FUNCTIONS
Objectives: To recall the concept of a function.
Suppose that A and B denote two sets of objects. They may be sets of numbers, sets of books, sets of humans, etc. Then a rule that assigns exactly one member of the set B to each member of the set A is called a function from A to B. For example, if A represents the set of books, and B represents the set of whole numbers, then the rule which assigns to every book, the number of pages in that book is a function from the set of books to the set of whole numbers.

A
B
Although the function concept is general
enough to
be defined for arbitrary sets A and
B
as
described above, in mathematics it is normally used when both
A
and
B are sets of numbers. We will apply the
function
concept to the situation where both A and
B are
the set of real numbers.
Recall 2: (The domain of a function)
The domain of a
function from A to B
is the members of A that have corresponding
members of B that are
assigned by the rule that the function
describes. In the book example
above, the domain is all books.
However, the domain may
not always be the whole set A.
Take, for example, the function from
the set of real numbers
the square root of that number.
That is,
etc.
The domain of this function is not the set of all real numbers because the negative numbers cannot be assigned a value. The square root of a negative number does not exist! (That is if we don't count the complex numbers!). In this case, the domain is the set of all non negative numbers.
The range of a function from A to B is the set of members of B that are assigned to members of A by the function. In the case of the books, the range is the numbers that are numbers of pages of books. Negative numbers are not in the range of this function, because a book cannot have a negative number of pages. Zero is not in the range, because a book with 0 pages is not considered a book (or is it?) Moreover, not all the positive whole numbers are in the range, because it may be that there are no books with (say) 13 pages; and, it is hardly likely that there are books with more than (say) 106 pages.
When the sets A and B are sets of numbers (as they will be in most cases,) the members of the domain are called the independent variables while the members of the range are called the dependent variables. This makes sense, given the fact that the members of B depend on the members of A.
There are several forms of notation used in illustrating particular functions. Most common is the single letter notation like f or g We denote the function as f(x), meaning that the function f assigns the value f(x) to the value x. For example, if f is the squaring function (the function that takes each real number to its square), then it is denoted f(x) = x2, meaning
f(-2) =
(-2)2 =
4,
f(2) = 22 =
4,
f(3) = 32 = 9,
f(4) = 42 = 16
Another notation that is commonly used in higher mathematics is f:A-> B, which is a more descriptive illustration, suggesting that f takes the values of A and sends them to values of B.
Through most of the
previous topics, we have
worked with equations that involve only one
variable.
Equations like
x2 - x + 40 = 0 or
These equations are useful in solving many problems. The word problems are examples of the usefulness of such equations. Many problems will involve more than one variable and it will become increasingly important to know how to treat equations of more than one variable, say an x and a y. Some of these equations may be put in the form where the y can be isolated and put on one side of the equation, with the remaining terms on the other side of the equation. Such is the case with the equation
In this case, if we may
assume that x is
different from 0, the equation can be put in the form y
equals
This is an equation
that may also be viewed
as a function. Namely, the function
that takes real numbers
to real numbers in the following
way:
Take the real number
x,
square
it,
subtract
60
and divide the
result by 2 times
x.
Let y be the
number that you get from this
recipe.
Mathematically, it is clear that y must
be the same as
Recall 7: (How the function
concept makes
equations more dynamic.) From Recall 6 an equation
like
can be thought of as a
function from
the set of real numbers to the set of real
numbers and
this function may be denoted
as
f(x) =
You may not think that there is much
difference
between thinking of the equation
and the function f(x) =
; but, the fact is that with the
function concept we may find f(x+2), which
is
This will become extremely useful in the Workout when we learn how to graph functions.
Example 1. Find the domain of the function that takes real numbers to real numbers.
f(x) =
.
Solution: Pick any number x in the interval where -5 < x < 5. This function cannot handle such numbers because the expression under the root sign is negative for such numbers; and when we take the square root, to find the value that the function takes x to, we find that we are are attempting to computethe square root of a negative number. The square root of a negative number is not a real number --there is no real number whose square is negative. Therefore, we must exclude the real number between -5 and 5 from the domain. The natural domain of this function is all real numbers greater than or equal to 5 and all real numbers less than or equal to -5.
Example 2. Find the domain of the function.
f(x) =3x/[(x-3)(x+2)].
Solution: Notice that when x = 3 or when x = -2, the rational expression is not defined as a finite real number (they are poles). Since functions must take real numbers to real numbers (and since every real number must be finite), we must exclude the values x = 3 and x = -2 from the domain. So the domain includes all real numbers except -2 and 3.
Example 3. Find the range of the function
f(x) =(x + 3)2 + 4
Solution: If x = -3, the term (x + 3)2 vanishes. Also this term can never be less than zero. So f(x) is smallest when x = -3 and the value of f(-3) is 4. As x takes values greater than -3, the function takes values that are greater than f(-3). In fact any value greater than 4 is achieved by some value different from x = -3. Take any large value, say 10,000, and ask if there is some value of x where f(x) = 10,000. This amounts to solving the equation
10,000 = (x + 3)2 + 4.
We may solve this equation by subtracting 4 from both sides to get
9,996 = (x + 3)2.
This means
that x + 3 = SQRT(9,996),
or,
x = -3 +SQRT(9,996).
We now know that f(-3+SQRT(9,996)) = 10,000.
Clearly, we may play the same game with any number greater than 4. So, the range of this function is the set of all real numbers greater than or equal to 4.
Example 4. Let f(x) be
Find f(x + 1).
Solution: The
function f itself is the rule
which says that if you give it a real number
x then it will first
cube that number, add 2
times the square of that number,
subtract 13 times that number and add
10,
then, divide the result by that number minus
1.
To find f(x + 1) simply follow the rules replacing that number by x+1. So, we have f(x+1) given by
EXERCISES
1. Write the following equation as a function of x
3x2 - xy = 2y - 4
2. Find the domain of the function
f(x) = (3x2+4)/(x+2).
3 Find the domain of the function
f(x) given by
4 Find the domain of the function given by the expression
/(x+2)
5 Find the range of the function f(x)
given by the
expression
6 Find
the range of the function f(x) given by the expression
7 For the function below, find f(-2), f(0),
f(1), and
f(2)
f(x)
=(4x3-x)/(2x-1)
8 For the function below, find f(x + 2) and
simplify your
answer.
f(x)
=(4x2-16x)/2x,
x different from 0.
9 For the
function below, find g(x2) .
g(x)
=(4x2-16x)/(2x-1), x different from 1/2
10. For the function below, find g((x
-1)(x-1))
11 Find
f(g(1/2)) and g(f(1/2)), where
and g(x) =2/x .
12 Find f(g(4)) and g(f(4)),
where
f(x) =1/(4-sqrt(x)) and g(x)
=4/(x2) .
13 Find
f(x)/g(x)
and simplify it,
where
14 Find
[f(x+2)-f(2)]/x and simplify it for the
function
f(x) = 3x2 + 2x - 1.
15
Find [f(x+2)-f(2)]/x and simplify it for the function
f(x) =1/(2x+1) .
2. All real numbers except x = -2.
4 -2/3 <= x <= 2/3
5 All real numbers greater than or equal to 2.
6 real numbers y such that 0 <= y <= 2.
7 f(-2) = 6, f(0) = 0, f(1) = 3, f(2) = 10.
8 f(x + 2) = 2x - 4.
12 f(g(4)) = 2/7 and g(f(4)) = 16.
13 -2/3 <= x <= 2/3
14 3x + 14.