9 May 1999

### asic Concepts

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

• The set X is called the domain of f and is denoted by Df .
• 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 X } is called the range of f and is denoted by Rf .� It is the set of all images of x under f.
• Range of f is a subset of Y, ie, Rf Y.
• From the graph of f, f is a function if every vertical line x = a for each a X cuts the graph of f at only one point.
 Note: In defining a function, the rule and domain must be given.� If the domain is not given, it is taken to be the largest possible domain for which the function is defined.� If the codomain is not specified, it is taken to be the range of the function.

### 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)� x1 = x2� for x1, x2 X.

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

y = f(x)� x = f-1(y).
• 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(x) = g(f(x)).
• gof is a function if Rf Dg.
• Dgof = Df .
• Rgof Rg.

### xamples

Example 1:  Two functions are defined as follows:

 f : x � x2 - 2x, x � R, x � 0; g : x � e2x, x � R.

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 2.
\� f is not one-one.

Rg = R+ = (0, ).�

g is one-one,�
as every horizontal line y = b, b 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 R, x 0.

 Let y �=� e2x ln y �= 2x x �=� �ln y
\� g-1 : x �ln x, x R+.

Example 2:  The functions f and g are defined by

 f : x � ln x, �x > 0; g : x � 1 - x, �x < 1.

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

Solution:

Rg = R+, Df = R+,
\� Rg Df� fog is a function.

 fog(x) �=� f(1 - x) �=� ln (1 - x)

\� fog : x ln (1 - x), x < 1.

Rfog = R.

Rf = R, Dg = (-�, 1),
\� Rf Dg� gof is not a function.

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