Up Your Score (38 page)

A
relation
is a set of ordered pairs (
x, y
). The
domain
is the set of all possible
x
values, while the
range
is the set of all possible
y
values.

A
function
is a relation where each
x
has its own
y
, but each
y
can correspond with a variety of
x
’s. Confused? Well, just remember that each student (
x
) in math class gets a grade (
y
). Each student gets only one grade, but a specific grade can be given to more than one person. You can also just do the
vertical line test
. If you can draw a vertical line through any part of the graph and it touches only one point, then it is a function (e.g., the sine graph). But if it touches more than one point, it is not (e.g., a circle).

Functions are usually defined by
f
(
x
) = something. Just remember that for such an equation,
f
(
x
) is always equal to the
y
value.

Example 1: Let the function
f
be defined by
f
(
x
) =
x
²
+ 2.

What does
y
equal when
x
= 3?

Plug in 3 for
x
:

f
(
x
) = 3
²
+ 2.

y
= 9 + 2 = 11

The answer is 11.

Example 2: The function
f,
where
f
(
x
) = (1 +
x
)
²
, is defined for –2 ≤
x
≤ 2. What is the range of
f
?

(A) 0 ≤
f
(
x
) ≤ 4

(B) 0 ≤
f
(
x
) ≤ 9

(C) 1 ≤
f
(
x
) ≤ 4

(D) 1 ≤
f
(
x
) ≤ 5

(E) 1 ≤
f
(
x
) ≤ 9

Since
f
is only defined for −2 ≤
x
≤ 2, then that is the domain of the function. To get the range, plug in random numbers between −2 and 2. You won’t always have to plug in every number, but if it’s a small set, it’s probably a good idea. No matter what the size of the set, though, it’s usually also a good idea to check the end points and the values of
x
that create zeros.

Clearly, the lowest number is 0, and the highest is 9. So (D) is the right answer. Notice that if you had plugged in only −2 and 2, you would have gotten (E) as the answer.

You get a function’s
inverse
by switching the
x
’s and the
y
’s. If the inverse also turns out to be a function, the original function is called
“one-to-one”
and passes both the vertical and horizontal line tests.

There are five special functions you will probably want to be familiar with:

Okay, you can stop quivering at the disturbing imagery.

When you combine functions, all that happens is that you insert whatever
x
is into the first equation. You could end up with a number or an expression. Then, for the second function, whenever you see
x
, you insert whatever you ended up with from the first function.

Example: Let
f
(
x
) =
x
²
+ 3 and
g
(
x
) = 3
x
. What does
f
(
g
(
x
)) equal when
x
= 5?

Since
g
(
x
) gets substituted for
x
in
f
(
x
), that means that you do
g
(
x
) first.
g
(5) = (3 × 5 = 15

So
g
(
x
) = 15. Plug that in for
x
in
f
(
x
).
f
(15) = (15
²
+ 3 = 228

If the problem had asked you to find
g
(
f
(
x
)), the answer would have been 84. Try it for yourself.

A linear function is anything that can be graphed as a straight line. They are always in the form
f
(
x
) =
mx
+
b
.

You solve linear equations just like you solve any other kind of equation.

Example: 7
x
+ 3 = 5
x
+ 6,
x
= ?
You have to subtract 3 from both sides (or you can subtract 6, it really doesn’t matter, but most people find it easier to work with positive numbers).
7
x
+ 3 – 3 = 5
x
+ 6 – 3
7
x
= 5
x
+ 3
Then you subtract 5
x
from both sides (or you can subtract 7
x
).
7
x
– 5
x
= 5
x
+ 3 – 5
x
2
x
= 3
Finally, divide both sides by 2, and you’re left with
x
=

If you aren’t confident and you have time, you can check your answer by plugging in
for all the
x
’s in the original equation. If both sides are the same, then you’ve got yourself an answer.

Quadratic equations, or equations in the form a
x
²
+ b
x
+ c something, can be solved by factoring the trinomial (expression with three terms) into two binomials (expressions with two terms). We factor by using the wise, age-old, foolproof system of . . . trial and error.

Let’s look at an example. We have the following three terms: 2
x
²
–
x
– 6 = 0. In order to solve it, first we should ask, Does it factor? It looks likes it doesn’t, because there’s nothing that all three terms have in common, and they aren’t cubes or squares or anything.

But looks can be deceiving.

Let’s see . . .

When trinomials are factored into binomials, they form something like (a
x
+ b)(c
x
+ d). So you write down your trinomial, and the binomials, with blanks for missing numbers:

2
x
²
–
x
- 6 = 0
Factors:
( __
x
+ __ )( __
x
+ __ )

And then you ask yourself, what two numbers multiplied together would give me 2, the first coefficient?

Well, this one’s pretty easy. So you fill in the first two numbers:

(2
x
+ __ )(
x
+ __ )

And then you take a look at the constant (the last number). What two numbers multiplied would give me 6?

Well, there’s 3 and 2, and then there’s 6 and 1. So try all four combinations.

(2
x
+ 6)(
x
+ 1)
(2
x
+ 1)(
x
+ 6)
(2
x
+ 3)(
x
+ 2)
(2
x
+ 2)(
x
+ 3)

Which one works? To check, FOIL them. You multiply the
F
irst terms of each binomial together, then multiply the
O
utside terms, then the
I
nside terms, and finally the
L
ast terms.

After FOILing all four sets of binomials, you might say, “Oh drat! None of them work.”

To which we say, “Hmm, you’re right. Guess this method doesn’t work.”

Just kidding. The thing is, the original trinomial has two negative signs in it. This means its binomial factors have to have a negative somewhere in there, too. So off you go again, this time trying various combinations of negative second terms.

Sooner or later, you’ll come up with: (2
x
+ 3)(
x
– 2) = 0.

So now, to solve for
x
, try setting each of the binomials equal to zero, because if A × B = 0, then either A or B
must
equal zero. In other words, you can separate these binomials into two equations of their own: