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Section 1.2: Functions and Their Graphs
Practice HW from Larson Textbook (not to hand in)
p. 17 # 1-23 odd, 33, 47, 48, 49, 50, 51
Interval Notation
Interval Notation provides a shorthand way of representing particular sets on the real
number line. Many of the answers for the problems given in your textbook use interval
notation. We review some of the basic intervals.

Closed Interval: [a, b]  {x | a  x  b} represents all numbers between and including
the endpoints a and b.
Graphical Representation:
x

Open Interval: (a, b)  {x | a  x  b} represents all numbers between a and b not
including a and b.
Graphical Representation:
x

Half Open Interval: [a, b)  {x | a  x  b}
x

Unbounded Intervals:
- [a, )  {x | x  a}
x
- [, b)  {x | x  b}
x
- (, )  {x |    x  }
x
2
Example 1: Write the inequalities 0  x  3 and x  1 in interval notation.
Solution:
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Functions
Definition: A function is a rule that assigns to each number x in one set (called the domain
of the function) a unique (one and only one) number y in another set (called the range).
Notation: A function is represented by y  f (x) .
Note: Graphically, the functional value f (x) gives the y coordinate of the ordered pair
( x, f ( x)) corresponding to the input value x.
Informally, we define the domain and range of a function as follows:
Domain - the set of x values (inputs) of which a function is defined.
Range – set of y values output by the function.
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Representing Functions Graphically
Example 2: Given the following graph of a function:
Answer the following:
a. Find f (3), f (1) and f (2)
b. For what values of x does f ( x)  2 ?
c. State the domain and range of f.
Solution:
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Representing Functions Algebraically
Example 3: Given f ( x)  2x 2  x  1. Find
a. f (1)
b. f (2)
c. f (t 2 )
d. f ( x  h)
e. f ( x  h)  f ( x)
Solution:
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Finding the Domain of a Function Algebraically
Recall that the domain of function is the set of input (x values) for which a function is
defined or makes sense. We demonstrate how to determine the domain of a function
algebraically in the next few examples.
x
Example 4: State the domain of the function f ( x)  2
.
x 4
Solution:
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Example 5: State the domain of the function f ( x) 
x2 .
Solution: For this problem, the domain must be values of x that, when input, give a nonnegative term under the square root. Thus, we solve the inequality
x2 0
Adding 2 to both sides solves this inequality.
x2
Thus, the domain is the following:
Domain: {x | x  2} or in interval notation this answer is [2,  )
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Example 6: State the domain of the function f ( x) 
3
.
1  sin x
Solution: The domain of this function will be restricted for values of x where the
denominator of this function is equal to zero, that is, where
1  sin x  0
Solving this equation for x gives
 sin x  1
sin x  1
Now, one angle value were sin x  1 is x 

. However, the sine function is periodic,
2
which means there is an infinite number of solutions to this equation. Since the period of
the sine function is 2 , we can express the solution in terms of integer multiples of 2 ,
that is
x

2
 2n 

2

2n (2)  4n 4n   (4n  1)
 



1 (2) 2
2
2
2
Thus, all solutions can be expressed by the formula. Thus, the domain is the set:

(4n  1) 




 x | x   2n  or  x | x 
2
2




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Example 7: State the domain of the function f ( x)  x 3  2x  1 .
Solution:
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Piecewise Functions
Piecewise functions are functions that have different output formulas for different parts of
the domain.
2 x  1 if
Example 8: Given the piecewise function f ( x)   2
if
 x
a. Find f (1), f (1), and f (2)
b. Sketch the graph.
x 1
x 1
Solution:
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Note: Not all equations represent that of a function.
Vertical Line Test: a graph is that of a function if no vertical line touches the graph more
than once.
Example 11: Determine whether the following graphs represent functions.
y
x
y
x
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Composition of Two Functions
The composition of two functions f and g is denoted by f  g . The composition ( f  g )( x)
takes an input value x, performs g on this input to obtain g (x ) , then performs f to obtain
f ( g ( x)) .
( f  g )( x)
x
g(x)
f (g(x))
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Example 12: Given
Find:
a. ( f  g )(0)
b. ( g  f )( 4)
Solution:
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Note: Algebraically, to compute ( f  g )( x)  f ( g ( x)) , substitute the function g (x ) for
every occurrence of x in the function f.
Example 13: For the functions f ( x)  x  1 and g ( x)  x 2  4 , find
a. f  g
b. g  f
Solution:
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Example 14: For the functions f ( x)  cos( x2 ) and g ( x)  x  1 , find
a. f  g
b. g  f
Solution: We calculate these two compositions as follows:

2
Part a: ( f  g )( x)  f ( g ( x))  f ( x  1) )  cos( x  1 )  cos( x  1)
Part b: ( g  f )( x)  g ( f ( x))  g (cos( x 2 ))  cos( x 2 )  1
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