Functions and Relations
Relation
- is a set of ordered pairs.
-mapping or pairing from the domain to the
range is one way to show correspondence in
a relation
algebraically, or graphically.
Function notation
𝑦 = 𝑓(𝑥)
Ex. A = {(-1, 3), (2, 0), (-3, 2)}
Domain is the set of independent variables.
Range is the set of dependent variables.
Function
Function
- is a relation in which each of the elements of
the domain is paired with exactly one
element of the range.
- is used to describe how one variable depends
on another.
- it can have the same value as domain.
- it
may
be
represented
numerically,
Function
Function
Vertical line test
- is a test that determines whether a
relation is a function or not by drawing
a vertical line through the graph of its
ordered pairs.
Operations on Functions
Inverse Function
Sum
-The inverse is usually shown by putting a little "-1"
after the function name, like this:
𝑦 = (𝑓 + 𝑔) 𝑥
𝒇−𝟏(𝒙)
𝑦 = 𝑓(𝑥) + 𝑔(𝑥)
Difference
𝑦 = (𝑓 − 𝑔)𝑥
𝑦 = 𝑓(𝑥) − 𝑔(𝑥)
Product
𝑦 = (𝑓𝑔)(𝑥)
𝑦 = 𝑓(𝑥)𝑔(𝑥)
Find the inverse function of f(x)=3x+2.
f (x)=3x+2
y = 3x + 2
Let f(x) =y
y - 2 = 3x
Transpose the constant to the left side
𝑦−2 3𝑥
3
=
3
Divide both sides by 3
Quotient
𝑓
𝑦 = ( ) (𝑥)
𝑔
𝑓(𝑥)
𝑦=
𝑔(𝑥)
So if 𝑓−1(𝑦) =
𝑦−2
3
Since the choice of the variable is arbitrary, we can
write this as:
𝑥−2
𝑓−1(𝑥) =
3
Composite Functions
(𝑓 ∙ 𝑔)(𝑥) 𝑜𝑟𝑓[𝑔(𝑥)]
(𝑔 ∙ 𝑓)(𝑥) 𝑜𝑟 𝑔[𝑓(𝑥)]
Rational Functions
Evaluation of Functions
- is defined as the quotient of polynomials in which
the denominator has a degree of at least 1.
𝑃(𝑥)
𝑓(𝑥) =
, 𝑤ℎ𝑒𝑟𝑒 𝑄 ≠ 0
𝑄(𝑥)
To evaluate the output for “f(x)” you just need to
substitute the given value to “(x)”.
Example.
Example.
Given 𝑓 (𝑥) = 6 − 𝑥2. Find 𝑓 (1)and 𝑓(−2)
Solution
1.
𝑓(1) = 6 − 12
𝑓(1) = 6 − 1
𝑓(1) = 5
2.
𝑓(−2) = 6 − (−2)2
𝑓(−2) = 6 − 4
𝑓(−2) = 2
(𝑥2 − 3𝑥 − 2)
(𝑥2 − 4)
The domain of this function includes all values of x,
except where 𝑥2 − 4 = 0
𝑓(𝑥) =
We can factor the denominator to find the
singularities of the function: 𝑥2 − 4=(x-2) (x+2)
Setting each linear factor equal to zero, we have
𝑥 − 2 = 0 𝑎𝑛𝑑 𝑥 + 2 = 0.. .
Solving each of these yields solutions𝑥 = 2 𝑎𝑛𝑑
𝑥 = −2; thus, the domain includes all 𝑥 not equal to
2 or -2