1.2-Basics of Functions

• A relation is any set of ordered pairs.

• The set of all first components of the ordered pairs is called the domain of the

relation.

• The set of all second components is called the range of the relation.

Example 1

Solution The domain is the set of all first components. Thus, the domain is:

{1994,1995,1996,1997,1998}.The range is the set of all second components. Thus, the range is:

{56.21, 51.00, 47.70, 42.78, 39.43}.

Find the domain and range of the relation:{(1994, 56.21), (1995, 51.00), (1996, 47.70),

(1997, 42.78), (1998, 39.43)}

Definition of a Function

• A relation becomes a function is a correspondence between two sets X and Y

that assigns to each element x of set X exactly one element y of set Y.

Example 2

Solution The elements in the domain and in blue.The elements in the range are in yellow.

Determine whether each relation is a function.a. {(1, 6), (2, 6), (3, 8), (4, 9)}

1234

689

Domain Range

(a) Figure (a) shows that every element in the domain corresponds to exactly one element in the range. No two ordered pairs in the given relation have the same first component and different second components. YES- the relation is a function.

689

1234

Domain Range

(b) Figure (b) shows that 6 corresponds to both 1 and 2. NO-This relation is not a function; two ordered pairs have the same first component and different second components.

b. {(6,1),(6,2),(8,3),(9,4)}

Example: Determine whether the equation defines y as a function of x:

2 4x y

Solve for y: 24y x

Since each value of x corresponds to ONE value of y, itIS a funciton!

Example: Determine whether the equation defines y as a function of x:

2 2 4x y

Solve for y: 2 24y x 24y x

Since each value of x will correspond to TWO values of y, itIs NOT a function!

Substitution: f (x) = x2 + 3x + 5.

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

a. We find f (2) by substituting 2 for x in the equation.

f (2) = 22 + 3 • 2 + 5 = 4 + 6 + 5 = 15

Thus, f (2) = 15.

Functions are usually given as equations using “function” notation:

Text Example cont.

Solutionb. We find f (x + 3) by substituting x + 3 for x in the equation.

f (x + 3) = (x + 3)2 + 3(x + 3) + 5

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

Equivalently,

f (x + 3) = (x + 3)2 + 3(x + 3) + 5= x2 + 6x + 9 + 3x + 9 + 5f (x + 3) = x2 + 9x + 23.

Square x + 3 and distribute 3 throughout the parentheses.

Text Example cont.

Solutionc. We find f (-x) by substituting -x for x in the equation.

f (-x) = (-x)2 + 3(-x) + 5

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

Equivalently,

f (-x) = (-x)2 + 3(-x) + 5 f (-x) = x2 –3x + 5.

“Piecewise” Functions are “split” into different pieces depending on the x-value

2( )

4

xf x

if 2

if 2

x

x

( 2)f (2)f (3)f

Graph this function:

21( ) 2

2 1

xf x

x

if 1

if 1

x

x

(0)f (1)f (6)f

Graph this function:

Ex. 3 Find the domain of each function:

a) 2( ) 3 4 5f x x x Domain is the set of all real numbers.

b)1

( )5

g xx

c) Volume of a sphere: 34

3V r

Since this is a radius, all r values must be positive. r > 0

d) ( ) 4 3h x x

Since what’s under the radical must be positive, solve the inequality:

4 3 0x 4

3x

All real numbers except -5

( , )

( , 5) ( 5, )

4( , ]

3

Ex. Find the domain and range of the function using your graphing calculator: 29y x

: 3 3D x : 0 3R y

Definition of a Difference Quotient

• The expression:

is called the difference quotient.

f (x h) f (x)

h0h

This is used extensively in Calculus!

Example: Find and simplify the difference quotient for:

2( ) 2 3f x x x

f (x h) f (x)

h

2 2[2( ) ( ) 3] (2 3)x xx h x h

h

2 2 22( 2 ) 3 2 3x xh h x h x x

h

2 2 22 4 2 3 2 3x xh h x h x x

h

24 2 (4 2 1)xh h h h x h

h h

4 2 1x h