Let's explore the two cases to described in the right hand frame. Why must we be concerned about a denominator containing x?
The denominator must not take on the value of zero.
Why is that?
When a denominator is zero, the expression is not defined because we don't know how to handle division by zero.
Does this mean that the domain must exclude x = 0?
Not necessarily. We have to look at the whole denominator, because it is the whole denominator which must not be zero.
What about square roots? Why do we have to be careful with them?
Since we are insisting on y values which are real numbers, we have to require that the square root be applied to a positive number, or zero. (The square root of a negative number is not a real number.)
Let's look at some examples.
Example 1.
What is the domain of y = 5x - 2?
Since there is no denominator or square root, the domain is all of the real numbers.
Example 2.
Find the domain of 
.
Here we have x in a denominator. What is the restriction on the denominator?
5x must not take on the value of 0. So we must avoid any value of x which makes 5x = 0.
What is the restriction on x?
x must not be 0.
Express the domain for .
{ x | x
¹ 0 }Example 3.
Now, consider 
. Can x = 0 be in the domain now?
Let's try it: With x = 0, we get
So we see that x = 0 does not lead to a denominator of 0.
How do we find the value of x that does lead to a denominator = 0?
We set the denominator to 0 and solve for the corresponding value of x.
Do it.
5x - 2 = 0. Add 2 to both sides.
5x = 2. Divide by 5 on both sides.
x = 2/5.
What is the interpretation of this result?
This is the value of x which must be avoided in the domain because 2/5 makes the denominator become zero.
Express the domain of .
{ x | x
¹ 0 } is the domain.Example 4.
Consider 
. Is there anything to be cautious about here?
Yes, the square root.
Why?
In order to produce a real number for y, the quantity under the square root sign must be positive, or zero.
How does that apply here?
5x - 2 must be positive or zero.
Express this as an inequality.
5x - 2
³ 0.Solve this for the related restriction on x.
5x - 2
³ 0. Add 2 to each side:5x ³ 2. Divide by 5 on each side.
x ³ 2/5.
Is 2/5 in the domain?
Yes.
What about 1/5, 0, etc?
These values also must be excluded from the domain because they are smaller than 2/5.
Express the domain.
The domain is { x | x ³ 2/5 }.
Example 5.
Consider 
. Is there anything to be cautious about here?
Yes, here we have both the square root and a denominator. As above, x = 2/5 results in
. This was OK in the numerator in Example 4.
Is it OK in this example?
No.
Why?
Because the square root is in the denominator and a zero in the denominator is an undefined situation.
Are the positive values of (5x - 2) still OK?
Yes.
Express the domain for .
The domain is { x | x > 2/5 }.
Note: the operator here is just "greater than", not "greater than or equal" as in Example 4.
