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 .

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

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.

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 

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:

For this function, the only thing to which you need to pay heed is the denominator of the fraction, which cannot be equal to zero. Thus, you can just look at the polynomial found in the denominator and solve for the values of  equal to

This is solved by factoring:

This indicates that 

Otherwise, all values are fine!

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:

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 

Looking at our numerator shows us that  can never be negative, as it would result in a nonreal solution for the numerator. The denominator solves to a real value for all positive integers until , where it causes us to divide by zero, and all numbers past  result in a negative radical (and thus a nonreal solution) for the denominator.

Testing  itself produces , which is a fine answer. Therefore, our domain is inclusive of  and exclusive of , and should be written as .