Lesson Objectives

- Learn how to write a function using function notation
- Learn how to evaluate a function using function notation
- Learn how to change between the implicit form and explicit form of a function

## How to Write a Function Using Function Notation

Up to this point, we have worked with functions in the following format:

y = 2x - 9

y = x

y = x

Where y is the dependent variable and x is the independent variable. When we have a function, we normally name the function as: f, g, or h. When working with functions, we will replace our familiar "y" with f(x), g(x), h(x),...etc

f(x) = 2x - 9

f(x) = x

f(x) = x

f(x) is read as "f of x", which means we have a function "f" that depends on the independent variable "x".

This notation can immediately be used to evaluate a function for a given value of x. Let's suppose we had the following function.

y = x - 3

We have previously seen that we could state that x = some value, and solve for y.

Let's suppose x = 7:

Plug in a 7 for x:

y = 7 - 3

y = 4

y is 4 when x is 7. This can be expressed as the ordered pair: (7,4)

We can show this using function notation by replacing the "x" in "f(x)" with the x-value of 7.

f(x) = x - 3

f(7) = x - 3

This is read as "f of 7"; what is the functions value, when x is 7?

f(7) = 7 - 3

f(7) = 4

The answer is the same either way, we are just learning new notation.

Let's look at a few examples.

Example 1: Find f(0), and f(1)

f(x) = 2x

f(0) = 2(0)

f(0) = 5

f(1) = 2(1)

f(1) = 2 - 1 + 5

f(1) = 6

Example 2: Find f(-2), and f(5)

f(x) = x

f(-2) = (-2)

f(-2) = -8 - 4

f(-2) = -12

f(5) = (5)

f(5) = 125 - 4

f(5) = 121

In some cases, we will be asked to plug in more complex expressions for our variable. Let's look at a few examples.

Example 3: Find f(a + 7), and f(a - 1)

f(x) = x - 12

Just like when we had a single number, we plug in whatever is inside of the parentheses for x in our function:

f(a + 7) = (a + 7) - 12

f(a + 7) = a - 5

f(a - 1) = (a - 1) - 12

f(a - 1) = a - 13

Example 4: Find f(z + 2), and f(z - 1)

f(x) = x

f(z + 2) = (z + 2)

f(z + 2) = z

f(z + 2) = z

f(z - 1) = (z - 1)

f(z - 1) = z

f(z - 1) = z

Example 5: Find f(-3), and f(10)

x

Let's start by solving for y:

y = x

Now we can replace y with f(x):

f(x) = x

f(-3) = (-3)

f(-3) = 9 - 20

f(-3) = -11

f(10) = (10)

f(10) = 100 - 20

f(10) = 80

y = 2x - 9

y = x

^{2}+ 12y = x

^{3}- 1Where y is the dependent variable and x is the independent variable. When we have a function, we normally name the function as: f, g, or h. When working with functions, we will replace our familiar "y" with f(x), g(x), h(x),...etc

f(x) = 2x - 9

f(x) = x

^{2}+ 12f(x) = x

^{3}- 1f(x) is read as "f of x", which means we have a function "f" that depends on the independent variable "x".

This notation can immediately be used to evaluate a function for a given value of x. Let's suppose we had the following function.

y = x - 3

We have previously seen that we could state that x = some value, and solve for y.

Let's suppose x = 7:

Plug in a 7 for x:

y = 7 - 3

y = 4

y is 4 when x is 7. This can be expressed as the ordered pair: (7,4)

We can show this using function notation by replacing the "x" in "f(x)" with the x-value of 7.

f(x) = x - 3

f(7) = x - 3

This is read as "f of 7"; what is the functions value, when x is 7?

f(7) = 7 - 3

f(7) = 4

The answer is the same either way, we are just learning new notation.

Let's look at a few examples.

Example 1: Find f(0), and f(1)

f(x) = 2x

^{2}- x + 5f(0) = 2(0)

^{2}- 0 + 5f(0) = 5

f(1) = 2(1)

^{2}- 1 + 5f(1) = 2 - 1 + 5

f(1) = 6

Example 2: Find f(-2), and f(5)

f(x) = x

^{3}- 4f(-2) = (-2)

^{3}- 4f(-2) = -8 - 4

f(-2) = -12

f(5) = (5)

^{3}- 4f(5) = 125 - 4

f(5) = 121

In some cases, we will be asked to plug in more complex expressions for our variable. Let's look at a few examples.

Example 3: Find f(a + 7), and f(a - 1)

f(x) = x - 12

Just like when we had a single number, we plug in whatever is inside of the parentheses for x in our function:

f(a + 7) = (a + 7) - 12

f(a + 7) = a - 5

f(a - 1) = (a - 1) - 12

f(a - 1) = a - 13

Example 4: Find f(z + 2), and f(z - 1)

f(x) = x

^{2}+ x - 1f(z + 2) = (z + 2)

^{2}+ (z + 2) - 1f(z + 2) = z

^{2}+ 4z + 4 + z + 2 - 1f(z + 2) = z

^{2}+ 5z + 5f(z - 1) = (z - 1)

^{2}+ (z - 1) - 1f(z - 1) = z

^{2}- 2z + 1 + z - 1 - 1f(z - 1) = z

^{2}- z - 1### Functions in Implicit Form

When a function is solved for the dependent variable "y" or "f(x)", it is said to be in "explicit" form. When a function is not solved for the dependent variable "y", it is said to be in "implicit" form. In some cases, we will have a function that is in implicit form and we will be asked to find the value of the function, given a certain input. When this occurs, we will first solve the equation for y. We then will replace y with "f(x)". Lastly, we will evaluate the function for the given value. Let's look at an example.Example 5: Find f(-3), and f(10)

x

^{2}- y = 20Let's start by solving for y:

y = x

^{2}- 20Now we can replace y with f(x):

f(x) = x

^{2}- 20f(-3) = (-3)

^{2}- 20f(-3) = 9 - 20

f(-3) = -11

f(10) = (10)

^{2}- 20f(10) = 100 - 20

f(10) = 80

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