Functions are equations that usually involve 2 variables. They are most often x and y. One can generally say that one of the variables is the input (usually x) and the other is considered the output (usually the y).

 

The domain of a function is all the possible values that can plug into the equation.

The range is, all the numbers that are the outputs from the domain.

 

A function is more precisely defined by saying that it is an algebraic expression in which each element of the domain produces, or is matched, with only one element of the range. In other words, for every x there is only one y.

An example of a function might look like

 

f(x)= 3x –1.

 

The f(x) at the beginning of the function is read as “a function of x” or “f of x.” It can be replaced with a y.

 

f(x) = 3x –1→ y = 3x – 1

 

Now say that one is using the above function f(x)= 3x –1 or y = 3x – 1. For understanding purposes, assign a domain to it. Remember the domain is the possible numbers that can plug into the function or the x’s.

Domain = {-3,-1,0,2}

 

Note- All those funny looking brackets mean is that it is a set of numbers. Anytime writing a set of numbers use those brackets.

 

 

 

All the possible values for x are -3, -1, 0, and 2. Plug each of these in for x in the equation and solve for y.

 

When x = -3

y = 3(-3) –1= -10

 

x = -1

y = 3(-1) –1= -4

 

x = 0

y = 3(0) –1= -1

 

x = 2

y = 3(2) –1= 5

 

The results of each of these equations, creates the range. Always write the range in order from least to greatest.

Range = {-10, -4, -1, 5}

 

Furthermore, as one can tell with the previous calculations the –3 for x will produce a y value of –10. That means that if one were to graph the function on an x-y axis the point (-3,10) would be on the graph.

 

Using that idea, to graph this function recognize that the following points would be on the graph.

(-3,10)

 

(-1,-4)

 

(0,-1)

 

(2,5)

 

 

Sometimes it is easier to understand functions and the definition of a function by mapping it. Creating two bubbles, one with the Domain and one with the Range does this. In between these two bubbles draw arrows, which represent how each element of the Domain is matched up with each element of the Range. Here is what mapping looks like using the previous function.

-3→10

-1→-4

0→-1

2 →5

 

Domain→Range

 

Remember that the definition of a function is an algebraic expression in which each element of the Domain produces, or is matched, with only one element of the Range. This is perhaps easier to see with mapping. Notice that each element of the Domain goes to one element of the Range.

 

Look at a couple of other maps and determine whether or not they are functions.

 

Example

Determine whether the following map is that of a function.

 

-5→2

0→0

6

 

 

Domain→Range

 

The fact that the lines cross means nothing. What one is concerned with is if every element of the Domain goes to one and just one element of the Domain. In this case it is a function even though the –5 and 6 are both going to 2. Perhaps an easier way to understand this is ask this question for each element of the Domain, “If I plug in a –5 for the x, do I know exactly what the answer will be?” The answer to this question for each element of the Domain is yes. Therefore, this is a function.

 

 

 

 

Example

Determine whether the following map is that of a function.

1→2

0→0

3

 

 

 

Domain→Range

 

Starting with the first element (1), ask the question, “If I plug in a 1 for x, do I know exactly what y will be?” The answer to this is no. Plug in a 1 for x, and one does not know if the result will be a 2 or 3 for y. Therefore, this is not a function.

 

One may ask “How can an equation ever be mapped to give a situation to plug in a value for x and get 2 values for y?” Here is one like that.

y² = x

 

Plug in a 4 for x and one can either have 2 or –2 for y. This would be an example of an equation that is not a function.

 

It may be asked to determine if a set of ordered pairs is a function. All one has to do is to see if each x is paired up with only one y.

 

Example

Determine if the set of ordered pairs is a function

 

{(0, 1),(1, 3),(2, 5)}

 

Each x has only 1 y. In other words, for the value of 0 know that y is 1, and so on for each x.

 

Yes, is a function

 

Answer!

 

 

 

 

 

 

Example

Determine if the set of ordered pairs is a function

 

{(-1,2),(2,3),(-1,4)}

 

In this case the –1 for x can go to either 2 or 4 for y. Thus, this is not a function.

Not a function

 

Answer!

 

 

 

 

 

 

 

Example

Determine if the set of ordered pairs is a function

 

{(-3,1),(0,4),(2,4)}

 

Each x goes with one y. So this is a function. It does not matter that 0 and 2(both x’s) both go to 4(y).

 

Yes, is a function

 

Answer!

 

 

 

 

 

 

Example

Determine if the set of ordered pairs is a function

 

{(-2, 1),(0, 1),(1, 3)}

 

Each x goes with only one y. This is a function.

Yes, is a function

 

Answer!

 

 

 

 

 

 

Example

Determine if the set of ordered pairs is a function

 

{(0,1),(0,1),(2,3),(3,-5)}

 

Even though the 0 is repeated for x it goes to 1 in both cases. So plug in a 0 for x and the y would be 1. Thus, this is a function.

 

There is another way to test and see if a graph is a function. It is called the vertical line test. The vertical line test says that if one can draw a vertical line anywhere on a graph and have it cross the graph more than once, then the graph is not a function.

 

 

 

 

 

Example

 

Is the graph a function?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

To see if this is a function, draw many vertical lines to see if any of them cross the graph in more than one place.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

From looking at the graph, none of these lines cross the graph more than once. Therefore, this is a function.

 

 

 

Example

Is the graph a function?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Draw some vertical lines to see if it crosses the graph more than once.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

It is easy to tell that the vertical line crosses the graph twice. Thus, this is not a function.

 

 

Example

Is the graph a function?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Regardless of where one draws a vertical line on this graph it will only cross it once.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This is a function.

 

 

Example

Is the graph a function?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Draw a vertical line right on top of this one and it would cross it in infinitely many places. Since that is more than once, this is not a function.

 

Remember to write a function as f(x) = something. Here are some examples of functions.

f(x) = 3x + 2

 

f(x) = -2x +3

 

f(x) = x²

 

g(x) = 2x

 

In the first three, one would start off by reading them as “f of x”. In the last example one would say “g of x”. It means nothing different that the other ones. The g(x) is typically used when there is more than one function in the problem. Sometimes h(x) will be used. Anything can be used for the first letter, all it does is name the function. Some examples of that will be shown later.

 

Now, work with a function and learn how to plug values into functions, and work with them.

 

 

 

 

Example

Let f(x) = 2x + 6

Find f(4).

What this is saying is find out what the value of the function is when x = 4. This can be very tricky when working with variables and negative numbers so it is very important to use parentheses when trying to evaluate a function this way. Plug the 4 in for the x everywhere it is in the function.

f(4) = 2(4) + 6

f(4) = 8 + 14

f(4)= 22

 

Answer!

 

Look at one that is a little more complicated and evaluate it for many different values. Remember, whatever is in the parentheses plug it in everywhere there is an x in the function.

 

 

 

 

 

Example

Let f(x) = x² + 3x – 2

Find f(0).

 

Plug in a 0 everywhere there is an x.

 

f(0) = (0)² + 3(0) – 2

 

Order of operations says to do the parentheses first.

 

f(0) = 0 + 3(0) – 2

 

Multiply.

 

f(0) = 0 + 0 – 2

 

Do the addition and subtraction.

 

f(0) = – 2

 

-2

 

Answer!

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(-1).

 

Plug in a –1 wherever there is an x.

 

f(-1) = (-1)² + 3(-1) – 2

 

Do the exponents.

 

f(-1) = 1 + 3(-1) – 2

 

Multiply.

 

f(-1) = 1 + -3 – 2

 

Add and subtract.

 

f(-1) = -4

 

-4

 

Answer!

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(2).

 

Plug in the 2 for x.

 

f(2) = (2)² + 3(2) – 2

 

Do the exponent.

 

f(2) = 4 + 3(2) – 2

 

Multiply.

 

f(2) = 4 + 6 – 2

 

Add and subtract.

 

f(2) = 8

 

8

 

Answer!

 

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(-3).

 

Plug in a –3 for every x.

 

f(-3) = (-3)² + 3(-3) – 2

 

Do the exponent.

f(-3) = 9 + 3(-3) – 2

 

Multiply.

f(-3) = 9 + -9 – 2

 

Add and subtract.

 

f(-3) = -2

 

-2

 

Answer!

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(a).

 

Plug in the a for every x.

 

f(a) = (a)² + 3(a) – 2

 

Do the exponent and multiply to get rid of all parentheses.

 

f(a) = a² + 3a – 2

 

This will not simplify anymore since there are no like terms.

 

a² + 3a – 2

 

Answer!

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

Find f(-b).

 

Plug in –b for every x.

 

f(-b) = (-b)² + 3(-b) – 2

 

Do the exponent.

 

f(-b) = b² + 3(-b) – 2

 

Multiply.

 

f(-b) = b² –3b – 2

 

This will not simplify any further.

 

b² –3b – 2

 

Answer!

 

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(x+2).

 

This one may be confusing at first but follow the same rules. Plug in x + 2 for every x.

 

f(x+2) = (x+2)² + 3(x+2) – 2

 

Do the exponent. Keep in mind, multiply x + 2 by x + 2. It is not x² + 4.

 

f(x+2) = (x+2)(x+2) + 3(x+2) – 2

 

Multiply the x + 2 by x + 2.

 

f(x+2) = x² + 4x + 4 + 3(x+2) – 2

 

Now go ahead and multiply the 3 and x + 2.

 

f(x+2) = x² + 4x + 4 + 3x + 6 – 2

 

Simplify by combining all like terms.

 

f(x+2) = x² + 7x + 8

 

x² + 7x + 8

 

Answer!

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(b - 1).

 

Plug in b-1 for every x.

 

f(b-1) = (b-1)² + 3(b-1) – 2

 

Do the exponent.

 

f(b-1) = (b-1)(b-1) + 3(b-1) – 2

 

Multiply the b - 1 and b – 1.

 

f(b-1) = b² –2b +1 + 3(b-1) – 2

 

Multiply the 3 and b-1.

 

f(b-1) = b² –2b +1 + 3b - 3 – 2

 

Combine all like terms.

 

f(b-1) = b² + b – 4

 

b² + b – 4

Answer!

 

 

 

 

 

 

Example

 

Let f(x) = x² + 3x – 2

 

Find f(x²).

 

Plug in x² for every x.

 

f(x²) = (x²)²+ 3(x²) – 2

 

Do the exponent.

 

f(x²) = x⁴+ 3(x²) – 2

 

Multiply.

 

f(x²) = x⁴+ 3x² – 2

 

x⁴+ 3x² – 2

 

Answer!

 

 

 

 

 

When dealing with more than one function there are certain operations that can be done, here is a list of those.

 

Addition (f + g)(x)= f(x) + g(x)

Subtraction (f – g)(x)= f(x) – g(x)

Multiplication (f * g)(x) = f(x) * g(x)

 

Division

 

 

Here are some examples explaining these operations.

 

 

 

 

Example

Let f(x) = x² + 1 and g(x) = 3x –4

Find (f + g)(2).

 

Apply the upper rule with addition.

(f + g)(2) = f(2) + g(2)

 

Find f(2) and g(2).

f(2)= (2)² +1 = 5

g(2) = 3(2) – 4 = 2

 

Applying the rule of addition add the results.

(f + g)(2) = f(2) + g(2)

 

5 + 2 = 7

 

Answer!

 

 

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 1 and g(x) = 3x –4

Find (f - g)(3).

 

Apply the definition for subtraction.

(f – g)(3) = f(3) – g(3)

 

Find f(3) and g(3).

(f – g)(3) = 10 – 5

(f – g)(3) = 5

 

Answer!

 

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² + 1 and g(x) = 3x –4

Find (f * g)(-2).

 

Use the rule for multiplication.

 

(f * g)(-2) = f(-2)*g(-2)

 

Find f(-2) and g(-2).

f(-2) = (-2)² + 1 = 4 + 1 = 5

g(-2) = 3(-2) – 4 = -6 – 4 = -10

 

Plug in the results into the formula.

 

(f * g)(-2) = 5(-10)

 

(f * g)(-2) = -50

 

Answer!

 

 

 

 

 

Example

 

Let f(x) = x² + 1 and g(x) = 3x –4

 

Find .

 

Apply the rule of division.

 

Find f(0) and g(0).

f(0) = (0)² + 1 = 0 + 1 = 1

g(0) = 3(0) – 4 = 0 – 4 = -4

 

Plug into the formula.

 

.

 

 

Answer!

 

 

 

Here are some more examples of applying the rules of functions.

 

 

 

 

 

 

Example

Let f(x) = x² – 4 and g(x) = x + 1.

Find (f + g)(x).

 

Apply the rule of addition.

 

(f + g)(x) = f(x) + g(x)

 

Since it is already known what f(x) and g(x) is, plug them in here.

 

(f + g)(x) = (x² – 4) + (x + 1)

 

The parentheses can drop since there is only an addition in between them.

 

(f + g)(x) = x² – 4 + x + 1

 

Simplify by combining like terms.

 

(f + g)(x) = x² + x – 3

 

Answer!

 

 

 

 

 

 

 

Example

 

Let f(x) = x² – 4 and g(x) = x + 1.

 

Find (f - g)(x²).

 

Apply the rule of subtraction.

(f - g)(x²) = f(x²) – g(x²)

 

Find f(x²) and g(x²).

f(x²) = (x²)² – 4 = x⁴ – 4

g(x²) = (x²) + 1 = x² + 1

 

Plug into the formula.

 

(f - g)(x²) = (x⁴ – 4) – (x² + 1)

 

Get rid of the parentheses. Instead of just dropping them, the second one has – in front of it. That is saying the same thing as –1, multiply it by everything on the inside of the parentheses.

(f - g)(x²) = x⁴ – 4 – x² – 1

 

Combine like terms.

 

(f - g)(x²) = x⁴ – x² – 5

 

Answer!

 

 

 

 

 

 

 

 

 

Example

Let f(x) = x² – 4 and g(x) = x + 1.

Find (f * g)(3x).

 

Apply the rule of multiplication.

(f * g)(3x) = f(3x)*g(3x)

 

Find f(3x) and g(3x).

f(3x) = (3x)² – 4 = 9x² – 4

g(3x) = (3x) + 1 = 3x + 1

 

Plug into the formula.

(f * g)(3x) = (9x² – 4)*(3x+1)

 

Multiply.

(f * g)(3x) = 27x³ +9x² –12x – 4

 

Answer!

 

 

 

 

 

 

 

Example

Let f(x) = x² – 4 and g(x) = x + 1.

Find .

 

Apply the rule of division.

 

 

f(x) and g(x) are already known so just plug them in here.

 

 

 

Answer!

 

Note- In the previous answer, there was a situation with a limited domain. That means that not any value will work for x. If one were to plug in a –1 for x there would be a zero on bottom. This would get a result of ₯ (infinity). Since this is not an amount that one can clearly state, say that the domain of this function would be all real numbers Ή -1. Which is read as “all real numbers not equal to –1”.

 

 

 

 

 

 

 

 

Example

 

Let f(x) = x² – 4 and g(x) = x + 1.

 

Find .

 

Apply the rule of division.

 

 

g(x) and f(x) are already known so plug them in.

 

 

Answer!

 

This function also has a limited domain. To find out what values of x would result in a 0 on bottom, set the bottom expression equal to 0. x² – 4 = 0. Solving this, one finds out that when x is 2 or –2 results in as zero on bottom. Thus the domain would be all real numbers Ή ±2.

 

Function composition means to combine more than one function into an expression. Here is the main rule for function composition.

(f ° g)(x) = f(g(x))

 

This is read as “ f of g of x”.

 

 

 

 

 

 

Example

Let f(x) = 2x + 1 and g(x) = x² + 2x – 3

Find (f ° g)(1).

 

Apply the rule of composite functions.

 

(f ° g)(x) = f(g(1))

 

Treat this like any other algebraic equation, the first thing to do is to find g(1), since it is the innermost parentheses.

 

g(1) = (1)² + 2(1) – 3

 

g(1) = 1 + 2(1) – 3

 

g(1) = 1 + 2 – 3

 

g(1) = 0

 

Plug this value into the equation for g(1).

 

(f ° g)(x) = f(g(1)) = f(0)

 

This means find f(0).

 

f(0) = 2(0) + 1

 

f(0) = 0 + 1

 

f(0) = 1

 

Answer!

 

Note- f(g(x)) and g(f(x)) are not the same thing and will rarely produce the same results. Here is the same problem worked out but only trying to find g(f(1)).

 

 

 

 

 

 

 

Example

Let f(x) = 2x + 1 and g(x) = x² + 2x – 3

 

Find (g ° f)(1).

 

Apply the rule of composite functions. This time the g and f are in different places.

 

(g ° f)(x) = g(f(1))

 

Find f(1).

 

f(1) = 2(1) + 1

 

f(1) = 2 + 1

 

f(1) = 3

 

Plug this value in for the f(1).

 

(g ° f)(x) = g(3)

 

Find g(3).

g(3) = (3)² + 2(3) – 3

g(3) = 9 + 2(3) – 3

g(3) = 9 + 6 – 3

g(3) = 12

 

Answer!

 

 

 

 

 

 

 

Example

Let f(x) = 2x + 1 and g(x) = x² + 2x – 3

 

Find (f ° g)(x).

 

Apply the rule of composite functions.

 

(f ° g)(x) = f(g(x))

 

Normally one might try to find g(x) at this point but g(x) = x² + 2x – 3 is already known.

 

(f ° g)(x) = f(x² + 2x – 3)

 

Now find f(x² + 2x – 3). Do this by plugging in x² + 2x – 3 everywhere there is an x in f(x).

 

f(x² + 2x – 3) = 2(x² + 2x – 3) + 1

 

Simplify.

 

f(x² + 2x – 3) = 2x² + 4x – 6 + 1

 

f(x² + 2x – 3) = 2x² + 4x – 5

 

Answer!

 

 

 

 

 

Example

Let f(x) = 2x + 1 and g(x) = x² + 2x – 3

 

Find (g ° f)(x-1).

 

Use the rule of composite functions.

 

(g ° f)(x-1) = g(f(x-1))

 

Find f(x-1).

 

f(x-1) = 2(x-1) + 1

 

f(x-1) = 2x-2 + 1

 

f(x-1) = 2x-1

 

Plug this into the equation.

 

(g ° f)(x-1) = g(2x-1)

 

Find g(2x-1).

 

g(2x-1) = (2x-1)² + 2(2x-1) – 3

 

Simplify.

 

g(2x-1) = 4x² – 4x + 1 + 4x – 2 – 3

 

g(2x-1) = 4x² – 4

 

Answer!