve meets the x-axis (y = 0)
Repeated root	A factor (x − a)² gives a touch at a. A factor (x − a)³ gives a flat crossing at a
Asymptote	A line the curve approaches but never reaches
Modulus
Proportional	y ∝ x means y = kx for a constant k. The constant k is found from one pair of values
Composite function	gf(x) = g(f(x)). Apply f first
One-one function	Each output comes from only one input. Only these have inverses
Inverse function	f⁻¹ undoes f: f⁻¹f(x) = x and ff⁻¹(x) = x

B7: sketching in steps

Polynomial (B7)

1. Factorise and mark the roots. Note which are repeated.
2. Find the y-intercept by putting x = 0.
3. Use the highest power and its sign for the ends. A positive x³ goes from bottom left to top right. A negative x³ goes from top left to bottom right. An even power with a positive coefficient has both ends up.
4. Join up: cross at single roots, touch at double roots.

Worked reminder. y = x(x − 4)(x + 1) has roots −1, 0 and 4, a positive cubic shape, and passes through the origin. At x = −2, y = (−2)(−6)(−1) = −12, which is below the axis, as expected for the far left.

Modulus of a linear function (B7)

1. Sketch the line y = ax + b.
2. Reflect any part below the x-axis into the upper half-plane.
3. The vertex is at x = −b/a on the x-axis. The y-intercept is |b|.

Worked reminder. y = |2x − 1| has its vertex at (1/2, 0), meets the y-axis at (0, 1), and has arm gradients +2 and −2.

Solving a modulus equation. Solve u = v and −u = v separately, then check both roots in the original. Reject any root that makes the right-hand side negative.

Reciprocal curves (B7)

Curve	a > 0	a < 0	Asymptotes
y = a/x	Quadrants 1 and 3	Quadrants 2 and 4	x = 0 and y = 0
y = a/x²	Both branches above the x-axis	Both branches below the x-axis	x = 0 and y = 0

Neither curve meets either axis. y = a/x is unchanged by a half-turn about the origin. y = a/x² is symmetric about the y-axis.

B7: graphs and equations

• The real solutions of f(x) = g(x) are the x-coordinates of the intersections of y = f(x) and y = g(x).
• To count roots of a hard equation, rearrange into two curves you can sketch, for example x³ + 2x − 6 = 0 as x² + 2 = 6/x.
• Algebraic and graphical methods must agree. Use the sketch to check you have not lost a solution (for example by dividing by x).
• A line is tangent to a quadratic curve when the combined equation has a repeated root, so its discriminant is zero.
• For an inequality, read off the x-values where one graph is above (or below) the other.

B7: proportion

Relationship	Equation	Graph
y ∝ x	y = kx	A line through the origin
y ∝ xⁿ	y = kxⁿ	Curve through the origin
y ∝ 1/x	y = k/x	Hyperbola
y ∝ 1/xⁿ	y = k/xⁿ	Curve with both axes as asymptotes

• Test y = kxⁿ by plotting y against xⁿ. A straight line through the origin confirms it, and the gradient is k.
• Percentage effects: if y ∝ x² and x rises by 10 per cent, y is multiplied by 1.1² = 1.21, a 21 per cent rise.
• If y ∝ 1/√x and x is multiplied by 4, y is multiplied by 1/2.

B8: composite functions

• gf(x) = g(f(x)). In general gf ≠ fg.
• Domain of gf: x must be in the domain of f, and f(x) must be in the domain of g.
• Range of gf: the output of g applied to the range of f.
• ff means apply f twice. For f(x) = 1/(x − 1), ff(x) = 1/(1/(x − 1) − 1) = (x − 1)/(2 − x), for x ≠ 1 and x ≠ 2.

Method for a composite equation. Build gf(x) in full, simplify, then solve. Do not substitute numbers until the composite is a single expression, unless the question asks for a value.

Worked reminder. f(x) = 2x + 1 and g(x) = x³. gf(x) = (2x + 1)³ and fg(x) = 2x³ + 1. At x = 1, gf(1) = 27 and fg(1) = 3.

B8: inverse functions

Method in steps

1. Check f is one-one on its domain, restricting the domain if not.
2. Write y = f(x) and rearrange to make x the subject.
3. Replace y by x to write f⁻¹(x).
4. The domain of f⁻¹ is the range of f.

Must-know facts

• Graph: mirror the curve in the line y = x. A point (a, b) on f becomes (b, a) on f⁻¹.
• If f(a) = b then f⁻¹(b) = a.
• Inputs of f⁻¹ are outputs of f, and the other way round.
• (fg)⁻¹ = g⁻¹f⁻¹: the order reverses. For f(x) = 2x + 1 and g(x) = x³, fg(x) = 2x³ + 1 and (fg)⁻¹(x) = ∛((x − 1)/2).
• Self-inverse: f⁻¹ = f, so ff(x) = x. Example: f(x) = (x + 3)/(x − 1) for x ≠ 1.
• Where f meets f⁻¹: if f is increasing, solve f(x) = x.

Worked reminder (domain restriction). f(x) = (x − 1)² + 2 for x ≥ 1 has range f(x) ≥ 2. Then y − 2 = (x − 1)², and x ≥ 1 gives x = 1 + √(y − 2). So f⁻¹(x) = 1 + √(x − 2) with domain x ≥ 2.

Must-know distinctions

Compare	Rule
gf and fg	gf applies f first. They usually differ
f⁻¹(x) and 1/f(x)	The inverse function is not the reciprocal
Touching and crossing	Even power of a factor: touch. Odd power: cross
Vertical and horizontal asymptote	x = 0 comes from division by zero. y = 0 is the limit of the curve for large
Domain of f and of f⁻¹	Domain of f⁻¹ is the range of f

Quick self-test

1. For y = (x + 1)(x − 3)², state where the curve meets each axis and say whether it crosses or touches the x-axis at each root.
2. Solve |x + 2| = 3 − x.
3. State the asymptotes of y = −5/x² and say which side of the x-axis the curve lies on.
4. y varies inversely with x, and y = 8 when x = 4. Find y when x = 16.
5. f(x) = 2x + 1 and g(x) = x³. Find gf(1) and fg(1).
6. Find the inverse of f(x) = (x + 3)/(x − 1) for x ≠ 1.
7. f(x) = (x − 1)² + 2 for x ≥ 1. State the domain of f⁻¹.
8. If f(x) = 2x + 1 and g(x) = x³, find (fg)⁻¹(55).
9. The point (3, 7) lies on y = f(x). Write down a point on y = f⁻¹(x) and the value of f⁻¹(7).
10. y = kx². By what percentage does y change when x increases by 10 per cent?

Answers

1. y-intercept (0, 9), since (1)(9) = 9. Crosses the x-axis at (−1, 0), touches it at (3, 0).
2. Right arm: x + 2 = 3 − x gives x = 1/2. Left arm: −x − 2 = 3 − x gives −2 = 3, impossible. So x = 1/2. Check: |5/2| = 5/2.
3. Asymptotes x = 0 and y = 0. Both branches lie below the x-axis.
4. y = k/x, so k = 32, and y = 32/16 = 2.
5. gf(1) = g(3) = 27. fg(1) = f(1) = 3.
6. y(x − 1) = x + 3, so x(y − 1) = y + 3 and x = (y + 3)/(y − 1). So f⁻¹(x) = (x + 3)/(x − 1), x ≠ 1. The function is self-inverse.
7. The range of f is f(x) ≥ 2, so the domain of f⁻¹ is x ≥ 2.
8. fg(x) = 2x³ + 1, so (fg)⁻¹(55) = ∛((55 − 1)/2) = ∛27 = 3. Check: fg(3) = 55.
9. (7, 3) lies on f⁻¹, and f⁻¹(7) = 3.
10. y is multiplied by 1.21, a 21 per cent increase.

Where marks are usually lost

• Writing gf when the working is fg, or applying the functions in the wrong order.
• Giving f⁻¹(x) with no domain, when the question asks for the inverse function as a function.
• Drawing a double root as a crossing, or leaving out the y-intercept.
• Leaving out the asymptotes on a sketch of y = a/x or y = a/x², or drawing a curve that touches them.
• Keeping a modulus root without checking the original equation.
• Cancelling a factor of x and so dropping the root x = 0.
• Choosing the wrong square root when inverting a restricted quadratic. Use the sign that matches the domain.
• Calling a relationship proportional when the graph does not pass through the origin.
• Reflecting the inverse in the x-axis or y-axis instead of the line y = x.

Official syllabus

AQA, A-level Mathematics (7357) specification, version 1.3, 31 January 2018, content B7 and B8 and Appendix A notation for functions.