Relation  set of ordered pairs
set of all elements of the ordered pairs  domain
set of all second elements  range
relation can be satisfied into 4 categories:
1.
2.
3.
4.
One-to-one
One-to-many
Many-to-one
Many-to-many
A function f, or a mapping f from a set A to a set B is a rule that assigns to each element x in A a
unique element y in B
Image of x under f is represented by the notation f(x)
A function can only be one-to-one or many-to-one
How to test if relation is a function
1. Vertical line test
2. Check if there is at least one x (input value) that corresponds to more than one y(output
value)
Reciprocal function
F(x) = 1/x where 𝑥 ∈ ℝ
When x = 0, f(x) is undefined)  no finite value
When x > 0, f(x) is +ve; when x < 0, f(x) is -ve  graph only lies in quadrants I and iii
Sign diagram:
As x approaches 0 from values less than 0, f (x)  -∞
As x approaches 0 from values reater than 0, f(x)  ∞
As x tends to infinity, f(x)  0
As x tends to negative infinity, f(x)  0
x=0 (vertical asymptote)
y=0 (horizontal asymptote)
f(x) is asymptotic to 2 axes
𝑝(𝑥)
rational function  function which can be written in the form 𝑞(𝑥) where p(x) & q(x) are polynomials,
q(x) is not a constant nor 0
vertical asymptote  when the function be undefined (e.g. /0, sqrt(-ve numbers),…)
for a function
𝑎𝑥 𝑐
𝑏𝑥 𝑔
if c > g  no horizontal asymptote
𝑎
if c = g  horizontal asymptote: y = 𝑏
if c < g  horizontal asymptote: y = 0
decomposing original rational function into sum of two algebraic fractions is known as partial
functions decomposition
long division  find oblique asymptote if x is not in the remainder
absolute value function (modulus function)
𝑓(𝑥) = |𝑥| = 𝑥, 𝑥 ≥ 0; −𝑥, 𝑥 < 0
Draft the graph of each absolute value sign
1. Do cases (e.g. for x>?  x+1 …)
2. Draw sign diagram
3. Usually get a mx+c form  draft the graph
|x| represents the magnitude (modulus) of x
|x-y|  absolute difference between a and b or graphically distance
Solving equations involving absolute values
1. Do cases (separate in which case numbers inside || will become negative or positive)
e.g.
2. When it is in the format of |?| = |?|
square both sides and solve the equation
let a > 0
if |x| ≤ a  -a ≤ x ≤ a
if |x| ≥ a  x ≥ a or x ≤ -a
let b > 0
if |x| ≤ b  no solutions
if |x| ≥ b  𝑥 ∈ ℝ
solving inequalities involve in absolute values
-
Do by cases
Restrict the domain that was set at the beginning
Can square both sides but must ensure that both sides is +ve
Onto function  function that maps an element x to every element y
Even function  f(x) = f(-x) reflectional symmetry about y-axis
Odd function  f(-x) = -f(x) 180degree rotational symmetry about the origin
Determining whether the function are even, odd, or neither by sub -x into the function and see the
relationship between them
For g(f(x)) to exist, range of f(x) ≤ domain of g
Identity function: i(x) = x
f(i(x)) = f(x)
i(f(x)) = f(x)
given a function f, if there exists a function g that f(g(x)) = g(f(x)) = x  g is inverse function of f
both functions are images of each other about line y=x
inverse of f  f-1
find the inverse function
1. Write x as the subject in terms of y, then rewrite x as f-1(x) and y as x
2. The f(n) with the mirror image graph about the line y=x is the inverse f(n)
only one-to-one function have an inverse FUNCTION as for many-to-one function, the inverse
relation won’t be a function since one input has to be mapped to multiple outputs
however inverse function can be changed to function with inverse function by splitting the function
(restrict domain) [e.g. quadratic function, use the axis of symmetry to split the graph into 2 parts]
in order the inverse of f exists
domain of f-1 = range of f
domain of f = range of f-1
self-inverse function  the function itself is symmetrical about the graph y=x
for polynomial long division, the dividend must be a polynomial of the same or higher degree than
that of denominator