Basic Concepts Injective, Surjective, Bijective Inverse Function Composite Function Examples

asic Concepts

A function f : X ® Y is a rule which associates each element x Î X with a unique element y Î Y such that y = f(x).

• The set X is called the domain of f and is denoted by D_f .
• The set Y is called the codomain of f.
• f(x) is called the image of x under f or f-image of x.
• The set { f(x) : x Î X } is called the range of f and is denoted by R_f .  It is the set of all images of x under f.
• Range of f is a subset of Y, ie, R_f Í Y.
• From the graph of f, f is a function if every vertical line x = a for each a Î X cuts the graph of f at only one point.

njective, Surjective, Bijective

• A function f : X ® Y is said to be injective or one-one if no two distinct elements of X have the same f-image, ie, f(x₁) = f(x₂)  Þ  x₁ = x₂  for x₁, x₂ Î X.

• A function f : X ® Y is said to be surjective or onto if for every y Î Y, there is at least one x Î X such that y = f(x), ie, R_f = codomain of f.

• A function f : X ® Y is said to be bijective if it is both one-one and onto.

nverse Function

For every bijective function f : X ® Y, there exists an inverse function f^-1 : Y ® X such that

• Domain of f^-1 = Range of f.
• Range of f^-1 = Domain of f.
• The graph of the inverse y = f^-1(x) is the reflection of the graph of y = f(x) in the line y = x.  Same scale must be used for both axes.

omposite Function

Let f and g be functions.
Then the composite gof, or simple gf, is defined by

• gof is a function if R_f Í D_g.
• D_gof = D_f .
• R_gof Í R_g.

xamples

Example 1:  Two functions are defined as follows:

For each of the functions, state the range and determine whether or not the function is one-one.
Give, in the same form, the definition of the functions gof and g^-1.

Solution:

	x2 - 2x  =  x(x - 2)  =  (x - 1)2 - 1. Rf = [-1, ¥).  f(0) = 0 = f(2)  but  0 ¹ 2. \  f is not one-one.
	Rg = R+ = (0, ¥).  g is one-one,  as every horizontal line y = b, b Î R, cuts the graph of y = g(x) at most once.

gf(x)	=	g(x2 - 2x)
	=	2x2 - 4x e

\  gof : x ®

2x2 - 4x e

, x Î R, x ³ 0.

Let y	=	e2x
ln y	=	2x
x	=	½ln y

\  g^-1 : x ® ½ln x, x Î R⁺.

Example 2:  The functions f and g are defined by

Show that fog is a function.  Define fog and state its range.  Explain why the function gof does not exist.

Solution:

R_g = R⁺, D_f = R⁺,
\  R_g Í D_f  Þ  fog is a function.

fog(x)  =  f(1 - x)  =  ln (1 - x) \  fog : x ® ln (1 - x), x < 1. Rfog = R.

R_f = R, D_g = (-¥, 1),
\  R_f Ë D_g  Þ  gof is not a function.