Functions

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 2 —

★ Exam Tip: The chapter’s review questions are all 9709 Paper 11 past-paper questions. The same moves keep coming back: complete the square → range, smallest value of k for one-one, find f⁻¹ and its domain, composites like fg(x) = k, and describe a sequence of transformations.

Why study functions? Real things are modelled as functions — a drink cooling with time, the height of a bicycle-tyre valve as the wheel rolls, depth of water in a cone filling from a tap, bacteria count (“when will it pass 5 billion?”). This chapter goes deeper: domain, range, one-one, composite, inverse, graph transformations.

✓ Quick check (prerequisites)

Skill	Try it	Answer
output of a function	f(x)=3x−2, find f(4)	10
composite function	f(x)=2x+1, g(x)=1−x, find fg(x)	2(1−x)+1 = 3−2x
simple inverse	f(x)=5x+4, find f⁻¹(x)	x−45
complete the square (Ch 1)	write 2x²−12x+5 as a(x+b)²+c	2(x−3)²−13

1. What is a Function?

Definition: a function is a relation that gives each member of one set exactly one member of another set (each input ⇒ one output). Another name: a mapping.

One-one vs many-one

one-one: a flat line cuts it ONCE

many-one: a flat line cuts it TWICE

NOT a function: a vertical line cuts it twice

	each input →	each output ←	example
one-one function	one output	only one input produces it	x ↦ x+2, x∈ℝ
many-one function	one output	can come from several inputs	x ↦ x², x∈ℝ
NOT a function	some input has two or more outputs		y² = x (x=3 gives y=±√3)

Tricks: Vertical line hits graph >1 time ⇒ not a function.
Horizontal line hits >1 time ⇒ many-one (no inverse).

Notation

Two ways to write the same functionf : x ↦ x + 2 (x ∈ ℝ) or f(x) = x + 2 (x ∈ ℝ)

“f : x ↦ x+2” reads “the function f is such that x is mapped to x+2” or “f maps x to x+2”.
f(x) = output of f when the input is x. e.g. f(x)=x+2 ⇒ f(5) = 5+2 = 7.
x ∈ ℝ means “x belongs to the set of real numbers”; “x ∈ ℝ, x ≥ 0” is often shortened to just x ≥ 0.

Domain and rangeDomain = set of input values (x) | Range (codomain / image set) = set of output values (f(x))

Remember: when you define a function you must also give its domain — the domain decides the range and whether the function is one-one.

Finding the range — method

Sketch the graph over the given domain (mark both end-points).
Range = the y-values covered: read from the lowest to the highest point.
Quadratic? Complete the square to find the vertex: (…)² ≥ 0 so the smallest value of the bracket is 0.
Check the end-points as well as the vertex — the other extreme is at an end.

Worked Ex 2.1range from a line

f(x) = 5 − 2x, x ∈ ℝ, −4 ≤ x ≤ 5. a domain b sketch c range.

a domain: −4 ≤ x ≤ 5. b straight line, gradient −2, y-intercept 5.

end-pointsx = −4: y = 5 − 2(−4) = 13x = 5: y = 5 − 2(5) = −5

domain = x-values used · range = y-values reached

c) range: −5 ≤ f(x) ≤ 13 (line slopes down, so highest at x=−4, lowest at x=5)

Worked Ex 2.2range from a quadratic

f(x) = (x − 3)² + 8, −1 ≤ x ≤ 9. Sketch and find the range.

Positive quadratic ⇒ ∪-shape. (x−3)² is a square so it is always ≥ 0; smallest is 0 at x = 3 ⇒ minimum point (3, 8).

end-pointsx = −1: y = (−1−3)² + 8 = 24x = 9: y = (9−3)² + 8 = 44

max is at an END of the domain (44, not 24)

range: 8 ≤ f(x) ≤ 44

Common mistake: stopping at the vertex, or taking 24 as the top — 9 is further from x=3 than −1, so the max is 44.

2. Composite Functions

Most functions are combinations of simpler ones. e.g. x ↦ 3x − 7 is “multiply by 3 then subtract 7” = g : x ↦ 3x (multiply by 3) then f : x ↦ x − 7 (subtract 7) ⇒ “first do g, then do f”.

fg(x) = f( g(x) ) — the one next to x acts first

Definition: when one function is followed by another, the result is a composite function.

KEY POINT 2.1fg(x) means: g acts on x first, then f acts on the result ⇒ fg(x) = f( g(x) )

KEY POINT 2.2 — three things to remember:

fg exists only if the range of g is inside the domain of f.
In general fg(x) ≠ gf(x) (order matters!).
ff(x) = apply f twice.

Explore 2.1 — who is right?

f(x) = 2x − 5, g(x) = 3x − 1 (x ∈ ℝ). Find gf(x). Students: A (3x−1)(2x−5) = 6x²−17x+5 · B 2(3x−1)−5 = 6x−7 · C 3(2x−5)−1 = 6x−16.

Answer: Student C ✓

gf(x) = g( f(x) )= g(2x − 5)= 3(2x − 5) − 1= 6x − 16

A ✗ multiplied the two functions (composite ≠ product). B ✗ did f(g(x)) = fg, wrong order — gf means f first.

How to find fg(x) — method

Trick: replace every x in the OUTER function by the whole inner function — use brackets! e.g. f(x)=(x−4)²−1 → f(11/2) = (11/2 − 4)² − 1.

Worked Ex 2.3value of a composite

f(x) = (x−4)² − 1 (x∈ℝ); g(x) = 2x+3x−2 (x∈ℝ, x>2). Find fg(4).

g firstg(4) = 2(4)+34−2 = 112fg(4) = f( 112 )= ( 112 − 4 )² − 1= 1¼

Worked Ex 2.4algebraic composites

f(x) = 2x + 3, g(x) = x² − 1 (x∈ℝ). Find a fg(x) b gf(x) c ff(x).

a (g first: x²−1)fg(x) = f(x² − 1)= 2(x² − 1) + 3= 2x² + 1

b (f first: 2x+3)gf(x) = g(2x + 3)= (2x + 3)² − 1= 4x² + 12x + 9 − 1= 4x² + 12x + 8

c (f twice)ff(x) = f(2x + 3)= 2(2x + 3) + 3= 4x + 9

Note fg(x) ≠ gf(x) — compare answers a and b.

Worked Ex 2.5a fraction composite

f : x ↦ 5x−2 (x∈ℝ, x≠2); g(x) = 3 − x² (x∈ℝ). Find a fg(x) b ff(x).

afg(x) = f(3 − x²)= 5(3 − x²) − 2= 51 − x²

bff(x) = f( 5x−2 )= 55x−2 − 2×(x−2) top & bottom: 5(x−2)5 − 2(x−2)= 5x − 109 − 2x

Trick in b: multiply numerator and denominator by (x − 2) to clear the little fraction.

Worked Ex 2.6fg(x) = k has real solutions

f(x) = x² + 4x, g(x) = 3x − 1 (x∈ℝ). Find the values of k for which fg(x) = k has real solutions.

fg(x) = (3x−1)² + 4(3x−1) = 9x² + 6x − 3fg(x) = k ⇒ 9x² + 6x + (−3 − k) = 0real solutions: b² − 4ac ≥ 06² − 4(9)(−3−k) ≥ 0144 + 36k ≥ 0k ≥ −4

Insight: fg(x)=k has a real solution exactly when k is in the range of fg — here fg(x) ≥ −4.

3. Inverse Functions

Definition: the inverse of f(x) is the function that undoes what f(x) has done. Written f⁻¹(x) (not 1/f(x)!).

f⁻¹ undoes f

KEY POINT 2.3

f f⁻¹(x) = f⁻¹ f(x) = x
domain of f⁻¹ = range of f
range of f⁻¹ = domain of f

KEY POINT 2.4: an inverse f⁻¹(x) exists if, and only if, f(x) is a one-one mapping. (Many-one ⇒ no inverse — unless you restrict the domain.)

Finding the inverse — 3 steps

Exampleinverse of f(x) = 3x − 1

y = 3x − 1x = 3y − 13y = x + 1y = x + 13

f⁻¹(x) = x + 13

Self-inverse functions

If f and f⁻¹ are the same function, f is self-inverse: f(x) = f⁻¹(x), so ff(x) = x. e.g. f(x) = 1x (x ≠ 0): f⁻¹(x) = 1x (x ≠ 0) ⇒ self-inverse.

Explore 2.2 — f(x) = (x−2)² + 1, x ∈ ℝ

restrict to one side of the vertex

type: many-one
vertex (2, 1)
domain ℝ · range f(x) ≥ 1
no inverse (many-one: e.g. f(0)=f(4)=5)
fix: restrict the domain to x ≥ 2 (or x ≤ 2) ⇒ one-one ⇒ inverse exists
for x ≥ 2: f⁻¹(x) = 2 + √x − 1, domain x ≥ 1

Worked Ex 2.7inverse with a root, then solve

f(x) = √x + 2 − 7 (x∈ℝ, x ≥ −2). a find f⁻¹(x) b solve f⁻¹(x) = f(62).

ay = √(x+2) − 7x = √(y+2) − 7x + 7 = √(y+2)(x+7)² = y + 2y = (x+7)² − 2

bf(62) = √64 − 7 = 1(x+7)² − 2 = 1(x+7)² = 3x + 7 = ±√3x = −7 ± √3

range of f is f(x) ≥ −7 ⇒ domain of f⁻¹ is x ≥ −7, so reject x = −7 − √3.

f⁻¹(x) = (x+7)² − 2 | solution: x = −7 + √3

Worked Ex 2.8smallest k for an inverse

f(x) = 5 − (x−2)², x∈ℝ, k ≤ x ≤ 6. a smallest k for which f has an inverse. b for that k find f⁻¹(x) and its domain & range.

a vertex (2, 5); f(6) = 5 − 16 = −11. Must be one-one ⇒ use only the right-hand side of the vertex ⇒ smallest k = 2.

f(x)=5−(x−2)² for 2≤x≤6

y = 5 − (x−2)²x = 5 − (y−2)²(y−2)² = 5 − xy − 2 = √(5 − x)y = 2 + √(5 − x)

f⁻¹(x) = 2 + √(5 − x)

domain of f⁻¹ = range of f ⇒ −11 ≤ x ≤ 5; range of f⁻¹ = domain of f ⇒ 2 ≤ f⁻¹(x) ≤ 6.

Remember (choosing ±): when you take a square root while finding f⁻¹, the range of f⁻¹ (= domain of f) tells you which sign to keep. Here y ≥ 2 so y − 2 = +√.

Restrict-domain rule: vertex at x = h ⇒ domain x ≥ h or x ≤ h gives one-one. “Least k” for x ≥ k ⇒ k = h.

Typical Paper-11 pattern (own example):

f : x ↦ x² − 6x + k, x ≥ p. Complete the square: (x−3)² + k − 9.

range: f(x) ≥ k − 9
smallest p for one-one: p = 3 (the vertex)
f⁻¹(x) = 3 + √(x − k + 9), domain x ≥ k − 9

4. Graph of a Function and its Inverse

Take f(x) = 2x + 1, −4 ≤ x ≤ 2. f(−4) = −7, f(2) = 5 ⇒ domain −4 ≤ x ≤ 2, range −7 ≤ f(x) ≤ 5.

f⁻¹ = mirror image of f in y = x

f⁻¹(x) = x − 12

domain of f⁻¹ = range of f ⇒ −7 ≤ x ≤ 5
range of f⁻¹ = domain of f ⇒ −4 ≤ f⁻¹(x) ≤ 2
points swap: (−4, −7) on f ↔ (−7, −4) on f⁻¹; (2, 5) ↔ (5, 2)

KEY POINT 2.5 The graphs of f and f⁻¹ are reflections of each other in the line y = x (because f f⁻¹(x) = x = f⁻¹ f(x)). When f is self-inverse, its graph is symmetrical about y = x.

Worked Ex 2.9draw f and f⁻¹

f(x) = (x−1)² − 2, x∈ℝ, 1 ≤ x ≤ 4. Draw f and f⁻¹ on the same axes.

(x−1)² ≥ 0, smallest 0 at x=1 ⇒ vertex (1, −2). f(4) = 9 − 2 = 7. One-one on this domain ⇒ inverse exists.

Method: sketch f, then reflect it in y = x (swap every x and y).

f: 1≤x≤4 and f⁻¹: −2≤x≤7

Algebra check: f⁻¹(x) = 1 + √(x+2); domain −2 ≤ x ≤ 7, range 1 ≤ f⁻¹ ≤ 4.

Worked Ex 2.10self-inverse and symmetry

f : x ↦ 2x + 7x − 2 (x∈ℝ, x≠2). a find f⁻¹(x). b what does it say about the graph?

a (collect y terms, factorise)y = 2x+7x−2x = 2y+7y−2xy − 2x = 2y + 7y(x − 2) = 2x + 7y = 2x+7x−2

f⁻¹(x) = f(x) ⇒ f is self-inverse

b graph of y = f(x) is symmetrical about the line y = x.

self-inverse ⇒ symmetric in y = x

Explore 2.3 (Ali’s claim): Ali says the two diagrams show a function and its inverse function. Check each with the tests: a curve with a hump (or a sideways curve) is many-one / not a function, so the red mirror image is only a reflection, not an inverse function. Reflecting in y = x always gives a graph; it is an inverse function only if the original is a one-one function.

5. Transformations of Graphs

From : translations, reflections, rotations, enlargements. Here: translations, reflections and stretches of y = f(x), and combinations.

Explore 2.4 — what you should discover
y = …	what happens to y = f(x)	acts on
f(x) + a	translate by vector (0, a) — up a	y (vertical)
f(x + a)	translate by vector (−a, 0) — left a	x (horizontal)
−f(x)	reflect in the x-axis	vertical
f(−x)	reflect in the y-axis	horizontal
a f(x)	stretch, factor a, parallel to the y-axis	vertical
f(ax)	stretch, factor 1/a, parallel to the x-axis	horizontal

① Translations

up a: vector (0, a)

f(x−3): RIGHT 3 · vector (3, 0)

KEY POINTS 2.6 – 2.8

y = f(x) + a : translation of y = f(x) by vector (0, a)
y = f(x − a) : translation by vector (a, 0)
y = f(x − a) + b : translation by vector (a, b)

Why does x → x − 3 move RIGHT? (y = x²−2x+1 → y = (x−3)²−2(x−3)+1). The curves are at the same height when x_new − 3 = x_old, i.e. x_new = x_old + 3 — the new curve is 3 units to the right of the old one.

Rule of thumb: number outside f(…) acts as it looks (+ up). Number inside the bracket does the opposite (x+5 ⇒ left 5).

Worked Ex 2.11translate 2 units right

y = x² + 5x translated 2 units to the right. Give the answer as y = ax² + bx + c.

y = x² + 5xreplace every x by (x − 2):y = (x−2)² + 5(x−2)y = x² − 4x + 4 + 5x − 10y = x² + x − 6

Worked Ex 2.12translate by a vector

y = √(2x) translated by (−5, 3).

y = √(2x)x → x + 5 (left 5), then add 3:y = √(2(x+5)) + 3y = √(2x+10) + 3

② Reflections

−f(x): flip in the x-axis

f(−x): flip in the y-axis

KEY POINTS 2.9 & 2.10 y = −f(x) is a reflection of y = f(x) in the x-axis (y → −y). y = f(−x) is a reflection in the y-axis (x → −x).

Worked Ex 2.13turning points after reflection

Quadratic y = f(x) has a minimum at (5, −7). Give the vertex and say max/min for a y = −f(x), b y = f(−x).

a reflect in x-axis: (5, 7) — now a maximum.

b reflect in y-axis: (−5, −7) — still a minimum (not turned over).

③ Stretches

2f(x): vertical ×2 · roots stay put

f(2x): horizontal ×½ · y-intercept stays

KEY POINTS 2.11 & 2.12

y = a f(x) : stretch of y = f(x), stretch factor a, parallel to the y-axis (y-values × a; x-axis invariant).
y = f(ax) : stretch of y = f(x), stretch factor 1/a, parallel to the x-axis (x-values ÷ a; y-axis invariant).

Other ways to say “stretch factor 2 parallel to the y-axis”: scale factor 2 with the line y = 0 invariant · stretch with the x-axis invariant · stretch relative to the x-axis · a vertical stretch, factor 2.
If a < 0, y = a f(x) = a stretch with negative factor, or a stretch with positive factor then a reflection in the x-axis.

Worked Ex 2.14stretch ×4 parallel to y

y = 5 − ½x² is stretched, factor 4, parallel to the y-axis.

let f(x) = 5 − ½x²4 f(x) = 20 − 2x²y = 20 − 2x²

Worked Ex 2.15identify the stretch

Single transformation mapping y = x² − 3x − 5 onto y = 4x² − 6x − 5?

f(x) = x² − 3x − 54x² − 6x − 5 = (2x)² − 3(2x) − 5= f(2x)

stretch parallel to the x-axis, stretch factor ½

6. Combined Transformations

Vertical (act on y)
f(x) + a	translation (0, a)
−f(x)	reflection in x-axis
a f(x)	vertical stretch, factor a

Horizontal (act on x)
f(x + a)	translation (−a, 0)
f(−x)	reflection in y-axis
f(ax)	horizontal stretch, factor 1/a

Explore 2.5 (apply to triangle T with vertices (0,0), (2,1), (3,0)): e.g. translate (0,3) then stretch ½ ≠ stretch ½ then translate (0,3); reflect in x-axis and translate (−2,0) gives the same image either way.

KEY POINT 2.13

Two vertical (or two horizontal) transformations: the order may change the result.
One horizontal + one vertical: the order does not matter.

KEY POINT 2.14 Vertical transformations follow the normal order of operations (as in arithmetic).
y = a f(x) + k: × a first, then + k.

KEY POINT 2.15 Horizontal transformations follow the opposite order.
y = f(bx + c): replace x by x+c (translate) first, then x by bx (stretch).

Worked Ex 2.16y = 2f(x) − 3

Given the graph of y = f(x), sketch y = 2f(x) − 3. Two vertical transformations ⇒ normal order: ① stretch vertically ×2 → ② translate (0, −3).

given graph

① stretch ×2

② then shift down 3

Each key point (x, y) → (x, 2y − 3). e.g. (1, −1) → (1, −2) → (1, −5); (2, 3) → (2, 6) → (2, 3).

Worked Ex 2.17describe the transformations

The graph of y = x² has image y = g(x) (graph below). a two ways to describe it b write g(x).

a way 1: translation (4, 0), then horizontal stretch, factor ½.

way 2: horizontal stretch, factor ½, then translation (2, 0).

x² → (x−4)² [translate (4,0)] → (2x−4)² [stretch ½]

b (way 1)translate (4,0): x → x − 4y = x² becomes y = (x − 4)²stretch ½: x → 2xy = (x − 4)² becomes y = (2x − 4)²g(x) = (2x − 4)²

(Using way 2 gives the same answer: (2x)² then x→x−2: (2(x−2))² = (2x−4)².)

Own example — y = f(2x − 6):

Write f(2x−6) = f(2(x−3)). Horizontal ⇒ opposite order: translate (6, 0) [x → x−6] then stretch ½ [x → 2x]. Equivalent: stretch ½ then translate (3, 0).

Common mistakes: ① f(x+a) moves left, not right. ② fg means g first. ③ f⁻¹ ≠ 1/f. ④ f(½x) is a stretch of factor 2 (1/a), not ½. ⑤ y = f(a − x) = reflect in y-axis then translate (a, 0). ⑥ Forgetting to restrict the domain before finding an inverse of a quadratic. ⑦ Missing the end-point when finding a range.

7. Full Chapter Cheat-Sheet

✓ Functions

function: each x ↦ exactly one y · one-one or many-one (a relation with 2 outputs for one input is NOT a function)
domain = inputs · range (image set) = outputs

✓ Composite functions

fg(x) = f(g(x)) — g acts first, then f
fg exists only if range of g ⊆ domain of f · fg(x) ≠ gf(x) in general · ff = f twice

✓ Inverse functions

f f⁻¹(x) = f⁻¹ f(x) = x · y = f(x) ⇔ x = f⁻¹(y)
Steps: ① y = … ② swap x, y ③ make y the subject
domain f⁻¹ = range f · range f⁻¹ = domain f
f⁻¹ exists ⇔ f is one-one
graphs of f and f⁻¹: reflections in y = x
f(x) = f⁻¹(x) ⇒ self-inverse ⇒ ff(x) = x and graph symmetric about y = x

✓ Transformations of y = f(x)

y = …	transformation
f(x) + a	translation (0, a)
f(x + a)	translation (−a, 0)
−f(x)	reflection in x-axis
f(−x)	reflection in y-axis
a f(x)	stretch factor a, parallel to y-axis
f(ax)	stretch factor 1/a, parallel to x-axis

✓ Combining transformations

two vertical or two horizontal: order can matter · one of each: order doesn’t matter
vertical → normal order of operations · horizontal → opposite order

Not included: the chapter’s practice exercises (2A–2H) and the 13 end-of-chapter past-paper questions are for you to try — use the methods above. Re-draw the graphs yourself for each transformation!