Additional Mathematics notes, chapter 1

Functions

Textbook Additional Mathematics Form 4 KSSM (DLP), pages 1–33

What this chapter covers

Explain functions using graphs and notation, and determine whether a relation or graph is a function (1.1.1)
Determine the domain, codomain and range of a function, including absolute value functions (1.1.2)
Find the image of a function when the object is given and the object when the image is given (1.1.3)
Describe and determine composite functions, find their images and objects, and find a function from a composite function (1.2.1–1.2.4)
Solve problems involving composite functions (1.2.5)
Describe the inverse of a function, use the horizontal line test and find inverse functions and their graphs (1.3.1–1.3.3)

1.1Functions

A function is a special relation from set X to set Y in which each element of X is mapped to one and only one element of Y. Example: the pay for a cashier at RM80 a day depends on the number of days worked.

f : x → 2x + 1. The object x is mapped to its image 2x + 1.

Notation	How to read it
f : x → 2x + 1	"f maps x to 2x + 1"
f(x) = 2x + 1	"2x + 1 is the image of x under f" or "f of x equals 2x + 1"
f(1) = 3	1 is the object, 3 is its image

A relation is not a function if an object has two images (one-to-many) or if an object has no image. An element in set Y with no object is allowed.

Function means one-to-one or many-to-one. One-to-many is never a function.

Vertical line test (graphs)

Draw a vertical line across the graph. If it cuts the graph at only one point, the graph is a function. If it cuts at no point or at two or more points, it is not a function.

A parabola with a vertical line cutting it at one point — One point: a function

A sideways parabola with a vertical line cutting it at two points — Two points: not a function

Test described in textbook p. 4.

Absolute value function

|x| is the numerical value of x: |x| = x when x ≥ 0 and |x| = −x when x < 0. The graph of f(x) = |x| is a V with its vertex at (0, 0).

Example 5 (textbook p. 8): f : x → |2x − 1| for −1 ≤ x ≤ 2

x	−1	0	12	1	2
f(x)	3	1	0	1	3

V shaped graph of y = |2x − 1| with vertex at (1/2, 0) and end points (−1, 3) and (2, 3)

Range: 0 ≤ f(x) ≤ 3. The part of the line y = 2x − 1 below the x-axis is reflected upwards.

Domain, codomain and range

Term	Meaning
Domain	the set of objects x that can be put into f
Codomain	the whole set Y that the images may come from
Range	the images that are actually obtained; the range is a part of the codomain

Domain = {1, 2, 3, 4}. Codomain = {3, 5, 7, 9, 11}. Range = {3, 5, 7, 9}.

A discrete function has separate points (domain is a list of values). A continuous function is a line or curve over an interval, for example 1 ≤ x ≤ 4.

Do not put every element of the codomain into the range. 11 is in the codomain but no object maps to it.

The function f(x) = 3x − 1 has domain {0, 1, 2}. State the range.

Show answerHide answer

f(0) = −1, f(1) = 2, f(2) = 5. Range = {−1, 2, 5}

Finding the image or the object

Put the object into the formula to find the image. To find the object, put the image equal to f(x) and solve.

Example 6 (textbook p. 9): f : x → 3x + 5x, x ≠ 0

f(5) = 3(5) + 55 = 16.
Find x when f(x) = 8: 3x + 5x = 8, so 3x² − 8x + 5 = 0, (3x − 5)(x − 1) = 0, x = 53 or x = 1.

Example 7 (textbook p. 10): absolute value equations

f(x) = |2x − 3|.
f(x) = 5: 2x − 3 = 5 or 2x − 3 = −5, so x = 4 or x = −1.
f(x) < 1: −1 < 2x − 3 < 1, so 1 < x < 2.
f(x) > 3: 2x − 3 > 3 or 2x − 3 < −3, so x > 3 or x < 0.

|expression| < k gives one "sandwich" inequality (−k < … < k). |expression| > k gives two separate parts.

If f(x) = |x − 4|, find the values of x such that f(x) = 6.

Show answerHide answer

x − 4 = 6 or x − 4 = −6, so x = 10 or x = −2

1.2Composite functions

A composite function is made by putting one function inside another. fg(x) = f[g(x)]: do g first, then f. Also f²(x) = f[f(x)].

f(x) = x + 2 and g(x) = x². Apply g first: fg(x) = x² + 2.

Example 8 (textbook p. 14): f : x → 2x and g : x → x² − 5

fg(x) = f(x² − 5) = 2(x² − 5) = 2x² − 10
gf(x) = g(2x) = (2x)² − 5 = 4x² − 5
f²(x) = f(2x) = 4x
g²(x) = g(x² − 5) = (x² − 5)² − 5 = x⁴ − 10x² + 20

fg is not the same as gf, and fg is not f × g. The function nearest to x is done first.

Image, object and the missing function

Example 9 (textbook p. 15): f : x → x − 1, g : x → x² − 3x + 4

g(2) = 2² − 3(2) + 4 = 2, then f(2) = 1, so fg(2) = 1.
fg(x) = x² − 3x + 3. If fg(x) = 7 then x² − 3x − 4 = 0, (x + 1)(x − 4) = 0, x = −1 or x = 4.

Example 10 (textbook p. 16): f : x → x − 2, given fg : x → 8x − 7

f[g(x)] = g(x) − 2 = 8x − 7, so g(x) = 8x − 5.
If gf(x) is given instead, let y = x − 2 so x = y + 2, rewrite in y, then change y back to x.

Order matters. To find f, substitute into g first. When the outer function is known, write g(x) − 2 = result and solve for g(x).

f(x) = 2x + 1 and g(x) = x². Find (a) fg(x) (b) gf(3)

Show answerHide answer

(a) fg(x) = f(x²) = 2x² + 1
(b) f(3) = 7, g(7) = 49

Problems with composite functions

In a word problem, name the functions, decide which one is applied first, then form the composite function and substitute.

Example 12 (textbook p. 17), idea only

Production q depends on workers n, and revenue r depends on q. Revenue as a function of n is r[q(n)], the composite function rq(n). Substitute n to get the revenue.

Repeated functions (Example 11, textbook p. 16)

If f(x) = 1/x² = x⁻², then f²(x) = f[f(x)] = x⁴, f³(x) = x⁻⁸ and f⁴(x) = x¹⁶. The power is (−2)ⁿ: positive for even n and negative for odd n. Hence f²²(x) = x^2²² and f³³(x) = x^−2³³.

1.3Inverse functions

The inverse function f⁻¹ undoes f: f : x → y ⇔ f⁻¹ : y → x. f⁻¹ is read "f inverse". It does not mean the reciprocal 1f(x).

Property	Meaning
Only a one-to-one function has an inverse	each image has exactly one object
ff⁻¹(x) = f⁻¹f(x) = x	use this to verify that two functions are inverses
Domain of f⁻¹ = range of f	and range of f⁻¹ = domain of f
(a, b) on f ⇒ (b, a) on f⁻¹	graph of f⁻¹ is the reflection of the graph of f in y = x

Horizontal line test

If every horizontal line cuts the graph of f at no more than one point, f is one-to-one and has an inverse. If a horizontal line cuts at two or more points, f has no inverse.

A rising curve cut once by a horizontal line — One point: inverse exists

A U shaped curve cut twice by a horizontal line — Two points: no inverse

Finding f⁻¹ algebraically

Write y = f(x).
Make x the subject.
Write x as f⁻¹(y), then replace y with x.

Example 19 (textbook p. 28): f : x → 5x + 2

y = 5x + 2, so x = y − 25. Hence f⁻¹(x) = x − 25 and f⁻¹(7) = 7 − 25 = 1.

Finding f⁻¹(3) without the formula (Example 14, textbook p. 21)

f(x) = xx − 4, x ≠ 4. Let a = f⁻¹(3), so f(a) = 3: a = 3(a − 4), a = 6. Hence f⁻¹(3) = 6.

f⁻¹(k) is the object whose image is k. Solve f(x) = k. It is often quicker than finding the whole formula.

Example 17 (textbook p. 26): f : x → x² − 6, 0 ≤ x ≤ 4

Graphs of f(x) = x² − 6 from (0, −6) to (4, 10) and its reflection f⁻¹ in the line y = x from (−6, 0) to (10, 4)

Range of f is −6 ≤ f(x) ≤ 10, so the domain of f⁻¹ is −6 ≤ x ≤ 10. The point (4, 10) on f gives (10, 4) on f⁻¹.

Check the domain. f(x) = x² − 6 for all real x has no inverse. A restricted domain such as 0 ≤ x ≤ 4 makes it one-to-one.

f(x) = x + 43. Find f⁻¹(x).

Show answerHide answer

y = x + 43, so x = 3y − 4. Hence f⁻¹(x) = 3x − 4