The domain is the set of all where the function is defined. Since we can't divide by 0, we need to find the values that would make the denominator zero. Set the denominator equal to zero and solve for to find:

Thus can take all values except for positive and negative 2, so the domain is:

You know that the square root function has real roots only for positive values. Therefore, you know that . From this, you can solve for your domain:

The two potential concerns for domain in a standard function are negative numbers inside even-powered radicals, and dividing by zero. In this case, the radical we have is odd-powered, so having a negative result underneath the radical is fine. All we need to do, then, is avoid dividing by zero, which means avoiding a result which sums to zero under the radical. In this case, only will create this situation, so we must avoid it.

Thus, our domain is

If a value of x makes the denominator of a equation zero, that value is not part of the domain. This is true, even here where the denominator can be "cancelled" by factoring the numerator into

and then cancelling the from the numerator and the denominator.

This new expression, is the equation of the function, but it will have a hole at the point where the denominator originally would have been zero. Thus, this graph will look like the line with a hole where , which is

.

Thus the domain of the function is all values such that

The domain of the function is all real numbers except x = –3. When x = –3, f(–3) is undefined.

The domain of the function is , because the result of taking the square root of a negative number is imaginary.

Therefore, for the function given, , when .

So the domain must be .

To find the domain we want to find any areas for which an x value will not work in our function. The potential problem area would be that under the radical.

Therefore, for the function given, , for all real numbers .

So the domain must be .

In other words, there is no real value of whose absolute value is less than let alone

To find the domain, you must find the values for which you can plug in for . Based on the given function, you know that you can not have a demoninator equal to . So, by setting the denominator equal to , you can find out the values that can not be in the domain, and then simply remove them from your set of numbers from negative infinity to positive infinity.

Thus,

Therefore, for our function, can not equal or it will be undefined. Thus, we get the set of numbers excluding positive .

Using our properties of exponents, we could rewrite as

This means that when we input , we first subtract 2, then take this to the fourth power, then take the fifth root, and then add three. We want to look at these steps individually and see whether there are any values that wouldn’t work at each step. In other words, we want to know which values we can put into our function at each step without encountering any problems.

The first step is to subtract 2 from . The second step is to take that result and raise it to the fourth power. We can subtract two from any number, and we can take any number to the fourth power, which means that these steps don't put any restrictions on .

Then we must take the fifth root of a value. The trick to this problem is recognizing that we can take the fifth root of any number, positive or negative, because the function is defined for any value of ; thus the fact that has a fifth root in it doesn't put any restrictions on , because we can add three to any number; therefore, the domain for is all real values of .

When x = 0 or x = 3, the function is undefined due to its denominator.

Thus the domain is all real numbers x, such that x is not equal to 0 or 3.