Section 4: Introduction to Functions
It is important that you watch the video first.
Define Relation, Domain, and Range
A relation describes a correspondence between two or more variables. Each of the following is an example of a relation.
x + 2y = 10
x² + y² = 25
{(x, y) | x > y }
{(4, 1), (2, –3), (0, 4), (6, 1)}
The domain of the relation is the set of all possible first coordinates. When we graph, we think of the first coordinate in terms of x.
The range of the relation is the set of all possible second coordinates. When we graph, we think of the second coordinate in terms of y.
Example 1 :
Find the domain and range of the relation:
{(7, 4), (–3, –3), (1, –2), (4, –6)}
Solution:
The domain consists of the set of all of the first coordinates: {7, –3, 1, 4}.
The range consists of the set of all of the second coordinates: {4, –3, –2, –6}.
Identify Functions
A function is defined as a relation in which each member of the domain is matched to exactly one member of the range. In other words, no two ordered pairs can have the same first coordinate and different second coordinate.
The relation shown in Example 1 is an example of a function since no ordered pairs have the same first coordinate.
Example 2:
Determine which of the following relations represents a function. If the relation does represent a function, then give the domain of the function.
a) {(10, 6), (23, –1), (38, 6), (10, 4), (59, 4)}
b) 
c) x = y² – 3
d) 
Solution:
- The set {(10, 6), (23, –1), (38, 6), (10, 4), (59, 4)} does not represent a function since the first coordinate, 10, corresponds to both 6 and 4
- does represent a function since every x-value that you substitute into the equation will produce a different y-value.
In order to determine the domain of the function, we need to consider the square root. The domain of a function is the set of all real numbers that are meaningful replacements for x. To find the domain, first decide if there are any values of x that are not meaningful replacements.
In this course, the types of expressions that we will consider which have values for x that are not meaningful are fractions, since they are undefined when the denominator is equal to zero, and square roots, since the square root of a negative number is imaginary.
In this case, we need to determine for which values for x, the expression is real.
To do this, set the expression under the radical sign (the radicand) to be grater than or equal to zero, and solve for x.
The domain will be the solution set to x + 5 > 0. Therefore, the domain of is x > –5. In interval notation, the domain is given as [–5, ∞). - x = y² – 3 does not represent a function since it is possible to find an x-value that corresponds to two different y-values.
If you select x = 1, and substitute it into the equation, then the equation becomes 1 = y² – 3, which has two solutions. Both y = 2 and y = –2 are solutions.
1 = (–2)² – 3 and 1 = 2² – 3 - does represent a function. Every x-value that you substitute into the equation will produce a different y-value.
In order to determine the domain of the function, we need to consider the fraction. The domain of the function will represent all meaningful values of x that can be substituted into this equation.
In this case, x cannot equal 2 since the number 2 will make the denominator equal to zero and the fraction will be undefined.
Function Notation
Function notation, such as y = f(x), illustrates the input and output process of a function.
y = f(x) is read “f of x,” where f is the name of a function and x is the value that is used as input into the function.
The value of a function is the output of the function, or the y -value that corresponds to an input value, x.
If f (x) = 5x – 7, then f (4) = 5(4) – 7 = 20 – 7 = 13. f (4) = 13.
To calculate the value of a function for any given x -value, just input a number into the function and simplify.
Example 3 :
Given f (x) = x² – 3x + 2, find
a) f (0)
b) f (2)
c) f (t )
d) f (3k )
e) f (x + 9)
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Solution: |
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| a. | f (0) = (0)²– 3(0) + 2 = 0 – 0 + 2 = 2 | f (0) = 2 |
| b. | f (2) = (2)²– 3(2) + 2 = 4 – 6 + 2 = 0 | f (2) = 0 |
| c. | f ( t ) = ( t )²– 3( t ) + 2 = t²– 3 t + 2 | f ( t ) = t²– 3 t + 2 |
| d. | f (3 k ) = (3 k )²– 3(3 k ) + 2 = 9 k²– 9 k + 2 | f (3 k ) = 9 k²– 9 k + 2 |
| e. | f ( x + 9) = ( x + 9)²– 3( x + 9) + 2
= ( x² + 18 x + 81) – 3 x – 27 + 2 = x² + 15 x + 56 |
f ( x + 9) = x² + 15 x + 56 |
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