Functions
Basic Concepts
Value (Image), Domain, Range,
Graph
Examples of Real functions
1. :
a. f(x) = 9
Constant function
b. Power functions f(x)=xn , where n is a natural number
Examples:
f(x) = x The identity function
f(x) = x2 The squaring function
f(x) = x3 The cubing function
The domain of each of these functions is R
b. Power functions f(x)=xn , where n is a negative integer
number.
Examples:
f(x) = 1/x The inverting function
f(x) = 1/x2 Inverting the square function
f(x) = 1/x3 Inverting the cube function
The domain of each of these functions is R – {0}
Polynomial Functions
Examples:
1. f(x) = any power function
2. f(x) = 8x9 + 6x5 -7x2 - x + 12
3. f(x) = 2x3 - 1
Rational Functions
A rational function is a function of the form p(x)/q(x), where
p(x) and q(x) are polynomials
Examples of rational functions are all polynomial functions
and hence all power functions.
The domain of a rational functions is the set of all real
Numbers which are not zeroes (roots) of the polynomial in
the denominator.
Example
3x  1
f ( x)  2
x  5x  4
domain f  the set of all real numbers satisfying :
x 2  5x  4  0
That ' s the set of all real numbers satisfying :
( x  1)( x  4)  0
That ' s the set of all real numbers satisfying :
x  1 , x  4
 domain f  R  {1 ,4}
Examples I
Determine domain f, and the
values: f(1), f(-2), f(0) and f(4) for
the given function f
Examples (1)
Determine domain f, and the values: f(1), f(-2), f(0)
and f(4) for the given function f
a.
f ( x)  2 x 2  3 x 
f (1)  2(1) 2  3(1) 
6
x4
6
1 4
 2  3  2  3
6
f (2)  2(2)  3(2) 
24
 8  6  1  13
2
f is defined for all x satisfying the condition :
x40
x4
 x  R  {4}
 x  (,4) (4, )
domain f  R  {4}  (,4) (4, )
f (4) does not exist , because 4 does not belong to domain f
b.
f ( x)  10
f (1)  f (2)  f (0)  f (4)  10
dom f  R
Range f  {10}
y
15
10
5
0
-5
-2.5
0
2.5
5
x
-5
-10
-15
c.
f ( x)  x
f (1)  1  1
f 2  2
f 0 0
f 4 4
dom f  R
Range f  [0, )  R 0
y
5
3.75
2.5
1.25
-5
-2.5
0
2.5
5
x
d.
f ( x)  3 2 x  8
f (1)  3 2(1)  8  6  8  2
f (2)  3 2(2)  8  12  8  4
f (0)  3 2(0)  8  0  8  8
f (2)  3 2(4)  8  16  8  8
domain f  R
Rangef  [0, )
e.
f ( x)  2 x  4
domain f  R
f (1)  2(1)  4  6
f ( 0)  2( 0)  4  4
f ( 2)  2( 2)  4  0
f (4)  2( 4)  4  12
Graph of f(x) = 2x + 4
y
10
5
0
-5
-2.5
0
2.5
5
x
-5
f.
f ( x)  x  3
domain f  R
f (1)  1  3  4
f ( 0)  0  3  3
f ( 2 )  2  3  1
f ( 4)  4  3  7
Graph of f(x) = x + 3
y
7.5
5
2.5
0
-5
-2.5
0
2.5
5
x
g.
f ( x)  2 x  6
f is defined for all x satisfying the inequality :
2x  6  0
 2x  6
 x3
 x  [3, )
domain f  [3, )
f (1), f (2) and f (0) do not exist
f (4)  2(4)  6  2
h.
f ( x)  6  2 x
f is defined for all x satisfying the inequality :
6  2x  0
 6  2x
3 x x 3
 x  (,3]
domain f  (,3]
f (4) does not exist
f (1)  6  2(1)  4  2
f (0)  6  2(0 )  6
f (2)  6  2(2)  10
i.
6
f ( x) 
( x  4)( x  2)
f is defined for all x satisfying the condition :
( x  4)( x  2)  0
 x  4 , x  2
 x  R  {4 ,  2}
 x  (,2) (2,4) (4, )
6
f ( x) 
( x  4)( x  2)
f (2) and f (4) do not exist
6
6
2
f (1) 


(1  4)(1  2) (3)(3)
3
6
6
f (1) 

6
(1  0)(1  0) (1)(1)
j.
6
f ( x)  2
x  2x  8
6
f ( x) 
( x  4)( x  2)
This is the same problem as i.
Case-Defined Function
(Piecewise Defined Function)
k.
f ( x) 

5 x 1
x 2  3 x 1
domain f  R
f (1)  5
f (2)  (2) 2  3  7
f (0)  (0) 2  3  3
f ( 4)  5
Case-Defined Function
(Piecewise Defined Function)
l.
f ( x) 

2 x  4 1 x  2
x  3 x 3
domain f  [1,2] [3, )
f (1)  2(1)  4  6
f ( 4)  4  3  7
f (0) does not exist
f (2) does not exist
m.
 3
; x  2
f ( x)   2
; x 1
x
domain f  (,2] (1, )
f (1) does not exist
f (4)  (4) 2  16
f (0) does not exist
f (  2)   3
3
f ( ) does not exist
2
f (1000)  3
How to graph this function?
y
5
y
25
20
2.5
15
0
-5
-2.5
0
2.5
5
10
x
5
-2.5
0
-5
-5
3
; x  2
f ( x)   2
; x 1
x
range f  (1, ) {3}
-2.5
0
2.5
5
x