{(1, 5), (2, 6), (3, 7), (4, 8)}

We need to ask ourselves, does every first element (or input) correspond with

**So, this relation would be an example of a
function.**

**Domain**

We need to find the set of all input values. In terms of ordered
pairs, that correlates with the first component of each one. So,
what do you get for the domain?

**If you got {1, 2, 3, 4}, you are correct!**

**Range**

We need to find the set of all output values. In terms of ordered
pairs, that correlates with the second component of each one. SO,
what do you get for the range?

**If you got {5, 6, 7, 8}, you are absolutely right!**

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

We need to ask ourselves, does every first element (or input) correspond with

**So, this relation would not be an example of
a function.**

**Domain**

We need to find the set of all input values. In terms of ordered
pairs, that correlates with the first component of each one. So,
what do you get for the domain?

**If you got {-1, 2}, you are correct!**

Note that if any value repeats, we only need to list it one time.

**Range**

We need to find the set of all output values. In terms of ordered
pairs, that correlates with the second component of each one. So,
what do you get for the range?

**If you got {2, 3, 4}, you are absolutely right!**

To check if *y* is a function of *x*,
we need to solve for *y *and then check to see
if there is only one output for every input.

***Inverse of squaring is taking the sq. root**

***Solved for y**

At this point we ask ourselves, would we get one value for *y* if you plug in any value for *x*?

If you answered no, you are correct.

For example, if our input value *x* is 0,
then our output value *y* could either be 1 or -1. Note that
I could have picked an infinite number of examples like this one.
You only need to show one example where the input value is associated with
more than one output value to disqualify it from being a function.

**This means that at least one input value is associated with more
than one output value, so by definition, y is not a function of x.**

To check if *y* is a function of *x*,
we need to solve for *y* first and then check
to see if there is only one output for every input.

At this point we ask ourselves, would we get one value for *y* if you plug in any value for *x*?

If you answered yes, you are right on. For example, if we plugged in
a 1 for *x*, then* y* would only equal one value, 5. Note that ANY value you would plug
in for *x* would produce only one value for *y*.

Note that since it is solved for *y*, *y *is
our output value and *x* is our input value.

**Since our answer to that question is yes, that means by definition, y is a function of x.**

Again, think of functional notation as a fancy assignment statement.
For example, when we are looking for *f*(0),
we are going to plug in 0 for the value of *x *in
our function *f *and so forth.

***Plug in 2 for x and evaluate**

Don't let the fact that we need to plug in an *a* throw you. You plug it into the function just like you do a number.
Everywhere you have an *x* in your function,
replace it with an *a*.

Don't let the fact that we need to plug in the expression *a* + *h* throw you. You plug it into the function
just like you do a number. Everywhere you have an *x* in your function, replace it with an *a + h*.

:

**Putting it all together we get:**

***Combine like terms**

***Divide out h from
every term**

***Plug in 4 for x**

Since we can plug in any value for *x* (input
variable) and get a real number answer for our function (output variable),
then there are no restrictions. **Therefore the domain is all real
numbers.**

Our restriction here is that the denominator of a fraction can never
be equal to 0. So to find our domain, we want to set the denominator
equal to 0 and restrict those values.

This time we do have a restriction on our domain. We can only
take the square root of a number greater than or equal to 0. So we must
set the radicand (the inside of the radical) greater than or equal to 0.

***Inv. of mult. 2 is div. 2**

***This is our domain**

Only values of *x *that are greater than
or equal to -1/2 make the original radicand greater than or equal to 0, **so
x **__>__ **-1/2 is our domain for this
function.**

Last revised on March 29, 2010 by Kim Seward.

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